Centripetal Press Textbook Sampler by Classical Academic Press

Centripetal Press Textbook Sampler A unique, 7th-12th grade curriculum built on the principles of

wonder, integration, and mastery.

WWW.CENTRIPETALPRESS.COM

In science instruction today, challenges abound, from determining correct teaching methods to deciding how best to ensure students’ retention of content beyond the course. These challenges are compelling thoughtful science educators to continually search for the best solution to encourage a love of science and increase student achievement. Introducing Centripetal Press Centripetal Press is an imprint of Classical Academic Press that focuses on the highest-quality science curriculum for charter schools. Founded by educator John D. Mays, Centripetal Press offers a completely unique approach to secondary science curriculum. In 1999, John began teaching at a classical school in Austin, Texas. During his first years there, he had three educational challenges on his mind. The first was the absence of literature that addressed how to teach science classically, and the fact that science-teaching methods utilized at classical schools throughout the country seemed no different from those utilized at any other school. The second was the near-universal under-performance of students in American schools in science and math. The third challenge was that science texts were often ugly, inaccurate, and enormous. John began developing and implementing teaching methods designed specifically to address the first two of these two challenges. Then, in 2010, he began developing a full secondary science curriculum to support the new teaching methods he had perfected and refined. Centripetal Press (for charter schools, public schools, and secular home schools) and Novare Science (for Christian schools and home schools) are the two product lines resulting from his work. The curriculum and principles used in textbook design are described below. The classical, mastery-based teaching methodology is fully described in John’s book, From Wonder to Mastery: A Transformative Model for Science Education.

Our Emphasis on Wonder, Integration, and Mastery The goal at Centripetal Press has always been to transform the way science is taught. If you want science instruction in your school to realize the full potential of students’ abilities, teaching science the way it has been taught for the past several decades will not do. New methods and new points of emphasis are required. We summarize our approach in terms of three categories— wonder, integration, and mastery.

WONDER The study of science should always begin with wonder. The world is a stunning place, full of surprises and jaw-dropping phenomena. The innate tendency of all humans to marvel at the natural world is one of the science teacher’s most valuable resources because our sense of wonder compels us to study and learn more about those things that amaze us. Our sense of wonder also plays a key role in making science fun. Stimulating this sense of wonder should be part of every science teacher’s daily classroom practices. One of the most profound truths about science education is that wonder leads to increased knowledge and increased knowledge, in turn, leads to an increased sense of wonder. Another reason wonder is so central for science education is that helping students to know the natural world has never been more important than it is today. Only if they know the world will students begin to love it, and only then will they be motivated to help take care of it. To nurture this love, we begin with the natural wonder we feel when we encounter the natural world. 3

INTEGRATION INTEGRATION

One of the hallmarks of classical education is that instruction is integrative. In science instruction, four key areas of integration are essential: • Epistemology—Understanding the nature of scientific knowledge is just as important for students as the scientific content itself. Further, it is crucial that students understand how scientific knowledge progresses and develops over time, a process we call the Cycle of Scientific Enterprise, represented by the figure below.

Hypothesis

Theory

THE CYCLE OF SCIENTIFIC ENTERPRISE

FACT FACT NEW FACT YES

Experiment

Analysis

Review

• Mathematics—As Galileo said, mathematics is the language of nature. Frequent use of grade-appropriate mathematical skills in science classes is essential for both comprehension and application of scientific content. • History—Developing key historical connections enhances understanding of science as a process. The history of science should not be relegated to sidebars but should instead be part of the “main thing.” • Language—The use of language is central to all human thought. Maximizing opportunities for students to develop skill at written expression is one of the educator’s highest priorities. 4

MASTERY Mastery essentially means proficiency and long-term retention of course content. The first step toward mastery learning is to change how we define success. The broken default pattern is what we call the Cram-Pass-Forget Cycle: students cram for tests, pass them, and forget most of what they crammed in just a few weeks. Success in such an environment is a matter of jumping through assessment hoops. Students are not only cheated by this regimen, they are bored with it, and teachers are demoralized by the results.

CRAM PASS FORGET TM

By contrast, Centripetal Press advocates methods and curriculum designed to promote proficiency and long-term retention using a Learn-Master-Retain cycle. This first involves culling the content scope to an amount that truly can be mastered in the course of a school year. Many educators unthinkingly prioritize quantity over quality. But we believe students should be presented with an amount of material they can learn deeply rather than a bloated scope of content they will neither comprehend nor remember. Even with a reduced scope, students who study for mastery typically outperform their peers as they move to higher-level classes. Second, leading students to mastery and retention requires teaching methods designed to RETAIN produce these results. The standard approach used today involves teaching a chapter, giving a test on the chapter, and moving on. By contrast, pedagogy designed for mastery and retention inMASTER volves continuous review, ongoing accountability for previously studied material, and embedding of basic skills into new material. Our proven method makes ample use of innovative strategies to enable LEARN students to master course content. TM

Our Textbook Design Philosophy We believe in the power of beauty to help cultivate the minds of our students. Accordingly, we take great pride in producing texts that are beautiful as well as practical. Each of the texts published by Centripetal Press is designed according to the following principles: • The texts are elegant. We use attractive color pallets, tasteful fonts, and a simple, uncluttered page layout to produce texts that are as pleasing to look at as they are to read. • The texts are designed to be read, not to distract. It seems popular today to design text pages to look like website pages, full of eye-popping features and pop-ups. The trouble with this is that websites are designed to distract the viewer and shift the viewer’s attention from one place to another—not an appropriate design goal for textbooks. By contrast, we use a clean and simple page layout that encourages reading and does not distract. • Graphics are chosen to serve a purpose. We believe graphics should serve and be relevant to the text, while also beautifying the page. We avoid the common practice of including irrelevant images just to liven things up. We believe that studying the natural world is inherently interesting and does not need to be supported by texts that try to make science look “cool.” • Our texts treat students with dignity. We believe the use of silly or childish content is demeaning to students, particularly at the secondary level. Most adolescents look forward eagerly to the day when they can be treated as adults. Instead of trying to amuse students with childish content, we seek to draw them upward to the adult world of scientific study. • Centripetal Press texts are slim and light-weight. Our mastery-learning approach entails focusing on essential content and going deep rather than wide. One of the happy consequences of this approach is that our texts have about 350–400 pages of content instead of the more typical 600–800, and they weigh in at just over two pounds instead of six or eight. These texts are slim, easy to carry around, and easy on the backpack.

The Centripetal Press Curriculum Our curriculum is designed to support a “physics-first” approach in which all students take an introductory physics course in 9th grade. Additionally, we advocate supporting two separate pathways (i.e., “tracks”) of programming in high school—one for grade-level students and one for accelerated students.* Our ideas about these course sequencing issues are explained in a blog post here, as well as in John D. Mays’s book, From Wonder to Mastery: A Transformative Model for Science Education.

MATH

prealgebra

algebra

geometry

algebra 2

AP statistics

precal

calculus

adv precal

AP calculus

gen chem

A & P or env sci

prealgebra algebra

geometry

physics

SCIENCE

physical

algebra 2

biology

earth

adv physics accel phys/chem

adv chem

adv biol molec biol

The table below lists the texts Centripetal Press has available currently or in the near future: TITLE

SUBJECT

TARGET GRADE

AVAILABILITY

Physical Science

currently available

Planet Earth: Land, Water, Sky

Earth Science

currently available

Introductory Principles in Physics

Physics

9 (grade-level)

currently available

Accelerated Studies in Physics and Chemistry*

Physics/Chemistry

9 (accelerated)

currently available

Principles of Chemistry

Chemistry

11 (grade-level)

currently available

Accelerated Chemistry*

Chemistry

10 (accelerated)

currently available

General Biology

Biology

10 (grade-level)

planned for 2022

Physics: Modeling Nature*

Physics

12 (accelerated)

planned for 2021

*Titles marked with an asterisk are materials within the Accelerated Track.

Support Resources Centripetal Press has created and developed unique resources for each course. Common elements include: • Student Edition To promote mastery and long-term retention, Centripetal Press textbooks feature subject matter that has been carefully curated and limited to the central, most important material for each course. Each text also includes specific learning objectives (supplied at the beginning of each chapter); clear, proven problem-solving strategies incorporated into all example problems; and answer keys to all computations in the text. The student editions integrate connections to history, mathematics, language, and philosophy, while regularly inviting students to pause and consider the sophistication of design in nature, the regularity of natural law, the human ability to model nature mathematically, and the fitness of the earth as a dwelling place for mankind. • Digital Resources A variety of digital resources are available to help teachers conduct class, including chapter exams and quizzes, review guides, semester exams, answer keys to all computation questions on the quizzes and tests, sample answers to verbal questions, recommendations and schedules for teaching the course, and more. (Contents vary per course.) • Experiment Manuals Centripetal Press aims to help educators implement an excellent science education program, which includes providing students with a robust lab experience. Using the experiment manuals, high school students will learn the background and objectives of an experiment, how to analyze results, and how to write a premier lab report. (For middle school texts, the experiment instructions are supplied as part of the digital resources.) • Solutions Manuals A useful resource for students, the solutions manuals feature stepby-step solutions, showing how each problem or calculation from the text is solved. • The Student Lab Report Handbook: 2nd Edition Enhance your student’s studies with this concise guide to creating premier lab reports from scratch. This handbook covers syntax and word choice, technical expression, data analysis, error prediction, creating graphs in Excel and Numbers, and more, making it an excellent tool to use with any Centripetal Press high school science course. We recommend supplying a copy of this resource to each student in their freshman year so they can use it during all four years of high school.

Contact Us Our team is here to assist as you choose curricula that will be the best fit for your classroom. We encourage you to take advantage of the following resources: Questions & Guidance We look forward to answering any questions you might have regarding our curriculum, placement, or pricing. Please contact us! Email: info@classicalsubjects.com Phone: (717) 730-0711 Quotes and Vendor Applications We are able to provide formal quotes for all of our materials, and can also submit applications required to become an approved vendor for your school district. Inquire at info@classicalsubjects.com. Get the Latest Information from Centripetal Press and Classical Academic Press For more information on Centripetal Press products, visit www.centripetalpress.com. Get the latest information on new products and support information at the QR link below, or at www.classicalacademicpress.com/pages/schools-charter.

Physical Science

PUBLISHED BY CENTRIPETAL PRESS Grade Level: 6-8th Grade Yearlong Course

CLICK TO VIEW IN OUR ONLINE CATALOG

PHYSICAL SCIENCE There are a great many physical science texts on the market today, but Physical Science from Centripetal Press is remarkably different. Here we highlight a few distinctive features of this text. •

Physical Science supports our signature philosophy of science education based on wonder, integration, and mastery. We always seek to build on and stimulate the student’s innate sense of wonder at the marvels found in nature. Integration refers to the epistemology, mathematics, and history embedded in the text, and a curriculum designed to help develop the student’s facility with using language to communicate scientific concepts. Physical Science is designed to be used in a mastery-learning environment, using the mastery teaching model John developed and describes in From Wonder to Mastery: A Transformative Model for Science Education. When teachers use this mastery-learning model, the result is high student achievement and exceptional long-term retention for all students.

• The chapter topics are presented in a unique and specialized sequence and structure. The chart below illustrates the relationship between the book’s 14 chapters. At the most basic level, the design philosophy behind the sequencing of topics in Physical Science is to first treat the most fundamental topics in a broad overview, and then to come back to these topics to study the details (including computations at an age-appropriate level). 10 Force and Motion

9 Properties

11 Compounds 12 Waves, Sound 13 Electricity and Reactions and Light

14 Magnetism

8 Measurement and Units 7 Science, Theories and Truth 5 Forces and Fields 1 Matter

6 Substances 2,3 Energy

4 Order in Nature

1.1 The Three Most Basic Things

In addition broadly double-looped structure, there are several key In additiontotothis this broadly double-looped structure, there are aother several structural elements in the book’s design. One of these pertains to the presentation other key structural elements in the book’s design. One of these pertains of Since explaining the definition of energy iswhat essentially (at any to energy. the presentation of energy. Since explaining energyimpossible is is essentially age level), we present it through a discussion that relates the major forms of enerimpossible (at any age level), we present it through a discussion relating the gy to our major sources of energy. This places an important technology application major of energy to our major of sources of energy. This places an imright in forms the middle of the broad overview the “three most basic things” covered in portant instance of technology application right in the middleand of interesting. the broad the first four chapters, and makes the energy presentation practical ofkey thestructural “three most basic things” covered in the fi4rst chapters, overview Another element is the positioning of Chapters andfour 7. These chapters crucial understanding the nature of the world live in and the nature and are makes thefor energy presentation very practical andweinteresting. of scientific knowledge (epistemology). we could callof these “meta-chapters,” Another key structural elementAsissuch, the positioning Chapters 4 and 7. by which we mean that their topics inform all of our thinking about standard science These chapters are crucial for understanding the nature of the world we live

in, and the nature of scientific knowledge (epistemology). As such, we could call these “meta-chapters,” by which2 we mean that their topics inform all of our thinking about standard science content. The topic of Chapter 4 is the third of the trilogy of “basic things”—the natural orderliness we call the laws

PHYSICAL SCIENCE content. The topic of Chapter 4 is the third of the “three most basic things”—the natural orderliness we call the laws of nature. The material on Science, Theories, and Truth (epistemology) is placed in Chapter 7 strategically to follow the broad overview of Chapters 1–6 , and provide reference for subsequent chapters and course discussions With the broad overview of fundamentals and meta-topics covered in the fall term, the spring term opens with an entrance into the use of mathematics in science. Chapter 8 on Measurement and Units is another meta-chapter that lays the foundation for computational applications. An additional benefit of waiting until the spring term to introduce math is that students will have had an entire semester of prealgebra by then and will be prepared to tackle the metric prefixes and unit conversions. The structural principles of this text are illustrated in the graphic above, with fundamental topics at the bottom and more detailed topics toward the top. While Chapters 4 and 7 influence and support all levels, Chapter 8 points to later chapters with mathematical content. •

The text includes 12 experiments, nine of which are quantitative. Quantitative experiments—those involving measurements, data, and quantitative analysis—are a hallmark of the physical sciences. We assume that students in 7th grade have little prior experience tabulating and graphing data, so the text includes a primer on graphing data immediately before the first experiment. The design of these experiments enables students to engage in quantitative laboratory work (measurement, data collection, graphing, analysis) well before the mathematical topics are introduced in the text (Chapter 8). All the experiments were specifically designed for this course and many make use of everyday materials such as Hot WheelsTM cars, golf and ping pong balls, magnetic compasses, StyrofoamTM cups and coolers, batteries, and lightbulbs.

Support Resources Three major support resources are available to accompany Physical Science. They include: Digital Resources The following materials are accessible exclusively through Centripetal Press: • a full year of weekly, cumulative quizzes • two semester exams • an 87-page resource manual for the 12 experiments • a document containing all answer keys • a full year of Weekly Review Guides • a document with recommendations for teaching the course • a lesson list and example calendar Special Parts Kit This experiment kit is a handmade tool that makes use of four special jigs made of wood or metal. We encourage educators to get the students involved in making the jigs if required facilities and expertise are available, and full construction details are included in the experiment resource manual. However, if making the jigs is not an option, sets may be purchased from Classical Academic Press here.

PHYSICAL SCIENCE Tips and Tools A variety of tips and tools are available at www.centripetalpress.com. Here, teachers will find scores of images and short videos organized chapter by chapter. Teachers are free to use any of this content as they wish to assemble vivid and fascinating lessons.

About Centripetal Press Centripetal Press is an imprint of Classical Academic Press that focuses on the highest-quality science curriculum for charter schools. Founded by educator John D. Mays, Centripetal Press offers a completely unique approach to secondary science curriculum. For more information on our vision for secondary science education, please refer to the Overview section on pg. 2 in this series of samples.

Sample Chapters The following pages contain samples from the text. The Table of Contents is shown, as well as Chapters 1, 3, and 4, the graphing primer and the first two experiments.

Physical Science A Mastery-Oriented Curriculum

John D. Mays

CENTRIPETAL PRESS Austin, Texas 2017

© 2017 Novare Science & Math LLC All rights reserved. Except as noted below, no part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by information storage and retrieval systems, without the written permission of the publisher, except by a reviewer who may quote brief passages in a review. All images attributed to others under any of the Wikimedia Commons licenses, such as CC-BY-SA-3.0 and others, may be freely reproduced and distributed under the terms of those licenses. Published by Centripetal Press centripetalpress.com

CENTRIPETAL PRESS

Printed in the United States of America Second printing, with new preface, 2019 ISBN: 978-0-9981699-4-1 Centripetal Press is an imprint of Novare Science & Math LLC. Cover design by Nada Orlic, http://nadaorlic.info/ For a catalog of titles published by Centripetal Press, visit centripetalpress.com.

Contents

Teacher Preface

Chapter 0

Start Reading Here

Chapter 1

Matter and Atoms

The Three Most Basic Things Atoms Electrons The Development of Atomic Theory

3 5 8 9

1.1 1.2 1.3 1.4

Chapter 2 2.1 2.2 2.3

Sources of Energy

What Is Energy? Where Is the Energy? Sources and Forms of Energy Electromagnetic Radiation Kinetic Energy Potential Energy Gravitational Potential Energy Chemical Potential Energy

xvi

14 15 16 20 20 24 26 27 28

Contents

Thermal Energy Nuclear Energy

32 33

Getting Started with Experiments

Experimental Investigation 1: Kinetic Energy

Chapter 3

3.1 3.2 3.3 3.4 3.5 3.6

Conservation of Energy

The Law of Conservation of Energy Mass-Energy Equivalence Heat and Heat Transfer Work Internal Energy Summary: Where Is the Energy?

45 48 49 52 55 57

Experimental Investigation 2: Heat Transfer by Conduction

Chapter 4

4.1 4.2 4.3

Why Are There Laws of Nature? Order and Structure in Nature The Remarkable Universe Conditions Necessary for Life Other Remarkable Observations Conditions Necessary for Exploration

Chapter 5 5.1

5.2

Order in Nature

Forces and Fields

The Four Forces The Gravitational Force The Electromagnetic Force The Strong Nuclear Force The Weak Nuclear Force Three Types of Fields The Gravitational Field The Electric Field The Magnetic Field

63 65 69 69 71 73 76 77 78 80 81 81 84 84 85 88

Experimental Investigation 3: Electrostatic Forces

Chapter 6

6.1 6.2 6.3

Substances

Atoms, Molecules, and Crystals Molecules Crystals The Substances Family Tree Elements

97 97 100 102 103 vii

Physical Science

6.4 6.5

Compounds Mixtures and Solutions

108 110

Experimental Investigation 4: Growing Crystals

116

Chapter 7

Science, Theories, and Truth

118

Science is Mental Model Building The Cycle of Scientific Enterprise Theory Case Study: Part 1 Hypothesis Case Study: Part 2 Experiment Analysis Case Study: Part 3 Review Case Study: Part 4 Facts and Theories Experiments and the Scientific Method Truth Ways of Knowing Truth Direct Observation Valid Logic Divine Revelation Relating Scientific Knowledge and Truth Summary: The Nature of Scientific Knowledge

119 122 122 123 123 124 124 124 125 125 126 127 130 132 134 134 134 135 136 137

7.1 7.2

7.3 7.4 7.5 7.6

7.7 7.8

Chapter 8 8.1 8.2 8.3 8.4 8.5

Measurement and Units

Science and Measurement The International System of Units Metric Prefixes Unit Conversions Calculating Volume

138 139 141 144 148 157

Experimental Investigation 5: Determining Volume

162

Chapter 9

Properties of Substances

164

Physical Properties Temperature, Pressure, and Volume Temperature Pressure Volume Phases of Matter

165 168 168 170 172 175

9.1 9.2

9.3 viii

Contents

9.4 9.5

Calculating Density Chemical Properties

179 185

Experimental Investigation 6: Determining Density

190

Experimental Investigation 7: Heat of Fusion

192

Chapter 10 Force and Motion

194

10.1 10.2 10.3 10.4

A Brief History of Motion Theory Velocity Acceleration Inertia and Newton’s Laws of Motion Inertia State of Motion Newton’s First Law of Motion Newton’s Second Law of Motion Newton’s Third Law of Motion

195 198 203 209 210 210 211 211 216

Experimental Investigation 8: Inertia and Force

220

Chapter 11 Compounds and Chemical Reactions

222

11.1 Tools for Chemistry Chemical Equations Ions and Polyatomic Ions Acids and Bases 11.2 How Compounds Form Main Goal #1 Main Goal #2 11.3 Chemical Reactions Salts Combustion Reactions Oxidation Reactions Redox Reactions Precipitation Reactions Acid-Base Reactions

223 223 225 227 230 230 232 233 233 234 235 236 238 240

Experimental Investigation 9: Observing Chemical Reactions

244

Chapter 12 Waves, Sound, and Light

248

12.1 What is a wave? 12.2 Types of waves 12.3 Common Wave Phenomena Reflection Refraction Diffraction

249 252 256 256 257 260 ix

Physical Science

12.4 Sound and Human Hearing

261

Experimental Investigation 10: Refraction

266

Chapter 13 Electricity

268

13.1 13.2 13.3 13.4 13.5

269 270 275 279 285

The Nature of Electricity Static Electricity How Electric Current Works DC Electric Circuits and Ohm’s Law Series and Parallel Circuits

Experimental Investigation 11: Series and Parallel Circuits

292

Chapter 14 Magnetism and Electromagnetism

296

14.1 Magnetism and Its Cause 14.2 Ampère’s Law 14.3 Faraday’s Law of Magnetic Induction

297 300 304

Experimental Investigation 12: Magnetic Field Strength

310

Glossary

312

Appendix A Making Accurate Measurements

327

A.1 A.2 A.3 A.4

327 328 329 329

Parallax Error and Liquid Meniscus Measurements with a Meter Stick or Rule Measurements with a Triple-Beam Balance Measurements with an Analog Thermometer

Appendix B Percent Difference

330

Appendix C References

332

Image Credits

333

Index

336

Chapter 1 Matter and Atoms

The drawing above is a depiction of a lithium atom. In the center is the atomic nucleus, containing four neutrons and three protons. In the much larger spherical regions surrounding the nucleus are the atom’s three electrons, two in the inner region and one in the larger outer region. To make the nucleus visible at this scale, it is drawn about 2,500 times larger than it should be. If the nucleus were drawn to scale for a diagram of this size, it would be 1/300th the size of the period at the end of this sentence.

OBJECTIVES After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary.

1. Name and briefly explain the three basic things the universe is made of. 2. Describe how the particles in atoms are organized. 3. Describe each of the three basic subatomic particles. 4. Describe the atomic models developed by John Dalton, J.J. Thomson, Ernest Rutherford, and Niels Bohr. 5. Describe the key features that the quantum model of the atom added to correct and complete the Bohr model. 6. State the contributions of Democritus and James Chadwick to atomic theory.

VOCABULARY TERMS You should be able to define or describe each of these terms in a complete sentence or paragraph.

1. 2. 3. 4. 5.

1.1

atom charge electron energy ion

6. 7. 8. 9. 10.

mass matter neutron nucleus orbital

11. 12. 13. 14. 15.

order proton shell subatomic particle volume

The Three Most Basic Things

What are the pages of this book made of? Paper, of course, but what is paper made of? The answer is that paper is made of the fibers from various kinds of plants, including trees. But what are these fibers made of at the most basic level? You probably already know the answer—atoms. The material stuff in the every day world is made of atoms, parts of atoms and a few other strange particles we can’t see. Matter is just our word for substances made of particles that have mass and take up space (have volume). The matter we normally encounter is made of atoms. There are different ways the atoms can be arranged, such as crystals and molecules. There are also different forms matter can take, depending on how hot or cold it is. We will discuss these things in more detail later. Our point here is that matter is one of the three basic ingredients that form the universe we live in. All matter is substance that Matter is one of the basic things the physical has mass and volume. universe is made of. But matter is not all there is 3

Physical Science

in the physical world. Going back to the pages of this book—how did the pages get here? How were they fashioned, printed and bound? And thinking even more deeply, what holds the pages together? Why don’t their atoms fly apart, like spray paint coming out of a can? The answer to these questions relates to energy. The pages of this book were fashioned into their present form through the use of energy. The machines that cut the trees, the factory that made the paper, and all the people involved in making the paper and the book used energy to do their work. But thinking more deeply again, the atoms in the pages are sticking together because of the energy in their attractions for each other. The atoms themselves are held together by energy. Nothing anywhere can happen without energy being involved, and energy itself is what holds everything together. Some scientists are content to say that matter and energy are the two basic ingredients of which the universe is made. However, there is one more basic ingredient that I think should not be left out. I am going to call this ingredient order. For over 100 years, scientists have marveled at the orderliness present at every level in the structure of the universe. Stars aren’t just distributed randomly in space; they are organized into spiral galaxies. At lower temperatures, protons, neutrons, and electrons are tightly organized into very predictable structures in atoms. Molecules in our bodies are part of a colossal protein factory that processes our food, carries energy to our muscles, flushes out waste, operates our many different organ systems, and builds new cells as we need them. The orderliness of the universe allows us to model the behavior of nature using mathematics. We call these models the “laws of nature” or the “laws of physics.” (As we will see in Chapter 7, a better term for these laws is theories.) But the fact that we can model nature with mathematics is simply amazing. Some of the order around us comes from human intelligence. We build houses and machines, write books and Order is present music, craft paintings and sculptures. But as mentioned everywhere in nature. above, there is order at large in the universe, and there appears to be deep mathematical structure in the universe. Many believe—even some nonreligious scientists and philosophers—that the order observed throughout the universe indicates that there is a purpose for the universe and for creatures like us who live in it. Some believe that we create meaning and purpose for ourselves; but others believe that since we humans express desires, intentions, and purposes, these thoughts about purpose must come from a larger purpose for the universe that comes from outside the universe—a power that we speak of as transcendent. There is much food for thought here! The universe is a stunning and wonderful place. So far as we know, human beings are unique creatures. We know of no other species that engages in language, reason, humor, art, love, and religious belief the way we do. There is so much in the world around us to be excited about and

Energy holds everything together and enables any process to happen.

Chapter 1

Matter and Atoms

interested in. How can life ever be boring? There is so much to learn, so much to wonder about! To summarize this first section, all material objects—all matter—is made of atoms. Energy is present in nature, holding everything together and enabling everything to happen. And order is present in the laws of physics and chemistry and in the use of matter and energy to make things. We will consider energy and order in more depth in later chapters. Our task for the rest of this chapter is to consider atoms in more detail. Learning Check 1.1 1. Describe the three basic things the universe is composed of. 2. Give an example of how order is evident in nature. 3. In the list below, some things are clearly the result of the order resulting from human intelligence, and some are the result of the order embedded in the laws of nature. Explain which is the case for each item. › › › › › › ›

1.2

an artist’s painting the arrangement of pieces of confetti on the floor at a party the shape of a wadded up piece of paper the design of the pages in a book the arrangement of all the leaves in a tree the sound of a cello string when it is plucked the arrangement of the keys on a computer keyboard

Atoms

Chances are that you have learned about atoms before, so you may already have an idea of how they are put together. In this section, I am going to describe some basic scientific facts about atoms. Later, we will take electron a brief look at how scientists figured these nucleus things out. Atoms are much too small to see. In order to see an object with our eyes, the object has to be big enough to reflect light waves into our eyes. But atoms are much smaller than the waves of visible light, so they do not reflect the waves, and we cannot see the atoms. What we know about atoms we have inferred from thousands of experiments. To Figure 1.1. An atom with its nucleus of protons infer something is to figure it out from the and neutrons, and electrons in a much larger evidence. If I go out to my car and find the region surrounding the nucleus. The nucleus is

actually much smaller than shown in the picture. 5

Physical Science

back bumper smashed, I infer that someone hit my car, even though I was not there to see it. neutron heaviest Figure 1.1 is a diagram of a small atom, the way scienno charge tists currently understand them. In the center is the nucleus (which is actually much smaller than shown in the figure). Within an atom are three different types of subatomic particles, depicted in Figure 1.2. There are two kinds of subatomic particles in the nucleus: protons and neutrons. proton positive (+) These particles have almost the same weight, although neucharge trons are a tiny bit heavier. Protons have a property called charge. This property is responsible for electricity and everything electrical in nature. There are two kinds of charge, electron negative (–) which we call positive and negative. (Benjamin Franklin was the first to call charge by these terms, back in the 18th charge Figure 1.2. The three subatomic century.) The charge on protons is positive. Neutrons have particles. no charge. A third particle inside the atom is the electron. Electrons weigh about 2,000 times less than protons. This means that their weight almost does not matter. But what does matter is their charge, which is exactly the same strength as protons, but negative. The electrons buzz around in a sort of layered cloud around the nucleus. More on electrons in a moment. Other than the nucleus and the electrons, the rest of the atom is completely empty space. This is actually a bit mind boggling, so here is an example to help you visualize this. Figure 1.3 shows an engraving of the ancient sports stadium in Rome called the Coliseum. The tiny figures on the ground near the center of the Coliseum are people. Imagine that one of those people has a flower pinned to his lapel. If the nucleus of an atom were the size of the head of that pin, the cloud where the electrons are would be the size of the entire Coliseum! Everything else in the atom is empty space—nothing in there, not even air. (Of course, air is also made of atoms.) Finally, when we speak of an atom alone by itself, we typically assume the atom is electrically neutral. This means that there is no net charge on the atom. The only way this can be is if the atom has an equal number of protons and electrons so that their charges balance out.

Figure 1.3. Engraving of the Roman Coliseum by Giovanni Piranesi. 6

Chapter 1

Matter and Atoms

Scientists, Experiments, and Technology Although atoms are too small to be seen, there are technologies that can make images of atoms we can study. The image to the right was made by a scanning tunneling microscope (STM) and the individual atoms are imaged as little circles. The STM uses beams of electrons reflecting off objects to construct an image of the object’s structure at a very small scale. This image shows the surface of a sample of gold. The atoms inside a sample of gold are arranged in a regular, repeating pattern—rows of atoms without gaps. But at the surface, gold atoms can have a gap between the rows that occurs after every five rows of atoms. In recent years scientists have been learning how to construct materials with very specific arrangements of atoms. A fascinating example is carbon nanotubes, represented in the computer image to the right. These are hollow tubes of carbon atoms with walls one atom thick. The diameter of a nanotube is about seven times the diameter of the carbon atom itself, and the tubes are extremely strong. An STM image of a nanotube is shown to the right.

Physical Science

It’s pretty easy for most atoms to gain or lose electrons. When they do, they are not electrically neutral any more. They have a net charge, either positive or negative. Atoms with a net charge like this are called ions. Ions with opposite charges attract each other. In later chapter, we will see that this is one of the main reasons atoms stick together to form chemical compounds. Learning Check 1.2 1. Explain why scientists can claim that atoms exist, even though atoms cannot be seen. 2. Describe the locations of the three types of subatomic particles found in atoms. 3. Describe which subatomic particles have charge and which do not. For those that do, identify which kind. 4. Compare the weights of the three subatomic particles. 5. Explain what ions are and how they form.

1.3

Electrons

Let’s talk just a bit more about the electrons and how they are arranged inside the atom. First, you need to know that electrons are weird, and it is hard to say just exactly what they are. Even though we refer to them as particles, they are certainly not hard little things like pellets or B-Bs. Electrons sometimes act like tiny particles, but they also sometimes act like waves. This is hard for everyone to understand, but that’s just how it is. We all have to live with the strange properties of electrons and just try to understand them the best we can. Another thing about electrons is that there is no way to know precisely where they are and how fast they are going at the same time. This is why I drew the red streaks around the electrons in Figure 1.1. By showing them as sort of smeared, I am trying to show the uncertainty we have about where they are or how fast they are moving. In an atom, every electron has a very specific amount of energy. The electrons are arranged in the atom according to how much enFigure 1.4. The first five orbitals in an atom. ergy they have. The clouds they buzz around in Each one can hold up to two electrons. are called orbitals or shells. Electrons with the same amount of energy go in the same orbital, but only two electrons can go in each orbital. The large spheres in Figure 1.1 represent the first two orbitals that every atom has. Figure 1.4 shows the shapes of the first five orbitals every atom has, beginning with the two spherical ones. After the first two, the next three are shaped in a double arrangement that looks like a thick hamburger bun. The orbital shapes get even weirder after that, as you may learn later when you take chemistry in high school.

Chapter 1

Matter and Atoms

Learning Check 1.3 1. State at least five facts about electrons. 2. Describe the shapes of the first five electron orbitals in atoms.

1.4

The Development of Atomic Theory

Our theories about atoms have been under development for a long time. Over the centuries, there have been scores of important scientists who contributed key insights to our present theory—or model—of the atom. Here we will look at a few of the most important developments along the way. The ancient Greek philosopher Democritus is usually given credit for first imagining that matter is made of atoms (Figure 1.5). Democritus lived in the 5th century BCE, and proposed that everything was made of tiny, indivisible particles. The word atom comes from the Greek word meaning indivisible. For over 2,000 years after Democritus, nothing much happened to further our understanding of atoms. But then the scientific revolution began to take off, and major developments began to occur regularly. In 1803, English scientist John Dalton (Fig- Figure 1.5. Greek philosopher Democritus. ure 1.6) published the first fully scientific model of the atom. Dalton’s theory included the idea that everything was made of indivisible atoms, as Democritus had said. Dalton went on to say that atoms combine together in whole-number ratios to form the compounds that different substances are made of. Dalton also proposed that atoms are not created or destroyed during chemical reactions, and that every atom of a given element is identical. All the points in Dalton’s theory were either correct or partially correct, and Dalton’s 1803 Dalton’s model was a maatomic model: jor step forward. indivisible particles. Dalton’s model was not correct about the notion that atoms are indivisible. The first news Figure 1.6. English scientist John Dalton. 9

Physical Science

that atoms had smaller Thomson’s 1897 pieces inside them came atomic model: from the work of anoththe Plum pudding er English scientist, J.J. model—a cloud Thomson (Figure 1.7). In of positively 1897, Thomson performed charged material a brilliant series of exwith thousands periments that produced of negatively beams of electrons inside a charged particles glass tube. At the time, no embedded in it. one knew anything about electrons, but Thomson took the bold step of proposing that the beams he had produced were made of particles that came from within atoms. As a result of his work, Thomson proposed a Figure 1.7. English scientist J.J. Thomson. new atomic model, one that everyone now calls the plum pudding model. Now, most American students these days don’t know much about plum pudding, so you can think positively charged of Thomson’s model as the “watermaterial melon model.” As illustrated in Figure 1.8, Thomson modeled the atom as a cloud of positively charged material with thousands of negatively charged particles embedded in it, like a waternegatively charged melon with its many seeds. particles The next scientist to develop a Figure 1.8. Thomson’s “plum pudding” model of the new atomic model was New Zealandatom. er Ernest Rutherford, (Figure 1.9). In 1909, Rutherford was in England exRutherford’s 1909 atomic model: a tiny perimenting with firing small, positively nucleus containing the positive charge charged particles at a thin foil made of and almost all the mass; negative pure gold. This work led Rutherford to electrons surrounding the nucleus; conclude that all the positive charge in most of the atom is empty space. an atom is concentrated in the center, not spread out as Thomson had proposed. Rutherford called this central concentration of charge the nucleus. Rutherford also proposed that the electrons discovered by J.J. Thomson were outside the nucleus, surrounding it, and that most of the atom was empty space. As you can see, with Rutherford’s work we have come a long way towards understanding atoms, and we now have a general idea of how they are structured. The 10

Chapter 1

Matter and Atoms

next development was put forward by Danish physicist Niels Bohr in 1913 (Figure 1.10). Bohr was the first to propose that the electrons were orbiting the nucleus like planets orbiting the sun. As depicted in Figure 1.11 on page 12, Bohr theorized that the orbits represented different “energy levels” for electrons. Lower energies were closer to the nucleus and higher energies were farther out. The lower-energy orbits would fill up first. The lowest orbit could hold two electrons. Orbits two and three could each hold eight electrons, and there were higher-energy or- Figure 1.9. Physicist Ernest Rutherford, from New Zealand. bits after that. Bohr’s model was very successful at explaining atomic behavior. However, it soon became clear that the electrons aren’t exactly orbiting. Instead, an electron sort of zooms around—at extremely high speed—in a three-dimensional cloud defined by how much energy the electron has (as we saw back in Figure 1.4). And as we saw before, it is difficult even to think of electrons as particles at all, since they also have wave-like properties. One scientist said that since we don’t really know what electrons are, we should just call them slithy toves. And when we talk about what they do, we can just say they gyre and gimble in the wabe!1 Our short history of the atomic model would not be complete without mentioning the discovery of one final important piece to the puzzle. In 1932, English scientist James Chadwick discovered the neutron. Scientists Figure 1.10. Danish physicist Niels The quantum model: already knew that nearly all Bohr. Building on Bohr— of the atom’s mass was in the electrons reside in nucleus, along with all the positive charge. But what they orbitals of various knew about mass and charge didn’t match up until Chadshapes nested wick demonstrated that there were electrically neutral pararound the nucleus. ticles, also in the nucleus, that had almost the same mass The 1913 Bohr model: Building on Rutherford—electrons orbit the nucleus like planets. The energy of the electron determines its orbit, and only fixed energies are possible.

1 In case you have forgotten, these terms are from the poem “Jabberwocky,” in Lewis Carroll’s Through the Looking-Glass, and What Alice Found There. 11

Physical Science

h ig he

ro rbi

(slightly more) as the protons. With electrons Chadwick’s discovery our basic understanding of the atom was complete. Scientists learned much more about atoms from experiments conducted throughout the 20th century. nucleus Our current model of the atom is called the quantum model. As I describe at the beginning of the chapter, nucleus the quantum model places the electrons in orbitals, rather than orbits. The quantum model describes the shapes of all the orbitals, and the rules governing which electrons go where among an atom’s orbitals. These rules Figure 1.11. Bohr’s planetary atomic model. are at the heart of chemistry, which is all about how the electrons in atoms interact with each other. We will leave the rest of the details of the quantum model for another science course. Learning Check 1.4 1. Describe the atomic models proposed by Dalton, Thomson, Rutherford, and Bohr. 2. Describe the additional features included in the quantum model. 3. Why did one scientist use the silly language of “Jabberwocky” to describe electrons?

Scientists, Experiments, and Technology The particles used by Ernest Rutherford to explore the thin gold foil in his research are called alpha particles. Alpha particles (often written as α-particles) are a form of nuclear radiation. Each alpha particle contains two protons and two neutrons. Alpha particles are emitted naturally as radioactive substances go through the process called nuclear decay. The image on the opposite page depicts a large nucleus emitting an alpha particle during nuclear decay. When this decay happens, the alpha particles typically exit the atomic nucleus at a speed of 15,000,000 meters per second! Alpha particles are sometimes used in common technologies such as smoke detectors. But I would rather tell you about a new technology being

Chapter 1

Matter and Atoms

Scientists, Experiments, and Technology (continued) explored by researchers today. This technology is a cancer treatment called unsealed source radiotherapy. The idea is to make use of the fact that though alpha particles will damage tissue, they do not penetrate the tissue very deeply. In unsealed source radiotherapy, small amounts of a radioactive substance are introduced into the body and directed near the site of a cancerous tumor. As the radioactive substance decays, the alpha particles it emits bombard the tumor and destroy it. Of course, the healthy tissue surrounding the tumor is also hit by the alpha particles. But because the alpha-particles do not penetrate the surrounding tissue very far, the healthy tissue is damaged only slightly. The damaged tissue will heal. The important thing is that the life-threatening cancer is destroyed. Chapter 1 Exercises Answer each of the questions below as completely as you can. Write your responses in complete sentences. 1. Describe J.J. Thomson’s contributions to the development of the atomic model. 2. When scientists say that atoms are mostly empty space, what do they mean? (How empty are they, and what’s in the empty part?) 3. Describe the particles found in the nucleus of atoms. 4. What determines where the electrons are in an atom? 5. Why was John Dalton’s atomic model so important? 6. Describe the three basic ingredients the universe is made of. 7. What are some of the properties of electrons? 8. What are some examples from nature that demonstrate that order is an important aspect of the natural world? 9. Describe the atomic model proposed by Ernest Rutherford. 10. What are orbitals?

Physical Science

Getting Started with Experiments The Art of Experimental Science Experimental research is one of the things that makes science so interesting and so much fun. Science is a lot more than just learning things in books. Throughout the history of science, new discoveries have been made and new theories have been tested in the laboratory. If you are the type of person who loves fooling around with parts, wires, wood, and chemicals, then the experiments in this book will be right up your alley. If you are the type of person who would prefer to stay inside where it is air conditioned and drink tea, then the experiments will give you an opportunity to get your hands dirty. Who knows, you may find you love doing experiments! History is full of people—both men and women—who helped in a lab when they were 13, and ended up becoming experimental chemists or physicists. It could happen to you. In the next few pages we will look at some of the important things to keep in mind while doing experimental work. I will conclude this introduction with a tutorial on preparing scientific graphs. Safety Safety is a major concern in any science laboratory, whether that lab is in a classroom, a research facility, your kitchen, or your garage. Here are some standard safety rules you should know and always follow: 1. Always wear safety goggles or safety glasses when heating substances on a hot plate or over a flame. 2. When handling hot substances or apparatus, use tongs or wear thermally protective gloves. 3. Use great care when handling glassware. As I always say, there are three ways to break something—improper procedures, silliness, or carelessness— and all are bad in a lab! 4. Wear protective eyewear and gloves when handling hazardous materials. 5. Always work under the supervision of a responsible and knowledgeable adult when using sharp tools, hot plates, flames, or chemicals. 6. Make sure you have a phone in your work area in case you ever need to call for help.

Getting Started with Experiments

7. Always follow written procedures, and don’t take short cuts. Do not revise procedures to suit yourself without consulting with a responsible and knowledgeable person who knows about the kind of work you are attempting to perform. 8. Never taste things in a science lab unless your instructor directs you to. 9. Always keep long hair tied back out of the way, and don’t wear loose, blowsy, or baggy clothing while working in a lab. 10. Make sure you have adequate ventilation. 11. Make sure you have a fire extinguisher in your work area. 12. Exercise care in everything you do, pay attention, and avoid horseplay. Care and Accuracy It is common for those new to experimental work to underestimate just how careful one needs to be in order to get accurate results from a science experiment. Students’ results are sometimes so inaccurate that they are useless, often requiring the students to do the work over again. I have already mentioned that carelessness can be a safety hazard. But carelessness can also result in equipment that doesn’t work properly, results that don’t turn out correctly, or data that are useless. If your experimental data aren’t any good, then you have wasted your time, and possibly your teammates’ time as well. So from the very beginning, make it your goal to follow directions carefully, to assemble apparatus with care and patience, and to record measurements with as much accuracy as possible. Resist the temptation to consider hastily or carelessly performed work as “good enough.” Developing a passion for care and accuracy makes a big difference in the quality of your results. And when your results are superior, you learn more and you end up finding scientific experiments much more satisfying. Doing It Over Is Okay Sometimes, even when you are being as careful as you know how to be, experiments don’t turn out the way they are supposed to. Welcome to the world of the scientist! How many thousands of light bulbs was it that Thomas Edison made before he found one that worked? Life in the science lab means sometimes things don’t work. So let me offer you this advice and encouragement. When things don’t work out right, just try to figure out how to improve your method and do the experiment, or part of it, over again. This may not be convenient; everyone is busy. But it is the right thing to do. If you have no idea what went wrong, then get some advice from someone who can help you figure it out. Keeping a Lab Journal Every practicing scientist maintains a lab journal, a written record of everything he or she does in the lab. As a science student, it is very important that you learn how to maintain your own lab journal, and that you faithfully document your work in it. Here’s why lab journals are important in the real world:

Physical Science

• When work is being passed from one researcher to another, the journal is a record of what has been done in the past and how it was accomplished. • When particular methods, equipment, or procedures used in the past need to be used again, the details are all in the lab journal. • When people become famous, apply for patents, win awards, and so on, all the background information folks need in order to verify the work or write the scientist’s biography is all there in the lab journal! You may not invent something that needs to be patented, but you will need to write reports on your experiments, and to do that you need a record of what you did, who helped, when it happened, and what equipment you used. You also need a place to record your data. So when you do experimental work, always document everything in your lab journal. (Your teacher may even grade you on how well your journal is kept.) Here is some advice about lab journals: 1. Keep your lab journal very neat and well organized. Don’t doodle in it, draw in it, or mess it up. 2. Don’t use a spiral notebook. Use a bound composition book with quadrille (graph) paper. (Quadrille paper makes it easy to set up tables and graphs.) Acceptable journals are the National 53-108, Mead 09100, and others available online and at office supply stores. 3. Put your name on it in case you misplace it. 4. For every experiment you work on, enter the following information: • the date (always enter the date again every day you work) • the names of team members working with you (enter these also every day you work, so you have a record of who is there and who is not each time you meet) • a complete list of all equipment, apparatus, materials and supplies you use in conducting the experiment • the manufacturer and model number for any electronic equipment you use • tables with all your data, with the original units of measure • calculations or unit conversions you perform as part of the experiment • observations or notes about anything that happens that you may need to write about in your report or remember later, including records of work that has to be repeated and why • methods or procedures you use, and the reasons for using them There are other items you can enter in your journal that become more important as you get into high school and college (such as sources, contacts, and prices for special

Getting Started with Experiments

chemicals or parts you have to order), but the list above should cover the things that you need to worry about for now. Take pride in maintaining a thorough lab journal. Make it a habit always to have it with you when you work in the lab, and always to document your work in your journal. Scientific Graphs Graphs are extremely important in reporting scientific information. Often, a graph is the best way to present scientific information so that the reader can examine it and understand it most easily. Because scientific graphs are so important, many of the experiments in this text require you to prepare a graph for displaying your experimental results. Depending on what grade you are in, you may have already learned about using Cartesian coordinates in graphs in your math class. If so, then learning how to prepare a proper scientific graph will be easy. But here I assume that some of the students using this book haven’t studied graphing yet and I describe how to prepare scientific graphs in some detail. Even if you already learned how to make graphs in your math class, there are still many specific details about scientific graphs that you need to know. So read this section carefully. We begin by considering an example experiment and some data from that experiment. We will use these data to illustrate how to set up and format a scientific graph. You may know that automobiles use water to keep the engine cool. You may also know that to keep the water from boiling in the summer or freezing in the winter, a product called antifreeze is mixed in with the water Amount of Boiling point in the car’s engine. I tested mixtures of water antifreeze, by (°C) and antifreeze to see what the boiling point volume (%) would be with different amounts of antifreeze mixed in. What I found is shown in the table 0 101.0 to the right. (You probably know that plain wa10 102.8 ter boils at 100°C. But you may not know that 104.3 common thermometers are only accurate to 20 +/– 1°C. This means that the reading might be 30 105.1 off by one degree too high, or one degree too 40 106.6 low. That is probably the explanation for why 110.0 my boiling point was recorded as 101.0°C with 50 plain water.) In this experiment there are two quantities we call variables. These variables are the amount of antifreeze in the mixture, and the boiling point of the liquid. Notice that it makes sense to think of one of these variables as depending on the other. I select the amount of antifreeze I put in the mixture and the boiling point that results depends on my selection. In the context of graphing scientific information, the variable the scientist selects values for is called the independent variable. In my example experiment, the amount of antifreeze in the mixture is the independent variable. The variable that depends on the scientist’s selection is called the dependent variable, and in the example this is the boiling point. A basic graph is a grid with a horizontal line across the bottom and a vertical line down the left side, as shown at the top of the next page. The two lines are called axes, and they are used as scales for locating the values of the variables for each data point (each row in the data table).

Physical Science

boiling point (°C)

On a graph, we associate the independent variable with a scale marked out on the horizontal axis of the graph. We associate the dependent variable with the vertical axis of the graph. The first thing we have to do to set up the graph is decide what scales to use. Look again at the data values in the data table. As you see, the values for the independent variable, the antifreeze concentration, range from 0 to 50. We want to pick a scale that comfortably covers this range of values without having a great deal of excess space on either end. I am going to choose a scale that starts at 0 and goes up to 60, just a bit higher than the highest value. From the table, the values for the dependent variable, the boiling point, range from 101.0 to 110.0. So I will scale my vertical axis from 100 to 112. When I label the axes with these two scales, the graph looks like the illustration below. (Most of the time in math your axis scales probably start at zero. Well here’s a little secret: they don’t have to!) The next thing we need to do is 112 label each of the axes. The label must contain the variable associated with 110 axis for dependent variable values that axis, and the units of measure 108 that go with it. In the third illustration I have added these labels. Notice that 106 I placed the units of measure in pa104 rentheses. This is one of the standard axis for independent variable values 102 methods of formatting the units and is the method you should use for now. 100 0 10 20 30 40 50 60 Notice also that there are no capital letters in my labels. This is traditional for scientific graphs, and is the formatting I prefer for my students (although some scientific publications are now capitalizing the variable names). Now we are ready to locate each of the data points from the 112 table on the graph. To do so, do 110 the following for each of the data 108 points (rows) in the data table. First, find the value of the inde106 pendent variable on the horizontal 104 axis, and using a ruler and pencil, draw a light vertical line from there 102 up into the graph. Next, locate the 100 value of the dependent variable on 0 10 20 30 40 50 60 amount of antifreeze, by volume (%) the vertical axis, and, using a ruler, draw a light horizontal line from there into the graph. Where these two lines meet is where you place some kind of symbol to represent that data point. In the next illustration, I depict this process for the third data point using dashed red lines. The first two data points are already shown on the graph.

Getting Started with Experiments

boiling point (°C)

Repeat the data point location process until all the data points 112 from the data table are accurately 110 located on the graph. Then connect each of the data points with 108 a straight line, drawn with a ruler. 106 The completed graph looks like the final illustration below. 104 There are a few final points 102 to make. First, the type of graph I have shown how to make in this 100 0 10 20 30 40 50 tutorial is called an x-y scatter plot, amount of antifreeze, by volume (%) or simply scatter plot. There are many other types of graphs, but this one is the most important to know about. Scatter plots are used all the time in science to present data and other types of scientific information. Second, if you are going to place your graph into a report, then your graph needs a title. Standard formatting for titles is to capitalize only the first letter 112 of the first word, and to place a pe110 riod at the end. Third, until you get to some 108 very fancy math in college, chances 106 are that the scales on your graphs need to be linear. This means 104 that the scales need to be marked 102 in round numbers and regularly spaced. On my graph, the hori100 0 10 20 30 40 50 zontal axis scale is marked in tens, amount of antifreeze, by volume (%) spaced two lines apart. The vertical graph is marked in twos, two lines apart. The regular spacing is crucial in order for the graph to display the correct relationships between the data points. Fourth, you don’t always have to connect the dots in a graph, but it is common to do so. Fifth, if you are displaying more than one data set on the same graph, you need to use different symbols or colors for the dots and lines of the different data sets. We run into this in Experimental Investigation 2. Finally, in high school you should begin learning how to prepare graphs on a computer. For now, your instructor may prefer that you concentrate on learning how to draw nice looking graphs by hand. (I think that is appropriate for middle school students.) But when drawing graphs by hand, make sure you a) use a ruler for the axes and for connecting data points, b) make the graph as neat as you are capable of making it, and c) locate your data points very accurately. Now it’s your turn! boiling point (°C)

Physical Science

Experimental Investigation 1: Kinetic Energy Overview • Using a Hot Wheels car and a ramp of Hot Wheels track, release the car from a standard height repeatedly, adding weights to the car each time to increase its mass. Measure the car’s mass before each run. • The goal of this experiment is to determine how the kinetic energy of the car varies as its mass changes, assuming the car’s speed at the bottom of the ramp stays the same. • Use an assembly of friction flaps and a measuring rule to slow the car and measure how far it travels while stopping. Use the stopping distance as a measure of the kinetic energy the car has at the bottom of the ramp. • Verify that the speed of the car is not affected by mass by releasing two cars with different masses together. • Collect mass and distance data, and prepare a graph of stopping distance versus mass. Basic Materials List • • • •

Hot Wheels cars and track lead weights (split-shot fishing sinkers) mass balance and measuring rule friction flap assembly of 3 × 5 index cards, and apparatus for adjusting its height • apparatus for mounting the track In the discussion of kinetic energy, you learned that the kinetic energy of a moving object depends on both the object’s mass and its speed. In this investigation, we examine how increasing an object’s mass increases the energy the object is carrying in its motion. This experiment will be fun because we get to use Hot Wheels cars! Your setup for this investigation is similar to the one shown above. We start the car rolling by allowing it to roll freely down a track formed into a ramp. This is how the car gets its kinetic energy—the gravitational potential energy at the top of the ramp is converted into kinetic energy at the bottom of the ramp. Our setup allows us to increase the mass—and the kinetic energy—of the car without increasing its speed. We do this by releasing the car over and over from the same height, so that its speed at the bottom of the ramp is always the same (more on this in a moment). But each time we release the car, we increase its mass just a bit by adding small lead weights to the car. To enable this, we use a car with a cargo bed to hold the weights.

Note to Instructors Full details on materials, preparations, and procedures for all the Experimental Investigations are available from the publisher (see Teacher Preface).

Experimental Investigation 1

Kinetic Energy

To get an idea of how much kinetic energy the car has at the bottom of the ramp, we make a device that uses friction to slow the car in a stopping zone. Our friction stopper uses flaps of index card to slow the car, and we gauge the amount of kinetic energy the car has by measuring how far into the stopping zone the car goes before the friction flaps bring it to a stop. The photo to the left shows a car being stopped by the friction flaps. The measuring rule next to the track allows the experimenter to measure how far the car travels into the stopping zone. Let’s go back for a minute to my comment that every time the car is released from the same height it will be going the same speed at the bottom of the ramp. Back in the 17th century, Galileo demonstrated that falling objects all accelerate at the same rate. This means that any object released from a given height above a table will be going the same speed when it reaches the table, regardless of its mass. Accordingly, our car will always be going the same speed at the bottom of the ramp, even as we add weight to it. But rather than take Galileo’s word for it, as part of this experiment you need to verify this for yourself. In the photo, you can see two ramps side by side. Before placing the friction flaps over the track, make a few trials with two cars released from the same height. Use two identical cars, and place weights in one of them to make it about 10–15% heavier than the other. As you do this, see if you can verify Galileo’s discovery. In order to make this verification, you need to work out a way to release the two cars at exactly the same time. For collecting experimental data, you use a single car. Adjust the height of the friction flap assembly so it is level, and so it stops the car before the car reaches the end of the track, even when the car is full of lead weights. Release it 3–5 times from the same height, adding another bit of weight to it each time. Vary the weight from a minimum (car empty), up to at least 130% of the car’s empty weight. For each run, measure the car’s mass and how far it travels into the friction/stopping zone, and record these data in a table in your lab journal. Analysis An important part of the analysis is to prepare a graph of stopping distance (vertical axis) versus mass (horizontal axis). Your instructor will help you with setting up your graph and labeling it properly. In your report for this experiment, address the following questions. 1. Were you able to verify that the car’s speed at the bottom of the ramp didn’t change, even as mass was added to the car? Describe how you did this. 2. How does the stopping distance relate to the car’s kinetic energy? Use your graph to help address this question. Explain what caused the stopping distance to increase. 3. Where does the kinetic energy of the car go as the car stops? 4. What does the shape of your graph tell you about the relationship between kinetic energy and mass?

Chapter 3 Conservation of Energy

Sir Benjamin Thompson, also known as Count Rumford, was an American-born British physicist and inventor. In the late 18th century, Thompson studied the heat produced during the process of boring cannon (drilling the hole down the center of a cannon after it is cast), and concluded that the heat was produced by the motion and friction of the drilling machine. His theory about heat was completely opposed to the prevailing view at the time, which held that heat was a substance inside objects. His ideas were foundational for the formulation of the law of conservation of energy a century later. 44

OBJECTIVES After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary.

1. State the law of conservation of energy and give examples of its application. 2. Explain how friction on a moving object affects the forms of energy present in a given situation. 3. Explain Einstein’s principle of mass-energy equivalence, and use the principle to explain the source of the energy in nuclear reactions. 4. Describe the three ways heat can transfer energy from one substance to another. 5. Explain how the heat radiating from a warm object can be used to determine the object’s surface temperature. 6. Define work, and give examples of situations when work is or is not being performed. 7. Use the concept of internal energy to explain how a substance retains energy within its atoms or molecules.

VOCABULARY TERMS You should be able to define or describe each of these terms in a complete sentence or paragraph.

1. conduction 2. conservation of energy 3. convection 4. fluid 5. heat

3.1

6. heat transfer 7. internal energy 8. mass-energy equivalence 9. mechanical equivalent of heat

10. radiation 11. thermal equilibrium 12. work

The Law of Conservation of Energy

In our energy study last chapter, we saw many examples of energy in one form being converted into another form. Nuclear energy produced by fusion reactions in the sun produces electromagnetic radiation. The gravitational potential energy in dammed-up water turns into kinetic energy as the water falls, and then electrical energy as the water spins a turbine-generator. Chemical potential energy in the molecules of fuel is converted into thermal energy in steam, and then kinetic energy in a spinning steam turbine, and finally into electrical energy by a generator. There are dozens of examples like this of energy in one form being transformed into another. It was during the 19th century that scientists began to solidify the 45

Physical Science

theory known as the mechanical equivalent of heat. According to this theory, mechanical forms of energy (such as kinetic energy) and heat are the same thing—energy. One can be converted into the other, but they are simply different forms of the same thing. At the time, this was a big discovery because for centuries scientists thought that heat was some kind of weightless gas—called caloric—that passed from warmer substances to cooler ones. As usual in scientific research, many experiments exploring this issue were conducted by many scientists, but over time the caloric theory just didn’t hold up. Heat was not a substance or gas flowing out of hot objects. Later in this chapter, we will explore the nature of heat in more detail—how energy is stored inside substances, and how it is transferred from one substance to another. For now all we need to note is that heat is just a form of energy, like kinetic energy and the other forms of energy we have studied. Once the experimentation of the 18th and 19th centuries had made evident the principle of the mechanical equivalent of heat, scientists soon discovered the law of conservation of energy, one of the most fundamental laws in physics. This important law is stated in the following box. The law of conservation of energy: Energy can be neither created nor destroyed, only changed in form.

To illustrate this principle, let’s look at two examples. The first example is a kid sliding down a metal slide, as depicted in Figure 3.1. At the top of the slide, the kid is at rest. His body has gravitational potential energy because of his elevated position in the gravitational field of the earth. As he slides down, the gravitational potential energy converts into kinetic energy as the kid picks up speed. Also, the friction between the kid’s pants and the slide causes heating, and heat is released into the atmosphere. This is because friction on moving objects always causes heating, which always releases energy into the environment. When he reaches the bottom, he has no gravitational potential energy left. The energy he started with has all been converted into heat and kinetic energy. gravitational potential energy kinetic energy

gravitational potential energy

heat from friction

Figure 3.1. Energy transformations as a kid slides down a slide. 46

kinetic energy

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According to the law of conservation of energy, none of the original gravitational potential energy is lost or destroyed, and no new energy is created. The only thing that happens is that the gravitational potential energy the kid starts with converts into different forms of energy. We can even express this in the form of an equation, like this: original gravitational potential energy

final kinetic energy

heat produced by friction

This is a sort of accounting equation. It states the kind of energy we start with and then accounts for where it all goes. The equation says that all the gravitational potential energy the kid has at the beginning is equal to the sum of the kinetic energy he has at the end and the energy released as heat on the way down. Our second example is to track the energy involved in an exploding firecracker, as illustrated in Figure 3.2. Before the explosion, there is chemical potential energy in the chemicals in the firecracker. With the explosion, this chemical potential energy is converted into several different forms of energy, including energy in the light, heat, and sound from the explosion, as well as the kinetic energy in the flying debris as the firecracker blows apart. acoustic energy in the sound of the bang

chemical potential energy in the explosive chemicals

kinetic energy in the flying debris electromagnetic radiation in the light and heat

Figure 3.2. Energy transformations in an exploding firecracker.

As in the previous example, when the firecracker explodes no energy is created or destroyed. The original energy is all there, just in different forms. Writing this in an equation, original chemical potential energy

kinetic energy in flying debris

heat and light

acoustic energy

With one exception, which we examine in the next section, current scientific theory now holds that the law of conservation of energy applies to every process and in every part of the universe. It is one of the fundamental laws of physics. In the next section, we take the law of conservation of energy one step further. 47

Physical Science

Learning Check 3.1 1. What is “caloric”? 2. State the law of conservation of energy. 3. Make up three of your own examples, different from the ones in the text, that demonstrate the law of conservation of energy. 4. When friction is present on a moving object, how does this affect the amount and forms of energy present in a given situation?

3.2

Mass-Energy Equivalence

From the previous chapter, you know that it was Albert Einstein that first theorized that light and energy are quantized. In 1905, when he published his famous paper on the subject, Einstein also published a paper demonstrating that mass and energy are related—equivalent, in fact—and that one can be transformed into the other! This theory is now called mass-energy equivalence. The equation Einstein discovered is one of the most famous equations in the history of science:

E = mc2 In this equation, E stands for energy, m stands for mass, and c stands for the speed of light, which is 300,000,000 meters per second. The equation basically says that there is an equivalent amount of energy associated with any given amount of mass. If you take a certain mass, in kilograms, and multiply it by the square of the speed of light (a very large number), the Mass and energy are result is the equivalent amount of energy associated with related, and one can that amount of mass. be converted into Remember our discussions about fission and fusion in the other. Chapters 1 and 2? In each case, energy is released. Einstein’s equation enables us to understand why. When two protons fuse together during fusion, their mass after they stick together is a tiny bit less than the sum of the masses of the two protons before they fuse together. What happens to the rest of the mass? It is transformed into energy, just as Einstein’s equation predicts. The same thing happens in fission. After a large atomic nucleus has been split apart, the sum of the masses of all the pieces is less than the original nuclear mass. The difference is converted to energy. This is truly amazing. It is also amazing that Einstein formulated this theory long before any experiments could be performed to confirm it. So far as we know, the only exception to the law of conservation of energy presented in the previous section is when mass is converted to energy during a nuclear reaction. But if we take the mass-energy equivalence into account using Einstein’s equation, then the law of conservation of energy applies across the board, without exception. Scientists refer to this as the conservation of mass-energy. 48

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Learning Check 3.2 1. Where does the energy come from in the fission reactors we use to produce electricity? 2. What does mass-energy equivalence mean?

3.3

Heat and Heat Transfer

In the context of science, heat is a technical term with a very specific meaning. The term heat applies specifically to energy that is in the process of being transferred from a warm object to a cooler one. In fact, it is incorrect to refer to an object or substance as containing or possessing heat. You can say that an object has kinetic energy, and you can refer to the internal energy of a substance (which I explain later in the chapter), but heat is a term that refers to energy that is in transit—on its way from one place to someplace else. In this section, we look at the three ways this happens. No doubt you already know that heat always flows from hotter objects to cooler objects. (We are not going to talk about thermodynamics much in this course, but in case you are wondering, it is the second law of thermodynamics that says heat always flows from the warmer object to the cooler one and never the other way around.) Heat continues to flow from the warmer object to the cooler one until they reach the same temperature. When they are the same temperature and heat flow between them ceases, the objects are in a state called thermal equilibrium. There are three ways heat can transfer from one substance to another. First, we have already seen (Section Objects are in thermal 2.3) how heat can travel as electromagnetic waves. Both equilibrium when the light and heat from the sun travel to earth as electrothey are at the same magnetic radiation. When we refer to heat transfer by temperature, and no radiation, this is what we mean. heat is flowing between We can see light, of course, but we cannot see the them. heat from a hot object like the sun because heat is in the infrared region of the electromagnetic spectrum. In Figure 3.3 is a sketch I drew to suggest how heat can radiate from a hot object, such as a horseshoe being heated in a blacksmith’s forge. The red wavy lines are supposed to suggest the heat radiating from that hot horseshoe, but as I said, we cannot see the heat. We sure can feel it on our Heat transfer by skin, though! radiation occurs The second way heat transfer happens is by a process when energy called conduction. Conduction occurs in solids, and of moves by infrared all solids metals conduct heat the most readily. electromagnetic waves. As with radiation, conduction is a very familiar process to all of us. You know that if you put a metal object, such as an iron frying pan, on a fire the handle of the object soon gets hot. 49

Physical Science

Scientists, Experiments, and Technology

radiant intensity (arbitrary units)

All warm objects emit infrared radiation into the environment. In fact, by detecting the wavelength where the strongest radiation is, we can calculate the temperature of the object. This UV visible IR is because at any temperature, a warm object radiates a range of wavelengths, and the radiation is strongest at a par5000 K ticular wavelength for a given temperature. The hotter an object is, the shorter the wavelength of the peak in the radiation the object emits. 4000 K In the graph shown to the left, the black curves show the radiation emitted by a hot object at three different tempera500 1000 1500 2000 2500 3000 tures. The 5000 K curve is at about the wavelength (nm) surface temperature of the sun. The 3000 K curve is a bit below the temperature of the glowing metal filament in a traditional light bulb. Note that the peak in the curves moves to longer—more infrared—wavelengths as the temperature gets lower. 3000 K

In the images below, the upper photo was made with visible light. The lower photo was made with infrared light, and then the different infrared wavelengths were translated to visible wavelengths so we can see the radiation patterns. The temperature scale in the lower photo associates the colors in the infrared photo to the temperature. Notice that the black bag is opaque to visible light, but transparent to infrared. The reverse is true of the man’s glasses. The color of the bag indicates that it is nice and cool—about 75°F. Notice also that the man’s facial hair is the same temperature as his face, so it appears as the same color in the infrared photo. 93.4 90 85 80 75 73.6

The relationship between the surface temperature of an object and the radiation it emits gave birth to a great new temperature measurement technology a few years ago. The hand-held device uses a lens to capture the infrared radiation from an object and display the object’s surface temperature.

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Conservation of Energy

These devices have built-in laser pointers to make it easy to point the device at a distant object, but the laser is not part of the temperature sensing at all. In the photo to the left, a maintenance employee is measuring the temperature of the walls in a building to help assess how the building’s air conditioners are performing. You can see the laser spot on the wall in the background. This happens even if the handle isn’t over the fire. The reason is that when an object gets hot, the atoms inside the object vibrate very vigorously. In a metal, these atomic vibrations in the hot part of the object are transferred to other nearby atoms so that they begin vibrating vigorously, too, which means they are hot, too. The sketch in Figure 3.4 illustrates this process. In the sketch, the blue spheres represent the atoms in the metal. I drew them in a regular, geometric pattern because in metals that’s how the atoms are arranged. Underneath the metal, there is a heat source, such as a gas flame on a stove. As Figure 3.3. Heat transfer by radiation. the heat from the flame warms the atoms in the metal, they begin vibrating much faster. These vibrations are transferred to neighboring atoms, which warms them so they begin vibrating faster, too. By this process of transferring vibrations, the heat spreads throughout the solid metal, although the atoms near the flame are always vibrating the most vigorously, and so are the hottest. The third way heat transfer can occur is by convection. Convection occurs in fluids atoms (liquids and gases). In fluids, atoms are free to move around, mix and mingle, and bump into each other. In hot fluids, the atoms are moving Heat transfer faster and in cool fluids the by conduction atoms are moving slower. occurs when As illustrated in Figure 3.5, vibrating atoms if a hot fluid mingles with in solids cause a cool fluid, the hot, fast heat source nearby atoms atoms begin colliding with to vibrate, too. the cooler, slower atoms. Figure 3.4. Heat transfer by conduction. 51

Physical Science

This mingling causes the hot atoms to transfer some of their kinetic energy to the cooler atoms. In the end, the cooler atoms have been warmed up by having energy transferred to them from the hot atoms. A good example of convection at work is in the old radiators commonly used to heat homes in the days before central heating systems (and still in use in many oldFigure 3.5. Heat transfer by convection. Red arrows represent fast, hot particles. Green arrows represent er homes, parslower, cooler particles. Heat transfer by ticularly in the convection occurs northern U.S.). when hot and cold Hot water is pumped through the inside of the radiator, makparticles in a fluid ing the iron radiator warm to the touch. When air molecules mix and collide, collide with the hot surface, they pick up kinetic energy from causing slower, the rapidly vibrating atoms in the metal. These hot, fast movcooler molecules to ing molecules then gradually work their way through the gain kinetic energy room, colliding with the slower, cooler molecules and exand speed up. changing energy so the cooler molecules begin moving faster, which means they get hotter. Learning Check 3.3 1. 2. 3. 4. 5.

3.4

Why is it incorrect to say that an object has heat in it? What is thermal equilibrium? How does conduction work? How does convection work? Write a paragraph explaining how we can determine the surface temperature of an object by sensing the infrared radiation the object is emitting.

Work

In the previous section, you learned that the term heat describes a process in which energy is transferred from one object to another. The term work is used to denote another such process. Work, like heat, is a technical term with a very specific meaning. Work is a mechanical process of transferring energy from one machine or person to another. When work is performed, a force is applied to an object, and the object is moved a certain distance in the direction of the force. This process always results in energy transfer. Figure 3.6 shows four examples of work being performed. First, when an archer stretches back a bowstring, he does work on the bow. This results in potential en52

Chapter 3

Conservation of Energy

ergy stored in the bent bow. When the archer releases the bow, the bow does work on the arrow by pushing the arrow very hard for a few inches. The result of this work is that the arrow gains kinetic energy. In the second example, in the upper right, some boys are pushing a car up a hill. We say that the boys are doing work on the car. The result is that the car now has gravitational potential energy, since it is at the top of the hill. In the third photo, a white car is giving a push start to a red race car. The race car was at rest, and now is moving, so the white car did work on the red car, and the red car now has kinetic energy. Finally, a construction crane hoists

archer

work

bow

bow now has potential energy

bow

work

arrow

arrow now has kinetic energy

white car

work

red car

red car now has kinetic energy

boys

crane

work

car

bucket

car now has gravitational potential energy

bucket now has gravitational potential energy

Figure 3.6. Four examples of work, in which energy is transferred from one machine or person to another by applying a force and moving a distance. 53

Physical Science

a bucket of concrete up in the air. The crane does work on the bucket, and now the bucket has gravitational potential energy. There is one more thing to notice about work. When work is done, energy transfer occurs. For this to happen, the When work is force doing the work and the distance involved must point performed, an in the same direction or along the same line. If they are at applied force causes right angles to one another, no work is done. an object to move in This is illustrated in Figure 3.7. When lifting a bucket, the same direction. a person does work on the bucket. Energy is transferred in this process because some of the energy in the person’s body (chemical potential energy) is transferred to the bucket, which now has gravitational potential energy. But during the process of carrying the bucket from one place to another, no work is done on the bucket. The bucket is not being raised, so it is not gaining gravitational potential energy. And it is not being accelerated, so it is not gaining kinetic energy (except a little nudge at first to get it going, but that is not what I am talking about here). The bucket is not gaining any energy because the upward force applied to hold up the bucket is at right angles to the distance the bucket is being moved.

F d

Figure 3.7. In raising the bucket (left), work is done on the bucket. In moving the bucket horizontally, no work is done on the bucket.

In this second case, does the person do any work at all? Yes, but not on the bucket. To perform the task of carrying the bucket, the energy transfer is heat from the person’s body into the air. (This is a very complex process, and we cannot say what the work is done on without getting into the biology of the human body, which we are not going to do.) In short, since the bucket gains no energy, no work is done on it. Learning Check 3.4 1. Make up four examples of your own that illustrate one machine or person doing work on another. 2. If you press as hard as you can against a brick wall until you begin breathing hard and sweating, did you do any work on the wall? Explain your answer.

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3.5

In the photo to the left, food items are slowly moving along on a conveyor belt. Is the conveyor belt doing work on the food? Explain your answer. At the right, baggage is moving up a conveyor belt into an airplane. The conveyer loads the baggage on the ground, and transports it up into the airplane. Is the conveyor belt doing work on the baggage? Explain your answer.

Internal Energy

When describing convection in Section 3.3, I noted that the particles in a hot fluid are moving faster than the particles in a cooler fluid. The fact that hotter particles move faster than cooler particles is very significant, and relates fundamentally to the energy in a substance. Every substance is made of atoms, and the atoms of every substance are always in motion. In a solid, the atoms vibrate. In a fluid, the atoms are moving around at high speed. This means the atoms in any substance always have kinetic energy. This kinetic energy in the atoms in a substance enables a substance to absorb energy by being heated. The hotter the substance is, the faster its atoms move and the more kinetic energy the atoms have. The internal energy of a substance is the sum of all of the kinetic energies of the individual particles (atoms or molecules) in the substance. Now, even though the number of particles in any ordinary quantity of matter is colossal, scientists have mathematical techniques that allow them to Internal energy is compute this sum, and thus to determine the internal energy in the sum of all the a substance. The hotter a substance is, the faster its particles are kinetic energies moving and the higher its internal energy is. of the particles in Now that you know about internal energy, let’s consider the a substance. entire picture about the energy in an object as it warms up, how the energy is stored in the object, how the energy is transferred into and out of the object, and so on. Figure 3.8 is a photograph of one side of a hollow brick wall, with bricks on one side of the wall warming in the sun. The other side of these bricks is the inside of the wall, where it stays cool and dark. Let’s use these warming bricks to consider how heat flows through the wall. Refer to the diagram in Figure 3.9 as we go. 55

Physical Science

First, the energy warming these bricks is in the infrared electromagnetic radiation arriving at the bricks from the sun. You might think that some of the energy warming the bricks would be from convection in the heated air, as the hot air molecules hit the bricks and transfer kinetic energy to the atoms in the bricks. But bricks in the direct sun become much hotter than the surrounding air, don’t they? This means that the bricks are actually warming the air, rather than the other way around. On the left of the diagram, the electromagnetic radiation from the sun is shown arriving at the outer surface of the bricks. Okay, so the energy in the infrared electromagnetic waves is absorbed by the atoms on the bricks’ surface, increasing their thermal energy. What does this mean the atoms do? They vibrate faster—because the Figure 3.8. Bricks in the sun on a warm day. radiant heat has been transferred to the atoms in the bricks. Since they are very hot now, they are vibrating vigorously. Then, since the bricks are solid, the atoms in the next layer of brick begin vibrating faster, too, and so on, one layer after another, as the energy works its way by conduction to the inside of the brick. The heat conducts its way to the cool side of the bricks, on the inside of the wall. Now, since the bricks are warmer than the cool air inside the wall, electromagnetic energy radiates from the bricks into the inside of the wall, warming it up a bit. I drew the waves inside the wall shorter (vertically) than the waves hitting the bricks from the sun to show that the heat ra-

outside in the sun

atoms inside the bricks

hollow space inside the brick wall

Figure 3.9. Energy transfer through a hollow brick wall. Energy arrives at the bricks, passes through the bricks by conduction, and radiates out the cooler side. (Note: Since the bricks are hotter than the surrounding air, some energy also radiates from the bricks to the left, back out into the air.) 56

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diating inside the wall is less intense. I also made the waves radiating inside the wall to have longer wavelengths than those arriving outside. The reason for this should be clear from our study of electromagnetic radiation earlier in this chapter (see the box on page 50). Lower temperature means longer wavelength radiation. As we end this section, there is one final point I wish to make about internal energy. If you are thinking that internal energy sounds a lot like the thermal energy we studied in Chapter 2, you are right. The two are very similar and some books treat them as synonyms. The energy in a substance from being heated, which is our definition for thermal energy, is simply the increase in the substance’s internal energy that occurs from the heating. But there is more than one way to increase the internal energy of a substance, so some authorities make this distinction between thermal energy and internal energy: the term internal energy refers specifically to the sum of all the kinetic energies in the particles of a substance. The internal energy can be increased two ways: by heating or by compressing the substance into a smaller volume. By contrast, the term thermal energy refers specifically to the portion of the internal energy that is due to heating. Learning Check 3.5 1. Imagine an unopened can of your favorite soft drink sitting in the sun. Explain what is meant when referring to the internal energy of the liquid inside that can. 2. Think about your can of soft drink again as it sits in the sun. Explain how the liquid in the can gets hot. Don’t be vague; use the terms and concepts we have encountered in the last two sections. 3. Think again about the brick wall we discussed in this section. Consider what changes in Figure 3.9 if the sun is suddenly covered up by a thick black storm cloud. Redraw the diagram, and explain what is going on in this new situation. 4. Building on the previous question and your new diagram, assume the sun never comes back out before nightfall. How long does the situation in your new diagram continue? What if it stays cloudy the next day and the overnight temperature holds steady with no sun for many days. What happens?

3.6

Summary: Where Is the Energy?

In this chapter and the last, we have covered the basics about energy. Now, based on what I have presented, if you think about it, there are really only a few basic ways that energy can be present in substances or in space. Just by way of reviewing and summarizing our study of energy, let’s review these possibilities. The first is in the random motion of the atoms in a substance. If it is a solid, they are vibrating. If it is a liquid or a gas, they are flying around (kinetic energy). The 57

Physical Science

Figure 3.10. Internal energy: The sum of the kinetic energies of the moving particles.

Figure 3.11. Kinetic energy (large scale): Energy in the overall motion of an object.

sum of all these kinetic energies is the substance’s internal energy. The internal energy of a substance depends on its temperature. The hotter it is, the faster its particles move, and the higher its internal energy is. The internal energy of a gas is suggested by Figure 3.10. The second way a substance can possess energy is if the substance as a whole is moving, so that as a whole it has kinetic energy. The substance could be moving in a line, rotating, orbiting or some combination of these. If so, the substance as a whole has kinetic energy. This is depicted in Figure 3.11. The third way a substance can have energy is if the substance is in a field, so that it has potential energy. As suggested by Figure 3.12, we studied the gravitational potential energy present with objects in a gravitational field. The object could also be in some other kind of field, such as a magnetic field or an electric field. If so, then the object may have some kind of potential energy associated with that kind of field. Chemical potential energy is actually due to the energy in electrical fields around protons and electrons in substances. (We will look more closely at electric and magnetic fields later.) Fourth, as Einstein discovered, there is energy associated with the mass in a substance, and during a nuclear reaction some of the mass in an atom is Figure 3.12. Gravitational potential energy: The energy that will be released by objects in a gravitational field if they are no longer held apart. Note: This is the famous and beautiful “Earthrise” photo taken by Apollo 8 astronaut Bill Anders. The moon is in the foreground. On Christmas Eve, during the mission, the Apollo 8 crew read the first 10 verses from the book of Genesis on a national television broadcast. At the time, that broadcast was the mostwatched TV program ever.

Chapter 3

Conservation of Energy

converted directly into energy, as depicted in Figure 3.13. If we were able to tear apart the nuclei in the atoms of a substance, massive quantities of energy would be released. But the only way to get at all this energy would be literally to destroy the substance by tearing apart its very atoms and breaking them down into individual protons and neutrons. Finally, energy is present in the electromagnetic radiation found throughout outer space. This includes the infrared, visible, and ultraviolet radiation coming from the sun and stars, as well as the Cosmic MicroFigure 3.13. Nuclear energy: Some of an atom’s wave Background radiation. mass is converted to energy during nuclear reactions.

Chapter 3 Exercises Answer each of the questions below as completely as you can. Write your responses in complete sentences. 1. Make up a situation in which you start with one form of energy, and some event occurs, causing the energy to be transformed into at least three other forms of energy. Use the law of conservation of energy to account for all the energy in your example. 2. Imagine you are camping, and you are sitting inside your tent in the sun on a hot day. Your tent is dark blue, so not much light comes in, and it is dark in there. Why does it get hot inside your tent? Explain how this happens. 3. Suppose there is a small gasoline engine in the top of a big warehouse. The engine is attached to some pulleys and is used to hoist cartons up to the second and third floors for storage. Explain how work is involved in getting a carton up to the third floor. 4. Consider again the previous question, and suppose you carry a gasoline can full of fuel up to the third floor and fill up the engine’s gas tank. To do this, you must do work on the can of fuel. Does this work energy get converted into the gravitational potential energy of the cartons hoisted to the upper floors? Why or why not? 5. Consider a glass of ice water sitting in a windowless room in a house. The temperature in the room is 72°F. Describe how energy from the room finds its way into the ice cubes in the ice water to melt them. 6. Consider again the glass of ice water in the previous question. Does the idea of thermal equilibrium apply at all here? Explain your answer.

Physical Science

Experimental Investigation 2: Heat Transfer by Conduction Overview • Assemble four containers of water. In a beaker, maintain boiling water at 100°C. In three Styrofoam cups place small amounts of room-temperature water. Place thermometers in each container. • Connect the heat source (hot water) to the room-temperature water with pieces of copper, aluminum, and plastic wire or cord all the same diameter and length. The goal of this experiment is to compare the ability of these three materials to conduct heat. The amount of heat a material can transfer by conduction is called its thermal conductivity. • Record the temperatures of the water in all four containers every 20 minutes until thermal equilibrium is reached. • Collect temperature and time data, and prepare graphs of temperature versus time for the water in the four different containers. Plot the three curves for the cups on the same set of axes, and the temperature for the hot water beaker on a separate graph. Basic Materials List • • • • • • • •

hot plate digital thermometers (4) Pyrex beaker, 600 mL graduated cylinder, 100 mL Styrofoam cups and lids duct tape Styrofoam ice chest (2) and small cardboard box samples of copper and aluminum wire, and plastic cord

We expect metals to conduct heat better than plastic. This is why metal tools like tongs and frying pans sometimes have plastic handles. But how much better do metals conduct heat? And is there a difference in the thermal conductivities of different metals? In this experiment, we examine these questions scientifically. To do this, we need a heat source, which will be a large beaker of water on a hot plate, maintained at a constant, hot temperature. Then we need a place for the heat to go. For this, we use three small Styrofoam cups of water. By monitoring the water temperature in the three cups, we have a way of quantifying how much heat is conducted into the water in each of the cups. Third, we need materials to conduct heat from the heat source to the water in the cups. For two of these materials, we use pieces of aluminum and copper wire. This way we can compare the thermal conductivities of these two common metals. For the third cup, we need a plastic material shaped like wire. The nylon monofilament line used in grass trimmers (“string trimmers”) works well. To prepare the heat conductors, cut equal lengths of copper wire, aluminum wire, and nylon line. You need four pieces of each material, and they need to be long enough to reach comfortably from the hot water beaker into the Styrofoam cups. Wrap the four pieces of each material together in duct tape to serve as a ther60

Experimental Investigation 2

Heat Transfer by Conduction

Analysis

acceleration, m/s/s

mal insulator, as shown in the first photo. On each end of each pair of conductors, leave one inch of exposed material. Make sure the exposed part is the same length on each material so we have a valid comparison of thermal conductivity. The setup is illustrated in the next two photos. A beaker containing about 500 mL of water is on a hot plate in a Styrofoam ice chest. This keeps the water in the beaker at a steady temperature of 100°C without warming up the room. Tape the insulating cups to a cardboard box inside a second ice chest. Use the graduated cylinder to measure out 100 mL of tap water into each of the cups, making sure that each cup has the same amount of water in it. Punch two holes in each lid, one for a thermometer and one to allow one of the conducting materials to enter the water. Label the cups so you don’t forget which one is which. (The cardboard box is there to elevate the cups to the same height as the hot water beaker, so the thermal conductors can be as short as possible.) To run the experiment, place the tap water in the cups 2–3 hours in advance so the water is at room temperature when the experiment begins. Insert the three conduct0.125 ing materials into the cups and the beaker through holes in the sides of the ice chests. Record the four starting temperatures in your lab journal. 0.1 Switch on the hot plate and continue to record all four temperatures every 20 minutes until thermal equilibrium is0.075 reached (about four hours). You will know the system is at thermal equilibrium when all temperatures are more or less holding steady. 0.05 0.025

An important part of the analysis is to prepare graphs of the four data sets, using the horizontal axis for the time and the vertical 0axis for the temperature. Plot the data for the 0 The water 60 in the cups 120 begins180 three small cups on the same axes (same graph). at around240 force, N 22°C, so you probably want to scale the vertical axis from 20°C up to a bit above the highest cup-temperature reading you have. Since you are plotting three sets of data on the same graph, you need to use a different symbol or color for each data set, as illustrated in the sample graph to the right. The hot water temperature is supposed to be steady at 100°C. This is a lot higher than the cup temperatures, so it needs to be on a graph by itself. Your instructor will help you with setting up your graphs and labeling them properly. In your report for this experiment, address the following questions: 1. From studying the graphs, how do the thermal conductivities of the three materials compare? 2. Why do you think the temperatures from the hot water beaker are recorded and plotted on a graph? What purpose does that serve? 3. Why are the three conductor materials wrapped in duct tape? 4. Why is it important that the three different materials are cut to the same length? 5. Imagine you are designing a part to go inside a machine, which we will call part U. Part U connects to three other parts, called a, b, and c. If you do not want heat to transfer between part U and parts a, b, and c, what kind of material should you consider using for making part U?

300

Chapter 4 Order in Nature

DNA is an enormous molecule found in the cells of all living organisms. DNA is referred to as a macromolecule, because it is assembled from groups of smaller molecules called sugars, phosphates, and nucleotides, all woven together in a spiraling shape known as a double helix. The chains of paired nucleotides in human DNA (shown in red in the image above) can be up to approximately 220 million pairs long. The sequencing of the molecules in DNA contains a sophisticated multi-layered code. Embedded in this code is the genetic information that specifies and governs the biological functioning and physical traits of the organism. 62

OBJECTIVES After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary.

1. 2. 3. 4.

Describe examples that illustrate the existence of “laws of nature.” Describe three possible explanations for why laws of nature exist. Give examples of mathematical structure in nature. Describe at least eight remarkable features of the universe, the galaxy, the solar system, or earth that illustrate the order necessary for complex life to exist on earth.

VOCABULARY TERMS You should be able to define or describe each of these terms in a complete sentence or paragraph.

1. 2. 3. 4.

4.1

atmosphere Big Bang fine-tuning galaxy

5. 6. 7. 8.

laws of nature magnetosphere Milky Way solar system

9. solar wind 10. steady-state universe

Why Are There Laws of Nature?

It is so common to hear people refer to the “laws of nature” that this phrase usually goes by without a second thought. But if you stop and think about it, the fact that laws of nature exist is actually quite remarkable. According to the standard scientific description of the origin of the universe, the entire universe began with an inconceivable explosion of energy, an event known as the Big Bang. From the moment the Big Bang happened, the laws of physics were present to shape the emerging matter and energy over 13.8 billion years into the stars, galaxies, and solar systems we see today. How could it be that from such an explosion a set of unwritten, mathematical principles would spontaneously emerge, and that these principles would consistently govern the behavior of matter and energy forever after? Let’s think about this for a moment. You know what explosions are like; you’ve probably seen thousands of them in movies. They are massively chaotic, with material flying everywhere and light and heat radiating out from the center of the explosion. But even in the midst of the chaos of an explosion, like the one shown in Figure 4.1, the motion of all the fragments, the patterns of light and heat, and the new compounds formed by the chemical (or nuclear) reactions involved are all governed by the laws of physics and chemistry. And apparently these laws were 63

Physical Science

there from the beginning of the universe, governing the expansion of electromagnetic radiation from the moment of the Big Bang. What do we mean by the “laws of nature”? Well, in the most general sense this phrase refers to everything in the entire domain of science, from the laws of electricity and magnetism, to the chemistry of exploding gunpowder, to the chemical system of detection and communication ants use to swarm a lump of sugar, to the patterns of behavior exhibited by a hunting tiger. The plain fact is, there are laws that govern how nature works. If there weren’t, the matter and energy spewing out from the Big Bang would have had no plan or Figure 4.1. Even the chaos of explosions is governed by the program for what to do next, and laws of physics and chemistry. there would have been no regulating principles to bring order to the expanding mess. It would be just a mess of stuff, nothing else. Consider also that in order for us to do science at all, there must be predictable laws governing the way nature works. Without predictable laws, nature would not submit itself to scientific analysis. But if that were the case, we wouldn’t be here to analyze it because the existence of life itself depends on predictable, orderly laws governing the way matter and energy interact. The title of this section is, “Why Are There Laws of Nature?” Philosophers and theologians have expressed three different views in answer to this question. All agree that the laws of nature are fascinating, beautiful, and wonderful. Those who love science also agree that the laws of nature are interesting to study. But people differ as to why the laws of nature exist. Let’s explore this question here briefly. Some people say there is no particular reason for the laws of nature. According to this first view, Science depends on the the laws of nature came into existence by chance. If regular behavior of the this is true, then we need not seek a reason for the world around us. If there existence of the laws; we just accept them and go weren’t laws of nature, there on to study them. A second view holds that there is would be no science. There some unknown natural principle that explains why would be no people, either. the laws of nature exist. According to this view, there is a natural explanation for the existence of the laws of nature, but we have not discovered this principle yet. At this time, no one knows what such an unknown natural principle might be. The third view is held by those who believe in a divine being (God) who created the universe. According to this view, the laws of nature exist because they were created

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Order in Nature

as part of nature by the creator, according to the purposes the creator had in mind at the creation. Our purpose in this chapter is to explore some of the many fascinating consequences of the fact that the laws of nature do exist. As I wrote in the first chapter, the order in nature is one of the three basic things that exists (along with matter and energy). We will not further explore the reason why the laws of nature exist. We must leave that question to the philosophers and theologians. I encourage you to ask your parents or teachers about this if you are interested in thinking more about it. For the rest of the chapter, we will focus on what we can say about the orderliness we find in nature. Learning Check 4.1 1. What is meant by the phrase “laws of nature”? 2. What are some examples of evidence for the existence of laws of nature? 3. Describe three possible explanations for why laws of nature exist.

4.2

Order and Structure in Nature

It has been said that all the basic laws of physics that govern the physical behavior of the known universe— ∇⋅D = ρ f which are expressible in mathematical equations—can be written on a single sheet of paper. Isn’t that astonishing? ∇⋅B = 0 Here are a few examples of these laws. In Figure 4.2 are the famous “Maxwell’s Equations.” These four equations ∇ × E = − ∂B were published by Scottish physicist James Clerk Maxwell ∂t in 1864, and they describe all of classical electricity and ∂D magnetism, including radio waves and light. ∇ × H = J + f Everything we know about gravity is summarized in ∂t the general theory of relativity, shown in Figure 4.3. German physicist Albert Einstein published this theory in Figure 4.2. Maxwell’s equations 1916. Maxwell’s equations are difficult to learn and under- describe all electricity and stand, but they are nothing compared to the difficulty of magnetism. the mathematics behind this equation of Einstein’s. 1 8π G Rab − Rg ab = 4 Tab And yet the main operating i! ∂ Ψ = ĤΨ principle of all the gravity 2 c ∂t in the universe can be exFigure 4.3. The primary equation of pressed in a single equation! Figure 4.4. The general Einstein’s general theory of relativity, a The behavior of par- Schrödinger equation, geometrical description of space and ticles such as protons and describing the behavior of time that accounts for gravity. electrons at the atomic or systems at the quantum (atomic) level.

Physical Science

quantum level is summarized by the “Schrödinger equation,” shown in Figure 4.4. This equation, published in 1926 by Austrian physicist Erwin Schrödinger, has dominated physics in the 20th and 21st centuries (along with Figure 4.5. An expression of Einstein’s equation). the Heisenberg uncertainty And the limits on what it is possible to know about principle, describing limits on what observers like us can know any quantum system are summarized by the Heisenberg about objects at the quantum uncertainty principle, shown in Figure 4.5. level. Of course, there are a few thousand other equations that come up in the study of physics, but it is amazing that they really all boil down to those shown here and a few others. The breathtaking complexity of the matter and energy we see in nature can be summarized so simply and elegantly! Not only that, but we can discover these laws and understand them! It is wonderful that the universe—and the powers of the human mind—are this way. Now, just as amazing as the simplicity and comprehensibility of nature’s mathematical structure is the fact that nature Many scientists has any mathematical structure at all. Many scientists have have agreed that wondered about this. Several have written comments such as the mathematical this statement by a Nobel Prize winning physicist: “the mirastructure of nature cle of appropriateness of the language of mathematics for the is a mystery that formulation of the laws of physics” is “something bordering seems miraculous. on the mysterious.” I have shown you some of the equations in physics that illustrate the order and structure embedded in nature. Now I want to show you some beautiful examples how this mathematical structure is evident just by looking at nature. Just check out the stunning, three-dimensional arrangement of cubes in the pyrite crystal of Figure 4.6. This is not just mathematical; it looks like someone has been playing with mathematics! Now, we all know about the planets in our solar system. The very creative image in Figure 4.7 shows them to scale, next to one another for size comparison (and for just gazing at their beauty.) Thousands of astronomers have labored at working out the mathematics of the planetary movements. All the planets travel in elegant, predictable orbits around the sun, and their orbits are all elliptical, which means , and can be they are shaped like this characterized with equations that look Figure 4.6. Pyrite crystals exhibiting an effect known

! σ xσ p ≥ 2

as “twinning.” 66

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Figure 4.7. The beautiful planets (left to right, back to front): Jupiter, Saturn, Uranus, Neptune, Earth, Venus, Mars, and Mercury.

like ax2 + by2 = 1. All the planetary moons travel in ellipses around their companion planets as well. (Some cool asides: the famous Great Red Spot on Jupiter is about three times the size of the earth. It is a giant storm, and we have known about it here on earth for 350 years. Saturn’s rings are made almost entirely of water ice, and average about 20 meters thick. The outer diameter of the rings is about the same as the distance from earth to earth’s moon.) Now let’s talk about the fruit fly, commonly used in biology classes all over the world to study the mathematical laws of genetic inheritance (which we won’t go into). It is such a simple little bug, common and easily squished. But check out the magnificent geometry built into the eye of a fruit fly shown in Figure 4.8, an image captured by a scanning electron microscope (SEM). Figure 4.9 is another image of an eye captured by an SEM, the eye of an antarctic krill. (Krill are small ocean crustaceans related to shrimp, with bodies about 3/4 inch long.) Perfect hexagons— imagine that!

Figure 4.8. A scanning electron microscope image of the eye of a fruit fly.

Figure 4.9. A scanning electron microscope image of the eye of an antarctic krill. 67

Physical Science

We could go on endlessly with these examples, but let’s look at just two more beauties. Figure 4.10 shows another SEM image, this time of a single-celled member of the algae family. If you can believe it, this thing is only about 10 micrometers in diameter, which is only about 20 times the wavelength of visible light. I just can’t get over the rich, velvety beauty of this image. And mathematical structure? This thing looks like it was built in a watch factory! (By the way, scientists don’t yet know what those round things are for, but I am confident they serve some purpose.) Figure 4.10. A scanning electron microscope image of a Finally, in Figure 4.11 we have coccolithophore, a single-celled algae that is abundant one of the world’s cutest marine aniin the oceans. mals, the bobtail squid! (I’m sure you will love this: they are also known as dumpling squid, or stubby squid.) This little guy is only a couple of inches long. Don’t you love the spiraling pattern of colors? (Why do the colors spiral?) But listen to what this critter has going on: it can create its own “invisibility cloak.” No, really. This squid has a mantle called the “light organ,” which is home to a glow-in-the-dark species of bacteria that live underneath it. By means of a fantastic optical system composed of a color filter, lens, and reflector built-in to the Figure 4.11. This bobtail squid is only a few centimeters squid’s mantle, the bacteria emit light long, but its light emitting capabilities allow it to hide its beneath the animal. From below, this own silhouette. light mimics the appearance of the light hitting the top of the squid. The net effect of this superb optical imaging system is to mask the squid’s own silhouette and protect the squid from predators! Imagine a squid and some bacteria having such elegant control over electromagnetic radiation. (You might also be interested to know that in exchange for the lighting service, the squid feeds the bacteria on sugar and amino acids.)

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Order in Nature

Oh wait—I almost forgot about the Fibonacci sequence, found all over the place in pine cones, sunflowers, flower stems, honeybees, and nautilus shells. On second thought, I will leave this one for you to look up yourself. Learning Check 4.2 1. Describe some equations scientists use to model the laws of nature. 2. Describe some examples in which the mathematical structure in nature is evident from its appearance.

4.3

The Remarkable Universe

For several decades now, scientists have been aware of quite a number of coincidences associated with the existence of complex life on planet earth. Let’s take a look at a few of these remarkable features.

Conditions Necessary for Life There are many conditions that must be exist. Here are some of them.

satisfied for complex life as we know it to

1. We need a lot of water to regulate temperature and transport nutrients. The earth’s surface is over 70% water (Figure 4.12). The water cycle involves a complex system of oceans, evaporation, cloud transport, rainfall, lakes, and rivers. This large amount of water is necessary to regulate the temperature of the planet. Also, because minerals can dissolve in water, the water cycle works as a global transport system to distribute the nutrients needed by plant and animal life. Current scientific thinking is that without the oceans, the temperature swings from day to night would be so drastic that Figure 4.12. 71% of the earth’s surface is covered in water, one of the main necessities for complex life to exist. complex life could not exist. 2. We have to be just the right distance from the sun. Within our solar system, earth is positioned in what some scientists have called the “Goldilocks zone”: not too hot, not too cold, but just right. If earth’s orbit were a bit smaller or a bit larger, the oceans would be unstable and life probably could not exist. 69

Physical Science

3. We need an atmosphere to trap heat. Our atmosphere (Figure 4.13) is important because we breathe it. But it serves another purpose as well: to trap heat from the sun and help regulate the surface temperature of the earth. Without it, daily temperature fluctuations would be so high that complex life could not exist. 4. We need a large moon.

Figure 4.13. The earth’s atmosphere, as seen during sunset

The tilt of the earth’s rota- from the International Space Station. tional axis is the cause of our seasonal climate variation, which helps keep the equator from getting too hot and the poles from getting too cold. Moderate temperatures are not only necessary for complex life to exist, but moderate temperatures also make sure earth’s water does not get too hot and boil away, creating the kind of harsh environment found on Venus. Our large moon stabilizes the tilt of earth’s axis, preserving the balance necessary for life. 5. Numerous physical constants must have just the right values.

There are numerous constants found in the equations of the laws of physics. These include the gravitational constant, the masses of the fundamental particles, the electrical charges on the fundamental particles, and others. Scientists have noted that for many of these constants, if their values were only slightly different the physics and chemistry of the universe would not support complex life such as ourselves. Many scientists refer to this as the fine-tuning of these physical constants. 6. We need a magnetic field around the planet to shield us from the harmful solar wind. The liquid iron core of the earth creates a magnetic shield around the earth called the magnetosphere. The magnetosphere (Figure 4.14) provides protection for our planet from the “solar wind,” a spray of Figure 4.14. This artist’s rendering of Earth’s magnetosphere illustrates the way it protects us from the harmful effects of the solar wind.

Chapter 4

Order in Nature

high-energy protons and electrons coming from the sun. Without the iron core and the resulting magnetic shield, it is likely that the solar wind would rob earth of its atmosphere, its water, and its life.

Other Remarkable Observations

As we have seen, the conditions on earth are wonderfully appropriate for supporting life. But there are other remarkable observations that should be included on any list of things that make our world amazing. Here are a few more notable features of our world. 1. The universe had a beginning. Current scientific theory holds that the universe had a beginning. But science has not always held this. Big Bang theory was not formulated until after 1929, when astronomer Erwin Hubble discovered that the galaxies are all accelerating away from one another. Prior to that, many scientists held a “steady-state” view of the universe, in which the universe has no beginning and has always existed. Today, scientists recognize that the universe began 13.8 billion years ago, it has been expanding ever since, and its rate of expansion is increasing. In other words, the galaxies are accelerating away from each other. 2. All nature exhibits exquisite balance. It is unfortunate that many of us these days spend most of our time indoors. But those who have spent a great deal of time outdoors studying the forests, mountains, rivers, and oceans are often struck by the delicate balance maintained by the ecosystems of the planet and the life that lives in them. The creatures are all exquisitely adapted to their environments, displaying peculiar physical features that enable them to flourish. Further, the creatures all seem to depend on one another and on the proper functioning of earth’s water cycle, day-night cycle, tides, and seasons. The balance in nature is one of its most remarkable features. 3. Our bodies exhibit incredible functionality. I hope the human body amazes you, because it is indeed amazing. Have you seen images showing the astonishing way babies develop in the womb, with a beating heart after 21 days? Have you thought about how wonderful it is that when we receive a minor wound, our blood coagulates (a highly complex process) to prevent us from losing too much blood? Aren’t you amazed at the stereoscopic vision provided by our eyes that enables us to perceive in three dimensions, in color, and over a range of light intensity that no camera can match? Aren’t you dazzled by the ability of the human immune system to fight off disease and infection? Consider that the human brain is understood to be the most complex object in the entire universe, enabling our amazing capabilities in mathematics, music, language, sculpture, athletics, engineering, medicine, and on and on. Sometimes when viewing beautiful works of art, such as Michelangelo’s sculpture called the Pietà, or Hans Holbein’s extraordinary 71

Physical Science

sketches of the hands of Erasmus (Figure 4.15), one can find oneself actually weeping from sheer amazement. Thinking about the incredible ability people have to create such things is mind-blowing. 4. DNA is an extremely sophisticated, multi-level coding system. We are still in the early stages of learning about human DNA, but already we know that the genetic coding in DNA conveys the entire set of instructions that specifies and regulates how the human body—the most sophisticated system in the universe—develops and functions. Amazingly, the coding in DNA operates at more than one level. There are codes embedded in the matrix of nucle- Figure 4.15. Our brains and hands are wonderful, and otides in DNA, but there are also enable us to create beautiful works of art, such as this codes on top of those codes! Un- sketch by Hans Holbein the Younger (c. 1523). raveling the coding embedded in human DNA is one of the hottest fields of scientific research today. 5. We perceive that the earth and the heavens are beautiful. Why do we perceive mountains, sunsets, forests, flowers, animals, and human beings as beautiful? Why do we perceive works of art as beautiful? In a universe that seems mostly hostile to life, it is remarkable that life exists in such abundance on earth, and that we humans with all our mental and emotional faculties are here in the midst of it. In fact, our perception of beauty is just one of the many emotions humans feel. The human range of emotion is unique. While scientists debate the degree to which animals experience emotion, there is no doubt about the amazing range of emotions humans experience. The fact that we perceive things as beautiful again points to the wonderful order amidst the complexity of the world around us. If our surroundings were not ordered in a harmonious way, we might still feel emotions such as fear, anger, and frustration, but we would hardly be expected to perceive beauty, love, peace, and joy. We are well adapted to the world we live in, and the orderliness of the laws of nature allows us to perceive many things around us as beautiful and good.

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Order in Nature

Conditions Necessary for Exploration We have looked at order and

structure in nature. We have noted many other remarkable features of nature, including our wonderful human characteristics. One notable human characteristic is our thirst for knowledge and our desire to understand the world—that is, our yearning to explore. Well, as it turns out, many of the features of earth that enable complex life to exist also happen to make exploration of the universe possible. We conclude this section with a few examples of this amazing circumstance. 1. The moon is just the right size and distance from earth to produce solar eclipses. I wrote above about how important the size of the moon is for enabling complex life to exist on the earth. The moon’s size is also exactly right for producing perfect solar eclipses, like one shown in Figure 4.16. Because of the moon’s size and distance from earth, during an eclipse the main disk of the sun is obscured, shutting off its Figure 4.16. Our moon is just the right size to produce total solar eclipses, like this one observed from the blinding light but still allowing us Apollo 12 spacecraft in 1969. to see the sun’s corona, a sort of atmosphere around the sun. Studies of the corona using images from eclipses were the key to discovering the elements of which the sun is composed, thus unlocking our understanding of the solar fusion process, solar history, the history of the universe, and in fact the entire field of astrophysics. 2. Our galactic location provides a great view of the heavens. Our solar system is in a galaxy called the Milky Way. As shown in the artwork of Figure 4.17, the Milky Way Galaxy is shaped like a bulging disk with four main spi- Figure 4.17. The position of earth in the galaxy is not ral arms extending out. I did not only safe, but it is perfect for astronomical exploration. 73

Physical Science

mention previously that our solar system’s location in the galaxy happens to be one of the safest places in the galaxy, but so it is. The main disk area is a very hazardous region because lots of star formation is going on, generating a lot of radiation hazardous to life. We are outside the hostile environment of the main disk, in a relatively safer place. But we also happen to be located in between two of the galaxy’s main spiral arms. Inside the arms there is a lot of dust. If our solar system were located there, our view of the heavens outside our solar system and outside our galaxy would be obscured. But since we are located in between two arms, we have a great view outside the galaxy. As a result, we are able to see distant galaxies and study their features. 3. Our atmosphere is transparent. If our atmosphere were like the atmosphere of Venus, we wouldn’t be doing much astronomical research, at least not from the planet’s surface. Venus is all clouds, and if you happened to be on the ground on Venus you would see nothing but thick cloud cover, 24/7. Fortunately for us, our atmosphere not only helps regulate the temperature by trapping the sun’s heat, but it is transparent as well, enabling us to gaze out into the heavens. The layers in our lovely and very cooperative atmosphere are on fine display in Figure 4.18, an image captured by Expedition 22, the 22nd long-term endurance crew of the International Space Station. The image elegantly captures the silhouette of Space Shuttle Endeavor, just before docking.

Figure 4.18. This beautiful image of Space Shuttle Endeavor was captured by the crew of the International Space Station. Behind the shuttle the layers of earth’s atmosphere are visible. 74

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So it just so happens that natural features essential for the existence of complex life also just happen to make our world suitable for scientific exploration. What an amazing coincidence! We are ready now to dig more deeply into the physics and chemistry of the world around us. But studying science is a lot more fun when we are aware of just how wonderful our world is. Studying and learning is always hard work, but studying science is a lot more rewarding when we pause to take note of the wonders that surround us, wonders we often overlook. This world is truly an amazing place and the best part of studying science is that we get to study these wonderful things every day! Learning Check 4.3 1. Make a list of six necessary conditions for complex life, and briefly describe each of them in your own words. 2. Make a list of at least five aspects of life on earth that are amazing or remarkable. On your list you may include some of the features discussed in this section, or some other features that you have thought of yourself. In your own words, explain why each of the features you list makes earth a remarkable place. 3. List three features of earth that facilitate our efforts to explore the universe around us.

Chapter 4 Exercises Answer each of the questions below as completely as you can. Write your responses in complete sentences. 1. Write a paragraph explaining the different views on why the laws of nature exist. 2. Write a paragraph explaining what we mean when we refer to the “order and structure” found in nature. 3. Imagine that one of your friends says to you that she doesn’t find the world to be a very remarkable place at all, and that, in fact, she finds the world to be quite ordinary and even boring. Write a paragraph or two in response to this friend. In your paragraph, make use of some of the material from this chapter. 4. Back at the beginning of Chapter 1, I write that there are three basic things in nature—matter, energy, and order. Why is it appropriate to say that order is one of the most basic things in nature? Write a paragraph explaining how you would respond to someone who claims that only matter and energy are basic, and that order is not part of nature.

Planet Earth

PUBLISHED BY CENTRIPETAL PRESS Grade Level: 8th Grade Yearlong Course

CLICK TO VIEW IN OUR ONLINE CATALOG

PLANET EARTH Planet Earth is an earth science text designed for middle school. The text is typically used in an 8th-grade course. Here we highlight a few distinctive features of this text. •

Planet Earth supports our signature philosophy of science education based on wonder, integration, and mastery. We always seek to build on and stimulate the student’s innate sense of wonder at the marvels found in nature. Integration refers to the epistemology, mathematics, and history embedded in the text, and a curriculum designed to help develop the student’s facility with using language to communicate scientific concepts. Planet Earth is designed to be used in a mastery-learning environment, using the mastery teaching model John developed and describes in From Wonder to Mastery: A Transformative Model for Science Education. When teachers use this mastery-learning model, the result is high student achievement and exceptional long-term retention for all students.

•

The text includes eight experiments. Three of these involve the use of topographic maps, a feature students and teachers love.

Support Resources Two major support resources are available to accompany Planet Earth. They include: Digital Resources The following materials are accessible exclusively through Centripetal Press: • a full year of weekly, cumulative quizzes • two semester exams • a 33-page resource manual for the eight experiments • a document containing all answer keys • a full year of Weekly Review Guides • a document with recommendations for teaching the course • a lesson list and example calendar Tips and Tools A variety of tips and tools are available at www.centripetalpress.com. Here, teachers will find scores of images and short videos organized chapter by chapter. Teachers are free to use any of this content as they wish to assemble vivid and fascinating lessons.

About Centripetal Press Centripetal Press is an imprint of Classical Academic Press that focuses on the highest-quality science curriculum for charter schools and secular home schools. Founded by educator John D. Mays, Centripetal Press offers a completely unique approach to secondary science curriculum. For more information on our vision for secondary science education, please refer to the Overview section on pg. 2 in this series of samples.

PLANET EARTH Contact Us Our team is here to assist as you choose curricula that will be the best fit for your classroom. We encourage you to take advantage of the following resources: Questions & Guidance We look forward to answering any questions you might have regarding our curriculum, placement, or pricing. Please contact us! Email: info@classicalsubjects.com Phone: (717) 730-0711 Quotes and Vendor Applications We are able to provide formal quotes for all of our materials, and can also submit applications required to become an approved vendor for your school district. Inquire at info@classicalsubjects.com. Get the Latest Information from Centripetal Press and Classical Academic Press For more information on Centripetal Press products, visit www.centripetalpress.com.

Sample Chapters The following pages contain samples from the text. The Table of Contents is shown, as well as Chapters 1, 2, and 6, and the first experiment.

Planet Earth Land, Water, Sky

Kevin Nelstead

Camp Hill, Pennsylvania 2020

Planet Earth: Land, Water, Sky © Classical Academic Press®, 2020 Edition 1.0 ISBN: 978-1-7326384-2-6 All rights reserved. Except as noted below, this publication may not be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior written permission of Classical Academic Press. All images attributed to others under any of the Wikimedia Commons licenses, such as CC-BY-SA-3.0 and others, may be freely reproduced and distributed under the terms of those licenses. Classical Academic Press 515 S. 32nd Street Camp Hill, PA 17011 centripetalpress.com Cover design: Scarlett Rugers, scarlettrugers.com IS.08.20

Contents Preface For Teachers

Our Emphasis on Wonder, Integration, and Mastery Considerations for Science Programming Lab Journals and Lab Reports Conducting Experiments Companion Book and Digital Resources Notes on Using this Text

Preface For Students

Stewarding Our Precious Planet Mastery Lab Journals

Chapter 1 Earth In Space

1.1 An Introduction to Earth Science 1.1.1 Earth Systems 1.1.2 Subdivisions of Earth Science 1.1.3 Further Specializations 1.2 Earth in the Solar System, Galaxy, and Universe 1.2.1 Earth in the Solar System 1.2.2 Earth—A “Just Right” Planet 1.2.3 Earth in the Galaxy and Universe 1.3 Earth’s Orbit and the Seasons 1.3.1 Earth’s Orbit 1.3.2 Solstices and Equinoxes 1.3.3 The Tropics and Polar Regions 1.4 Phases of the Moon 1.5 Eclipses 1.5.1 Solar Eclipses 1.5.2 Lunar Eclipses 1.6 Calendars 1.6.1 The Hebrew Calendar 1.6.2 The Islamic Calendar 1.6.3 Western Calendars Chapter 1 Exercises

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xiv xvi xvi xvi xvi xvii

xx xxi xxii

4 4 6 7 8 8 10 12 14 14 15 17 18 20 20 22 23 23 24 25 27

Contents

Chapter 2 Visualizing Earth

Chapter 3 Thinking About Earth

Chapter 4 Matter and Minerals

2.1 Locations on Earth 2.1.1 Latitude and Longitude 2.1.2 Satellite Navigation Systems 2.2 Map Projections and the Shape of Earth 2.2.1 Map Projections 2.2.2 The Shape of Earth 2.3 Remote Sensing 2.4 Mapping Earth 2.4.1 Geographic Information Systems 2.4.2 Topographic Maps 2.4.3 Map Margin Information 2.4.4 Gradient and Percent Slope Chapter 2 Exercises Experimental Investigation 1: Interpreting Topographic Maps

3.1 The Cycle of Scientific Enterprise in Earth Sciences 3.1.1 Scientific Facts 3.1.2 Scientific Theories 3.1.3 Hypotheses and Scientific Research (Experiments) 3.1.4 Analysis 3.2 Experimental Science and Historical Science 3.3 Challenges in Studying Earth 3.4 Stewardship of Earth 3.4.1 Earth Resources 3.4.2 Sustainability 3.4.3 The Degradation of Nature Chapter 3 Exercises

4.1 Atoms, Elements, and Crystals 4.1.1 Atomic Structure 4.1.2 The Periodic Table of the Elements 4.1.3 Compounds and Crystals 4.1.4 Chemical Bonds 4.2 Minerals 4.2.1 Definition of Mineral 4.2.2 Formation of Minerals 4.2.3 Mineral Groups 4.2.4 Gems vi

29 30 33 34 35 37 38 42 43 43 46 47 49 52

57 58 59 60 61 62 66 68 69 69 71 73

75 75 77 79 81 83 83 85 86 89

Contents

4.3 Mineral Properties and Identification 4.3.1 Luster, Color, and Streak 4.3.2 Crystal Form, Hardness, Cleavage, and Density 4.3.3 Other Mineral Properties 4.4 Mineral Resources 4.4.1 Ores 4.4.2 Mines Chapter 4 Exercises 4.4.3 Minerals as Nonrenewable Resources Experimental Investigation 2: Identifying Minerals

92 92 93 96 97 97 99 100 100 102

Chapter 5 Rocks and the Rock Cycle

104

Chapter 6 Plate Tectonics and Mountain Building

132

5.1 The Rock Cycle 5.2 Igneous Rocks 5.2.1 Magma and Lava 5.2.2 Types of Magma and Lava 5.2.3 Intrusive and Extrusive Igneous Rocks 5.2.4 Identifying Igneous Rocks 5.3 Sedimentary Rocks 5.3.1 Clastic Sedimentary Rocks 5.3.2 Organic and Chemical Sedimentary Rocks 5.3.3 Identifying Sedimentary Rocks 5.4 Metamorphic Rocks 5.4.1 Contact Metamorphism and Regional Metamorphism 5.4.2 Identifying Metamorphic Rocks 5.5 Energy Resources 5.5.1 Coal 5.5.2 Petroleum and Natural Gas 5.5.3 Environmental Concerns with Fossil Fuels 5.5.4 Alternative Energy Resources Chapter 5 Exercises Experimental Investigation 3: Identifying Rocks

6.1 Continental Drift 6.2 The Ocean Floor 6.2.1 Continental Margins 6.2.2 Mid-Ocean Ridges 6.2.3 Seamounts and Marine Sediments 6.3 Seafloor Spreading

106 108 109 110 111 111 114 115 116 118 119 119 120 122 123 124 125 127 128 130

133 136 136 138 139 142

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Contents

6.4 Plate Tectonics 6.4.1 Plate Boundaries 6.4.2 Tectonic Plate Movements in the Past 6.5 Mountains 6.5.1 Folding 6.5.2 Faulting 6.5.3 Mountain Building Chapter 6 Exercises

145 146 149 149 150 152 154 157

Chapter 7 Volcanoes and Earthquakes

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Chapter 8 Weathering, Erosion, and Soils

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7.1 Volcanoes and Plate Tectonics 7.1.1 Volcanism at Divergent Plate Boundaries 7.1.2 Volcanism at Convergent Plate Boundaries 7.1.3 Volcanism Within Tectonic Plates 7.2 Volcanoes 7.2.1 Volcanic Materials 7.2.2 Types of Volcanoes 7.2.3 Caldera Eruptions 7.2.4 Fissure Eruptions 7.3 Igneous Intrusions 7.4 Earthquakes 7.4.1 Elastic Deformation 7.4.2 Faults 7.4.3 Seismic Waves 7.4.4 Measuring and Locating Earthquakes 7.5 Earthquake Damage 7.5.1 Earthquake Intensity 7.5.2 Tsunamis 7.6 The Interior of the Earth 7.6.1 Crust 7.6.2 Mantle 7.6.3 Core Chapter 7 Exercises Experimental Investigation 4: Studying Volcanoes with Topo Maps

8.1 Mechanical and Chemical Weathering 8.1.1 Mechanical Weathering 8.1.2 Chemical Weathering 8.1.3 Rates of Weathering

viii

160 160 163 164 167 167 171 174 175 177 179 179 180 181 184 187 188 189 191 191 192 193 195 196

199 200 202 204

Contents

8.2 Erosion 8.2.1 Agents of Erosion 8.2.2 Transportation and Deposition 8.2.3 Differential Weathering and Erosion 8.3 Soils 8.3.1 Soil Composition 8.3.2 Soil Horizons and Profiles 8.3.3 Formation of Soils 8.3.4 Soil Types 8.3.5 Soil Conservation Chapter 8 Exercises Experimental Investigation 5: Modeling Weathering

205 205 208 209 211 211 214 215 217 220 223 224

Chapter 9 Surface Water and Groundwater

228

Chapter 10 Landforms

262

9.1 The Hydrologic Cycle 9.2 Streams 9.2.1 Drainage Networks 9.2.2 Streamflow in Channels 9.2.3 Stream Profiles and Base Level 9.2.4 Erosion and Transport of Sediments by Streams 9.2.5 Stream Deposition 9.3 Stream Landforms 9.3.1 Stream Valleys 9.3.2 Floods and Floodplains 9.4 Groundwater 9.4.1 Porosity and Permeability 9.4.2 Aquifers 9.4.3 Springs and Wells 9.4.4 Groundwater as a Resource 9.5 Caverns and Groundwater Landforms Chapter 9 Exercises Experimental Investigation 6: The Stream Table

10.1 Landforms Caused By Mass Movement 10.1.1 Variables That Influence Mass Movement 10.1.2 Types of Mass Movement 10.2 Desert Landforms 10.2.1 Streams in the Desert 10.2.2 Wind Erosion and Transport 10.2.3 Wind Deposits

229 232 232 234 237 238 239 243 243 245 249 250 250 252 254 255 257 258

263 264 266 269 269 270 272

Contents

10.3 Glaciers 10.3.1 Glacial Ice 10.3.2 Alpine Glaciers 10.3.3 Ice Sheets 10.4 Glaciation in Earth’s past 10.4.1 Evidence for an Ice Age 10.4.2 The Timing and Worldwide Effects of Glaciation Chapter 10 Exercises Experimental Investigation 7: Studying Glaciers with Topo Maps

274 275 277 280 283 283 285 287 288

Chapter 11 Unraveling Earth History

292

Chapter 12 Oceanography

330

11.1 Geologic Time 11.1.1 The Development of the Concept of Geologic Time 11.1.2 Hutton and Lyell 11.2 Relative Age 11.2.1 Rock Layers 11.2.2 Principles of Relative-Age Dating 11.2.3 Unconformities 11.2.4 Putting it all Together 11.3 Fossils 11.3.1 Types of Preservation 11.3.2 Trace Fossils 11.3.3 Uses of Fossils 11.4 Absolute Dating 11.4.1 Isotopes and Radioactivity 11.4.2 Radiometric Dating 11.4.3 Types of Radiometric Dating 11.5 Sedimentary Environments 11.6 An Overview of Earth History 11.6.1 Divisions of Earth History 11.6.2 The Precambrian 11.6.3 The Paleozoic Era 11.6.4 The Mesozoic Era 11.6.5 The Cenozoic Era Chapter 11 Exercises

12.1 The Oceans 12.1.1 History of Ocean Exploration 12.1.2 Oceans and Seas

293 294 294 296 296 298 300 302 303 303 306 306 307 307 308 310 311 317 317 320 322 323 326 328

331 332 334

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12.2 Seawater 12.2.1 Salinity 12.2.2 Physical Properties of Seawater 12.3 Currents and Waves 12.3.1 Ocean Surface Currents 12.3.2 Waves 12.4 Tides 12.5 Marine Life 12.5.1 Types of Marine Organisms 12.5.2 Basic Requirements for Ocean Life 12.5.3 Ocean Environments 12.5.4 Ocean Food Resources 12.6 Shorelines 12.6.1 Erosional Landforms along Coastlines 12.6.2 Depositional Landforms along Coastlines 12.6.3 Humans and Coastlines Chapter 12 Exercises

336 336 340 341 342 343 345 348 348 350 351 354 355 355 356 359 361

Chapter 13 The Atmosphere

362

Chapter 14 Weather

384

13.1 The Composition and Structure of Earth’s Atmosphere 13.1.1 Composition of the Atmosphere 13.1.2 Layers in the Atmosphere 13.2 Properties of the Atmosphere 13.2.1 Temperature 13.2.2 Atmospheric Pressure 13.2.3 Wind 13.2.4 Humidity 13.2.5 Precipitation 13.3 Energy and Water in the Atmosphere 13.3.1 Solar Radiation and Transfers of Energy 13.3.2 Evaporation, Condensation, and Energy 13.4 Circulation of the Atmosphere 13.4.1 Global Wind Systems 13.4.2 Jet Streams Chapter 13 Exercises

14.1 Clouds and Precipitation 14.1.1 Formation of Clouds 14.1.2 Types of Clouds 14.1.3 Precipitation

363 363 365 367 368 370 372 374 375 376 376 378 379 379 382 383

386 386 387 389

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14.2 Air Masses and Fronts 14.2.1 Air Masses 14.2.2 Fronts 14.2.3 Weather Maps 14.3 Weather Forecasts 14.3.1 Making a Basic Weather Forecast 14.3.2 Analog Weather Forecasting 14.3.3 Numerical Weather Forecasting 14.4 Severe Weather 14.4.1 Thunderstorms, Severe Thunderstorms, and Tornadoes 14.4.2 Tornadoes 14.4.3 Hurricanes Chapter 14 Exercises Experimental Investigation 8: Weather Maps

392 392 394 395 399 399 400 400 402 403 407 409 413 414

Chapter 15 Climate and Air Pollution

416

Glossary

448

Appendix Minerals

477

References

480

Image Credits

481

Index

492

15.1 Climate 15.1.1 Climate Charts 15.1.2 Factors that Determine Climate 15.2 Classification of Climates 15.2.1 The Köppen Climate-Classification System 15.2.2 A Closer Look at the Climate Groups 15.2.3 Weather Extremes 15.3 Climate Change 15.3.1 Natural Climate Changes 15.3.2 Human-Caused Climate Change 15.4 Air Pollution 15.4.1 Air Pollutants 15.4.2 Factors that Affect Air Pollution 15.4.3 Indoor Air Pollution Chapter 15 Exercises

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417 418 418 422 422 424 431 434 434 438 442 443 444 445 446

Chapter 1 Earth In Space

In December 1968, three astronauts went into orbit around the Moon in the Apollo 8 spacecraft. This was the first time humans had gone further from Earth than just a few hundred kilometers above the surface. As they circled the Moon, the Apollo 8 crew were the first people to have a direct, close-up view of the desolate craters, plains, and mountains of another world. Since radio signals cannot go through or around the solid sphere of the Moon, they were out of radio contact with Earth for about thirty minutes each time they went behind the Moon. As they came back from their fourth journey behind the Moon, they saw something that caught their attention far more than the unexplored surface of the Moon—they saw Earth rising above the lunar horizon. The Moon was gray and barren; Earth on the other hand hung brilliantly in the sky with its blue oceans, white clouds and ice caps, and variously-hued continents. They journeyed 384,000 kilometers to the Moon, but then realized that the most important object in their view was not the Moon, but Earth.

Objectives After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary.

Define and describe the four major Earth systems. Give examples of specialties practiced by different kinds of Earth scientists. Describe Earth’s location in the solar system, galaxy, and universe. Define what a habitable zone is, and apply this definition to Earth’s place in the solar system and the solar system’s place in the Milky Way galaxy. 5. Give examples of ways in which Earth seems to be “just right” for complex life. 6. Describe Earth’s orbit around the Sun. 7. Explain how the tilt of Earth’s axis causes the seasons. 8. Explain why it is hot in the summer and cold in the winter in the Northern Hemisphere. 9. List, in order, the phases of the Moon throughout the lunar cycle. 10. Describe the position of the Sun, Earth, and Moon for each phase of the lunar cycle. 11. Explain what causes partial and total solar eclipses and explain why they do not occur every month. 12. Explain the cause of lunar eclipses. 13. Explain how solar, lunar, and lunisolar calendars work and give an example of each. 14. Compare the Julian and Gregorian calendars, explaining how the Gregorian calendar corrected a flaw in the Julian calendar. 1. 2. 3. 4.

Vocabulary Terms You should be able to define or describe each of these terms in a complete sentence or paragraph. 1. Antarctic Circle 2. aphelion 3. Arctic Circle 4. atmosphere 5. autumnal equinox 6. biosphere 7. ecliptic 8. first quarter 9. full Moon 10. galaxy 11. geocentric model 12. geology 13. Gregorian calendar 14. habitable zone 15. heliocentric model

16. hydrosphere 17. Julian calendar 18. last quarter 19. lithosphere 20. lunar calendar 21. lunar phase 22. lunisolar calendar 23. meteorology 24. Milky Way galaxy 25. new Moon 26. oceanography 27. partial solar eclipse 28. penumbra 29. perihelion 30. solar calendar

31. solar eclipse 32. summer solstice 33. total lunar eclipse 34. total solar eclipse 35. Tropic of Cancer 36. Tropic of Capricorn 37. umbra 38. universe 39. vernal equinox 40. waning crescent 41. waning gibbous 42. waxing crescent 43. waxing gibbous 44. winter solstice

Chapter 1

1.1

An Introduction to Earth Science

The image on the opening page of this chapter is a picture taken by astronauts on the spacecraft Apollo 17, the final mission to the Moon in 1972. From space, one can see that Earth is not a monotonous place. Some of its land surface is brown and barren, and other parts are covered with lush vegetation. Greater than 70% of the surface is covered by oceans. Both land and the oceans near the poles are covered with water in a different form: snow and ice. Forming a thin layer on top of all these is the atmosphere, with its ever-changing patterns of clouds.

1.1.1 Earth Systems What you see from space can be categorized into different Earth systems, illustrated in Figures 1.1 through 1.5. Scientists think of these systems as concentric spheres, with the solid Earth at the center, then water, then the air. Living organisms are present in all three. This gives us four primary Earth systems: the lithosphere, the hydrosphere, the atmosphere, and the biosphere. The lithosphere is the rigid outer layer of Earth, composed mostly of solid rock. The lithosphere includes Earth’s crust and the upper part of the underlying mantle. The rocks of the crust are exposed in many places at the surface of Earth. There are two basic types of crust: continental crust and oceanic crust. The continental crust is composed largely of a lighter-colored rock called granite and averages about 30 to 40 km (20–30 mi) in thickness. On the other hand, the oceanic crust is typically about 5 km (3 mi) thick and composed of dark, dense igneous rocks (that is, rocks formed from molten rock) called basalt and gabbro. Beneath the lithosphere is the rest of Earth’s rocky mantle, and beneath the mantle is Earth’s iron core. We discuss Earth’s crust, mantle, and core in Chapter 7. The hydrosphere is the part of Earth that is made out of water. This water is present as a liquid in the oceans, seas, lakes, and rivers, and as groundwater in pores in rocks and soil. Additionally, water is present as a solid— snow and ice—on land and the polar oceans, and as a gas in the atmosphere. Most of Earth’s water is present as salt water in the oceans. The greatest amount of fresh water—that is, nonsalty water—is contained in ice, primarily in the ice caps that cover Greenland and Antarctica. Compared to the oceans and ice caps, Figure 1.1. Lithosphere—the solid, rigid outer layer of Earth. This quarry is in Australia.

Earth In Space

there is only a tiny amount of water in lakes and streams. Earth is surrounded by a layer of gases called the atmosphere. The thickness of the atmosphere compared to the rest of the planet is like the peel of an apple compared to the rest of the fruit. The most abundant gas in the atmosphere is nitrogen (78%), followed by oxygen (21%). The remaining 1% is made up of argon, carbon dioxide, and a number of gases that are present in small proportions. The atmosphere also contains a variable amount of water vapor—water in its gaseous state. The atmosphere serves a number of functions. The oxygen in the atmosphere is necessary for respiration for most living things, and the carbon dioxide is necessary for photosynthesis. The atmosphere also helps to maintain the temperature of the surface of Earth in a range that is suitable for advanced organisms such as plants and animals. Figure 1.2. Hydrosphere—Earth’s water. Havasu Falls is In addition, the gases of the atmo- in the Grand Canyon in Arizona. sphere help prevent various types of harmful radiation from the Sun and deep space from reaching Earth’s surface. The biosphere is made up of all organisms that live on Earth, together with the environments in which they live. Life exists in almost every environment on Earth:

Figure 1.3. Atmosphere—The gaseous layer that surrounds Earth. This thunderstorm occurred over New Mexico.

Figure 1.4. Biosphere—All living things on Earth. This temperate rainforest is in Redwood National Park in California.

Chapter 1

the surface, the soil, the air, the deep sea, hot springs, ice, and even cracks in hot rocks thousands of meters beneath the surface. These four systems all interact with each other. It is obvious that organisms in the biosphere are dependent on the lithosphere, hydrosphere, and atmosphere for oxygen, water, nutrients, and space to live. However, not only does the biosphere use resources from the other systems, the biosphere in turn affects those systems as well. Plants have changed the atmosphere by producing oxygen. Plants also help to break down minerals in the soil by removing nutrients and exchanging water with the hydrosphere. Likewise, there are interactions between the lithosphere, hydrosphere, and atmosphere. For example, water— Figure 1.5. All four Earth systems—lithosphere, coming from the atmosphere—falls hydrosphere, atmosphere, and biosphere—are interacting on Earth and causes erosion of the along this Atlantic Ocean coastline in Maine. soil, which is part of the lithosphere. Many of the interactions between the solid Earth, water, air, and life are extraordinarily complex and are not fully understood.

1.1.2 Subdivisions of Earth Science

Figure 1.6. A geologist sampling 1150°C (2100°F) lava at Kilauea, a volcano in Hawaii.

Earth scientists work in a number of specialties. Traditionally, these are broken down into three major subdivisions: geology, oceanography, and meteorology, illustrated in Figures 1.6 through 1.8. Geology is the study of the materials that make up Earth and the processes that change Earth over time. Geologists, of course, study rocks, but they also study a range of materials and processes that will be covered in this book, such as volcanoes, earthquakes, fossils, streams, glaciers, water resources, mineral resources, and energy resources.

Earth In Space

Figure 1.7. Oceanographers use submersibles such as DeepWorker to explore the ocean’s depths.

Figure 1.8. Meteorologists using radar to study a tornado.

Oceanography is the study of the oceans. An oceanographer might study ocean currents, processes that occur on beaches, mineral resources on or beneath the ocean floor, or organisms that live in the ocean. Meteorology is the study of the atmosphere. Meteorologists not only make weather observations and forecasts. They are also interested in studying air pollution and understanding long-term trends and changes in climate at various places.

1.1.3 Further Specializations Due to the explosive growth of scientific knowledge, it is impossible for a geologist, oceanographer, or meteorologist to have complete knowledge of his or her subject. Usually, Earth scientists have a broad knowledge of the sciences and a specialty that is the focus of their work. Some specialties of Earth sciences include: • Climatology

The study of climate, which is the long-term average of weather conditions in an area.

• Ecology

The study of the interactions between organisms and their environments. Ecology is often considered to be a topic in biology, but it is also an important topic in Earth sciences. Organisms affect Earth, and Earth affects the organisms that live on it.

• Geochemistry

The use of chemistry to understand Earth processes.

• Geophysics

The use of physics to understand Earth, including its shape, interior, magnetism, and surrounding space.

• Hydrology

The study of the movements and quality of water, either on Earth’s surface or under it.

• Marine biology

The study of life and ecosystems in the oceans.

• Mineralogy

The study of the formation, composition, and distribution of minerals.

• Paleontology

The study of past life on Earth and how it has changed over time.

• Petroleum geology

The study of the location, migration, and production of oil and gas resources. 7

Chapter 1 • Petrology

The study of rocks.

• Planetary geology

The application of geological principles to other worlds, such as planets, moons, and asteroids.

• Volcanology

The study of volcanoes.

This list represents only some of the many specializations within the Earth sciences. Even within these specialties, a scientist usually focuses on an even narrower topic. A climatologist might be most interested in desert climates, a paleontologist might specialize in coral fossils of the Jurassic Period, or a volcanologist might focus on the chemical composition of volcanic rocks produced from volcanoes like the ones in Hawaii. However, even within these specializations the best scientists are those who can relate their data to work being done by workers in other specializations. Because of this, scientists often work in teams and attend meetings with other scientists so they can exchange ideas and look for interactions and relationships between their work and that of others. Learning Check 1.1 1. Distinguish among the lithosphere, hydrosphere, atmosphere, and biosphere. 2. Suggest two ways that the biosphere interacts with each of the other Earth systems. 3. Give a definition for each of the three major subdivisions of Earth science that you will learn about in this course.

1.2

Earth in the Solar System, Galaxy, and Universe

We cannot completely understand Earth without having an understanding of Earth’s place in the solar system, galaxy, and universe. After all, things that happen at a great distance from Earth can greatly influence our planet. Energy from the Sun, produced by nuclear fusion of hydrogen and helium in the Sun’s core, constantly bathes Earth’s land, water, and air in the form of electromagnetic radiation. There is gravitational attraction between Earth and the Sun, Moon, and other planets in the solar system. Rare astronomic events can influence Earth as well, such as the collision of large meteorites or even asteroids with Earth.

1.2.1 Earth in the Solar System Until the 16th and 17th centuries, most scientists believed that Earth was at the center of the physical universe. It was thought that the Sun, Moon, and five planets known at the time (Mercury, Venus, Mars, Jupiter, and Saturn) all orbited around Earth in perfectly circular orbits, as illustrated in Figure 1.9. In this model, the stars are points of light that also revolve around Earth. This Earth-centered picture of the universe is known as a geocentric model. The story of how scientists changed their minds about this model of the universe is fascinating and was an important 8

Earth In Space

turning point in the history of science, but it is a topic for another course. It took about Stars one hundred years from the work of Nicolaus Copernicus Saturn Sun (1473–1543) until after the death of Galileo (1564–1642) for most scientists to abandon Moon Venus the geocentric model of the universe. Earth Jupiter Today we have a very difMercury ferent picture of the place of Earth in the solar system and Mars of the universe as a whole, illustrated in Figure 1.10. We now understand that the Sun, not Earth, is at the center of the solar system. This model of the solar system is known as Figure 1.9. In the geocentric model, Earth is at the center of the a heliocentric model. Earth is universe, and all other bodies orbit Earth. one of eight planets that orbit the Sun. The four innermost planets—Mercury, Venus, Earth, and Mars—are com-

Mars

Neptune

Moon Uranus

Earth

Venus

Saturn

Sun Mercury

Jupiter

Figure 1.10. In the heliocentric model, planets, asteroids, comets, and other solar system bodies orbit the Sun. Our Sun is just one of over 100 billion stars in the Milky Way galaxy. This drawing is highly not to scale.

Chapter 1

posed largely of rock and are known as the terrestrial (Earth-like) planets. As Figure 1.11 suggests, we can apply what we know about Earth to the study of the other terrestrial planets because they have similar compositions. Similarly, we can apply what we learn about Mercury, Venus, and Mars back to our study of Earth. The outer four planets—Jupiter, Saturn, Uranus, and Neptune—are much more massive. They are composed largely or entirely of gas and known as the gas giants.

1.2.2 Earth—A “Just Right” Planet Most places in the solar system—and in the universe as a whole—are quite hostile to complex life. Complex life is life that is more sophisticated than bacteria, and includes all plants and animals. In most places in the universe, the temperatures are too hot or too cold, or there isn’t water, or the right elements— such as carbon—aren’t present, or there is too much damaging electromagnetic radiation for living organisms to thrive. Mars

Figure 1.11. The Curiosity rover on Mars, one of the four terrestrial planets. Many of the geologic features that occur on Earth, such as volcanoes, sand dunes, and stream channels, can also be studied on Mars.

Within the solar system, Earth occupies a special location. If Earth ITABLE ZONE HAB were somewhat closer to the Sun, the temperature would be hot enough to boil the oceans. Living Earth organisms, from bacteria to humans, are dependent on liquid water to exist. Even on Venus, the next planet closer to the Sun, the temperature on the surface is around sun 460°C (860°F)! On the other hand, Mercury Venus if Earth were farther away from the Sun, it would be cold enough to freeze all water on the surface. As an example, consider Mars, with an average surface temperature of around –55°C (–67°F), though Figure 1.12. Our solar system’s habitable zone, tinted in blue. temperatures can rise above freezThere is debate about exactly where the inner and outer limits of the habitable zone are. The Sun and planets are not ing (0°C) on a summer day near the Equator. So far as we know, Mars is drawn to scale. 10

Earth In Space

currently a lifeless planet. It certainly cannot support the abundance of life that exists on Earth. The region around a star in which liquid water can exist on the surface of planets—and which therefore can support advanced life—is known as the habitable zone, illustrated in Figure 1.12. Determining the inner and outer limits of the habitable zone is not a straightforward task, but certainly Venus is too close to the Sun, and Mars is at or near the outer edge of our solar system’s habitable zone. Sometimes astronomers refer to planets in the habitable zone as “Goldilocks planets,” where the temperature is neither too hot, nor too cold, but just right. There may be places outside of the habitable zone where conditions are just right for life—such as in warm water beneath the surface of some of the moons of Jupiter—but it is considered unlikely by most scientists that these environments could support life more sophisticated than microscopic bacteria. Earth is also “just right” in other ways: 1. Earth seems to be an ideal size. If it were considerably smaller, it would not have strong enough gravity to hold onto most of its atmosphere. Mars has only 10.7 percent of the mass of Earth, and has a very thin atmosphere. If, on the other hand, Earth were considerably larger, it would probably have a much denser atmosphere due to stronger gravity. This would likely lead to much higher surface temperatures. 2. Earth has a good amount of water. There is enough water to support life, but not so much as to completely cover Earth with water. 3. Earth seems to have just the right chemical composition. For example, Earth has a small amount of carbon, an essential element for all the primary molecules of life, such as proteins, sugars, and DNA. But if it had considerably more carbon, the composition of both the solid Earth and the atmosphere would be radically different, and inhospitable to life. Earth also has an iron core, which causes Earth’s magnetic field and helps to protect the surface from harmful radiation from space. The chemical composition of Earth’s crust also seems to be just right for the long-term maintenance of life. 4. Many scientists believe that gravitational interactions between Earth and the Moon—the same interactions that cause ocean tides—keep Earth’s axis tilted at a fairly constant angle near 23.5°. If Earth didn’t have such a large Moon—all other moons in the solar system are small compared to the size of their parent planets—occasional changes in the angle of Earth’s tilt could cause catastrophic changes to the climate, which would make the continued existence of complex life difficult. We take a closer look at Earth’s axis in the next section. 5. Earth has plate tectonics, which, as we discuss in Chapter 6, is the process that moves continents and other parts of the lithosphere around on Earth’s surface. It turns out that plate tectonics is a process which helps to make Earth a suitable home for the flourishing of life. Geologists believe that a number of factors have 11

Chapter 1

to be just right for plate tectonics to occur, such as the presence of oceans, and having the right amount of radioactivity in the rocks of the lithosphere. There are many more ways in which the universe as a whole, and our planet in particular, seem to be fine-tuned to support complex life—dozens more, in fact! Scientists believe that if all these factors were not fine-tuned the way they are, the complex life on Earth would not exist.

1.2.3 Earth in the Galaxy and Universe On a very dark night, one can see up to two or three thousand stars, only a very tiny fraction of the total number of stars in the galaxy. A galaxy is a massive system of stars gravitationally bound to one another. Our Sun is but one of perhaps 200 billion (200,000,000,000) stars in the Milky Way galaxy, the spiral galaxy in which our solar system is located. If we were able to travel outside of the Milky Way galaxy— which would take millions of years to do with our current spacecraft—we would see that it looks something like the galaxy shown in Figure 1.13. As Figure 1.14 illustrates, our solar system is located at the edge of a spiral arm, roughly half way from the center of the galaxy to the outer edge. Our Sun is orbiting the center of the galaxy with a velocity of about 800,000 kilometers per hour, but even moving that fast, it takes over 200,000,000 years for the Sun to make one revolution around the galactic core. Just as there is a habitable zone in the solar system, where conditions are right for life to flourish, so there seems to be a galactic habitable zone in spiral galaxies where conditions are right for life to exist. Near the galactic core, stars are closer to each other Figure 1.13. Our Milky Way galaxy is a spiral galaxy, like the Andromeda galaxy pictured here. than they are in our neighborhood of the galaxy, and it is believed that catastrophic events, such as stars passing close to one another and disrupting planetary orbits due to gravitational attraction, would make the entire central region of the galaxy inhospitable for complex life. On the other hand, it appears that solar systems near the outer edges of spiral galaxies do not contain a high enough proportion of atoms of elements heavier than hydrogen and helium to have terrestrial planets—worlds made of heavier elements such as iron, silicon, and oxygen.

Earth In Space

The Milky Way galaxy is just part of the much larger universe. The universe is composed of all the matter and energy that exists, as well as space and time themselves. This includes everything we can see with our most powerful telescopes, such as the distant galaxies shown in Figure 1.15. The Milky Way galaxy is just one of hundreds of billions of galaxies. If there are 200,000,000,000 stars in our galaxy, and at least 100,000,000,000 observSun able galaxies in the universe, then how many stars are there in the entire universe? At a minimum, that number is 20,000,000,000,000,000,000,000 stars. Figure 1.14. An artist’s conception of the Milky Way This enormous number is virtu- galaxy showing the location of our Sun and solar ally incomprehensible to us. It would system. take over 600 trillion years to count 20,000,000,000,000,000,000,000 stars, taking one second per star. Because the universe is so enormous, and our Sun and planet are so small in comparison, it would be easy to conclude that we humans are rather insignificant. Earth is our home,

Figure 1.15. The universe is made up of many billions of galaxies. Almost every object in this image, taken by the Hubble Space Telescope, is a galaxy composed of billions of stars.

Chapter 1

however, and it is an ideal home for both humans and the wide range of living organisms we share the planet with. So far as we know, planets as ideally suited for life are not common in the galaxy. If planets such as Earth are rare, then advanced, intelligent beings such as ourselves are perhaps even less common. We have a responsibility, therefore, to take care of the world we live in. Learning Check 1.2 1. Contrast the geocentric model of the solar system with the heliocentric model. 2. Describe Earth’s location in space, relative to the Sun and solar system, the galaxy, and the universe. 3. In what ways does Earth seem to be “just right” for complex life? 4. What is meant by a habitable zone, for both the solar system and galaxy as a whole? 5. How many stars do scientists estimate there are in the Milky Way galaxy?

1.3

Earth’s Orbit and the Seasons

1.3.1 Earth’s Orbit Earth takes 365.24 days to make one complete orbit around the Sun. This orbit, however, is not circular. Instead, Earth’s orbit around the Sun is slightly elliptical, meaning that Earth is slightly closer to the Sun at some times and farther away at others. At its closest, Earth is about 147,000,000 km from the Sun and at its farthest it is about 152,000,000 km from the Sun, for an average of about 150,000,000 km (93,000,000 mi). As illustrated in Figure 1.16, the point on the orbit of a planet, asteroid, or comet around the Sun where it is closest to the Sun is called the perihelion, and the point where it is farthest from the Sun is called the aphelion. The difference between the perihelion and aphelion for Earth’s orbit is not all that great and from a distance Earth’s orbit would appear to be nearly circular. One interesting thing about this elliptical orbit is that Earth is closest to the Sun (perihelion) during the first week of January and farthest from the Sun (aphelion) the first week of July. This perihelion aphelion means Earth is actually slightly closer to the Sun when it is winter in the Northern HemiSun sphere and slightly farther away from the Sun in the summer. This tells us that the distance from one of Earth’s hemispheres to the Sun is not the primary cause of seasons. This disFigure 1.16. For an elliptical orbit, the aphelion tance does have a minor effect on seasons— is where a planet is farthest from the Sun; the the Northern Hemisphere winter is slightly perihelion is where it is closest to the Sun.

Earth In Space

warmer than it would be if Earth’s orbit were circular, and Northern Hemisphere summer is slightly cooler than it would otherwise be—but the main cause of Earth’s seasons is the tilt of Earth’s axis.

1.3.2 Solstices and Equinoxes The plane in which Earth orbits the Sun is called the ecliptic. As Earth orbits the Sun, its axis of rotation is not perpendicular to the ecliptic; it is tilted at an angle of about 23.5° as Figure 1.17 shows. As Earth orbits the Sun, its axis always points in the same direction. In Figure 1.18, 23.5° you see that no matter where Earth is in its orbit, its axis is inclined 23.5° to the right. As Earth completes its annual trip N around the Sun, different parts of the planet receive different amounts of sunlight. At the far left of Figure 1.18, on about June 21, the Northern Hemisphere is tilted towards the Sun and receives more direct sunlight than the Southern Hemisphere does. This is the summer solstice—the moment of time Equ ato when the Sun is highest in the sky in the hemisphere and r regarded by many as the first day of summer. In general, the summer solstice is the day of the year with the most hours of S sunlight. The winter solstice occurs on about December 21 in the Northern Hemisphere. This is when the Sun is lowest in the sky in the hemisphere, and the day with the least Figure 1.17. Earth’s axis is hours of sunlight. At the time the summer solstice occurs in tilted 23.5° from vertical, the Northern Hemisphere, the winter solstice occurs in the relative to the ecliptic. Southern Hemisphere. About halfway between the solstices are the equinoxes, when Earth’s axis neither tilts away nor toward the Sun, and there is approximately an equal amount of daylight and nighttime. The vernal equinox is the equinox that occurs between the winter solstice and summer solstice, on about March 22 in the Northern Hemisphere. The autumnal equinox occurs between the summer solMarch 21 vernal equinox stice and winter solstice, on about September 22 in the Northern Hemisphere. Tropic of Cancer The temperature on Earth’s surface depends Tropic of Capricorn on the amount of heat June21 December 21 received from the Sun, summer solstice winter solstice which is received in the September 23 autumnal equinox form of electromagnetic radiation. Earth Figure 1.18. As Earth orbits the Sun, its axis is always pointed the same direction. This means that for part of the year, the North Pole is pointing is always gaining en- more towards the Sun; at the opposite part of the year, the North Pole ergy from the Sun, but is pointing away from the Sun. Date names apply to the Northern Earth also loses heat Hemisphere. 15

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energy into space at the same time. If Earth did not lose heat to balSun’s path December 21 ance out the energy it East receives from the Sun, it would get hotter and South hotter and eventually melt! The electromagnetic radiation from the Sun is primarily visible light, which is absorbed North West by Earth’s surface and atmosphere, leading to Figure 1.19. The path the Sun takes across the sky in the Northern Hemisphere at the latitude of the central United States on the summer warming. The warm solstice (June 21) and winter solstice (December 21). Earth, at the same time, re-radiates electromagnetic radiation back out into space, mainly as infrared radiation. We can see this balance swinging back and forth on a daily basis: the temperature rises as the Sun warms Earth’s surface during the day, but then drops at night. We also see this happening on a longer time scale, as the Northern Hemisphere warms up through the spring and summer and cools off through fall and winter. It is all about which is greater at the time: energy received from the Sun or energy radiated back into space from Earth. There are two primary reasons why it is hotter in the summer than in the winter. Both reasons are related to the path the Sun takes across the sky. We think of the Sun as rising in the east and setting in the west. But as Figure 1.19 illustrates, in the summer the Sun actually rises in the northeast and takes a long, high path across the sky before setting in the northwest (from a Northern Hemisphere perspective). In the winter, on the other hand, the Sun rises in the southeast, and takes a much shorter, lower path across the sky before setting in the southwest. From these observations, it becomes apparent why it is hot in the summer. First, the Sun is in the sky for a long time on a summer day, so there is a long period of time for Earth’s surface to absorb radiation from the Sun. Second, the Sun is higher in the sky on a summer day, so sunlight is concenFigure 1.20. The flashlights illustrate the difference trated on a smaller area than in the between having the Sun high in the sky (summer) winter. The difference in sunlight conand low in the sky (winter). In the summer, energy is centration is illustrated with flashlight concentrated in a smaller area, but in winter, energy from the Sun is spread out over a larger area, resulting in beams in Figure 1.20. The opposite is less heating of the ground. true in winter: the Sun is in the sky for Sun’s path June 21

Earth In Space

a shorter period of time, so there is less time for Earth’s surface to absorb sunlight and sunlight hits Earth’s surface at a lower angle, spreading out the energy from the Sun over a greater area.

1.3.3 The Tropics and Polar Regions Near Earth’s Equator, changes in the Sun’s position in the sky don’t have as great of an effect on temperatures. Whether it is summer solstice, winter solstice, or an equinox, the Sun climbs up high in the sky every day and this part of Earth is generally warm year-round. Between latitudes 23.5°N and 23.5°S, there is at least one day in the year when the Sun is directly overhead at noon.1 At latitude 23.5°N, the Sun climbs up to directly overhead at noon on the day of the Northern Hemisphere summer solstice. As Figure 1.21 illustrates, this latitude is known as the Tropic of Cancer. This latitude is the farthest north that the Sun can be seen directly overhead. The southern equivalent, at 23.5°S, is the Tropic of Capricorn. The region between the Tropic of Cancer and the Tropic of Capricorn is known as the tropics. Except in mountainous areas, this region is warm all year. In the United States, only Hawaii is in the tropics. In many tropical areas, the main difference in seasons is not between hot and cold, but between a wet season and a dry season. The situation is quite different in polar regions. On the day of the Northern Hemisphere summer solstice, the Sun does not set at any location farther north than the Arctic Circle, which is at about 66.5°N. These regions are sometimes referred to as “the land of the midnight Sun.” But the opposite situation occurs at the Earth axis

sun rays

Arctic Circle

Tropic of Cancer

Equator Tropic of Capricorn Antarctic Circle Figure 1.21. Earth at Northern Hemisphere winter solstice, showing the Tropic of Capricorn, Tropic of Cancer, Arctic Circle, and Antarctic Circle. 1 Latitude and longitude are discussed in Chapter 2.

Chapter 1

winter solstice, when the Sun does not rise above the horizon north of the Arctic Circle. The Southern Hemisphere equivalent of the Arctic Circle is the Antarctic Circle, which is at approximately 66.5°S. Despite having up to 24 hours of sunlight, polar regions do not get hot in the summer because the Sun is always low in the sky. Learning Check 1.3 1. Calculate the approximate speed at which Earth moves in its orbit around the Sun. You can simplify the problem by assuming that Earth’s orbit around the Sun is circular. 2. What effect does the elliptical orbit of the Sun have on Earth’s seasons? 3. Explain how the tilt of Earth’s axis causes seasons. 4. Speculate as to what seasons would be like if Earth’s axis were not tilted, and what seasons would be like if Earth’s axis were tilted at 45° instead of 23.5°.

1.4

Phases of the Moon

There are nights when the Moon shines bright in the sky, bright enough to walk outside without the need for artificial lighting. There are other nights when it is pitch black and there is no Moon above, but thousands of stars shine in the dark sky. As the Moon orbits Earth, its appearance, as well as the time it rises and sets, changes from day to day. The phase of the Moon, or lunar phase, is the shape of the Sunlit portion of the Moon’s face as seen from Earth. As the Moon orbits Earth, the same hemisphere of the Moon always faces us, but part of the near side of the Moon is in sunlight, and part of it is in darkness. The phases of the 29.5-day lunar cycle are outlined in Table 1.1. A helpful way of visualizing how lunar phases work is shown in Figure 1.22, which shows the Moon revolving around Earth in a counter-clockwise direction. The phases of the Moon are determined by the positions of the Sun, Earth, and Moon. The Moon emits no light of its own; we only see light that is reflected off it. Note that the side of the Moon facing the Sun is illuminated and the side facing away from the Sun is dark. As the Moon orbits Earth, varying amounts of light and dark are visible from Earth’s surface. When the Moon is in the position represented by the right side of Figure 1.22, the side of the Moon facing Earth is completely in darkness, while the side of the Moon facing away from Earth is fully illuminated. This is called the new Moon. When the Moon is new, its position in the sky is close to the position of the Sun, so it rises in the morning, sets in the evening, and is invisible to us. A couple of days after the new Moon, the Moon has advanced in its orbit around Earth, and a thin sliver is visible in the western sky just after sunset. This is called the waxing crescent; the term waxing means increasing or growing. If you observe the Moon from night to night, you see that the line that separates light from dark slowly moves across the Moon’s face from right to left. About a week after the new Moon, the 18

Earth In Space

Phase New Moon

Waxing Crescent

First Quarter

Waxing Gibbous

Full Moon

Waning Gibbous

Last Quarter

Moonrise

Moonset

Viewing

Sunrise

Sunset

Moon not visible

Just after sunrise

Just after sunset

Moon visible in the west just after sunset

Noon

Midnight

Moon visible in afternoon and early evening

Late afternoon

After midnight

Moon visible most of the night

Sunset

Sunrise

Moon visible all night

Before midnight

Mid-morning

Moon visible late night until midmorning

Midnight

Noon

Moon visible after midnight until noon

Waning Crescent Just before sunrise Just before sunset

Moon visible in the east just before sunrise

Table 1.1. Phases of the Moon, with moonrise and moonset times.

Moon has completed a quarter of its orbit around Earth and it appears half illuminated. Rather than calling this a half Moon, however, it is referred to as a quarter Moon—in this case, the first quarter (top of Figure 1.22). After the first quarter, the near side becomes more fully illuminated, with more than half but less than all of the near side showing, a phase called the waxing gibbous. A full two weeks after the new Moon, the Moon is fully illuminated when viewed from Earth, and we see a full Moon (far left of Figure 1.22). During the following two weeks, the process reverses (lower part of Figure 1.22). A few days after the full Moon, darkness starts to slowly creep across the Moon’s face from right to left, leading to the waning gibbous phase. Waning means decreasing, the opposite of waxing. The last quarter occurs a week after the full Moon, and a few days later, the waning crescent appears in the sky shortly before 19

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sunrise. After 29.5 days, the cycle repeats itself, starting sunlight Waxing Waxing First Quarter with a new Moon. Crescent Gibbous Something many people don’t realize is that the Full New Moon is in the sky during the day time just as often Waning Waning it is in the sky during the Gibbous Crescent Last Quarter night time. We don’t notice the daytime Moon as often for a couple reasons. First, Figure 1.22. As the Moon orbits Earth, varying amounts of the near the sky is bright during the side of the Moon become illuminated by the Sun, causing lunar day and it is easy to miss phases. a pale crescent or quarter Moon—or even a gibbous Moon—against the bright blue sky. Second, even though the Moon is in the sky when it is at or near the new Moon phase, there is simply not enough of the sunlit side of the Moon visible for us to see. The time of day that the Moon rises and sets goes hand in hand with the phases of the Moon. For example, the full Moon is on the opposite side of Earth from where the Sun is, so the full Moon rises roughly when the Sun sets and sets roughly when the Sun rises. Each day, the Moon rises about 50 minutes later; thus three days after the full Moon, the waning gibbous Moon rises about 2.5 hours (150 minutes) later. Table 1.1 summarizes when the Moon rises and sets for different phases. Learning Check 1.4 1. Explain why the Moon has phases. 2. Why does the full Moon rise at roughly the same time that the Sun sets? 3. If you were on the Moon, how would the appearance of Earth change throughout the 29.5-day lunar cycle?

1.5

Eclipses

In addition to phases, the orbit of the Moon around Earth causes another interesting phenomenon: eclipses. An eclipse occurs when one celestial body, such as Earth or the Moon, blocks the Sun from another celestial body.

1.5.1 Solar Eclipses A solar eclipse occurs when the Moon passes in front of the Sun, causing the Moon’s shadow to fall on Earth. In a partial solar eclipse, the Moon only covers part of the Sun’s disk, as shown in Figure 1.23. In the case of a total solar eclipse, the Moon completely covers the Sun, as in Figure 1.24. During a total solar eclipse, 20

Earth In Space

the sky becomes dark, but the total part of the eclipse lasts for only a few minutes. Solar eclipses can only happen when the Moon is directly between the Earth and the Sun, when there is a new Moon, illustrated in Figure 1.25. A shadow from a light source, such as the Sun or a light bulb, has multiple parts to it. The darkest, inner part of the shadow is called the umbra. During a solar eclipse, the Moon completely blocks the Sun within the umbra. The Figure 1.23. Partial solar eclipse. (Caution: looking penumbra is the part of the shadow that directly at a partial eclipse can damage your eyes!) is not completely dark. Within the penumbra, the Sun is not completely blocked, so a viewer sees a partial solar eclipse. The umbra and penumbra are easy to demonstrate, using a table lamp and a disk of some type, as shown in Figure 1.26. Total solar eclipses are fairly rare events, and most people have never seen one. (I have seen a total solar eclipse once, in 1979). Looking back at Figure 1.22 (lunar phases), one might think that a solar eclipse should occur every month. But the Moon orbits Earth in a plane that is about 5° tilted relative to the ecliptic, the plane in which Earth orbits the Sun. This means that in most months, the new Moon passes either above or below the Sun in the sky rather than right in front of it. In some years, there are no total solar eclipses. In other years, there may be up to five solar eclipses, although this only happens rarely. Because of the small size of the umbra on Earth, total solar eclipses are only visible in narrow bands on Earth’s surface. There is actually more to the Sun than just the bright disk we normally see in the sky. When that disk is blocked, we can still see the solar atmosphere, appearing as a thin ring of pink light around the eclipse and a faint white glow that extends farther out (Figure 1.24). The solar atmosphere is made of hot glowing Figure 1.24. Total solar eclipse. This is the only time that the solar atmosphere is visible from Earth. plasma, an ionized gas. 21

Chapter 1

Earth orbit Moon orbit penumbra umbra

penumbra umbra

Figure 1.26. Umbra and penumbra. An ant within the umbra would not be able to see the light bulb, and an ant in the penumbra would be able to see part of the light bulb.

Sun

The Sun is 400 times as large as the Moon, but it is also 400 times farther away. The result is that the Moon and the Sun appear to be almost exactly the same size in the sky. It is truly amazing that we live on a planet that has total solar eclipses in which the Moon almost perfectly covers the Sun. This requires a Moon that is just the right size, at just the right distance from Earth. Being able to see solar eclipses is not just something that is cool, but something that has been critical in the development of science. For example, total solar eclipses helped us to understand the composition of the Sun, and helped to confirm the validity of Albert Einstein’s general theory of relativity.

Figure 1.25. A solar eclipse occurs when the Moon blocks the light of the Sun. Viewers located in the umbra, which is very small on Earth, see a total solar eclipse; viewers located in the penumbra see a partial solar eclipse.

umbra penumbra Earth orbit Moon orbit

Sun

Figure 1.27. A total lunar eclipse occurs when the Moon passes through Earth’s umbra.

1.5.2 Lunar Eclipses Lunar eclipses occur when Earth’s shadow covers the face of the Moon. This occurs when the Moon is full, when Earth is between the Moon and the Sun, but does not happen every time the Moon is full. Again, because the Moon’s orbit around Earth is inclined at 5° to the ecliptic, the Moon usually misses Earth’s shadow as it orbits. Like all shadows, Earth’s shadow has an umbra and penumbra, illustrated in Figure 1.27. When the Moon passes through Earth’s penumbra, it becomes slightly

Earth In Space

darker, but this dimming can be rather subtle. A total lunar eclipse occurs when the Moon passes completely into Earth’s umbra. During a total lunar eclipse, the Moon does not become completely dark, even though Earth completely blocks the Sun. If you were standing on the Moon at this time, you would see Earth completely blocking the Sun, with a brilliant red sunset forming a ring around the entire Earth. This red light actually illuminates the Moon during a total lunar eclipse, giving the Moon the reddish glow shown in Figure 1.28. For this reason, the eclipsed Moon is often referred to as a “blood Figure 1.28. The Moon during a total lunar eclipse. The reddish glow is caused by sunlight bending Moon.” through Earth’s atmosphere. Earth’s shadow is much larger than the Moon’s shadow. Because of this, Earth’s umbra can completely cover the Moon (unlike the Moon’s umbra, which darkens only a small area on Earth). During a lunar eclipse, anyone on the night side of Earth can see the full Moon, and therefore can see the eclipse occurring. People on the day side of Earth miss the eclipse. Learning Check 1.5 1. Describe how the Sun, Earth, and Moon must be arranged for solar and lunar eclipses to occur. 2. During a solar eclipse, why do people in some places see a partial eclipse, while others see a total eclipse? 3. Explain why there aren’t solar and lunar eclipses every month.

1.6

Calendars

Closely related to Earth’s revolution around the Sun and the Moon’s orbit around Earth is the topic of calendars. The calendar we use in much of the world today, including in all of Western society, is based on Earth’s orbit around the Sun. However, before examining our Western calendar, we will take a brief look at some calendars that have been used in other parts of the world.

1.6.1 The Hebrew Calendar Most calendars are systems that divide the passage of time into years, months, and days, but the details vary. We will start by looking at the Hebrew calendar, used by the Jews in biblical times as well as in modern Israel.

Chapter 1

The Hebrew calendar is an example of a lunisolar calendar—one based on both the phases of the Moon and the orbit of Earth around the Sun. The Moon orbits Earth once every 29.5 days, so each month in the Hebrew calendar has either 29 or 30 days, giving a total of 354 days in twelve months. Each month starts with the new Moon. Earth does not completely orbit the Sun in 354 days, and the Hebrew calendar compensates for this by adding a thirteenth month in seven out of every nineteen years. This adjustment is necessary so the months occur in a consistent time of the year. Without this correction, the seasons—and associated religious holidays—would occur about eleven days earlier each year. For example, the original Passover, as recorded in the biblical book of Exodus, occurred in the spring. Without the addition of a leap month every few years, Passover would occur in April for about three years, then in March for a few years, and so on. The calendar was also used for agricultural purposes as well, such as determining times for planting and harvesting. The numbering of years according to the Hebrew calendar is based on a medieval calculation of the date for the creation of the world. The Jewish New Year celebration, Rosh Hashanah, usually occurs in September, with the Hebrew year 5781 beginning in September of 2020.

1.6.2 The Islamic Calendar The Islamic calendar is used by Muslims throughout the world for tracking religious holidays and as the day-to-day calendar for business and government in many Islamic countries. The Islamic calendar is a lunar calendar in which months are based strictly on the phases of the Moon. Each new day begins at sunset. Like the Hebrew calendar, each month in the Islamic calendar has either 29 or 30 days. A new month begins at the first sighting of the thin crescent Moon the day after the new Moon, shown in Figure 1.29. This determination is usually made by religious authorities for a given country, and if they do not see the crescent Moon for some Figure 1.29. Months in the Islamic calendar begin with the first sighting of the waxing crescent Moon.

Earth In Space

reason (such as cloud cover), then the new month doesn’t begin until the next evening. Traditionally, it has been difficult to create a calendar for future months of the Islamic year, or to make a firm appointment for a given date in a future month. A person could not predict in advance whether the current month would have 29 or 30 days; it depended on whether the crescent Moon was visible in the western sky at the end of the 29th day. Some Islamic countries now allow astronomical calculations to determine when new months will begin, but others continue to rely on actual observations of the sky. The Islamic calendar has twelve months that average 29.5 days long, for a typical total of 354 days in a year. Unlike the Hebrew calendar and other lunisolar calendars, there are no leap months. In North America and Europe, we associate certain months and holidays with certain seasons—Christmas is in winter, and July is a summer month. Islamic months and holidays, on the other hand, come eleven days earlier each year, so the months and holidays migrate through the seasons. For example, the month of Ramadan—the month in which adult Muslims are obligated to fast from food and water from sunrise to sunset every day—occurs in the summer for a few years, then in the spring for a few years, and so on. The Islamic year 1442 begins in August of 2020.

1.6.3 Western Calendars There have been two main calendars used in Western society over the past 2,000 years. They are both solar calendars, based primarily on Earth’s orbit around the Sun. The first of these is the Julian calendar, named after Julius Caesar, who introduced it in 46 BCE (46 BC). It was very similar to the calendar we use today. There were twelve months with either 30 or 31 days, except February which normally had 28 days, making a total of 365 days in a year. Every fourth year was declared to be a leap year, with a 29th day in February. Originally, the Julian calendar had no consistent way of numbering years. Sometimes a year would be indicated by the reign of an emperor, as in the sentence “In the fifteenth year of the reign of Tiberius Caesar.” The idea of numbering the years starting with the birth of Jesus—Anno Domini (Latin for “year of the Lord”), or AD—originated in about AD 525, and gradually spread throughout the Western world. Unfortunately, the calculations for determining the birth of Jesus were off by a few years—Jesus was probably born around 4–6 BC (BC means Before Christ). There was no year zero; 1 BC was followed by AD 1. It is now common to use CE (Common Era) and BCE (Before Common Era) instead of AD and BC. Either way, the calendar we use has its origin date as the birth of Jesus. There was a minor problem with the Julian calendar that grew with time. The average length of a year in the Julian calendar is 365.25 days, but Earth actually takes about 365.24 days to orbit the Sun. Over a period of many centuries, this means that months and religious holidays migrate relative to the seasons. Given enough time, the observance of Easter would eventually have become a winter, rather than a spring, event. In 1582, Pope Gregory XIII introduced a revision to 25

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the Julian calendar, known today as the Gregorian calendar. It took over three hundred years for all the countries of Europe to adopt the Gregorian calendar; Great Britain and its American colonies switched to the Gregorian calendar in 1752. It has become the international standard calendar, even in countries that use other calendars for religious or cultural purposes. The main difference between the Julian and Gregorian calendars is the reduced number of leap years in the Gregorian calendar. According to the Julian calendar, any year divisible by four is a leap year. The Gregorian calendar requires that years divisible by 100 must also be divisible by 400 in order to be a leap year. The years 1700, 1800, and 1900 were leap years according to the Julian calendar, but not leap years according the Gregorian calendar. The year 2000 was a leap year because it is divisible by 400. Some Christian holidays have fixed dates, such as Christmas, which falls on 25 December on the Gregorian calendar. Other church holidays, such as Easter, are actually based on a lunisolar calendar. Easter occurs on the first Sunday after the first full Moon following the March equinox—the vernal equinox in the Northern Hemisphere. Learning Check 1.6 1. Contrast the Hebrew and Islamic calendars, distinguishing between the lunar calendar and the lunisolar calendar. 2. Describe the features of the Julian and Gregorian calendars. Explain why it was necessary to switch from one to the other.

Earth In Space

Chapter 1 Exercises Answer each of the questions below as completely as you can. Write your responses in complete sentences unless instructed otherwise. 1. Give a definition for each of the four major Earth systems and describe how these systems interact with each other. 2. Draw a sketch of our solar system, showing the Sun, planets, and habitable zone. 3. What are some ways in which Earth appears to be specially designed for living organisms to thrive? 4. Describe what is meant by the term “galactic habitable zone.” 5. Describe Earth’s orbit around the Sun and explain why the shape of this orbit is not the primary cause of seasons. 6. Draw a sketch like Figure 1.18 and use it to explain the cause of Earth’s seasons. 7. Explain why it is that, outside the tropics, summer days are warm or hot and winter days are cool or cold. 8. Predict what Earth’s days and seasons would be like if Earth’s axis were tilted at close to 90° rather than 23.5°. 9. Draw a sketch of Earth, showing the location of the Tropics of Cancer and Capricorn, and Arctic and Antarctic Circles. 10. Describe the path the Sun would take across the sky viewed from the North Pole on the summer solstice, autumnal equinox, and winter solstice. 11. Draw sketches of the eight phases of the Moon, in order, from the new Moon through the waning crescent. 12. Draw a sketch like Figure 1.22 and use it to explain the cause of lunar phases. 13. When people on Earth see a quarter Moon, what would Earth look like from the Moon? 14. Draw a sketch of a total solar eclipse with labels for the Sun, Moon, Earth, umbra, and penumbra. 15. Explain why solar eclipses are only visible in limited geographic areas, but lunar eclipses are visible from the entire night side of Earth. 16. Describe the similarities between the Hebrew and Islamic calendars, and then explain what the biggest difference is between how they work. 17. Compare the Julian and Gregorian calendars and explain how the Gregorian calendar corrected a problem with the Julian calendar.

Chapter 2 Visualizing Earth

The ancient Greek scholar Eratosthenes of Cyrene, who lived from about 276 to 195 BCE, is considered by many to be the first true geographer. Other Greek philosophers, such as Aristotle, had demonstrated that Earth was a sphere, but Eratosthenes was able to go further and use Sun angles at different locations to determine the actual circumference of the spherical Earth. He also determined that Earth is tilted on its axis. Eratosthenes wrote books about Earth, and correctly divided our world into polar, temperate, and tropical climate zones. He recognized that cities and countries at the same distance from the Equator—what we now call latitude—had similar climates. He created a map that showed latitude lines running parallel to the Equator, as well as lines that ran perpendicular to them—lines we now recognize as lines of longitude.

Objectives After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary. 1. Read latitude and longitude values from a map, with a precision of minutes. 2. Explain why navigators have been able to determine latitude since ancient

3. 4. 5. 6. 7. 8. 9.

times, but it wasn’t until the 1700s that navigators could accurately determine longitude at sea. Describe how GPS receivers use satellites to determine locations on Earth. Explain what map projections are and why they are necessary for the creation of maps. Describe the shape of Earth. Define remote sensing and give examples of how remote sensing is used to study Earth. Distinguish between GPS and GIS. Interpret features on a topographic map, especially contour lines. Calculate gradients and percent slopes from a topographic map.

Vocabulary Terms You should be able to define or describe each of these terms in a complete sentence or paragraph. 1. 2. 3. 4. 5. 6. 7. 8. 9.

2.1

azimuthal projection cartographer cartography conic projection contour interval contour line coordinate system cylindrical projection echo sounding

10. electromagnetic spectrum 11. Geographic Information System (GIS) 12. Global Positioning System (GPS) 13. gradient 14. index contour line 15. latitude 16. longitude

17. map legend 18. map projection 19. map scale 20. oblate spheroid 21. percent slope 22. prime meridian 23. remote sensing 24. scale bar 25. topographic map

Locations on Earth

We do not have an original copy of the ancient world map made by Eratosthenes, shown at the beginning of the chapter. The map shown in the image is a reconstruction based on written descriptions in the writings of Eratosthenes and other ancient scholars. But we know Eratosthenes had a good understanding of the size and shape of our planet, as well as of the distribution of climate and vegetation zones. He also introduced the general concepts of latitude and longitude, which have been refined in the centuries since. You may have already discovered that maps are fascinating. It is easy to pick up a map to look up a location, only to find yourself studying all the details for an hour or more! Cartography is the art and science of making maps, and a person who makes maps professionally is a cartographer. Cartography is an art, since a 29

Chapter 2

cartographer must have creativity and a sense of what makes a map pleasing to the eye. It is also a science, because it involves measurements, analysis of data about Earth, and an understanding of topics such as ecology, landforms, and electromagnetic radiation. Obviously, modern cartographers have many more tools available to them than Eratosthenes did, such as computers, satellites, and advanced surveying equipment.

2.1.1 Latitude and Longitude Locations on Earth are specified using a numerical coordinate system. You are familiar with coordinate systems in mathematics. The Cartesian coordinate system locates points with an x-y ordered pair on a graph. The most widely used coordinate system for specifying locations on Earth North Pole is the latitude-longitude geographic coordinate 60°N system, in which locations are given as angles. 60° 30°N Latitude is the angle between lines drawn from 30° the center of Earth to the surface, as illustrated in Equator 0° Figure 2.1. The 0° line is the Equator and other 30° lines are measured north or south of the Equator. 30°S 60° Lines of latitude run parallel to the Equator and to 60°S each other, so they are always the same distance South Pole apart. Because of this, lines of latitude are sometimes called parallels. Navigators at sea have been able to determine latitude since ancient times. On a clear night in the Northern Hemisphere, it is fairly easy to determine the latitude by observing Polaris, the North Star. For an observer on Earth’s surface, the angle between the line of sight to the horizon and the line of sight to Polaris in the night sky is very close to the observer’s latitude. Polaris is almost directly above Earth’s North Pole, so if it is directly overhead, you know you are at 90°N (the North Pole). At a point half way between the Equator and the North Pole, Polaris is 45° above the hoFigure 2.1. Angles from Earth’s center are rizon and the latitude is 45°N. There is no clearly used to establish lines of latitude on the visible star directly above the South Pole, so there surface (top). Lines of latitude are parallel to each other and to the Equator, which is is no commonly used pole star in the Southern 0° (bottom). Hemisphere. Longitude is the angle east or west of a 0° line running from the North Pole to the South Pole. This 0° line is called the prime meridian. The lines of longitude, illustrated in Figure 2.2, are sometimes called meridians. Meridians are measured as angles east or west of the prime meridian, and longitude values range from 0° to 180°, for a total measurement of 360° around the entire planet. 30

Visualizing Earth

In the past, many countries had their own prime meridians; there was a meridian of Paris, one of London, another of Washington DC, and many others. In 1884, an agreement was reached at an international conference to recognize the meridian going through Greenwich, England, to be the international standard prime meridian. The straight north-south boundaries between many of the western states in the United States were originally defined as a certain number of degrees west of the Washington DC meridian, rather than in terms of the longitude west of Greenwich. Unlike lines of latitude, lines of longitude are Figure 2.2. Lines of longitude. The prime not parallel—they come together at the poles. At meridian runs through Greenwich, the Equator, one degree of latitude and one degree England, and is designated as 0°. of longitude cover very close to the same distance on Earth’s surface. As one travels closer to a pole, the lines of longitude get progressively closer to one another, as Figure 2.2 shows. Locations on Earth’s surface are expressed with latitude first, then longitude. In Figure 2.3, you see that New York City is located at 41°N, 74°W (stated as 41 degrees north, 74 degrees west), and Sydney, Australia, is at 34°S, 151°E. On a global scale, this might be sufficient to locate a place, but on a local scale, latitude and longitude may be expressed with a much higher degree of precision. Each degree of latitude or longitude is divided into sixty minutes, and each minute is divided into sixty seconds. Minutes are represented by the symbol ’ and seconds by the symbol ”. This enables us to state locations more precisely. For example, the Statue of Liberty in New York City is at 40°41’21”N, 74°2’41”W. (These coordinates locate the statue to within about 30 meters.)

Figure 2.3. Latitude and longitude of New York City and Sydney, Australia.

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Although mariners have known how to determine latitude at remote locations for many centuries, they did not have tools for accurately determining longitude until the mid-1700s. On land, people can use surveying techniques to measure distances and angles, and thus determine how far east or west a location is from a central meridian, but those methods cannot be used at sea. A common method for determining locations at sea was by dead reckoning, which was based on the estimated course of Figure 2.4. One of Harrison’s marine chronometers. a ship. A ship would leave from a known loThe invention of accurate clocks in the 1700s enabled navigators to make accurate plots of their cation and the navigator estimated how far positions at sea for the first time. the ship had travelled based on the speed of the ship through the water. Sometimes this worked well enough and ships arrived at their planned destinations. However, there were numerous instances when ships or entire fleets missed their destination and got lost at sea, crashed into rocks, or encountered other disasters. Because of this, seafaring nations sought more accurate means of determining longitude. A key step toward more accurately determining longitude at sea was the invention of highly accurate clocks. In the mid-1700s, British clockmaker John Harrison invented the marine chronometer, pictured in Figure 2.4. Harrison labored over a period of thirty years to produce a clock that was sufficiently accurate for navigation. The invention of the marine chronometer enabled navigators to use the positions of the Sun, Moon, planets, and stars at a specific time to make accurate determinations of both latitude and longitude. The navigator used a sextant like the one shown in Figure 2.5 to measure the angle between the horizon and an astronomical body—the Sun by day or Moon, planet, or star by night—and then consult books (called nautical almanacs) with printed tables of the positions of these bodies at various times. This allowed the navigator to determine a ship’s position to within a few miles, usually accurate enough to get the ship to its destination and avoid disasters. Figure 2.5. A sextant is used to measure the angle between the horizon and the Sun or other celestial body.

Visualizing Earth

2.1.2 Satellite Navigation Systems Today, locations on Earth can be precisely determined by using signals from satellite navigation systems. The most commonly used satellite navigation system is the Global Positioning System, or GPS. A GPS satellite is shown in Figure 2.6. The Global Positioning System always has a minimum of 24 satellites orbiting Earth. These satellites continuously transmit radio signals that are used to determine locations on Earth. Figure 2.7 shows how the satellites surround the planet. At any given moment, there are at least four satellites in the sky over any location. The radio signals Figure 2.6. A Global Positioning System (GPS) satellite, with solar panels to supply it with sent by the satellites contain the time the electricity. signal was sent and the position of the satellite when the signal was sent. Figure 2.8 shows a typical hand-held GPS receiver. The receiver determines the amount of time it has taken for the signal to travel from each of the satellites. Since the signals travel at the speed of light (300,000 kilometers per second), the receivers must be able to measure time intervals with a precision of billionths of a second. Some of the satellites are closer Figure 2.7. The 24 satellites of the Global Positioning System to the receiver (GPS) provide worldwide service. and others are farther away. The receiver calculates the distance to four different satellites with known positions and uses these four distances to calculate its own position, as illustrated in Figure 2.9. Satellite navigation now has hundreds of applications. Figure 2.8. GPS receiver with a map showing the These systems were originally developed for military use, but current location of the now are used for airplane, ship, and car navigation; surveying; receiver.

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studies of earthquakes and volcanoes; tracking wildlife; and recreational activities such as backpacking. Smart phones and tablet computers typically have GPS receivers built into them. The Global Positioning System is operated by the United States, but there are similar systems operated by Russia and China. Other satellite navigation Figure 2.9. The spheres represent different distances from GPS satellites. systems are being devel- The point where the surfaces of all four spheres intersect is the location oped by other countries. of the receiver. Learning Check 2.1 1. Compare lines of latitude to lines of longitude, explaining how they are similar and how they are different. 2. Explain how Harrison’s invention of the first marine chronometer enabled navigators to determine their longitude while at sea. 3. Describe how the GPS system is used.

2.2

Map Projections and the Shape of Earth

Imagine a balloon with a map of the United States drawn on it. If we were to pop that balloon, it would be a challenge to lay that map completely flat on a table. As shown in Figure 2.10, we would have to stretch and distort the rubber in order to get the United States or any other portion of the balloon to be flat. This is the same problem cartographers have when they take regions on the roughly spherical Earth and make maps on flat surfaces, such as a sheet of paper or computer monitor. The true sizes, angles, and shapes of features on Earth—continents, oceans, countries, and so forth—must be disFigure 2.10. The curved surface of the United States fits properly on torted in order to construct the inflated balloon, but the popped balloon has to be stretched for a flat map. These distortions the United States to fit on a flat surface. 34

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are obvious on maps that show large areas, like those in Figure 2.11, but distortions exist even on maps of small areas, such as on a map of a city.

2.2.1 Map Projections A map projection is a method by which the curved surface of Earth is portrayed on a flat surface. To visualize how projections work, imagine drawing lines from the center of the globe through features on Earth’s surface, and extending these lines onto a surface that can be flattened out. There are numerous projections in use, but most of them fall into a few categories. A common type of projection is a cylindrical projection, constructed by extending features on Figure 2.11. Both these maps distort the size and shape of the United States and Canada. Earth’s surface onto a cylinder wrapped around The only true portrayal of the size and shape Earth, as illustrated in Figure 2.12. There are of features on Earth’s surface is on a globe. several ways of doing this. A common example is the Mercator projection, which portrays all lines of latitude and longitude as parallel to each other. Because lines of longitude in reality meet at the poles, the Mercator projection distorts the sizes of features closer to the poles. If you look at a globe—on which features are portrayed in their true proportions—you see that Greenland is considerably smaller than South America. However, on a Mercator projection, Greenland looks as large or larger than South America, as on the world map shown in Figure 2.13. The advantage of a Mercator projection is that both Greenland and South America are preserved in roughly their true shapes. Another advantage is that straight lines drawn on a Mercator projection map follow a constant compass direction, and thus are easy to use for navigation. Mercator projections are frequently used for world maps and are also often used for maps of smaller areas, such the topographic map shown in Figure 2.14. A conic projection is constructed by extending features on Earth’s curved surface onto a cone, as prime meridian illustrated in Figure 2.15. This reduces distortion of features near the latitudes at which the imagiEquator nary cone intersects Earth’s surface. Maps of entire countries or continents are commonly drawn using Figure 2.12. For a cylindrical projection, features on a globe are projected conic projections. The onto a cylinder that is wrapped around the globe. 35

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Figure 2.13. A world map on a Mercator projection, from 82° S to 82° N.

projection is designed so there is a minimum amount of distortion of size or shape in the area of interest. Usually, features farther from the center of the map are more distorted, as is apparent on the north and south edges of the map in Figure 2.16. An azimuthal projection, shown in Figure 2.17, is created by projecting Earth’s surface onto a plane that touches the planet at only one point. There is no distortion of features at this point, and the farther features are from that point the larger they become. The most Figure 2.14. A topographic map of Mount Shasta, California, on a Mercator projection. For maps of small areas such as this, distortions due to the map projection are not significant for most purposes.

Visualizing Earth two standard parallels define the map layout

Figure 2.15. For a conic projection, features on a globe are projected onto a cone that is placed over a globe.

common type of map that uses an azimuthal projection is one centered on one of Earth’s poles. In Figure 2.18, the Arctic Ocean is portrayed much as it would be on a globe, with little distortion of size or shape. Features become increasingly distorted as the distance from the North Pole increases.

Figure 2.16. A map of North America on a conic projection. This projection preserves the size and shape of features near the central latitudes of the map, but distortions increase to the north and south.

2.2.2 The Shape of Earth There are many variations of each of these map projections, each with advantages and disadvantages. The only way to really portray Earth without significant distortions is on a globe. But a perfectly spherical globe still involves slight distortions. As I have described the Earth, I intentionally used phrases like “roughly spherical.” In fact, Earth actually has an equatorial bulge—it is slightly flattened at the poles and has a greater circumference when measured around the Equator than when measured from pole to pole. A more accurate term for Earth’s shape is oblate spheroid. An oblate spheroid, portrayed in Figure 2.19, is a flattened sphere where the Equator has a greater radius than the measurement from the center to a pole. If gravity were the only force acting on material in Earth, it would be a sphere, but Earth’s daily rotation on its axis

plane of projection Equator

Figure 2.17. For an azimuthal projection, features on a globe are projected onto a plane that touches the globe at one point.

Figure 2.18. A map centered on the North Pole with an azimuthal projection. Features near the North Pole have little distortion of size or shape, but areas at a greater distance from the pole are noticeably stretched out.

Chapter 2 Figure 2.19. Rather than being a perfect sphere, Earth is slightly flattened at the poles and has a slight bulge at the Equator. This shape is called an oblate spheroid. The amount of flattening is greatly exaggerated in this diagram.

N causes it to flatten at the poles relative to the Equator. Polar Radius The difference be6356.752 km tween Earth’s radius measured through the Equatorial Radius poles and through the Equator is less than 1%, 6378.137 km which is why Earth seems spherical. However, all cartographic and scientific measurements of Earth, such as latitude and longitude, are actually made by considering Earth as an oblate spheroid S rather than as a sphere. The most commonly acFigure 2.20. Earth’s radius measured cepted measurements for Earth’s radius are indi- through the Equator is slightly greater cated in Figure 2.20. than its radius measured through the poles.

Learning Check 2.2 1. Explain why it is impossible to make an accurate map of Earth on a flat sheet of paper without using a projection. 2. Describe the difference between cylindrical, conic, and azimuthal projections. 3. Describe the shape of Earth.

2.3

Remote Sensing

Remote sensing is the collection of information about Earth and the environment from a distance, usually by detecting and recording energy from the electromagnetic spectrum emitted from or reflected by Earth’s surface, vegetation, water, or atmosphere. The electromagnetic spectrum, illustrated in Figure 2.21, includes electromagnetic energy ranging from high-energy gamma rays and X-rays to lowenergy radio waves, with the visible light spectrum in the middle. All portions of the electromagnetic spectrum are used in remote sensing. Radio waves, on the left side of Figure 2.21, have long wavelengths but low energy. You can remember that radio waves have low energy when you think of how many radio signals are going through you right now—from every radio and TV station in your area, from cell phones, and from other sources—with no apparent effect. X-rays and gamma rays are on the other end of the spectrum. They have very short wavelengths—measured in billionths or trillionths of a meter—but very high energy. 38

Visualizing Earth

0.000000000001 m = 0.001 nm

0.00000000001 m = 0.01 nm

0.0000000001 m = 0.1 nm

0.000000001 m = 1 nm

0.00000001 m

0.0000001 m

0.000001 m = 1 μm

0.00001 m

0.0001 m

0.001 m = 1 mm

0.01 m = 1 cm

0.1 m

10 m

100 m

1000 m = 1 km

Data from remote sensing is VISIBLE often presented in RADIO WAVES microULTRAGAMMA the form of pictures INFRARED X-RAYS waves VIOLET RAYS AM Radio FM Radio referred to as imagery (or sometimes referred to as aerial or satellite photography). If you have used internet sites or apps on mobile devices to view an overhead picture of radiation wavelength shorter wavelength your house, school, longer wavelength lower energy higher energy or other location, Figure 2.21. The electromagnetic spectrum. you have viewed remote-sensing imagery. Some of the earliest aerial photography was taken using cameras mounted to balloons, kites, and even pigeons, as shown in Figure 2.22. Today, remote-sensing data are typically collected from above using airplanes or satellites, such as the NASA Landsat satellite shown in Figure 2.23. Much of this technology was initially invented for military and intelligence purposes, leading Figure 2.22. Before airplanes became common for to the development of what are commonly aerial reconnaissance, attempts were made to use referred to as “spy satellites,” but many oth- pigeons as aerial photographers. er uses have now been developed. Remote sensing satellites have provided regular coverage of most of Earth for over forty years. Imagery from these satellites is used to trace changes to forests, farmland, cities, water bodies, and other features. In addition to using visible light, many satellites record infrared radiation, which means the satellite detects wavelengths of light the human eye cannot see. This enables the sensor to distinguish changes in vegetation and soil moisture that are too subtle to be seen with visible light. Figure 2.24 shows the burn scar from a large forest fire Figure 2.23. Landsat satellites capture images of in Colorado. Healthy trees are shown with the entire Earth once every 16 days.

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Figure 2.24. Satellite imagery is used to analyze damage to forests due to wildfires. In this image, red is healthy vegetation, gray is bare ground or man-made features, and dark brown is the burn scar from a wildfire.

Figure 2.25. Remote sensing images from weather satellites give detailed and timely information about Earth’s atmosphere. In this image centered on North America, the highest clouds, which are associated with the tops of strong storm systems, are red and orange.

Figure 2.26. Echo sounding uses sound waves transmitted through water to determine water depth.

red to emphasize that these data were collected using infrared wavelengths, in addition to visible light. Infrared wavelengths are also useful for geological investigations, such as searching for mineral deposits. Remote-sensing images are important for many uses in oceanography and

Visualizing Earth

meteorology. In the past, weather observations were limited to those made at manned weather stations and there were large parts of Earth where storms could develop without anyone knowing about them. Now we have a continuous global picture of what is happening in the atmosphere. The first weather satellites sent back images that enabled forecasters to see the extent of cloud cover, but not much more. Instruments on modern satellites detect the temperatures of cloud tops, as shown in Figure 2.25, and measure wind speeds, precipitation, lighting strikes, concentrations of various pollutants, and much more. Since Earth’s land surface, oceans, and atmosphere emit Figure 2.27. A map of a submarine volcano in the infrared radiation at night as well as dur- Aegean Sea, with elevations determined using ing the day, weather satellites even gather echo sounding. The main crater is about 1.5 km (1 mi) across. The deepest part of the crater remote-sensing data in the dark. Not all remote sensing data is pro- (purple) is about 500 m (1640 ft) below sea level, the shallowest part of the crater rim (white duced using electromagnetic radiation. An and and dark red) is about 10 m (33 ft) below the example of a different technology is echo ocean surface. sounding, which uses sound waves rather than electromagnetic radiation to determine the depth of water in oceans or lakes. By measuring the time required for the sound waves to travel from the ship to the sea floor and back to microphones in the water, oceanographers are able to determine the depth of the water, as illustrated in Figure 2.26. It takes a large amount of echo-sounding data to map even a small part of the ocean floor, such as the map of a submarine volcano depicted in Figure 2.27. Sound waves are also transmitted into the solid Earth to determine the structure and layering of rocks in Earth’s Figure 2.29. Use of radar on NASA’s crust and deeper. This tech- Magellan probe allowed scientists nique is commonly used by to make a complete map of Venus. geologists exploring for pe- The colors are added by computer software. As in Figure 2.27, blue troleum deposits. areas are low in elevation and Probes sent to other red and white areas are high. The Figure 2.28. Because of its thick parts of the solar system also surface of Venus is extremely hot layers of clouds, it is impossible use remote sensing to gather and dry, with an average surface to see the surface of Venus from space using visible light. information, using the same temperature of 462°C (863°F). 41

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principles as those used for observing Earth. We would know very little about the surfaces of Mercury, Venus, Mars, or the moons of the outer planets if it were not for remote sensing. For example, Venus is enveloped by a thick layer of clouds, as shown in Figure 2.28, and its surface is never visible from space. However, the surface of Venus has been mapped in great detail using radar mounted on the Magellan probe. Radar determines distances using radio waves. The transmitter sends out radio waves that bounce off the surface of a planet and back to a receiver on the satellite. The Magellan probe was able to gather data to make an elevation map of almost the entire surface of Venus, as shown in Figure 2.29. Learning Check 2.3 1. List the six main regions of the electromagnetic spectrum, from low energy to high energy. 2. Describe a way that remote sensing is used to study each of the four Earth systems (lithosphere, hydrosphere, atmosphere, and biosphere).

2.4

Mapping Earth

A map is a model of the world or part of the world, showing natural and manmade features. Constructing a map involves mathematics, computer skills, geographic knowledge, and a lot of patience. A well-made map is not just a technical document; it must also be pleasing to the eye. The “art” side of cartography involves making decisions about colors, text placement, and map symbols. These choices make the difference between constructing a well-made map and one that is difficult to read or even misleading. In the past, maps were usually printed on paper. Now, people increasingly turn to digital maps—on computer screens or mobile devices— on a day-to-day basis. There are many types of maps—road maps, political maps, maps that show the spread of diseases, and on and on. Maps are used in Earth sciences to present information about all sorts of topics, ranging from weather maps and maps of ocean surface temperatures, to maps showing the topography of mountain ranges or the sea floor. You have heard the phrase, “a picture is worth a thousand words.” The same concept applies to cartography: “a map is worth a thousand words.” The use of maps powerfully and quickly communicates complex concepts to both scientific and general audiences. Consider a map such as the one in Figure 2.30 depicting the average wind speed 80 meters above the ground for the United States. This map was constructed as part of an effort to identify the best locations for placing wind turbines. Wind turbines are used to generate electricity. The purple areas are those with the highest average wind speeds and the yellow and green areas are the areas with lowest average wind speeds. You can quickly look at the map and see that the areas with strongest winds are in the Great Plains, from northwest Texas up through North Dakota.

Visualizing Earth

Figure 2.30. Map showing average wind speed, in meters per second, at 80 meters above the ground.

2.4.1 Geographic Information Systems All maps are prepared using map projections because the features portrayed on the map must be distorted in order to fit on a flat surface. In the past, this was a laborious process, but it is now done quickly by computers. Almost all maps, whether paper or digital, are now made using Geographic Information Systems. A Geographic Information System (or GIS) is computer software that analyzes data about Earth and presents it in the form of maps. The data contained in the GIS may include imagery; elevation data; and points, lines, and polygons that represent natural and man-made features such as towns, roads, streams, lakes, and political boundaries. Using GIS software, maps that would have taken months to construct in the past are typically produced in a matter of hours or days. There is much more to a GIS than just making maps. Cartographers or other GIS users also compare different sets of data to help make decisions. For example, a forester might use GIS to compare soil types, elevations, and sun angles in order to decide the best places planting tree seedlings after a forest fire. This was a very time-consuming task before the availability of GIS technology.

2.4.2 Topographic Maps An important type of map in Earth sciences is the topographic map. A topographic map represents the surface of Earth, showing elevations by the use of con43

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tour lines or color tints (e.g., the color bands shown in Figure 2.27). Topographic maps also show other natural and man-made features, such as streams, lakes, forests, roads, and buildings. They are used by geologists, foresters, wildlife biologists, engineers, planners, military personnel, campers, and backpackers. In the United States, detailed topographic maps have been made for the entire nation by government agencies such as the U.S. Geological Survey and the U.S. Forest Service. A section from a U.S. Geological Survey topographic map is shown in Figure 2.31. The topographic maps made by these agencies are referred to as quadrangles. Most other countries have similar topographic mapping programs. Most topographic maps portray elevation using contour lines—curves that connect points of equal elevation. The brown curves in Figure 2.31 are contour lines. Each point along the 900-foot contour represents an elevation of 900 feet above sea level, and each point along the 1,100-foot contour represents an elevation of 1,100 feet above sea level. The shapes of the contour lines on the map reflect the shape of the land surface. The starting point for measuring elevation is sea level, which is assigned a value of zero. Most topographic maps in the United States show el-

Figure 2.31. A part of a topographic map showing roads, buildings, water bodies, trees, and elevations.

Visualizing Earth

26 0

evations in feet, but military topographic maps and topographic maps produced by other countries indicate elevations in meters. Figure 2.32 illustrates the relationship between contour lines and landscape. In places where the land is relatively level, such as along the stream in the center of the diagram, the contour lines are far apart. Where the slope is steep, the contour lines are close to each other. There are four basic principles associated with contour lines, illustrated in Figures 2.33, 2.34, and Figure 2.32. The relationship between 2.35: landscape and contour lines.

275

1. As just stated, where contour lines are far apart the slope is gentle; where they are close together the slope is steep. 2. Contour lines never cross each other. On some topographic maps, they can touch each other where there are cliffs, but they do not actually cross each other. 3. Where a contour line crosses a stream on a topographic map, the contour line is Vshaped, with the V pointing upstream toward areas with higher elevation. 4. Contour lines form closed loops around elevated areas (e.g., hills) and depressed areas (e.g., craters). For example, a hill on a topographic map looks like a series of concentric, irregular loops. Often, the contour lines extend beyond the edges of an individual map so you don’t always see the complete closed loop.

Figure 2.33. Principles 1 and 2—Contour lines are far apart in flat areas, and close to each other in steep areas, but contour lines never cross.

The contour interval of a topographic map is the difference in elevation between adjacent contour lines. If the contour interval is 20 feet, then contour lines are shown for every multiple of 20 feet, such as at 100, 120, 140, and 160 feet. In flat areas, there is not a great elevation difference from one part of a map to another, so a topographic map of an area of plains might have a contour interval of only 10 feet. Most hilly arFigure 2.34. Principle 3—Contour lines make eas can be sufficiently portrayed with a 20-foot a V where they cross streams, with the V contour interval; mountainous areas typically pointing upstream. 45

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have contour intervals of 40 feet. If a mountainous area were portrayed with a 10-foot contour interval, the contours would be so close to each other that they would blend together. In order to make a map easier to read, every fourth or fifth contour line, known as an index contour line, is drawn with a heavier line weight and usually labeled to indicate its elevation. If the contour interval on a map is 20 feet, then every fifth contour line has a value that is a multiple of 100 Figure 2.35. Principle 4—Contour lines form feet and desigclosed loops around hills. nated as an index contour line. Contour interval and index contours are depicted in Figure 2.36.

2.4.3 Map Margin Information Some of the most important features of a top- Figure 2.36. This map has a contour ographic map (or other types of maps) are found interval of 40 feet, with index contours not in the main part of the map but in the margins every 200 feet. around the map’s edge. The margin gives information about the location of the map, publisher, date, and accuracy of the map. One important element of the map margin is the map scale. The map scale is the ratio between a length on a map and the corresponding horizontal distance on the ground. For example, a map scale of 1:100,000, means that features shown on the map are one-100,000th their true size. It also means that one unit of measurement on the map represents 100,000 of those same units on Earth. In this case, 1 cm on the map represents 100,000 cm on the ground. This is a convenient scale, because it means that 1 cm on the map also represents 1 km on the ground. (There are 100,000 cm in 1 km.) Another way of expressing map scale is with a scale bar. A scale bar graphically represents the scale of the map and consists of a line with marks like a ruler. The map scale and scale bar for a U.S. Geological Survey topographic map are shown in Figure 2.37. Another important part of the map margin is the legend, an example of which is shown in Figure 2.38. A map legend is a key that indicates what each symbol on the map repreFigure 2.37. Scale bar and map scale from a topographic map. sents. Many sym46

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bols on a map are designed to look like the features they symbolize. For example, trees are shown with a green tint and water bodies are depicted with blue. However, many other symbols are not the same color as the features they represent. Manmade features such as buildings are usually black, regardless of what color they really are.

2.4.4 Gradient and Percent Slope We can describe the slope of a geographic feature such as a hillside, stream, or road as gentle or steep, or we can actually measure a slope to determine its gradient. Gradient is a mathematical measurement of the rate of change of the elevation of Earth’s surface over a given horizontal distance. In algebra, the slope of a line is the ratio of “rise” over “run.” The gradient of Earth’s surface is measured in a similar way. One way to express gradient is as the rise or drop in elevation per horizontal distance. For instance, if a stream drops in elevation 50 ft in 2.0 mi, its gradient is calculated as

BUILDINGS AND RELATED FEATURES

CONTOURS 6000

RAILROADS AND RELATED FEATURES

RIVERS, LAKES, AND CANALS MINES AND CAVES

RIVERS, LAKES, AND CANALS – continued

ROADS AND RELATED FEATURES

SUBMERGED AREAS AND BOGS

VEGETATION

Figure 2.38. Simplified legend for U.S. Geological Survey topographic maps.

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gradient =

rise (or drop) 50 ft = = 25 ft/mi run 2 mi

percent slope =

rise feet × 100% run feet

0.47%

run = 10,560 ft

rise = 50 ft

When expressed this way, the units of gradient are always distance of rise or drop per distance of run, such as feet per mile (ft/mi) or meters per kilometer (m/km). Gradient is also often presented as a percent slope. Percent slope is the rise over run of Earth’s surface multiplied by 100%, as illustrated in Figure 2.39. When calculating percent slope, the units for the rise and run must be the same, so the calculation uses an equation such as

Let’s calculate the percent slope for the same stream. The stream drops 50 feet in 2.0 miles, but we percent slope = (rise/run) × 100% = (50 ÷ 10,560) × 100% must first convert the miles to feet. There are 5,280 = 0.47% feet in a mile, so the conversion is as follows: Figure 2.39. Calculating percent slope. (The sketch is not to scale; the slope is highly exaggerated to make it visible.)

5280 ft 2 mi ⋅ = 10,560 ft 1 mi The stream’s percent slope is calculated as follows: percent slope =

rise feet 50 × 100% = × 100% = 0.47% run feet 10,560

Note that because there are units of feet in both the numerator and denominator of this ratio, the length units cancel out and the result has units of percent; the answer is 0.47%, not 0.47 feet per foot. Gradient can be determined by interpretation of contour lines on topographic maps. To calculate the gradient of a hillside, for example, one determines the rise by calculating the difference between the elevation for the highest contour line and the lowest contour line. The run is determined by using the map scale to measure the horizontal distance between those points. A sample calculation based on a topographic map is shown in Figure 2.40.

Figure 2.40. Point A has an elevation of 8000 feet, and point B has an elevation of 7400 feet, a rise of 600 feet. The horizontal distance between A and B is 3600 feet. The percent slope is (rise/run) × 100% = (600 ft/3600 ft) × 100 % = 16.7%.

Visualizing Earth

Learning Check 2.4 1. Explain what a Geographic Information System is. 2. Explain how contours are used to portray Earth’s landscape on a topographic map. 3. What are the four basic principles of contour lines? 4. Explain what is meant by the term map scale.

Chapter 2 Exercises Answer each of the questions below as completely as you can. Write your responses in complete sentences unless instructed otherwise. 1. Use the map of the United States below to determine the name of the cities with the following coordinates: a. 40°N, 105°W

b. 42°N, 88°W

c. 26°N, 80°W

d. 46°N, 123°W

e. 41°N, 74°W

g. 42°N, 71°W

h. 30°N, 95°W

34°N, 118°W

2. Use the map of the United States above to determine the latitude and longitude, to the nearest degree, of the following cities: a. Albuquerque, New Mexico

b. Atlanta, Georgia

c. Dallas, Texas

d. Minneapolis, Minnesota

e. San Francisco, California

g. Toronto, Ontario, Canada

h. Washington, DC

Seattle, Washington

Chapter 2

3. Use the map of a portion of Yellowstone National Park below to determine the names of the features with the following coordinates: a. 44°59’N, 110°42’W b. 44°48’N, 110°27’W c. 44°38’N, 110°27’W d. 44°32’N, 110°50’W e. 44°25’N, 110°34’W

4. Use the map of a portion of Yellowstone National Park above to determine the latitude and longitude, to the nearest minute, of the following features: a. Fishing Bridge b. Lower Falls c. Norris Geyser Basin d. Old Faithful e. Tower Falls 5. Illustrate how features on a globe can be projected onto a cylinder, cone, and plane. 6. Describe the shape of Earth and explain why it is not spherical. 7. Describe an example of remote sensing using one of the non-visible parts of the electromagnetic spectrum. 8. What are some advantages of using satellites to observe Earth’s weather? 9. Explain why contour lines are far apart in flat areas and close together 50

Visualizing Earth

in steep areas. 10. Use the topographic map of High Top Mountain below to answer the following questions (elevation values are in feet). a. What is the contour interval of this map? b. What is the interval between index contours? c. Estimate an elevation value for the × at the top of High Top Mountain. d. What are the gradients between A—B and C—D, measured in feet per mile? e. What are the percent slopes between A—B and C—D?

Answers 1. a. Denver b. Chicago c. Miami d. Portland e. New York f. Los Angeles g. Boston h. Houston 2. a. 35°N, 107°W b. 34°N, 84°W c. 33°N, 97°W d. 45°N, 93°W e. 38°N, 122°W f. 48°N, 122°W g. 44°N, 79°W h. 39°N, 77°W 3. Mammoth Hot Springs b. Mount Washburn c. Mud Volcano d. Lower Geyser Basin e. West Thumb 4. a. 44°34’N, 110°23’W b. 44°43’N, 110°30’W c. 44°44’N, 110°42’W d. 44°28’N, 110°50’W e. 44°54’N, 110°23’W 10. a. 20 ft b. 100 ft c. between 1580 and 1600 ft, e.g., 1590 ft d. 1400 ft/mi, 480 ft/mi e. 26%, 9%

Chapter 2

Experimental Investigation 1: Interpreting Topographic Maps Overview Mount Shasta is a volcano in northern California, and is the second tallest volcano in the Cascade Range of western North America. Mt. Shasta last erupted somewhere between 200 and 300 years ago, and is considered to be a serious volcanic hazard. A good way to study and visualize a mountain like Mount Shasta is by interpreting a topographic map. In this investigation, you examine various features of a U.S. Geological Survey topographic Mount Shasta. A newer volcanic cone called Shastina is map of the Mt. Shasta area. You also forming on its west side (right side in this image). study how to construct topographic profiles, which are graphs showing cross sections along Earth’s surface. Basic Materials List • U.S. Geological Survey 1:24,000 topographic map of Mount Shasta, California, preferably 1998 edition. • Ruler • Graph paper (5 squares to an inch preferred) Part 1—Topographic Map Interpretation In your lab journal, record the latitude and longitude of each corner of the map, and draw a sketch of the map with labels as illustrated to the right. Then address the following questions: 1. What is the scale of this map? What does that mean?

45° 30’ N

113° W

44° 30’ N 112° W

2. What is the distance, in a straight line, between the summit of Mt. Shasta and Clarence King Lake? Give your answer in miles and meters. 3. What is the contour interval of this map? What does that mean? What is the interval between index contours? 4. What is the elevation of the summit of Mt. Shasta? How do you know? What is this value in meters? 52

Experimental Investigation 1: Interpreting Topographic Maps

5. What is the elevation of Sisson Lake? How do you know? What is this value in meters? 6. What is the elevation difference between the top of Bolam Glacier and the bottom of Bolam Glacier? What is the overall gradient of Bolam Glacier in feet per mile? What is the overall percent slope of Bolam Glacier? 7. What is the gradient along Road 43N21 in the northwest corner of the map? How does this compare to the gradient of Bolam Glacier? Describe how the contour lines relate to the gradients at these two locations. 8. The trail heading south from the Bolam trailhead (at the end of road 43N21 in Question 7) does not follow a straight line. Why not? 9. Where is the lowest elevation on the map and what is the elevation at this location? 10. Compare the size of the glaciers (blue contour lines) on the north side of Mt. Shasta with the size of the glaciers on the south side of Mt. Shasta. Form a hypothesis as to why there is a difference in the areas of these glaciers. Part 2—Constructing a Topographic Profile

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A topographic profile is a cross-section through Earth’s surface along a line. One can think of a topographic profile as a graph of what Earth’s surface would look like if one were to take a slice out of Earth’s crust and view it from the side. The first step in constructing a topographic profile is to lay a strip of paper along the profile line, and transfer contour values to the paper, as shown below.

The second step is to take the strip of paper and transfer the values to a graph on graph paper, as shown by the black dots below. For the vertical scale on your graph, use the same scale as shown on the map. On a 1:24,000 topographic map, one inch represents 24,000 in (2,000 ft) on Earth’s surface. In the topographic profile above, the horizontal and vertical scales are both 1:24,000, so 1 inch in either the horizontal or vertical directions represents 2,000 ft. Since the scales on the horizontal and vertical axes of the graph are the same, the graph visually represents the true slope of Earth’s surface. 53

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In some cases, such as in flat areas, it is preferable to have the vertical scale larger than the horizontal scale in order to emphasize subtle variations in the topography. This is called vertical exaggeration. Finally, draw a smooth curve to represent Earth’s surface along the profile line. Note that on the topographic profile, the curve is steeper where contour lines are closer together; the slope is more gentle where contour lines are farther apart. Construct topographic profiles from A to B for the three topographic maps shown on the next page.

Experimental Investigation 1: Interpreting Topographic Maps

Map 1: McCloud, California

Map 2: Rileyville, Virginia

Map 3: Renova West, Pennsylvania

Chapter 6 Plate Tectonics and Mountain Building

Mount Everest, in the Himalaya Mountains on the border between China and Nepal, is Earth’s highest mountain, with a summit 8,848 m (29,029 ft) above sea level. Like most mountain ranges, the Himalayas are a complex mountain range, with a mixture of igneous, sedimentary, and metamorphic rocks spanning a range of ages. The summit of Mt. Everest has a layer of limestone containing an assortment of marine fossils. This limestone overlays older metamorphic rocks. Based on the fossils and the arrangement of the rock layers, geologists have concluded that the sedimentary rocks at the summit of Mount Everest were formed in a shallow sea that existed between India and the rest of Asia. Since that time, India has moved northward—look back at Figure 3.2—squeezing the sedimentary layers that were deposited and resulting in the uplift of the marine rocks to their present elevation. India continues to move northward at about two centimeters per year and the Himalayas are still uplifting at a rate of up to a centimeter per year. The motion of the continents is now described by the theory of plate tectonics, which has replaced the older theory of continental drift.

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Objectives After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary. 1. Describe the evidence that first led people to consider that some of the

continents might have once been joined together.

2. State the details of Wegener’s four main arguments in support of his theory of

continental drift.

3. Explain why most geologists did not accept Wegener’s theory of continental

drift.

4. Summarize the evidence that led to the acceptance of seafloor spreading in

the 1960s.

5. Explain how seafloor spreading and plate tectonics are different than 6. 7. 8. 9.

Wegener’s theory of continental drift. Describe the three types of tectonic plate boundaries. Explain how convection in Earth’s mantle is related to plate tectonics. Describe Earth’s continental margins and seafloor. Discuss the relationships of compression and tension to folding and faulting in Earth’s crust.

Vocabulary Terms You should be able to define or describe each of these terms in a complete sentence or paragraph. 1. abyssal plain 2. anticline 3. asthenosphere 4. atoll 5. continental drift 6. continental margin 7. continental rise 8. continental shelf 9. continental slope 10. convergent boundary 11. deep-ocean trench 12. divergent boundary 13. fault

6.1

14. fault-block mountain 15. fold 16. geomagnetic reversal 17. graben 18. guyot 19. horst 20. hydrothermal vent 21. manganese nodule 22. mid-ocean ridge 23. monocline 24. normal fault 25. Pangea 26. plate tectonics

27. reverse fault 28. seafloor spreading 29. seamount 30. subduction 31. stress 32. strike-slip fault 33. syncline 34. tectonic plate 35. thrust fault 36. transform boundary 37. volcanic island arc

Continental Drift

Within a century after the voyages of Christopher Columbus to the New World, the creation of reasonably accurate maps of the Atlantic Ocean led some scholars to note how the continents on opposite sides of the ocean, particularly South America and Africa, seemed to fit together like pieces of a puzzle, as illustrated in Figure 6.1. Abraham Ortelius, writing in the late 1500s, suggested that North and South America had torn away from Europe and Africa through a process of earthquakes 133

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Figure 6.1. The fit between the coastlines of South America and Africa was noticed in the 1500s as soon as the first accurate maps of the Atlantic Ocean were created.

and flooding. Others speculated about whether these continents had once been next to each other, but it wasn’t until the early 20th century that scientists began seriously to think about this hypothesis. In 1915, a German meteorologist named Alfred Wegener published a book in which he presented multiple lines of evidence for a theory called continental drift. This theory proposed that not only the continents surrounding the Atlantic Ocean but all the continents were once joined together in a supercontinent called Pangea1, illustrated in Figure 6.2. This supercontinent then split into a number of smaller continents, such as North America, Australia, and Antarctica. These continents then slowly moved to their present positions over a period of

millions of years. Wegener outlined four main arguments supporting the theory of continental drift: 1. The fit of the continents. Like others before him, Wegener was intrigued by the similarity of the coastlines on opposite sides of the Atlantic Ocean.

2. Fossil evidence. By Wegener’s time, most paleontologists (scientists who study fossils and past life) believed that there must have been some sort of connection between Africa and South America in the past. In particular, fossils of an extinct aquatic reptile called Mesosaurus are found on both sides of the Atlantic, but nowhere else in the world. If Mesosaurus were a good enough swimmer to cross the Atlantic, then its fossils should be more widespread. Scientists had proposed that there had once been land connecting Africa and South America, but Wegener preferred to explain the Eurasia distribution of Mesosaurus by North continental drift. The distriAmerica bution of a fossil fern called Equatorial Tethys Sea Realm Glossopteris, found in all the region affected by Southern Hemisphere contiAfrica South Late Paleozoic glaciation America nents, also suggests that these India continents were once joined Australia to each other. The fossil eviAntarctica Figure 6.2. The arrangement of the continents about 250 million years ago. 1 Also spelled Pangaea. The name derives from the Greek pan, meaning “all” or “whole,” and Gaia, meaning “Mother Earth” or “land.”

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Fossil evidence of the Triassic land reptile Lystrosaurus.

Africa India

South America Figure 6.3. The distribution of fossils of the fern Glossopteris, the aquatic reptile Mesosaurus and other fossils are evidence that the continents were once together in a supercontinent.

Antarctica

Fossil remains of Cynognathus, a Triassic land reptile approximately 3 m long.

Australia

Fossil remains of the fern Glossopteris, found in all the southern continents, suggest they were once joined.

Fossil remains of the freshwater reptile Mesosaurus.

dence that the continents were once assembled together is summarized in Figure 6.3. 3. Similarity of rock types. On a jigsaw puzzle, not only must the shapes of the pieces to fit together, but the picture must be consistent from piece to piece as well. Wegener observed a similar matching of rock types across the Atlantic Ocean. The rocks of the Appalachian Mountains of eastern North America line up well with similar rocks in Great Britain and Scandinavia, as illustrated in Figure 6.4.

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NOR

4. Ice-age deposits. There is abundant evidence for an ancient ice age that covered parts of South America, Africa, India, Antarctica, and Australia. Geologists have studied scratches in the rocks to determine which way the glacial ice was moving. The directions make most sense if we assume these continents were once connected to each other. Figure 6.5 illustrates A how geologists theorize these continents were RT TNS IAN M O H C N LA connected during the ice PA AP age compared to their present positions. H

Despite multiple argu- Figure 6.4. If the Atlantic Ocean did not exist between North America and Europe, rocks formed in mountain belts (orange) on the two ments in favor of continen- continents would line up. 135

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AFRICA

NORTH AMERICA

EUROPE

ASIA

INDIA AFRICA

SOUTH AMERICA

AUSTRALIA

ANTARCTICA

SOUTH AMERICA

INDIA

AUSTRALIA

ANTARCTICA

Figure 6.5. According to geologists, there was an ice age in the Southern Hemisphere between 220 and 300 million years ago. The distribution of ice makes much more sense if the continents were once all together than if they were in their present positions.

tal drift, most geologists during Wegener’s lifetime were hostile to the idea that the continents had moved such great distances. The main objection to Wegener’s continental drift theory was that there was no credible explanation for how or why massive continents could move around on the face of the Earth. It seemed that the continents would somehow have to plow through the rocky crust of the ocean floor in order to move around. Wegener perished in 1930 while on an expedition on the Greenland ice cap and did not live to see his theory become widely accepted. Learning Check 6.1 1. What was the earliest evidence that suggested that Earth’s continents might have moved in the past? 2. How did glacial deposits in the southern continents such as Africa and Australia support the idea of continental drift? 3. How was fossil evidence used to support the theory of continental drift? 4. Why did most scientists initially reject Wegener’s theory of continental drift?

6.2

The Ocean Floor

6.2.1 Continental Margins The edges of the continents are known as continental margins. The east coast of North America, diagrammed in Figure 6.6, is typical of most continental margins around the Atlantic, Indian, and Arctic Oceans. The figure shows the major features of this type of continental margin. The shallowest part of the continental margin is the continental shelf, which starts at the shoreline and extends outward for a distance ranging from a few kilometers to several hundred kilometers. In most places, the continental shelf is a gently-sloping plain underlain by continental crust, so continental shelves are really a part of the continents themselves even though 136

Plate Tectonics and Mountain Building continental margin continental shelf

continental slope

continental rise

abyssal plain

sediments continental crust

oceanic crust

Figure 6.6. An Atlantic-type continental margin.

they are under water. The continental shelf contains some of the world’s greatest reserves of oil and gas, such as those along the Gulf of Mexico coast of Texas and Louisiana. It is also the home of some important fisheries, such as the Grand Banks off the coast of Newfoundland, Canada. The continental shelves extend out to where the depth reaches about 130 m (430 ft). Beyond this, the seafloor drops more steeply down to depths that are generally 3,000 m (10,000 ft) or greater. This sloping part of the seafloor is known as the continental slope. The continental slope is the true edge of the continents because it is where the granitic continental crust transitions to basaltic oceanic crust. In places, the continental slope is cut by deep canyons carved by currents carrying sediments from the top of the slope down to deeper parts of the ocean. Along the abyssal plain

oceanic crust

mantle

deep-ocean trench

continental shelf

continental crust

sediments

mantle

Figure 6.7. A Pacific-type continental margin with an offshore deep-ocean trench.

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base of the continental slope is a more gently sloping region called the continental rise, composed of sediments eroded and transported off the continental shelf and slope. Beyond the continental slope lie the abyssal plains. These are blanketed by thick accumulations of fine mud and are among the flattest places on Earth. The continental margins around most of the Pacific Ocean, as well as a few other places, are much narrower than the Atlantic-type margins. In some places there is almost no continental shelf and the seafloor slopes steeply down to the deep-ocean trenches—the deepest parts of the oceans. A continental margin with a deep-ocean trench is illustrated in Figure 6.7. The deepest place in the oceans is the Marianas Trench, near Guam in the western Pacific Ocean, with a depth of about 10,994 m (36,070 ft). The deep-ocean trenches are typically less than 100 km (60 mi) wide. They generally run parallel to the edges of continents, such as along the western coast of South America, or along arc-shaped chains of volcanic islands, as in the case of the Aleutian Islands of Alaska.

6.2.2 Mid-Ocean Ridges A prominent feature of the ocean floor is the series of mid-ocean ridges, a mountain range that runs through all the oceans, and has a combined length of about 70,000 km (44,000 mi). One portion of the mid-ocean ridge is the Mid-Atlantic Ridge, running down the center of the Atlantic Ocean and mirroring the coastlines on either side of the ocean, as shown in Figure 6.8. For the most part, the mountains

Figure 6.8. Earth’s mid-ocean ridges. The ridges are the red and orange areas and the black lines mark the rift zone along the axis of the ridges. Note how the Mid-Atlantic Ridge mirrors the coastlines on the east and west sides of the Atlantic Ocean. Color bands reflect the ages of the oceanic crust. Red represents the youngest areas and dark blue the oldest.

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of the mid-ocean ridges fall well short of being tall enough to rise above sea level. One important exception is Iceland, which sits right on top of the MidAtlantic Ridge. Along the center of much of the mid-ocean ridge system, there is a valley called a rift, which can be over 1,000 m (3,300 ft) deep. This rift is the site of frequent volcanic activity. It is also the location of un- Figure 6.9. A “black smoker” is a type of hydrothermal vent on the derwater hot springs called ocean floor. The black cloud is composed of superheated water filled hydrothermal vents. These with tiny particles of sulfide minerals. vents, like the one shown in Figure 6.9, discharge superheated water with temperatures up to 350°C (660°F). This fluid, heated by magma beneath the rift, is rich in metal sulfides and oxides, which turn into crystals of minerals such as pyrite, chalcopyrite, and hematite. Underwater deposits formed by hydrothermal vents may someday be mined for copper, zinc, lead, and other metals. Some ores Figure 6.10. Ely Seamount in the Gulf of Alaska. now on the continents are believed to have Note the volcanic crater at the summit. This image and the one in Figure 6.11 were created from echo formed in similar oceanic environments.

6.2.3 Seamounts and Marine Sediments

sounding data, as described in Figure 2.27.

Scattered over much of the ocean floors are isolated volcanic mountains called seamounts. Figure 6.10 is an image of a seamount in the Gulf of Alaska. Seamounts number in the tens of thousands and many rise hundreds or thousands of meters above the seafloor without breaking the surface of the ocean. Examples of seamounts that do rise above sea level include the Hawaiian Islands and Tahiti in the Pacific Ocean, and Figure 6.11. Bear Seamount, in the Atlantic Ocean the Azores and St. Helena in the Atlantic off the coast of Massachusetts, an example of Ocean. Almost all seamounts have a vol- a guyot. The summit rises over 2000 m (6600 ft) above the seafloor.

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Fringing reef, Maldives 2

4 Figure 6.12. Progression from a volcanic island (1), to a fringing reef (2), barrier reef (3), to a coral atoll (4). If the island sinks further and coral growth stops, the next step is a guyot.

Barrier reef around Bora Bora, Polynesia

canic origin, and some are sites of active volcanism. A guyot is a flat-topped seamount, illustrated in Figure 6.11. The summits of guyots are hundreds or thousands of meters below the water surface, yet the sediments and fossils on their summits suggest that the summits were Bikini Atoll, South Pacific once at sea level. It is believed that Figure 6.13. Three islands illustrating progression of a sinking at one time these mountains were volcanic island through fringing reef, barrier reef, and atoll. 140

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Figure 6.14. The White Cliffs of Dover in England are made of chalk. The chalk is composed of microscopic skeletons of single-celled marine organisms.

above sea level, but were eroded to a flat surface by wave erosion. The remains of the volcano then subsided (sank) deeper into the ocean basin. Guyots seem to be the end product of a process that happens to many volcanic islands. At some point, volcanoes become extinct; that is, they stop erupting. In tropical waters, coral reefs form around the islands. Coral organisms live as shallow-water colonies composed of many billions of individual organisms and produce skeletons made of calcium carbonate (limestone). When the reef is adjoining the island, it is called a fringing reef, as illustrated in Figure 6.12 (2). Since the island no longer grows larger through volcanic eruptions, the island begins both to erode and sink. However, the coral is still able to grow, even as the rest of the island sinks. Eventually the reef makes a ring called a barrier reef (Figure 6.12 (3)), with the remnants of the volcano in the center. As the volcano continues to erode and subside, it reaches the point where the volcano is completely submerged beneath the waves (Figure 6.12 (4)), resulting in the formation of an atoll—a low-lying, ring-shaped island that surrounds a central lagoon. Figure 6.13 shows tropical islands at various stages in this process. Sediment accumulations on the seafloor are thickest near the continents and very thin at the mid-ocean ridges. There are three types of sediments found in deep ocean basins, and these formed in three different ways. The first is sediments derived from erosion of land. Sand and gravel tend to be deposited close to the shoreline, but extremely fine-grained clay can remain suspended in the water for many years. Scientists estimate that it can take between 5,000 and 50,000 years for just one centimeter of land-derived sediment to accumulate on the abyssal plains. Another source of sediments in the ocean basins is biological activity. Singlecelled organisms in shallow seawater form microscopic skeletons made of either calcium carbonate or silica. When the organisms die, the skeletons settle to the seafloor, forming fine-grained sediments called calcareous ooze or siliceous ooze. Chalk, shown in Figure 6.14, is a type of limestone made of calcareous ooze. The chalk used on chalkboards used to be made of the rock chalk, but is now commonly made of the soft mineral gypsum. 141

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Sediments formed directly from seawater represent a third type. An important example of this is manganese nodules—rocks about the size and shape of potatoes that lie scattered in many places on the ocean floor. Manganese nodules, shown in Figure 6.15, are made mostly of manganese and iron, but also contain significant amounts Figure 6.15. Manganese nodules on the seafloor. of valuable metals such as copper, nickel, chromium, molybdenum, and cobalt. Manganese nodules grow around some sort of nucleus, such as a sand grain or shark tooth, and appear to grow extremely slowly—at rates measured in millimeters per million years. The metals that go into these rocks come from ions dissolved in seawater. Manganese nodules have not been mined extensively to date, but if the price of the metals these nodules contain increases significantly, it might someday become economical to mine them. Learning Check 6.2 1. Describe the different parts of the typical continental margin found around most of the Atlantic Ocean. 2. How are the continental margins around much of the Pacific Ocean different from continental margins around the Atlantic Ocean? 3. Why are hydrothermal vents associated in many cases with the midocean ridges? 4. Why do guyots generally have flat tops? 5. What are three different ways in which seafloor sediments form?

6.3

Seafloor Spreading

Another significant discovery about Earth’s seafloor in the 1960s had to do with Earth’s magnetic field. Because Earth’s core is mostly iron, Earth behaves as a giant magnet. All magnets have a north pole and south pole, and Earth has a North Magnetic Pole and a South Magnetic Pole. (Earth’s North Magnetic Pole in the Arctic is actually physically a south magnetic pole, as shown in the left-hand globe in Figure 6.16, since the north pole of the magnetic needles in compasses point toward it.) An unexpected discovery made in the early 1960s was that Earth’s magnetic field occasionally changes polarity—the North Magnetic Pole becomes the South Magnetic Pole and the South Magnetic Pole becomes the North Magnetic Pole! This change in the direction of Earth’s magnetic field, illustrated in Figure 6.16, is called a geomagnetic reversal. The planet itself does not flip over (thankfully), but something happens in the core that reverses the polarity of Earth’s magnetic field. The history of geomagnetic reversals is recorded in volcanic rocks. When magma crys142

Plate Tectonics and Mountain Building Reversed Polarity

Normal Polarity (today)

Figure 6.16. Earth’s magnetic field reverses in polarity, typically every few hundred thousand years. magnetic stripes tallizes, iron-bearing minerals, such as magnetite, in the resulting rocks become oriented to Earth’s magnetic field. In other words, these minerals act as compasses, pointing north and locking in place the orientation of Earth’s magnetic field at the time the rock cooled. We now know that the last major reversal was about 700,000 years ago. Volcanic rocks formed since that time have “normal” polarity. From 2.4 million years ago to 700,000 years ago, most volcanic rocks have “reversed” magnetic polarity. 10 5 0 5 10 million years ago As oceanographers studied the seafloor in the 1960s, ridge axis they discovered that the seafloor has bands of normal and Figure 6.17. Map view of reversed magnetic polarity preserved in the rocks. The bands of normal (black) and bands run parallel to the mid-ocean ridges, as illustrated reversed polarity on the seafloor in Figure 6.17. The rocks right along the rift zone of the southwest of Iceland. mid-ocean ridges have normal polarity. Rocks somewhat down Normal A the flanks of the ridges have re- magnetic polarity versed polarity, and rocks even farther from the ridge crest have Reversed normal polarity. Geologists inter- magnetic polarity B preted this to mean that the rocks along the ridge are the youngest and the seafloor gets progressively older with distance from C the ridge. In the early 1960s, geologist lithosphere magma Harry Hess combined this knowlFigure 6.18. As magma forms new continental crust at the edge of seafloor topography and mid-ocean ridge, the newly-formed rocks have a magnetic seafloor magnetism into a theory polarity, either normal or reversed. Older crust moves away called seafloor spreading. Accord- from the ridge.

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ing to the theory of seafloor spreading, new oceanic crust forms at the mid-ocean ridges and then slowly moves away from the ridge crest. This new crust is formed by magma that intrudes up into fissures and crystallizes along the center of the ridge. As new crust moves away from the ridge—at a rate of just a few centimeters per year—room is made for more magma to rise up from the mantle. Figure 6.18 illustrates how this explains the magnetic banding of rocks on the seafloor. As new crust forms above a magma chamber beneath the mid-ocean ridge, it bears the current magnetic polarity, whether normal or reversed. As additional oceanic crust is formed at the mid-ocean ridge, older crust moves horizontally away from the ridge. Seafloor spreading involves not only Earth’s crust, but Earth’s mantle as well. For one thing, the upper mantle is the source for magma that gets pushed up into the mid-ocean ridges—a key element of seafloor spreading. Additionally, much of the mantle seems to be directly involved in the movements of Earth’s crust. The crust plus the upper mantle, as you recall from Chapter 1, is referred to as the lithosphere; the deeper mantle is the asthenosphere. The lithosphere is fairly rigid, while the asthenosphere, though composed of solid rock, is able to flow because of the immense pressures at great depths. You can think of oceanic lithosphere as resting upon a pair of conveyer belts that move oceanic lithosphere away from the mid-ocean ridge. This process is illustrated in Figure 6.19. As magma rises near the ridge, older oceanic lithosphere is carried away from the ridge. As new oceanic crust is formed at the mid-ocean ridges, old oceanic crust is recycled back into the mantle at the deep-ocean trenches. Seafloor spreading occurs at a rate of only a few centimeters per year, but it has been measured. Scientists use lasers to measure distances between continents. The light in a laser beam travels at the speed of light, about ridge 300,000 km per second. By lithosphere measuring the time required trench for laser beams to bounce off trench asthe nosp satellites in Earth orbit—meaher e surements with precision in trilmantle lionths of seconds—scientists have confirmed the motions of 700 km the continents. The motions of the continents have also been outer core confirmed by measurements made using sophisticated GPS inner receivers. There are additional core evidences for seafloor spreadFigure 6.19. According to the theory of mantle convection, ing, some of which we look at seafloor spreading occurs because of convection currents in in the next chapter on volcanoes the asthenosphere, the deeper part of Earth’s mantle. This and earthquakes. convection involves rising hot mantle material beneath midocean ridges and sinking mantle material near the deep-ocean trenches.

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Learning Check 6.3 1. Describe features of the mid-ocean ridges that support the theory of seafloor spreading. 2. How do patterns of geomagnetism recorded in rocks of the seafloor support the theory of seafloor spreading? 3. In what ways is the mantle believed to be involved in seafloor spreading?

6.4

Plate Tectonics

Seafloor spreading provides the needed mechanism to explain Alfred Wegener’s hypothesis of continental drift. Geologists no longer use the term continental drift to describe the motions in Earth’s crust because it is not just the continents that are in motion. Instead, geologists now speak of the theory of plate tectonics.2 According to the theory, Earth’s entire lithosphere—including the crust and uppermost part of the mantle—is composed of tectonic plates that are slowly moving around on Earth’s surface. Plate tectonics is one of the most important concepts in geology. The theory explains a wide variety of phenomena, such as most of the

Figure 6.20. Earth’s lithosphere is composed of seven major plates and a number of smaller plates. The red arrows indicate the direction plates are moving relative to one another. 2 “Tectonics” refers to processes that deform Earth’s crust.

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lithospheric plate

asthenosphere

lithospheric plate asthenosphere

world’s volcanoes and earthquakes, and the formation of mountain ranges and many ore deposits. According to the theory of plate tectonics, Earth’s lithosphere is divided into seven major plates and a number of smaller plates. As you see in Figure 6.20, the North American Plate includes not only most of the continent of North America, but also a significant amount of the Atlantic and Arctic Oceans. The North American Plate, like the other plates, moves around on Earth’s surface as a solid unit.

6.4.1 Plate Boundaries There are three types of tectonic plate boundaries, illustrated in Figure 6.21. The first of these, called divergent boundaries, are where C asthenosphere plates are moving apart from each other. Most Figure 6.21. The three main types of divergent boundaries are found at the midtectonic plate boundaries: A) divergent, B) convergent, C) transform. ocean ridges, where new oceanic crust is being formed by magma coming up from Earth’s mantle. However, not all divergent boundaries are found on the ocean floors. A prominent feature in Africa is the East African Rift, running north-south from Lake Malawi up to the Red Sea, as shown EURASIAN on the map in Figure 6.22. The East AfriPLATE can Rift is similar in some ways to the rift Persian Gulf that runs down the center of the mid-ocean ARABIAN ridges (Figure 6.21A) and is home to a PLATE Nile number of volcanoes. Over a period of milRiver lions of years, the rift could open up to the AFRICAN point where a new ocean basin forms, splitPLATE n Erta (Nubian) f Ade ting the African plate into two plates. This Gulf o Ale appears already to have happened along the INDIAN Red Sea, at the north end of the East AfriPLATE can Rift, where the African Plate and ArabiEquator an Plate are moving apart from each other. Oldoinyo Lengai A second type of plate boundary is the AFRICAN PLATE Lake (Somalian) convergent boundary, where two plates colVictoria Legend lide with each other. There are three types of plate convergent boundaries, based on the types boundaries East African Lake of crust involved. lithospheric plate

Malawi

rift zone

volcano 1. Oceanic-continental convergence. When Figure 6.22. The East African Rift, a divergent plate oceanic lithosphere collides with conti-

boundary on a continent.

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tre

nc h

volc anic arc

nental lithosphere, the oceanic lithosphere is forced or pulled downward into the mantle, usually at an angle of about 45°. The basaltic oceanic crust is forced downward because its density (3.3 g/cm3) is higher than the granitic continental crust density (2.7 g/cm3). This process, in which an oceanic plate dives down into the mantle beneath another plate, is called subduction. Subduction creates deep-ocean trenches, as illustrated in Figure 6.23, and is a process that recycles the oceanic crust created at mid-ocean ridges. As oceanic lithosphere subducts deeper into Earth’s mantle, it heats and eventually partially melts, providing magma for overlying volcanoes. One example of continentaloceanic convergence is along the west coast of South America, where the Nazca Plate is oceanic crust continental crust subducting beneath the South lithosphere lithosphere American Plate. This forms the Peru-Chile Trench, which runs along most of the western coast asthenosphere of South America. oceanic-continental convergence

3. Continental-continental convergence. When two continental plates collide, neither can be subducted into the mantle because the low density of granite makes continental crust buoyant, so it cannot be subducted. As illustrated in Figure 6.25, when continents collide the crust is buckled and deformed,

oceanic crust

an d

isl

tre

ar c

2. Oceanic-oceanic convergence. Figure 6.23. When an oceanic plate converges with a When oceanic crust converges continental plate, oceanic lithosphere is subducted into with oceanic crust, one of the Earth’s mantle and a deep-ocean trench forms. plates subducts beneath the other, as illustrated in Figure 6.24. Typically, one of the plates is older and cooler than the other, and this is the plate that subducts. The deep-ocean trenches associated with oceanic-oceanic convergent boundaries are typically curved and have a chain of volcanic islands parallel to the trench on the plate that is not subducted. This curved chain of islands is called an island arc. An excellent example of an island arc is the Aleutian Islands of Alaska, which run parallel to the deep-ocean trench formed where the Pacific Plate subducts beneath an oceanic portion of the North American Plate. Each island in this arc has a volcano (or multiple volcanoes) on it. lithosphere

continental crust lithosphere lithosphere

asthenosphere oceanic-oceanic convergence Figure 6.24. When an oceanic plate converges with another oceanic plate, one plate subducts beneath the other and a deep-ocean trench forms. A line of islands called a volcanic island arc forms parallel to the trench.

147

Chapter 6

A N I C E R A M

San

San Francisco

d An rea

L A

Faul t

s Los Angeles

Murray Facture Zone

A Molokai Fr acture Zon

148

Me n Fra doci ctu no re Z on

P L

The third type of plate boundary is the transform boundary, where two plates slide past each other. At divergent boundaries, new crust is created; at convergent boundaries, old oceanic crust is destroyed. But at transform boundaries, crust is neither created nor destroyed. Transform boundaries usually occur in oceanic crust and are common along offset portions of the mid-ocean ridges. Figure 6.26 locates some transform boundaries along the western coast of North America, where the boundary between the Pacific Plate and the smaller Juan de Fuca Plate is made up partly of mid-ocean ridges (red lines) and partly of transform faults. There are only a few places where transform boundaries occur in continental crust. One of these is in Califor-

I C I F P A C

Juan

de Fu

H N O R T

ca Rid

hig h

pla tea u

forming a mountain range. The ge an Himalaya Mountains, the highnr i a t un est mountain range on Earth, mo were formed by the collision of continental crust India into the south side of the continental crust Eurasian Plate, illustrated back lithosphere lithosphere in Figure 3.2. This process continues today. Another mountain asthenosphere range that formed by continentalcontinental-continental convergence continental convergence millions Figure 6.25. When continental plates converge, neither of years ago is the Appalachian plate can be subducted into Earth’s mantle. The Himalaya Mountains, formed when Africa and Appalachian Mountains formed in this fashion. collided with North America over a period of time between 325 and 260 million years ago. This collision was part of the events that joined the continents together to form the supercontinent PanCanada gea, shown in Figure 6.2. Africa Unit ed States JUAN and North America later split apart DE FUCA from one another, forming the AtPLATE lantic Ocean.

Mexico

Figure 6.26. The Pacific coast of western North America has divergent, convergent, and transform tectonic plate boundaries. The Juan de Fuca Ridge is part of the midocean ridge system. Subduction of the Juan de Fuca Plate beneath the North American Plate occurs without a visible deep-ocean trench off the coast of Washington and Oregon. The San Andreas Fault and the far eastern part of the Mendocino Fracture Zone are transform plate boundaries, where plates slide past one another.

Plate Tectonics and Mountain Building

nia, where the Pacific Plate slides past the North American Plate along the San Andreas Fault. The piece of California west of the San Andreas Fault is on the Pacific Plate and moving northward relative to the North American Plate. The Pacific coast of North America actually has convergent, divergent, and transform plate boundaries close to one another, as Figure 6.26 indicates.

6.4.2 Tectonic Plate Movements in the Past If we project plate motion millions of years into the past, we see that Earth was once a very different place. At present, the Atlantic Ocean is becoming 2 cm wider each year. Reversing this motion to some point in the past indicates that the Atlantic Ocean was completely closed—North and South America were united with Europe and Africa. Doing the same thing with the motion of other plates, such as the Indian, Australian, and Antarctic Plates, leads us back to Wegener’s idea of Pangea, the supercontinent that included most of Earth’s landmasses over 100 million years ago. We can apply similar reasoning to imagine the future. What might a map of Earth look like millions of years from now? Will the Atlantic Ocean continue to open wider, with the Pacific Ocean getting narrower? Geologists believe there is evidence that the widening of the Atlantic Ocean will eventually slow down and even reverse, leading to an eventual re-closing of the Atlantic. Learning Check 6.4 1. Describe the three types of plate boundaries. 2. What are the similarities and differences between the three types of convergent plate boundaries? 3. Describe what will happen to Africa in a few million years if divergence continues at the Great Rift Valley (the East African rift). 4. The Atlantic Ocean is becoming 2 cm wider each year. How much wider was it in the year 2020 than it was in 1492, the year Columbus made his first trip across the Atlantic?

6.5

Mountains

Mountains form in a number of different ways. For example, volcanoes such as Mt. Rainier in Washington and Vesuvius in Italy form when lava flows erupt from a central vent, growing to considerable height after dozens or even hundreds of eruptions. We look more closely at volcanoes and how they relate to plate tectonics in Chapter 7. Most mountain ranges are formed by the forces involved in plate tectonics, which are able to contort rock layers by squeezing them from the sides, or stretch rocks until they break apart. We begin our study of mountains by looking at the various ways mountain-building processes deform rocks.

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6.5.1 Folding When tectonic plates converge with one another, a tremendous amount of force is applied on solid rocks in Earth’s crust. This force is measured as stress—force per unit area. Low amounts of stress actually cause rock layers to bend rather than break, especially at the temperatures and pressures encountered deeper in Earth’s

stress

Figure 6.27. Compressional stress causes rocks to deform by folding or breaking (faulting), depending on the rocks and the conditions.

crust. If the stress is applied for long periods of time, rock layers become highly distorted. If rocks are subjected to high amounts of stress in a shorter period of time, it is more likely that the rocks will break rather than bend. When rocks are squeezed by stress coming from opposite directions, the stress is called compression. Compression of rocks results in either folding or breaking, as illustrated in Figure 6.27. A fold is a bend in a rock layer. Most sedimentary rocks are deposited in horizontal layers; if the layers are no longer horizontal, anticline we can assume the layers have been syncline folded. There are three basic types of folds: anticlines, synclines, and monoclines, illustrated in Figure 6.28. An anticline, like the one shown in Figure 6.29, is a fold where rock layers slope away from the central axis of the fold, a formation that resembles an upside monocline down “U”. In some areas, the folded layers of sedimentary rocks in anticlines act as petroleum traps, such as in the upper part of Figure 5.22. A syncline is a fold in which the rock layers slope inward toward the cenFigure 6.28. The three types of folds are anticlines, tral fold axis, a right-side-up “U”. Ansynclines, and monoclines. 150

Plate Tectonics and Mountain Building

ticlines and synclines are typically found next to each other or in alternating patterns of anticlines and synclines. The third type of fold is a monocline, commonly found in areas with mostly horizontal layers of sedimentary rocks. The monocline is like a step with horizontal rocks on either side of the fold. Folds can be large or small features. The Figure 6.29. An anticline in Lebanon. anticline in Figure 6.29 is large enough to form a mountain. On the other hand, the folds in Figure 6.30 are visible in a small outcrop of rock. Some folds are even smaller and are best studied in hand-sized samples or even with microscopes. It might seem strange to you to think of solid rocks bending. Part of this is because the small samples we hold in our hands seem rather inflexible. Another reason is that we are accustomed to observing rocks in the low-stress environment of Earth’s surface. At the high pressures and temperatures deep in Earth’s crust, rocks are much more likely to deform by folding than they are near the surface. But even near Earth’s surface, solid rocks can fold. Geologists studying thin sections of folded rocks with microscopes can see where sand grains have moved past one another as the rocks folded. Also visible are tiny fractures that cut through the rock, enabling the rock to Figure 6.30. These anticlines and synclines in southern Greece are each just fold. Many folded a few meters across. The size of folds ranges from many kilometers down to sedimentary rocks millimeters. 151

Chapter 6

also show evidence of fracturing that occurred along with the folding. Folding occurs in sedimentary, metamorphic, and igneous rocks.

6.5.2 Faulting Compressional stress causes rocks to fold, as we have seen. It also causes rocks to break, as illustrated in Figure 6.27. A fault is a break in rocks where movement has occurred. If the rock cracks, but there is no movement along the crack, the crack is considered to be a fracture, not a fault. The opposite of compression is tension, stress that stretches rocks apart instead of squeezing them together, as illustrated in Figure 6.31. Tensional stress occurs at divergent plate boundaries, such as at the East African Rift mentioned earlier, and also at other places within tectonic plates. Tension may cause rocks to break, creating faults.

stress

faults

Figure 6.31. Tensional stress pulls rocks apart, which can lead to faulting.

If a fault is not vertical in Earth’s crust, geologists refer to the rocks above the fault as the hanging wall and the rocks below the fault as the footwall, as illustrated in Figure 6.32. If you can imagine standing inside Earth’s crust with your feet right on the fault, your feet would be on the footwall, and the rocks hanging above your feet would be the hanging wall. If the hanging wall moves down relative to the footwall, the fault is called a normal fault. Normal faults are caused by tension in Earth’s crust; the faults in Figure 6.31 are normal faults. If the hanging wall moves up relative to the footwall, the fault is known as a reverse fault, the type shown in Figure 6.27. Reverse faults are caused by compression in Earth’s crust. Some reverse faults have a low angle relative to a horizontal plane. These are known as thrust faults. Like all reverse faults, thrust faults form because of compressional stress, such as the stress present at convergent plate boundaries. Thrust faults can even be fault hanging wall footwall horizontal, sliding older rocks over younger rocks. One of the most famous examples of this is found in Glacier National Park in Montana, where older rocks have been pushed Figure 6.32. The hanging wall and footwall of a fault. about 80 km (50 mi) horizontally To an ant on the fault, the foot wall would be beneath their feet, and the hanging wall would be hanging above over younger rocks, as illustrated in Figures 6.33 and 6.34. Normally, oldthem. 152

Plate Tectonics and Mountain Building

East

Chief Mountain

Precambrian rocks > 1 billion years old

West

horizontal force due to tectonic plate collision

Cretaceous rocks 80–100 million years old

Figure 6.33. Thrust faulting occurs when compression causes older rocks to slide over younger rocks. In this diagram of the main thrust fault in Glacier National Park, Montana, Precambrian sedimentary rocks over one billion years old have been thrust over Cretaceous sedimentary rocks, which are about 80 to 110 million years old. Solid lines show layering in the rocks; dashed lines indicate probable layering before erosion in the mountains.

Lewis Thrust Fault

Precambrian limestone (older) Cretaceous sandstones and shales (younger)

Figure 6.34. Chief Mountain in Glacier National Park, with older (Precambrian) rocks pushed over younger (Cretaceous) rocks along a nearly horizontal thrust fault.

er rocks are deeper in Earth’s crust and younger rocks are nearer the surface. A thrust fault can stack rocks so they are “out of order,” with the older rocks above younger rocks. Before the development of the theory of plate tectonics, geologists had no explanation for how such dramatic compressional forces could exist in Earth’s crust. The convergence of tectonic plates provides a source for these powerful compressional forces. Thrust faults are forming today in places like the Andes Mountains Figure 6.35. Movement along a strike-slip fault is horizontal. 153

Chapter 6

Piqiang Fault

2 km

Figure 6.36. A satellite image of a strike-slip fault in China. The rocks on the east side of the fault have moved horizontally about 4 km (2.5 mi) relative to the rocks on the west side of the fault. (Note that in this view the camera is pointing straight down, although it may at first appear otherwise.)

in South America, where oceanic-continental convergence is occurring, and in the Himalaya Range in Asia, the site of ongoing continental-continental convergence. Another important type of fault is the strike-slip fault, where movement along the fault is horizontal rather than vertical, as illustrated in Figure 6.35. The San Andreas Fault, source of many of California’s earthquakes, including the 1906 earthquake that leveled San Francisco, is a strike-slip fault. The San Andreas Fault is mapped in Figure 6.26. The San Andreas Fault is part of the transform boundary between the Pacific Plate and the North American Plate. All transform boundaries are strike-slip faults, but not all strike-slip faults occur at plate boundaries. Figure 6.36 shows a strikeslip fault with a few kilometers of movement in the middle of the Eurasian plate.

6.5.3 Mountain Building Most mountain ranges on Earth can be related one way or another to plate tectonics. The movement of tectonic plates provides the energy necessary to cause uplift and deformation of vast areas of Earth’s crust. The Rocky Mountains are near the western edge of the North American Plate, and the Alps are related to the boundary between the Eurasian Plate and the African Plate. Some mountain ranges, such as the Appalachian Mountains of the eastern United States, are not now near the edge of a tectonic plate, but were near plate boundaries millions of years ago. All mountain ranges were formed by a combination of processes, such as volcanism, or folding and faulting, but some mountain ranges are dominated by one of 154

Plate Tectonics and Mountain Building

these processes over the others. The Appalachian Mountains are a mountain range where folding was an especially significant part of their formation, as the satellite image in Figure 6.37 shows. In Figure 6.38, a syncline is exposed in a road cut along a highway in the Appalachian Mountains. Figure 6.37. A satellite image of a portion of the Appalachian Mountains Fault-block moun- in Pennsylvania. The darker colors are forested areas on ridges formed by tains form when fault- anticlines and synclines. ing causes large blocks of Earth’s crust to rise or sink relative to one another, usually along normal faults. In faultblock mountains, uplifted blocks are called horsts and downthrown blocks are called grabens, as pictured in Figure 6.39. Many of the mountain ranges of the western United States, such as the Sierra Nevada in California, the Wasatch Range of Figure 6.38. This road cut through one of the ridges of the Appalachian Utah, and the Basin and Mountains in Maryland provides an excellent view of a syncline. Range area of Nevada and surrounding states, are horst graben fault-block mountains. Figure 6.40 shows the Grand Tetons of Wyoming. There has been stress up to 9,000 m (30,000 ft) of upward movement along the Teton Fault at the base of the Grand Tetons. Most of the upnormal lifted block has been eroded fault away, and the graben has been Figure 6.39. Fault-block mountains with uplifted blocks called filled with thousands of me- horsts and downthrown blocks called grabens. 155

Chapter 6

Figure 6.40. The Grand Tetons of Wyoming. The mountain range is a horst and the valley in the foreground is a graben.

ters of sediments, leaving the tops of the mountains about 2,200 m (7,200 ft) above the valley floor. Other mountain ranges are formed by regional uplift, in which broad areas of Earth’s crust are lifted to higher elevations. The Colorado Plateau of the southwestern United States is an example of an area that has been uplifted in this manner. Much of the Colorado Plateau has been uplifted by over a kilometer with relatively minor folding and faulting of sedimentary rock layers, as evidenced by the nearly horizontal layers seen in places like the Grand Canyon, shown in Figure 3.8. Learning Check 6.5 1. Describe different ways in which stress can deform rocks. 2. Explain why compressional stress can cause rocks to fold or fault, while tensional stress generally causes only faulting. 3. Draw sketches of an anticline, a syncline, and a monocline. 4. Draw sketches of a normal fault, a reverse fault, and a strike-slip fault. 5. How are folding and faulting related to plate tectonics?

156

Plate Tectonics and Mountain Building

Chapter 6 Exercises Answer each of the questions below as completely as you can. Write your responses in complete sentences unless instructed otherwise. 1. The far eastern coast of Canada and the west coast of Ireland are about 3,000 km apart. With the Atlantic Ocean widening by about 2 cm each year, how many years did it take for the Atlantic Ocean to open? Assume a constant rate of opening (which may or may not be a reasonable assumption). 2. Which of Wegener’s arguments for continental drift seems strongest to you? 3. Draw and label a series of diagrams illustrating the development of an atoll. 4. The magnetic bands on the seafloor near the East Pacific Rise, part of the mid-ocean ridge system, are farther apart than the magnetic bands associated with the Mid-Atlantic Ridge. Formulate a hypothesis to explain the cause of this. 5. Explain why seafloor sediments are thinner near the mid-ocean ridge than they are in the abyssal plains. 6. Explain how mantle convection is related to mid-ocean ridges and deep-sea trenches. 7. Thinking again about our discussion of theories in Chapter 3, how would you respond to someone who says, “I don’t believe in plate tectonics; it’s just a theory”? 8. What types of plate boundaries are most often associated with mountain ranges on continents? Why? 9. What types of faulting would you expect to find in the Himalaya Mountains? 10. The portion of California west of the San Andreas Fault is on the Pacific Plate and moving north relative to the North American Plate at an average rate of 3.5 cm per year. At this rate, how many years would it take for western California to move up to the Gulf of Alaska, 2,500 km to the north?

157

Introductory Principles in Physics PUBLISHED BY CENTRIPETAL PRESS Grade Level: 9th Grade Yearlong Course

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INTRODUCTORY PRINCIPLES IN PHYSICS Introductory Principles in Physics is an introductory physics text designed for grade-level students in 9th grade. The text is also suitable for use in higher grades due to the inclusion of some extra chapters (see below). Here we highlight a few distinctive features of this text. •

Introductory Principles in Physics supports our signature philosophy of science education based on wonder, integration, and mastery. We always seek to build on and stimulate the student’s innate sense of wonder at the marvels found in nature. Integration refers to the epistemology, mathematics, and history embedded in the text, and a curriculum designed to help develop the student’s facility with using language to communicate scientific concepts. Finally, Introductory Principles in Physics is designed to be used in a mastery-learning environment, using the mastery teaching model John developed and describes in From Wonder to Mastery: A Transformative Model for Science Education. When teachers use this mastery-learning model, the result is high student achievement and exceptional long-term retention for all students.

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The text includes five unique experiments. These experiments are a departure from the norm. Students make pendulums with washers and string; make velocity predictions with Hot WheelsTM cars going down ramps; push real cars in the parking lot to measure acceleration; measure the density of aluminum and PVC; and build test circuits using precision resistors, a voltmeter, and an electronic breadboard.

•

The text includes two bonus chapters. When used in 9th grade, 11 chapters make up the full course of study. However, knowing that some schools would use this text with grade-level students in 11th grade, we added Chapter 8 (Pressure and Buoyancy) and Chapter 13 (Geometric Optics). The content and more advanced mathematics of these extra chapters make the text suitable for use in either 9th or 11th grade.

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INTRODUCTORY PRINCIPLES IN PHYSICS Support Resources A number of support resources are available to accompany Introductory Principles in Physics. They include: The Student Lab Report Handbook Students should begin writing their lab reports from scratch in 9th grade. This popular manual gives them everything they need. We recommend supplying this handbook to every freshman so they can refer to it throughout high school. Read more about this resource on pg. 6. Solutions Manual to Accompany Introductory Principles in Physics This book contains complete written solutions for all the computations in the chapter exercises. Favorite Experiments for Physics and Physical Science This reference book contains complete background information, procedures, and material lists for the experiments and demonstrations used in all three of our high school physics courses. Experiments for Introductory Physics and ASPC This reference book contains material for the experiments only for Introductory Principles in Physics, taken from Favorite Experiments listed above. (Also included is one additional experiment used only in the sister course, Accelerated Studies in Physics and Chemistry.) Digital Resources The following materials are accessible exclusively through Centripetal Press: • a full year of weekly, cumulative quizzes • two semester exams • a document containing answer keys for all the quizzes and tests and sample answers for all the verbal questions in the text • a full year of Weekly Review Guides • a document with recommendations for teaching the course • a lesson list and example calendar • a sample graded lab report Tips and Tools This resource is full of tips and tools are available at www.centripetalpress.com. Here, teachers will find scores of images and short videos organized chapter by chapter. Teachers are free to use any of this content as they wish to assemble vivid and fascinating lessons.

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Sample Chapters The following pages contain samples from the text. The Table of Contents is shown, as well as Chapters 1, 2, and 4, and the first two experiments.

Introductory Principles in Physics A Mastery-Oriented Curriculum

John D. Mays

Second Edition

CENTRIPETAL PRESS

Austin, Texas 2017

Introductory Principles in Physics © Classical Academic Press®, 2020 Edition 2.2 ISBN: 978-0-9972845-8-4 All rights reserved. Except as noted below, this publication may not be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior written permission of Classical Academic Press. All images attributed to others under any of the Wikimedia Commons licenses, such as CC-BY-SA-3.0 and others, may be freely reproduced and distributed under the terms of those licenses. Copyright for all images attributed to CERN belongs to CERN, for the benefit of the CMS Collaboration. Classical Academic Press 515 S. 32nd Street Camp Hill, PA 17011 www.ClassicalAcademicPress.com/Novare/ Cover design by Nada Orlic, http://nadaorlic.info/ ISP.04.20

Contents

Preface for Teachers

viii A Different Textbook Philosophy ix Mastery, Integration, and Wonder ix Mastery ix Integration ix Wonder x Learning vs. Not Learning x 2. Teaching With This Text xi Skills and Prerequisites xi Optional Chapter Content xi Assignments, Homework, and the Weekly Workload xii Experiments and Experimental Error xii 3. Companion Resources xiii Slaying the Cram-Pass-Forget Dragon xiii Experiments Resources xiii Teacher Resource CD xiv The Student Lab Report Handbook xiv Solutions Manual to Accompany Introductory Principles in Physics xiv 1.

Preface for Students A Solid Study Strategy

Chapter 1 The Nature of Scientific Knowledge

1.1 Modeling Knowledge 1.1.1 Kinds of Knowledge 1.1.2 What is Truth and How Do We Know It? 1.1.3 Propositions and Truth Claims 1.1.4 Truth and Scientific Claims 1.1.5 Truth vs. Facts 1.2 The Cycle of Scientific Enterprise 1.2.1 Science 1.2.2 Theories 1.2.3 Hypotheses 1.2.4 Experiments 1.2.5 Analysis 1.2.6 Review 1.3 The Scientific Method 1.3.1 Conducting Reliable Experiments 1.3.2 Experimental Variables Do You Know ... What are double-blind experiments? 1.3.3 Experimental Controls Chapter 1 Exercises Do You Know ... How did Sir Humphry Davy become a hero?

Chapter 2 Motion 2.1

Computations in Physics

xvi xviii 2 3 3 4 5 7 8 9 9 10 12 13 13 13 14 14 15 16 17 18 19 20 21 i

Contents

2.1.1 The Metric System 2.1.2 MKS Units 2.1.3 Converting Units of Measure Do You Know ... How is the kilogram defined? 2.1.4 Accuracy and Precision 2.1.5 Significant Digits 2.1.6 Scientific Notation 2.1.7 Problem Solving Methods 2.2 Motion 2.2.1 Velocity Universal Problem Solving Method 2.2.2 Acceleration 2.3 Planetary Motion and the Copernican Revolution 2.3.1 Science History and the Science of Motion 2.3.2 Aristotle 2.3.3 Ptolemy 2.3.4 The Ptolemaic Model 2.3.5 The Ancient Understanding of the Heavens 2.3.6 The Ptolemaic Model and Theology 2.3.7 Copernicus and Tycho 2.3.8 Kepler and the Laws of Planetary Motion 2.3.9 Galileo 2.3.10 Newton, Einstein, and Gravitational Theory Do You Know ... Who built the first monster telescope? Chapter 2 Exercises

Chapter 3 Newton’s Laws of Motion

3.1 Matter, Inertia, and Mass 3.2 Newton’s Laws of Motion 3.2.1 The Three Laws of Motion 3.2.2 Actions and Reactions 3.2.3 Showing Units of Measure in Computations 3.2.4 Weight 3.2.5 Applying Newton’s Laws of Motion Thinking About Newton’s Laws of Motion 3.2.6 How a Rocket Works Do You Know ... Does gravity cause elliptical orbits? Chapter 3 Exercises Do You Know ... Where is Isaac Newton’s tomb?

Chapter 4 Energy

4.1 What is Energy? 4.1.1 Defining Energy 4.1.2 The Law of Conservation of Energy 4.1.3 Mass-Energy Equivalence 4.2 Energy Transformations

21 23 23 24 27 28 32 34 34 34 36 38 41 41 42 43 43 46 48 49 52 55 57 58 60 64 65 66 66 71 72 73 74 76 78 79 80 83 84 85 85 86 86 86

Contents

4.2.1 Forms of Energy 4.2.2 Energy Transfer 4.2.3 The “Energy Trail” Do You Know ... What is dark energy? 4.2.4 The Effect of Friction on a Mechanical System 4.2.5 Energy “Losses” and Efficiency 4.3 Calculations with Energy 4.3.1 Gravitational Potential Energy and Kinetic Energy 4.3.2 Work Do You Know ... What is alpha radiation? 4.3.3 Applying Conservation of Energy 4.3.4 Conservation of Energy Problems 4.3.5 Energy in the Pendulum Chapter 4 Exercises Do You Know ... Why are there pendulums in clocks?

Chapter 5 Momentum

5.1 Defining Momentum 5.2 Conservation of Momentum 5.2.1 The Law of Conservation of Momentum 5.2.2 Elastic and Inelastic Collisions 5.2.3 Problem Solving Assumptions 5.2.4 The Directionality of Momentum 5.2.5 Solving Problems with Conservation of Momentum 5.3 Momentum and Newton’s Laws of Motion Do You Know ... What is angular momentum? Chapter 5 Exercises Do You Know ... What is a hydraulic jump?

Chapter 6 Atoms, Matter, and Substances

6.1 Atoms and Molecules 6.2 The History of Atomic Models 6.2.1 Ancient Greece 6.2.2 John Dalton’s Atomic Model 6.2.3 New Discoveries 6.2.4 The Bohr and Quantum Models of the Atom 6.3 Volume and Density 6.3.1 Calculations with Volume 6.3.2 Density 6.4 Types of Substances 6.4.1 Major Types of Substances 6.4.2 Elements 6.4.3 Compounds Do You Know ... What structures can carbon atoms make? 6.4.4 Heterogeneous Mixtures 6.4.5 Homogeneous Mixtures

86 87 88 89 90 91 92 92 96 97 98 100 102 104 107 108 109 112 112 112 113 115 115 120 121 122 125 126 127 129 129 129 130 135 137 137 137 141 141 141 143 144 146 147 iii

Contents

6.5 Phases and Phase Changes 6.5.1 Phases of Matter Do You Know ... Why are crystals so fascinating? 6.5.2 Evaporation Do You Know ... What is a triple point? Do You Know ... What causes the crystal structure of ice? 6.5.3 Sublimation Chapter 6 Exercises

Chapter 7 Heat and Temperature

7.1 Measuring Temperature 7.1.1 Temperature Scales 7.1.2 Temperature Unit Conversions 7.2 Heat and Heat Transfer 7.2.1 How Atoms Possess Energy 7.2.2 Internal Energy and Thermal Energy 7.2.3 Absolute Zero 7.2.4 Thermal Equilibrium Do You Know ... How fast are air molecules moving? 7.3 Heat Transfer Processes 7.3.1 Conduction In Nonmetal Solids 7.3.2 Conduction in Metals 7.3.3 Convection 7.3.4 Radiation 7.4 The Kinetic Theory of Gases 7.5 Thermal Properties of Substances 7.5.1 Specific Heat Capacity 7.5.2 Thermal Conductivity 7.5.3 Heat Capacity vs. Thermal Conductivity Chapter 7 Exercises Do You Know ... What is the temperature in outer space?

Chapter 8 Pressure and Buoyancy

8.1 Pressure Under Liquids and Solids 8.2 Atmospheric Pressure 8.2.1 Air Pressure 8.2.2 Barometers 8.2.3 Absolute Pressure and Gauge Pressure 8.3 Archimedes’ Principle of Buoyancy 8.4 Flotation Do You Know ... What was Archimedes’ “eureka” discovery? Chapter 8 Exercises

148 148 151 153 153 154 154 155 158 159 159 160 162 162 162 163 163 163 164 164 165 166 167 168 169 169 169 170 172 175 176 177 180 180 180 182 184 187 188 189

Contents

Chapter 9 Waves, Sound, and Light

9.1 Modeling Waves 9.1.1 Describing Waves 9.1.2 Categorizing Waves 9.1.3 Modeling Waves Mathematically 9.2 Wave Interactions 9.2.1 Reflection 9.2.2 Refraction 9.2.3 Diffraction 9.2.4 Resonance 9.2.5 Interference Do You Know ... Do skyscrapers have resonant frequencies? 9.3 Sound Waves 9.3.1 Pressure Variations in Air 9.3.2 Frequencies of Sound Waves 9.3.3 Loudness of Sound 9.3.4 Connections Between Scientific and Musical Terms 9.4 The Electromagnetic Spectrum and Light Do You Know ... What causes sonic booms? Chapter 9 Exercises

Chapter 10 Introduction to Electricity

10.1 The Amazing History of Electricity 10.1.1 Greeks to Gilbert 10.1.2 18th-Century Discoveries Intriguing Similarities between Gravity and Electricity 10.1.3 19th-Century Breakthroughs 10.2 Charge and Static Electricity 10.2.1 Electric Charge Do You Know ... Who made the first color photograph? 10.2.2 How Static Electricity Forms Do You Know ... Why are plasmas conductive? 10.3 Electric Current 10.3.1 Flowing Charge 10.3.2 Why Electricity Flows So Easily in Metals Chapter 10 Exercises Do You Know ... Whose pictures did Einstein have on his walls?

Chapter 11 DC Circuits

11.1 Understanding Currents 11.1.1 Electric Current 11.1.2 The Water Analogy 11.2 DC Circuit Basics 11.2.1 AC and DC Currents

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Contents

11.2.2 DC Circuits and Schematic Diagrams 11.2.3 Two Secrets 11.2.4 Electrical Variables and Units 11.2.5 Ohm’s Law 11.2.6 What Exactly Are Resistors and Why Do We Have Them? 11.2.7 Through? Across? In? 11.2.8 Voltages Are Relative 11.2.9 Power in Electrical Circuits 11.2.10 Tips on Using Metric Prefixes in Circuit Problems 11.3 Multi-Resistor DC Circuits 11.3.1 Two-Resistor Networks 11.3.2 Equivalent Resistance 11.3.3 Significant Digits in Circuit Calculations 11.3.4 Larger Resistor Networks 11.3.5 Kirchhoff ’s Laws 11.3.6 Putting it All Together to Solve DC Circuits Do You Know ... Was there really a War of Currents? Chapter 11 Exercises Do You Know ... What is an uninterruptible power supply?

Chapter 12 Fields and Magnetism

12.1 Three Types of Fields 12.2 Laws of Magnetism 12.2.1 Ampère’s Law 12.2.2 Faraday’s Law of Magnetic Induction 12.2.3 The Right-Hand Rule 12.3 Magnetic Devices 12.3.1 Solenoids 12.3.2 Motors and Generators 12.3.3 Transformers Do You Know ... Where were the first transformers made? Chapter 12 Exercises Do You Know ... What is Nikola Tesla’s claim to fame?

Chapter 13 Geometric Optics

13.1 Ray Optics 13.1.1 Light As Rays 13.1.2 Human Image Perception 13.1.3 Flat Mirrors and Ray Diagrams 13.1.4 Real and Virtual Images 13.2 Optics and Curved Mirrors 13.2.1 Concave and Convex Optics 13.2.2 Approximations In Geometric Optics 13.2.3 Spherical Mirrors 13.2.4 The Mirror Equation 13.3 Lenses

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Contents

13.3.1 Light Through a Lens 13.3.2 Single-Lens Applications 13.3.3 The Lens Equation 13.3.4 Multiple-Lens Systems 13.3.5 Imaging with The Eye Do You Know ... How are rainbows formed? Chapter 13 Exercises

302 304 306 310 312 313 314

Glossary

316

Appendix A Reference Data

336

Appendix B Chapter Equations and Objectives Lists B.1 Chapter Equations B.2 Chapter Objectives Lists

Appendix C Laboratory Experiments

C.1 Important Notes C.2 Lab Journals C.3 Experiments Experiment 1 The Pendulum Experiment Experiment 2 The Soul of Motion Experiment Experiment 3 The Hot Wheels Experiment Experiment 4 Density Experiment 5 DC Circuits

Appendix D Scientists to Know About Appendix E Making Accurate Measurements E.1 E.2 E.3 E.4 E.5

Parallax Error Measurements with a Meter Stick or Rule Liquid Measurements Measurements with a Triple-Beam Balance Measurements with an Analog Thermometer

338 338 339 345 345 345 346 346 349 354 356 358 365 366 366 367 367 368 368

Appendix F References

369

Appendix G Image Credits

371

Index

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CHAPTER 1

The Nature of Scientific Knowledge

Theory → Hypothesis → Experiment In 1915, Albert Einstein produced his general theory of relativity. In 1917, Einstein announced an amazing new hypothesis: according to the theory, light traveling through space bends as it passes near a star. In 1919, this hypothesis was confirmed by teams under the leadership of Sir Arthur Eddington, using photographs taken of stars positioned near the sun in the sky during a solar eclipse. The image above is a positive created from one of Eddington’s negatives.

OBJECTIVES After studying this chapter and completing the exercises, students will be able to do each of the following tasks, using supporting terms and principles as necessary: 1. 2.

Define science, theory, hypothesis, and scientific fact. Explain the difference between truth and scientific facts and describe how we obtain knowledge of each. 3. Describe the “Cycle of Scientific Enterprise,” including the relationships between facts, theories, hypotheses, and experiments. 4. Explain what a theory is and describe the two main characteristics of a theory. 5. Explain what is meant by the statement, “a theory is a model.” 6. Explain the role and importance of theories in scientific research. 7. State and describe the steps of the “scientific method.” 8. Define explanatory, response, and lurking variables in the context of an experiment. 9. Explain why experiments are designed to test only one explanatory variable at a time. Use the procedures the class followed in the Pendulum Experiment as a case in point. 10. Explain the purpose of the control group in an experiment. 11. Describe the possible implications of a negative experimental result. In other words, if the hypothesis is not confirmed, explain what this might imply about the experiment, the hypothesis, or the theory itself.

1.1

Modeling Knowledge

1.1.1 Kinds of Knowledge There are many different kinds of knowledge—propositional truth, scientific knowledge, emotional knowledge, aesthetic knowledge, and others. As students of science, it is quite important that we understand what we mean by scientific knowledge, and how this kind of knowledge differs from others. This is necessary in order to have a proper understanding of what science is and how it works. In this chapter, we focus on two of these basic categories of knowledge: scientific knowledge and truth. We leave emotional knowledge, aesthetic knowledge, and the others for you to pursue in another course of study. One kind of knowledge is truth. Humans have sought to know truth since ancient times, and still do. Although not everyone today agrees about truth, most people in the world today do still believe that truth exists; that there are some things that are true and others that are false. This is the position taken in this text. It is also the case that most people in the world—though clearly not all—hold to some form of religious faith. And although most religious faiths make truth claims of one sort or another, religion is not the only source of truth. There are other sources that we can agree about regardless of our various religious beliefs. Some people handle the distinction between the truth and scientific knowledge by referring to religious teachings as one kind of truth and scientific teaching as a different kind of truth. The problem here is that there are not different kinds of truth. There is only one truth, but there are different kinds of knowledge. Truth is one kind of knowledge, and scientific knowledge is a different kind of knowledge. 3

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We are going to unpack this further over the next few pages, but here is a taste of where we are going. Scientific knowledge is not static. It is always changing as new discoveries are made. On the other hand, the truth does not change. I have developed a model of knowledge that emphasizes the differences between what various sources of truth can reveal to us and what scientific investigations teach us. This model is not perfect (no model is), nor is it exhaustive, but it is very useful, as all good models are. Our main goal in the next few sections is to develop this model of knowledge. The material in this chapter is crucial if you wish to have a proper understanding of what science is all about. To understand science correctly, we need to understand what we mean by scientific knowledge. Unfortunately, there is much confusion among non-scientists about the nature of scientific knowledge and this confusion often leads to misunderstandings when we talk about scientific findings and scientific claims. This is nothing new. Misconceptions about scientific claims have plagued public discourse for thousands of years and continue to do so to this day. This confusion is a severe problem, one much written about within the scientific community in recent years. To clear the air on this issue, it is necessary to examine what we mean by the term truth, as well as the different ways we discover truth. Then we must discuss the specific characteristics of scientific knowledge, including the key scientific terms fact, theory, and hypothesis.

1.1.2 What is Truth and How Do We Know It? Epistemology, one of the major branches of philosophy, is the study of what we can know and how we know it. Both philosophers and theologians claim to have important insights on the issue of knowing truth, and because of the roles science and religion have played in our culture over the centuries, we need to look at what both philosophers and theologians have to say. The issue we need to treat briefly here is captured in this question: What is truth and how do we know it? In other words, what do we mean when we say something is true? And if we can agree on a definition for truth, how can we know whether something is true? These are really complex questions, and philosophers and theologians have been working on them for thousands of years. But a few simple principles will be adequate for our purpose. As for what truth is, my simple but practical definition is this: Truth is the way things really are. Whatever reality is like, that is the truth. If there really is life on other planets, then it is true to say, “There is life on other planets.” If you live in Poughkeepsie, then when you say, “I live in Poughkeepsie” you are speaking the truth. The harder question is: How do we know the truth? According to most philosophers, there are two ways that we can know truth, and these involve either our senses or our use of reason. First, truths that are obvious to us just by looking around are said to be evident. It is evident that birds can fly. No proof is needed. So the proposition, “Birds can fly,” conveys truth. Similarly, it is evident that humans can read books and that birds cannot. Of course, when we speak of people knowing truth this way we are referring to people whose perceptive faculties are functioning normally.

The Nature of Scientific Knowledge

The second way philosophers say we can know truth is through the valid use of logic. Logical conclusions are typically derived from a sequence of logical statements called a syllogism, in which two or more statements (called premises) lead to a conclusion. For example, if we begin with the premises, “All men are mortal,” and, “Socrates was a man,” then it is a valid conclusion to state, “Socrates was mortal.” The truth of the conclusion of a logical syllogism definitely depends on the truth of the premises. The truth of the conclusion also depends on the syllogism having a valid structure. Some logical structures are not logically valid. (These invalid structures are called logical fallacies.) If the premises are true and the structure is valid, then the conclusion must be true. So the philosophers provide us with two ways of knowing truth that most people agree upon—truths can be evident (according to our senses) or they can be proven (by valid use of reason from true premises). Believers in some faith traditions argue for a third possibility for knowing truth, which is by revelation from supernatural agents such as God or angels. Obviously, not everyone accepts the possibility of knowing truth by revelation! Specifically, those who do not believe in God do not accept the possibility of revelations from God. Additionally, there are some who accept the existence of a transcendent power or being, but do not accept the possibility of revelations of truth from that power. So this third way of knowing truth is embraced by many people, but certainly not by everyone. In this text, I make no religious claims. But I think it is important in this discussion to recognize that people can agree on the nature of scientific knowledge regardless of their religious views. I have close friends who are religious believers and close friends who are atheists, and we have discussed this question for many years. I have been very encouraged by the agreement we have found in our understanding about science. And in the unsettling times in which we live, when people seem to be quarreling about almost everything, I think it is important to know about places where people can agree, even when they differ on other issues. Few people would deny that knowing truth is important. This is why we started our study by briefly exploring what truth is. But this is a book about science, and we need now to move to addressing a different question: what does science have to do with truth? The question is not as simple as it seems, as evidenced by the continuous disputes between religious and scientific communities stretching back over the past 700 years. To get at the relationship between science and truth, we first look at the relationship between propositions and truth claims.

1.1.3 Propositions and Truth Claims Not all that passes as valid knowledge can be regarded as true, which I defined in the previous section as “the way things really are.” In many circumstances—maybe most—we do not actually know the way things really are. People do, of course, often use propositions or statements with the intention of conveying truth. But with other kinds of statements, people intend to convey something else. Let’s unpack this with a few example statements. Consider the following propositions: 1.

I have two arms.

My wife and I have three children.

I worked out at the gym last week.

My car is at the repair shop. 5

Chapter 1

Texas gained its independence from Mexico in 1836.

Atoms are composed of three fundamental particles—protons, neutrons, and electrons.

God made the world.

Among these seven statements are actually three different types of claims. From the discussion in the previous section you may already be able to spot two of them. But some of these statements do not fit into any of the categories we explored in our discussion of truth. We can discover some important aspects about these claims be examining them one by one. So suppose for a moment that I, the writer, am the person asserting each of these statements as we examine the nature of the claim in each case. I have two arms. This is true. I do have two arms, as is evident to everyone who sees me. My wife and I have three children. This is true. To me it is just as evident as my two arms. I might also point out that it is true regardless of whether other people believe me when I say it. (Of course, someone could claim that I am delusional, but let’s just keep it simple here and assume I am in normal possession of my faculties.) This bit about the statement being true regardless of others’ acceptance of it comes up because of a slight difference here between the statement about children and the statement about arms. Anyone who looks at me will accept the truth that I have two arms. It will be evident, that is, obvious, to them. But the truth about my children is only really evident to a few people (my wife and I, and perhaps a few doctors and close family members). Nevertheless, the statement is true. I worked out at the gym last week. This is also true; I did work out last week. The statement is evident to me because I clearly remember going there. Of course, people besides myself must depend on me to know it because they cannot know it directly for themselves unless they saw me there. Note that I cannot prove it is true. I can produce evidence, if needed, but the statement cannot be proven without appealing to premises that may or may not be true. Still, the statement is true. My car is at the repair shop. Here is a statement that we cannot regard as a truth claim. It is merely a statement about where I understand my car to be at present, based on where I left it this morning and what the people at the shop told me they were going to do with it. For all I know, they may have taken my car joy riding and presently it may be flying along the back roads of the Texas hill country. I can say that the statement is correct so far as I know. Texas gained its independence from Mexico in 1836. We Texans were all taught this in school and we believe it to be correct, but as with the previous statement we must stop short of calling this a truth claim. It is certainly a historical fact, based on a lot of historical evidence. The statement is correct so far as we know. But it is possible there is more to that story than we know at present (or will ever know) and none of those now living were there. Atoms are composed of three fundamental particles—protons, neutrons, and electrons. This statement is, of course, a scientific fact. But like the previous two statements, this statement is not—surprise!—a truth claim. We simply do not know the truth about atoms. The truth about atoms is clearly not evident to our senses. We cannot guarantee the truth of any premises we might use to construct a logical proof about the insides of atoms, so proof is not able to lead us to the truth. And so far as I know, there are no supernatural agents who have revealed to us anything about atoms. So we have no access to knowing how atoms really are. What we do have are the data from many experiments, which may or may not tell the whole story. Atoms may have other components we don’t know about yet. The best 6

The Nature of Scientific Knowledge

we can say about this statement is that it is correct so far as we know (that is, so far as the scientific community knows). God made the world. This statement clearly is a truth claim, and many people in the world believe it. Other people disagree on whether the statement is true. I include this example here because we soon see what happens when scientific claims and religious truth claims get confused—this was at the core of what the trial of Galileo was all about (even though that story is widely misunderstood). This confusion is also at the heart of today’s “faith vs. science” debate. But regardless of whether a person is a religious believer, the issue is important for us all. We all need to learn to speak correctly about the different claims people make. To summarize this section, some statements we make are evidently or obviously true. But for many statements, we must recognize that we don’t know if they actually are true. The best we can say about these kinds of statements—and scientific facts are like this—is that they are correct so far as we know. Finally, there are metaphysical or religious statements about which people disagree; some claim they are true, some deny the same, and some say there is no way to know.

1.1.4 Truth and Scientific Claims Let’s think a bit further about the truth of reality, both natural and supernatural. I think most people agree that regardless of what different people think about God and nature, there is some actual truth or reality about nature and the supernatural. Regarding nature, there is some full reality about the way, say, atoms are structured, regardless of whether we currently understand that structure correctly. So far as we know, this reality does not shift or change from day to day, at least not since the early history of the universe. So the reality about atoms—the truth about atoms—does not change. And regarding the supernatural, there is some reality about the supernatural realm, regardless of whether anyone knows what that is. Whatever these realities are, they are truths, and these truths do not change either. Now, I have observed over the years that since (roughly) the beginning of the 20th century, careful scientists do not refer to scientific claims as truth claims. They do not profess to knowing the ultimate truth about how nature really is, and often point out that knowing the ultimate truth about reality is not what science is about. Hang with me here for a couple of examples of this. One of my favorite examples of this is a statement made by Danish scientist Niels Bohr, pictured in Figure 1.1, one of the great physicists of the 20th century. Bohr said, “It is wrong to think that the task of physics is to find out how nature is. Physics concerns what we can say about nature.” Scientific claims are understood to be statements about our best understanding of the way things are. Most scientists believe that over time our scientific theories get closer and closer to the truth of the way things really are. But when they are speaking carefully, scientists do not claim that our present understanding of this or that is the Figure 1.1. Danish physicist Niels Bohr truth about this or that. (1885–1962). 7

Chapter 1

Examples of Changing Facts In 2006, the planet Pluto was declared not to be a planet any more. In the 17th century, the fact that the planets and moon all orbit the earth changed to the present fact that the planets all orbit the sun, and only the moon orbits the earth. At present we know of only one kind of matter that causes gravitational fields. This is the matter made up of protons, electrons, and neutrons, which we discuss in a later chapter. But scientists now think there may be another kind of matter contributing to the gravitational forces in the universe. They call it “dark matter” because apparently this kind of matter does not reflect or refract light the way ordinary matter does. (We also study reflection and refraction later on.) For the existence of dark matter to become a scientific fact, a lot of evidence is required, evidence which is just beginning to emerge. If we are able to get enough evidence, then the facts about matter will change. Another example is a statement made by the famous American chemist Gilbert N. Lewis. In a lecture he gave in 1925, Lewis said, “The scientist is a practical man and his aims are practical aims. He does not seek the ultimate but the proximate. The theory that there is an ultimate truth, although very generally held by mankind, does not seem to be useful to science except in the sense of a horizon toward which we may proceed.” Both Bohr and Lewis are pointing us toward a proper understanding of what scientific knowledge is all about. It is not about knowing the truth. It is about modeling nature, as we explore in some detail in Section 1.2.

1.1.5 Truth vs. Facts Whatever the truth is about the way things are, that truth is presumably absolute and unchanging. If there is life on other planets, then that’s the way it is, period. And if matter is made of atoms as we think it is, then that is the truth about matter and it is always the truth. But what we call scientific facts, by their very nature, are not like this. Facts are subject to change, and sometimes do, as new information comes becomes known through ongoing scientific research. Our definitions for truth and for scientific facts need to take this difference into account. As we have seen, truth is the way things really are. By contrast, here is a definition for scientific facts: A scientific fact is a proposition that is supported by a great deal of evidence. Scientific facts are discovered by observation and experiment, and by making inferences from what we observe or from the results of our experiments. A scientific fact is correct so far as we know, but can change as new information becomes known. So facts can change. Scientists do not put them forward as truth claims, but as propositions that are correct so far as we know. In other words, scientific facts are provisional. They are always subject to revision in the future. As scientists make new scientific discoveries, they must sometimes revise facts that were formerly considered to be correct. But the truth about reality, whatever it is, is absolute and unchanging.

The Nature of Scientific Knowledge

The distinction between truth and scientific facts is crucial for a correct understanding of the nature of scientific knowledge. Facts can change; truth does not.

1.2

The Cycle of Scientific Enterprise

1.2.1 Science Having established some basic principles about the distinction between scientific facts and truth, we are now ready to define science itself and examine what science is and how it works. Here is a definition: Science is the process of using experiment, observation, and logical thinking to build “mental models” of the natural world. These mental models are called theories. We do not and cannot know the natural world perfectly or completely, so we construct models of how it works. We explain these models to one another with descriptions, diagrams, and mathematics. These models are our scientific theories. Theories never explain the world to us perfectly. To know the world perfectly, we would have to have certain, exhaustive knowledge of the absolute truth about reality, which we do not. So theories always have their limits, but we hope they become more accurate and more complete over time, accounting for more and more physical phenomena (data, facts), and helping us to understand the natural world as a coherent whole.

Hypothesis An informed prediction, based on a theory

Scientific Fact

Theory

Experiment

Our best explanation at present

Putting the hypothesis to the test

Scientific Fact Scientific Fact

New Scientific Fact

Yes

Analysis Are the experimental results consistent with the theory we started with?

No Review

Reconsider experimental methods, appropriateness of hypothesis, adequacy of theory

Figure 1.2. The Cycle of Scientific Enterprise.

Chapter 1

Scientific knowledge is continuously changing and advancing through a cyclic process that I call the Cycle of Scientific Enterprise, represented in Figure 1.2. In the next few sections, we examine the individual parts of this cycle in detail.

1.2.2 Theories Theories are the grandest thing in science. In fact, it is fair to say that theories are the glory of science, and developing good theories is what science is all about. Electromagnetic field theory, atomic theory, quantum theory, the general theory of relativity—these are all theories in physics that have had a Theory Scientific profound effect on scientific progress and on the way we all live.1 Our best explanation Fact Now, even though many people do not realize it, all scienat present Scientific tific knowledge is theoretically based. Let me explain. A theory is Fact a mental model or explanatory system that explains and relates Scientific Fact together most or all of the facts (the data) in a certain sphere of knowledge. A theory is not a hunch or a guess or a wild idea. Theories are the mental structures we use to make sense of the data we have. We cannot understand any scientific data without a theory to organize it and explain it. This is why I write that all scientific knowledge is theoretically based. And for this reason, it is inappropriate and scientifically incorrect to scorn these explanatory systems as “merely a theory” or “just a theory.” Theories are explanations that account for a lot of different facts. If a theory has stood the test of time, that means it has wide support within the scientific community. It is popular in some circles to speak dismissively of certain scientific theories, as if they represent some kind of untested speculation. It is simply incorrect—and very unhelpful—to speak this way. As students in high school science, one of the important things you need to understand is the nature of scientific knowledge, the purpose of theories, and the way scientific knowledge progresses. These are the issues this chapter is about. All useful scientific theories must possess several characteristics. The two most important ones are: •

The theory accounts for and explains most or all of the related facts.

•

The theory enables new hypotheses to be formed and tested.

Theories typically take decades or even centuries to gain credibility. If a theory gets replaced by a new, better theory, this also usually takes decades or even centuries to happen. No theory is ever “proven” or “disproven” and we should not speak of them in this way. We also should not speak of them as being “true” because, as we have seen, we do not use the word “truth” when speaking of scientific knowledge. Instead we speak of facts being correct so far as we know, or of current theories as representing our best understanding, or of theories being successful and useful models that lead to accurate predictions. An experiment in which the hypothesis is confirmed is said to support the theory. After such an experiment, the theory is stronger but it is not proven. If a hypothesis is not confirmed by an experiment, the theory might be weakened but it is not disproven. Scientists require a great deal of experimental evidence before a new theory can be established as the best explanation for a body of data. This is why it takes so long for theories to become widely accepted. And since no theory ever explains everything perfectly, there are always phe1

The term law is just a historical (and obsolete) term for what we now call a theory.

The Nature of Scientific Knowledge

nomena we know about that our best theories do Examples of Famous Theories not adequately explain. Of course, scientists continue their work in a certain field hoping eventu- In the next chapter, we encounter Einally to have a theory that does explain all of the stein’s general theory of relativity, one of facts. But since no theory explains everything the most important theories in modern perfectly, it is impossible for one experimental physics. Einstein’s theory represents our failure to bring down a theory. Just as it takes a best current understanding of how gravlot of evidence to establish a theory, so it takes a ity works. large and growing body of conflicting evidence Another famous theory we address later before scientists abandon an established theory. is the kinetic theory of gases, our present At the beginning of this section, I state that understanding of how molecules of gas theories are mental models. This statement needs too small to see are able to create presa bit more explanation. A model is a representa- sure inside a container. tion of something, and models are designed for a purpose. You have probably seen a model of the organs in the human body in a science classroom or textbook. A model like this is a physical model and its purpose is to help people understand how the human body is put together. A mental model is not physical; it is an intellectual understanding, although we often use illustrations or physical models to help communicate to one another our mental ideas. But as in the example of the model of the human body, a theory is also a model. That is, a theory is a representation of how part of the world works. Frequently, our models take the form of mathematical equations that allow us to make numerical predictions and calculate the results of experiments. The more accurately a theory represents the way the world works, which we judge by forming new hypotheses and testing them with experiments, the better and more successful the theory is. To summarize, a successful theory represents the natural world accurately. This means the model (theory) is useful because if a theory is an accurate representation, then it leads

Key Points About Theories 1.

A theory is a way of modeling nature, enabling us to explain why things happen in the natural world from a scientific point of view.

A theory tries to account for and explain the known facts that relate to it.

Theories must enable us to make new predictions about the natural world so we can learn new facts.

Strong, successful theories are the glory and goal of scientific research.

A theory becomes stronger by producing successful predictions that are confirmed by experiment. A theory is gradually weakened when new experimental results repeatedly turn out to be inconsistent with the theory.

It is incorrect to speak dismissively of successful theories because theories are not just guesses.

We don’t speak of theories as being proven or disproven. Instead, we speak of them in terms such as how successful they have been at making predictions and how accurate the predictions have been.

Figure 1.3. Key points about theories.

Chapter 1

to accurate predictions about nature. When a theory repeatedly leads to predictions that are confirmed in scientific experiments, it is a strong, useful theory. The key points about theories are summarized in Figure 1.3.

1.2.3 Hypotheses Hypothesis A hypothesis is a positively stated, informed prediction An informed prediction, about what will happen in certain circumstances. We say a based on a theory hypothesis is an informed prediction because when we form hypotheses we are not just speculating out of the blue. We are applying a certain theoretical understanding of the subject to the new situation before us and predicting what will happen or what we expect to find in the new situation based on the theory the hypothesis is coming from. Every scientific hypothesis is based on a particular theory. Often hypotheses are worded as if-then statements, such as, “If Hypothesis various forces are applied to a An informed prediction, based on a theory pickup truck, then the truck accelerates at a rate that is in direct proportion to the net force.” Every scientific hypothesis is based on Scientific Fact Scientific Fact

Theory

Our best explanation at present

Scientific Fact

Key Points About Hypotheses 1.

A hypothesis is an informed prediction about what will happen in certain circumstances.

2. Every hypothesis is based on a para theory and it is the hypothesis that is directly ticular theory. tested by an experiment. If the experiment turns 3. Well-formed scientific hypotheses out the way the hypothesis predicts, the hypothmust be testable, which is what esis is confirmed and the theory it came from is scientific experiments are destrengthened. Of course, the hypothesis may not signed to do. be confirmed by the experiment. We see how scientists respond to this situation in Section 1.2.6. Figure 1.4. Key points about hypotheses.

Examples of Famous Hypotheses Einstein used his general theory of relativity to make an incredible prediction in 1917: that gravity causes light to bend as it travels through space. In the next chapter, you read about the stunning result that occurred when this hypothesis was put to the test. The year 2012 was a very important year for the standard theory in the world of subatomic particles, called the Standard Model. This theory led in the 1960s to the prediction that there are weird particles in nature, now called Higgs Bosons, which no one had ever detected. Until 2012, that is! An enormous machine that could detect these particles, called the Large Hadron Collider, was built in Switzerland and completed in 2008. In 2012, scientists announced that the Higgs Boson had been detected at last, a major victory for the Standard Model, and for Peter Higgs, the physicist who first proposed the particle that now bears his name. 12

The Nature of Scientific Knowledge

The terms theory and hypothesis are often used interchangeably in common speech, but in science they mean different things. For this reason you should make note of the distinction. One more point about hypotheses. A hypothesis that cannot be tested is not a scientific hypothesis. For example, horoscopes purport to predict the future with statements like, “You will meet someone important to your career in the coming weeks.” Statements like this are so vague they are untestable and do not qualify as scientific hypotheses. The key points about hypotheses are summarized in Figure 1.4.

1.2.4 Experiments Experiments are tests of the predictions in hypotheses, under controlled conditions. Effective experiments are difficult to perform. Thus, for any experimental outcome to become regarded as a “fact” it must be replicated by several different experimental teams, often working in different labs around the world. Scientists have developed rigorous methods for conducting valid experiments. We consider these briefly in Section 1.3.

Experiment

Putting the hypothesis to the test

1.2.5 Analysis In the Analysis phase of the Cycle of Scientific Enterprise, researchers must interpret the exYes Are the Analysis experimental results perimental results. The results of an experiment consistent with the theory we are essentially data, and data always have to be instarted with? terpreted. The main goal of this analysis is to deNo termine whether the original hypothesis has been confirmed. If it has, then the experiment has produced new facts that are consistent with the original theory because the hypothesis was based on that theory. As a result, the support for the theory has increased—the theory was successful in generating a hypothesis that was confirmed by experiment. As a result of the experiment, our confidence in the theory as a useful model has increased and the theory is even more strongly supported than before.

1.2.6 Review If the outcome of an experiment does not confirm the hypothesis, the researchers must consider all the possibilities for why this might have happened. Why didn’t our theory, which is our best explanation of how things work, enable us to form a correct prediction? There are a number of possibilities, beginning with the experiment and going backwards around the cycle: •

No Review

Reconsider experimental methods, appropriateness of hypothesis, adequacy of theory

The experiment may have been flawed. Scientists double check everything about the experiment, making sure all equipment is working properly, double checking the calculations, looking for unknown factors that may have inadvertently influenced the outcome, verifying that the measurement instruments are accurate enough and precise enough to do the job, and so on. They also wait for other experimental teams to try the experiment to see if they get the same results or different results, and then compare. 13

Chapter 1

(Although, naturally, every scientific team likes to be the first one to complete an important new experiment.) •

The hypothesis may have been based on a incorrect understanding of the theory. Maybe the experimenters did not understand the theory well enough, and maybe the hypothesis is not a correct statement of what the theory says will happen.

•

The values used in the calculation of the hypothesis’ predictions may not have been accurate or precise enough, throwing off the hypothesis’ predictions.

•

Finally, if all else fails, and the hypothesis still cannot be confirmed by experiment, it is time to look again at the theory. Maybe the theory can be altered to account for this new fact. If the theory simply cannot account for the new fact, then the theory has a weakness, namely, there are facts it doesn’t adequately account for. If enough of these weaknesses accumulate, then over a long period of time (like decades) the theory might eventually need to be replaced with a different theory, that is, another, better theory that does a better job of explaining all the facts we know. Of course, for this to happen someone would have to conceive of a new theory, which usually takes a great deal of scientific insight. And remember, it is also possible that the facts themselves can change.

1.3

The Scientific Method

1.3.1 Conducting Reliable Experiments The so-called scientific method that you have been studying ever since about fourth grade is simply a way of conducting reliable experiments. Experiments are an important part of the Cycle of Scientific Enterprise, and so the scientific method is important to know. You probably remember studying the steps in the scientific method from prior courses, so they are listed in Table 1.1 without further comment. We will be discussing variables and measurements a lot in this course, so we should take the opportunity here to identify some of the language researchers use during the experimental process. In a scientific experiment, the researchers have a question they are trying to answer (from the State the Problem step in the scientific method), and typically it is some kind of question about the way one physical quantity affects another one. So the researchers design an experiment in which one quantity can be manipulated (that is, deliberately varied in a controlled fashion) while the value of another quantity is monitored. A simple example of this in everyday life that you can easily relate to is varying the amount of time you spend each week studying for your math class in order to see what effect the time spent has on the grades you earn. If you reduce the time you spend, will your grades go down? If you increase the time, will they go up? A precise answer depends on a lot The Scientific Method 1.

State the problem.

Collect data.

Research the problem.

Analyze the data.

Form a hypothesis.

Form a conclusion.

Conduct an experiment.

Repeat the work.

Table 1.1. Steps in the scientific method.

The Nature of Scientific Knowledge

This is the quantity you adjust.

Study Time Grades in Math Experimental System

This is where you look to see the effect.

Figure 1.5. Study time and math grades in a simple experimental system.

of things, of course, including the person involved, but in general we would all agree that if a student varies the study time enough we would expect to see the grades vary as well. And in particular, we would expect more study time to result in higher grades. The way your study time and math grades relate together can be represented in a diagram such as Figure 1.5. Now let us consider this same concept in the context of scientific experiments. An experiment typically involves some kind of complex system that the scientists are modeling. The system could be virtually anything in the natural world—a galaxy, a system of atoms, a mixture of chemicals, a protein, or a badger. The variables in the scientists’ mathematical models of the system correspond to the physical quantities that can be manipulated or measured in the system. As I describe the different kinds of variables, refer to Figure 1.6.

1.3.2 Experimental Variables When performing an experiment, the variable that is deliberately manipulated by the researchers is called the explanatory variable. As the explanatory variable is manipulated, the researchers monitor the effect this variation has on the response variable. In the example of study time versus math grade, the study time is the explanatory variable and the grade earned is the response variable. Usually, a good experimental design allows only one explanatory variable to be manipulated at a time so that the researchers can tell definitively what its effect is on the response variable. If more than one explanatory variable were changing during the course of the experiment, researchers may not be able to tell which one was causing the effect on the response variable. A third kind of variable that plays a role in experiments is the lurking variable. A lurking variable is a variable that affects the response variable without the researchers being aware of it. This is undesirable, of course, because with unknown influences present the researchers may not be able to make a correct conclusion about the effect of the known Lurking Variable Researchers manipulate input here

Explanatory Variable Response Variable Experimental System

Researchers measure outcome here

Figure 1.6. The variables in an experimental system.

Chapter 1

explanatory variables on the response variable under study. So researchers have to study their experimental projects very carefully to minimize the possibility of lurking variables affecting their results.

Do You Know ...

What are double-blind experiments?

The human mind is so powerful that if a person believes a new medication might help, the person’s condition can sometimes improve even if the medication itself isn’t doing a thing! This is amazing, but in medical research it means that the researchers can have a hard time determining whether a person is helped by the new medication, or by feeling positively about the medication, or even by the attention given to him or her by the doctor. Pictured below is Lauren Wood, a clinician involved in vaccine research at the Center for Cancer Research, which is part of the National Cancer Institute. Just as with every other scientific researcher, Dr. Wood’s research is conducted according to methods that have been developed to ensure that people’s beliefs about the research don’t influence the outcome of the research. The approach is to divide the patients who will participate in testing a new medication into two groups, control and experimental. The experimental group is given the new medication. The control group is given a placebo—a fake medication such as a sugar pill— that has no effect on the person’s medical condition. Further, none of the patients know whether they are given the placebo or the real medication. This technique, called a blind experiment, allows the researchers to determine whether a new medication actually helps, as they compare the results of the control and experimental groups. But there’s more. It turns out that the researchers themselves can affect the results of the experiment if they know which patients are receiving a placebo and which ones are receiving the medication under study. How can this happen? Well, if the researchers know who is getting the real medication, they might subconsciously act more positively with them than with other patients. This might be because the researchers expect those getting the new medication to improve, and this expectation gets subconsciously communicated to the patients. The positive attitude might be perceived as more encouraging and patients might improve just because of the encouragement! The way around this dilemma is to use a double-blind experiment. In a double-blind experiment, neither the patients nor the researchers know which patients are getting the placebo and which are getting the real treatment. A team of technicians is in the middle, administering the medication and keeping records of who received what. The researchers are not allowed to see the lists until the research results are finalized. The double-blind experiment is the standard protocol followed today for new medical research. 16

The Nature of Scientific Knowledge

In our example about study time and math grades, there could be a number of lurking variables affecting the results of the experiment. Possible lurking variables include changes in the difficulty of the material from one chapter to the next, and variations in the student’s ability to concentrate due to fatigue from seasonal sports activities.

1.3.3 Experimental Controls The last thing we consider in this section is an important way researchers control an experiment to ensure the results are valid. You are probably aware that developing new medical treatments is one of the major goals of experimental research in the 21st century. Many experiments in the field of medical research are designed to test some new kind of treatment by comparing the results of the new treatment to those obtained using a conventional treatment or no treatment at all. This is the situation in medical research all the time for experiments testing new therapies, medications, or procedures. Clinical trials are experiments conducted by researchers on people to test new therapies or medications. In experiments like these, the people (patients) involved in the study are divided into two groups—the control group and the experimental group. The control group receives no treatment or some kind of standard treatment. The experimental group receives the new treatment being tested. The results of the experimental group are assessed by comparing them to those of the control group. Another example will help to clarify all these terms. Let’s say researchers have developed a variety of fruit tree that they believe is more resistant to drought than other varieties. According to the researchers’ theoretical understanding of how chemical reactions and water storage work in the biological systems of the plant, they hypothesize that the new variety of tree will be able to bear better fruit during drought conditions. To test this hypothesis by experiment, the scientists develop a group of the new trees. Then they place the trees in a test plot, along with other trees of other varieties, and see how they perform. Figure 1.7 shows a researcher working in an agricultural test plot. In our fruit tree example, the trees of the new variety are in the experimental group and the trees of the other varieties are in the control group. The response variable is the quality of the plant’s fruit. Researchers expect that under drought conditions the fruit of the new variety will be better than the fruit of the other varieties. The explanatory variable is the unique feature of the new variety that relates to the plant’s use of water. The trees are exposed to drought conditions in the experiment. If the new variety produces higher quality fruit than the control group, then the hypothesis is confirmed, and the theory that led to the hypothesis has gained credibility through this success. One can imagine many different lurking variables that could affect the outcome of this experiment without the scientists’ awareness. For example, the new variety trees could be planted in locations that receive different amounts of moisture or sun than the locations where the control group trees are, or, the nutrients in the soil in different locations might vary. In a good experimental design, researchers seek to Figure 1.7. An agricultural research identify such factors and take measures to ensure that assistant working in a test plot. 17

Chapter 1

they do not affect the outcome of the experiment. They do this by making sure there are trees from both the experimental group and the control group in all the different conditions the trees will experience. This way, variations in sunlight, soil type, soil water content, elevation, exposure to wind, and other factors will be experienced equally by trees in both groups.

Chapter 1 Exercises As you go through the chapters in this book, always answer the questions in complete sentences, using correct grammar and spelling. Here is a tip that will help improve the quality of your written responses: avoid pronouns! Pronouns almost always make your responses vague or ambiguous. If you want to receive full credit for written responses, avoid them. (Oops. I mean, avoid pronouns!)

Study Questions Answer the following questions with a few complete sentences. 1.

Distinguish between theories and hypotheses.

Explain why a single experiment can never prove or disprove a theory.

Explain how an experiment can still provide valuable data even if the hypothesis under test is not confirmed.

Explain the difference between truth and facts and describe the sources of each.

State the two primary characteristics of a theory.

Does a theory need to account for all known facts? Why or why not?

It is common to hear people say, “I don’t accept that; it’s just a theory.” What is the error in a comment like this?

Distinguish between facts and theories.

Distinguish between explanatory variables, response variables, and lurking variables.

10. Why do good experiments that seek to test some kind of new treatment or therapy include a control group? 11. Explain specifically how the procedure you followed in the Pendulum Experiment satisfies every step of the “scientific method.” 12. This chapter argues that scientific facts should not be regarded as true. Someone might question this and ask, If they aren’t true, then what are they good for? Develop a response to this question. 13. Explain what a model is and why theories are often described as models. 14. Consider an experiment that does not deliver the result the experimenters had expected. In other words, the result is negative because the hypothesis is not confirmed. There are many reasons why this might happen. Consider each of the following elements of the Cycle of Scientific Enterprise. For each one, describe how it might be the driving factor that results in the experiment’s failure to con18

The Nature of Scientific Knowledge

firm the hypothesis. a. the experiment b. the hypothesis c. the theory 15. Identify the explanatory and response variables in the Pendulum Experiment, and identify two realistic possibilities for ways the results may have been influenced by lurking variables.

Do You Know ...

How did Sir Humphry Davy become a hero? Sir Humphry Davy (1778–1829) was one of the leading experimenters and inventors in England in the early nineteenth century. He conducted many early experiments with gases; discovered sodium, potassium, and numerous other elements; and produced the first electric light from a carbon arc. In the early nineteenth century, explosions in coal mines were frequent, resulting in much tragic loss of life. The explosions were caused by the miners’ lamps igniting the methane gas found in the mines. Davy became a national hero when he invented the Davy Safety Lamp (below). This lamp incorporated an iron mesh screen around the flame. The cooling

from the iron reduces the flame temperature so the flame does not pass through the mesh, and thus cannot cause an explosion. The Davy Lamp was produced in 1816 and was soon in wide use. Davy’s experimental work proceeded by reasoning from first principles (theory) to hypothesis and experiment. Davy stated, “The gratification of the love of knowledge is delightful to every refined mind; but a much higher motive is offered in indulging it, when that knowledge is felt to be practical power, and when that power may be applied to lessen the miseries or increase the comfort of our fellow-creatures.”

CHAPTER 2

Motion

Orrery Orreries, mechanical models of the solar system, were well-known teaching tools in the 18th century, often forming the centerpiece of lessons on astronomy. They demonstrated Copernicus’ theory that the earth and other planets orbit the sun. This example, from around 1750, is smaller but otherwise similar to George II’s grand orrery. This photo of the orrery was taken in the British Museum in London. 20

OBJECTIVES Memorize and learn how to use these equations: v f − vi d a= t t After studying this chapter and completing the exercises, students will be able to do each of the following tasks, using supporting terms and principles as necessary: v=

1. 2. 3.

Define and distinguish between velocity and acceleration. Use scientific notation correctly with a scientific calculator. Calculate distance, velocity, and acceleration using the correct equations, MKS and USCS units, unit conversions, and units of measure. 4. Use from memory the conversion factors, metric prefixes, and physical constants listed in Appendix A. 5. Explain the difference between accuracy and precision and apply these terms to questions about measurement. 6. Demonstrate correct understanding of precision by using the correct number of significant digits in calculations and rounding. 7. Describe the key features of the Ptolemaic model of the heavens, including all the spheres and regions in the model. 8. State several additional features of the medieval model of the heavens and relate them to the theological views of the religious authorities opposing Copernicanism. 9. Briefly describe the roles and major scientific models or discoveries of Copernicus, Tycho, Kepler, and Galileo in the Copernican Revolution. Also, describe the significant later contributions of Isaac Newton and Albert Einstein to our theories of motion and gravity. 10. Describe the theoretical shift that occurred in the Copernican Revolution and how religious officials (both supporters and opponents) were involved. 11. State Kepler’s first law of planetary motion. 12. Describe how the gravitational theories of Kepler, Newton, and Einstein illustrate the way the Cycle of Scientific Enterprise works.

2.1

Computations in Physics

In this chapter, you begin mastering the skill of applying mathematics to the study of physics. To do this well, you must know a number of things about the way measurements are handled in scientific work. You also need to have a solid problem-solving strategy you can depend on to help you solve problems correctly without becoming confused. These topics are addressed in this chapter.

Chapter 2

the USCS (U.S. Customary System). You have probably studied these systems before meter m length and should already be familiar with some of kilogram kg mass the SI units and prefixes, so our treatment second s time here will be brief. If you think about it, you would probampere A electric current ably agree that the USCS is cumbersome. kelvin K temperature One problem is that there are many differcandela Cd luminous intensity ent units of measure for every kind of physimole mol amount of substance cal quantity. For example, just for measuring length or distance we have the inch, Table 2.1. The seven base units in the SI unit system. foot, yard, and mile. The USCS is also full of random numbers like 3, 12, and 5,280, and there is no inherent connection between units for different types of quantities. By contrast, the SI system is simple and has many advantages. There is usually only one basic unit for each kind of quantity, such as the meter for measuring length. Instead of having many unrelated units of measure for measuring quantities of different sizes, fractional and multiple prefixes based on powers of ten are used with the units to accommodate various sizes of measurements. A second advantage is that since quantities with different prefixes are related by some power of ten, unit conversions can often be performed mentally. To convert 4,555 ounces into gallons, we first have to look up the conversion from ounces to gallons (which is hard to remember), and then use a calculator to perform the conversion. But to convert 40,555 cubic centimeters into cubic meters is simple—simply divide by 1,000,000 and you have 0.040555 m3. (If you are not clear on the reason for dividing by 1,000,000, just hold on until we get to the end of Section 2.1.3.) Another SI advantage is that the units for different types of quantities relate to one another in some way. Unlike the gallon and the foot, which have nothing to do with each other, the liter (a volume) relates to the centimeter (a length): 1 liter = 1,000 cubic centimeters.1 For all these reasons, the USCS is not used much in scientific work. The SI system is the international standard and it is important to know it well. In the SI unit system, there are seven base units, listed in Table 2.1. (In this text, we use only the first five of them.) There are also many additional units of measure, known as derived units. All the derived units are formed by various combinations of base units. To illustrate, below are a few examples of derived units that we discuss and use in this book. Note, however, that we won’t be working much with the messy fractions; they are simply shown to illustrate how base units are combined to form derived units. Unit

Symbol

Quantity

•

the newton (N) is the SI unit for measuring force: 1 N = 1

•

the joule (J) is the SI unit for measuring energy: 1 J = 1

•

the watt (W) is the SI unit for measuring power: 1 W = 1

kg ⋅m s2

kg ⋅m 2 s2 kg ⋅m 2 s3

The liter is not actually an official SI unit of measure, but it is used all the time anyway in scientific work.

Motion

Using the SI system requires knowing the units of measure—base and derived—and the prefixes that are applied to the units to form fractional units (such as the centimeter) and multiple units (such as the kilometer). The complete list of metric prefixes is shown in Appendix A in Table A.1. The short list of prefixes you need to know by memory for use in this course is in Table A.2. Note that even though the kilogram is a base unit, prefixes are not added to the kilogram. Instead, prefixes are added to the gram to form units such as the milligram and microgram.

2.1.2 MKS Units

Variable

Variable Symbol

Unit

Unit Symbol

A handy subset of the SI system length d (distance) meter m is the so-called MKS system. The MKS L (length) system uses only base units—such as h (height) the meter, kilogram, and second (hence, r (radius), etc. “MKS”) as units for mass, length, and mass m kilogram kg time—along with other units derived time t second s from the base units. The mass, length, and time units, and the symbols and Table 2.2. The three base units in the MKS system. variables used with them, are listed in Table 2.2. Dealing with different systems of units can become quite confusing. But the wonderful thing about sticking to the MKS system is that any calculation performed with MKS units produces a result in MKS units. This is why the MKS system is so handy. The MKS system dominates calculations in physics and we use it almost all the time in this course. To convert the units of measure given in problems into MKS units, you must know the conversion factors listed in Appendix A in Tables A.2, A.3, and A.4. Table A.5 lists several common unit conversions that you are not required to memorize but should have handy when working problem assignments.

2.1.3 Converting Units of Measure One of the most basic skills scientists and engineers use is re-expressing quantities into equivalent quantities with different units of measure. These calculations are called unit conversions. Mastery of this skill is essential for any student in high school science, and you use it a lot in this course. You have studied unit conversions in your math classes for the past few years. But this skill is so important in science that we are going take the time in this section to review in detail how to perform unit conversions. Let’s begin with the basic principle of how this works. First, you know that multiplying any value by unity (one) leaves its value unchanged. Second, you also know that in any fraction if the numerator and denominator are equivalent, the value of the fraction is unity, which means one. A “conversion factor” is simply a fractional expression in which the numerator and denominator are equivalent ways of writing the same physical quantity. This means a conversion factor is just a special way of writing unity (one). Third, we know that when multiplying fractions, factors that appear in both the numerator and denominator may be “cancelled out.” So when performing common unit conversions, what we are doing is repeatedly multiplying our given quantity by unity so that cancellations alter the units of measure until they are expressed the way we wish. Since all we are doing is multiplying by one, the value of our original quantity is unchanged; it simply looks different because it is expressed with different units of measure. 23

Chapter 2

Do You Know ...

How are the base units defined?

The definitions of the base units all have interesting stories behind them. In the past, several units were defined by physical objects, such as a metal bar (the meter) or metal cylinder (the kilogram). But over time these definitions have been replaced. (The last one was the kilogram, replaced in 2019.) Now, each base unit is defined in terms of a physical constant that itself is defined with a specific, exact value. The definitions of all but two of the units also depend on other unit definitions, as the arrows in the graphic indicate. The official definition of the second is based on waves of light emitted by cesium atoms. The speed of light is defined as 299,792,458 m/s, and the meter is defined as the distance light travels in 1/299,792,458 seconds. Let me elaborate a bit more on the idea of unity I mention above, using one common conversion factor as an example. School kids all learn that there are 5,280 feet in one mile, which means 5,280 ft = 1 mi. One mile and 5,280 feet are equivalent ways of writing the same length. If we place these two expressions into a fraction, the numerator and denominator are equivalent, so the value of the fraction is unity, regardless of the way we write it. The equation 5,280 ft = 1 mi can be written in a conversion factor two different ways, and the fraction equals unity either way: 5280 ft 1 mi = =1 1 mi 5280 ft So if you have a measurement such as 43,000 feet that you wish to re-express in miles, the conversion calculation is written this way: 43,000 ft ⋅

1 mi = 8.1 mi 5280 ft

There are two important comments to make here. First, since any conversion factor can be written two ways (depending on which quantity is placed in the numerator), how do we know which way to write the conversion factor? Well, we know from algebra that when we have quantities in the numerator of a fraction that are multiplied, and quantities in the denominator of the fraction that are multiplied, any quantities that appear in both the numerator and denominator cancel. Most units of measure are mathematically treated as multiplied quantities that can be cancelled out.2 In the example above, we desire that “feet” in the given quantity (which is in the numerator) cancels out, so the conversion factor is written with feet in the denominator and miles in the numerator. Second, if you perform the calculation above, the result that appears on your calculator screen is 8.143939394. So why didn’t I write down all those digits in my result? Why did I round my answer off to simply 8.1 miles? The answer to that question has to do with the sig2

An example of a unit that cannot always be treated this way is the degree Fahrenheit.

Motion

nificant digits in the value 43,000 ft that we started with. We address the issue of significant digits later in this chapter, but in the examples that follow I always write the results with the correct number of significant digits for the values involved in the problem. There are several important techniques you must use to help you perform unit conversions correctly; these are illustrated below with examples. You should rework each of the examples on your own paper as practice to make sure you can do them correctly. As a reminder, the conversion factors used in the examples below are all listed in Appendix A. You should study Appendix A to see which ones you must know by memory and which ones are provided to you on quizzes. 1.

Use only horizontal bars in your unit fractions. Never use slant bars.

In printed materials, one often sees values written with a slant fraction bar in the units, as in the value 35 m/s. Although writing the units this way is fine for a printed document, you should not write values this way when you are performing unit conversions. This is because it is easy to get confused and not notice that one of the units is in the denominator in such an expression (s, or seconds, in my example), and the conversion factors used must take this into account. Example 2.1 Convert 57.66 mi/hr into m/s. Writing the given quantity with a horizontal bar makes it clear that “hour” is in the denominator. This helps you to write the hour-to-seconds factor correctly. 57.66

mi 1609 m 1 hr m ⋅ ⋅ = 25.77 hr mi 3600 s s

Now that you have your result, you may write it as 25.77 m/s if you wish, but do not use slant fraction bars in the units when you are working out the unit conversion.

The term “per” implies a fraction.

Some units of measure are commonly written with a “p” for “per,” such as mph for miles per hour or gps for gallons per second. Change these expressions to fractions with horizontal bars when you work out the unit conversion. Example 2.2 Convert 472.15 gps to L/hr. When you write down the given quantity, change the gps to gal/s, and write these units with a horizontal bar: 472.15

gal 3.785 L 3600 s L ⋅ ⋅ = 6, 434,000 s 1 gal 1 hr hr

Chapter 2

Use the × and ÷ keys correctly when entering values into your calculator.

When dealing with several numerator terms and several denominator terms, multiply all the numerator terms together first, hitting the × key between each, then hit the ÷ key and enter all the denominator terms, hitting the ÷ key between each. This way you do not need to write down intermediate results and you do not need to use any parentheses. Example 2.3 Convert 43.17 mm/hr into km/yr. The setup with all the conversion factors is as follows: 43.17

mm 1m 1 km 24 hr 365 day km ⋅ ⋅ ⋅ ⋅ = 0.378 hr 1000 mm 1000 m 1 day 1 yr yr

To execute this calculation in your calculator, enter the values and operations in this sequence: 43.17 × 24 × 365 ÷1000 ÷1000 =

When converting units for area and volume such as cm2 or m3, use the appropriate length conversion factor twice for areas or three times for volumes.

The unit “cm2 ” for an area means the same thing as “cm × cm.” Likewise, “m3 ” means “m × m × m.” So when you use a length conversion factor such as 100 cm = 1 m or 1 in = 2.54 cm, you must use it twice to get squared units (areas) or three times to get cubed units (volumes). Example 2.4 Convert 3,550 cm3 to m3. 3550 cm3 ⋅

1m 1m 1m ⋅ ⋅ = 0.00355 m3 100 cm 100 cm 100 cm

Notice in Example 2.4 that the unit cm occurs three times in the denominator, giving us cm3 when they are all multiplied together. This cm3 term in the denominator cancels with the cm3 term in the numerator. And since the m unit occurs three times in the numerator, they multiply together to give us m3 for the units in our result. Notice also that the denominator is 100∙100∙100 = 1,000,000. This is why I write in Section 2.1.1 that to convert from cm3 to m3 we just divide by 1,000,000. Pay attention to this and don’t make the common (and silly) mistake of dividing by 100! This issue only arises when you have a unit raised to a power, such as when using a length unit to represent an area or a volume. When using a conversion factor such as 3.785 L = 1 gal, the units of measure are written using units that are strictly volumetric (liters and 26

Motion

gallons), and are not obtained from lengths the way in2, ft2, cm3, and m3 are. Another common unit that uses a power is acceleration, which has units of m/s2 in the MKS unit system. Example 2.5 Convert 5.85 mi/hr2 into MKS units. 5.85

mi 1609 m 1 hr 1 hr m ⋅ ⋅ ⋅ = 0.000726 2 2 hr 1 mi 3600 s 3600 s s

With this example you see that since the “hour” unit is squared in the given quantity, the conversion factor converting hours to seconds must appear twice in the conversion calculation.

2.1.4 Accuracy and Precision The terms accuracy and precision refer to the limitations inherent in making measurements. Science is all about investigating nature and to do that we must make measurements. Accuracy relates to error, which is the difference between a measured value and the true value. The lower the error is in a measurement, the better the accuracy. Error can be caused by a number of different factors, including human mistakes, malfunctioning equipment, incorrectly calibrated instruments, or unknown factors that influence a measurement without the knowledge of the experimenter. All measurements contain error because (alas!) perfection is simply not a thing we have access to in this world. Precision refers to the resolution or degree of “fine-ness” in a measurement. The limit to the precision obtained in a measurement is ultimately dependent on the instrument used to make the measurement. If you want greater precision, you must use a more precise instrument. The precision of a measurement is indicated by the number of significant digits (or significant figures) included when the measurement is written down (see next section). Figure 2.1 is a photograph of a machinist’s rule and an architect’s scale set side by side. Since the marks on the two scales line up consistently, these two scales are equally accurate. But the machinist’s rule (on top) is more precise. The architect’s scale is marked in 1/16-inch increments, but the machinist’s rule is marked in 1/64-inch increments. It is important that you are able to distinguish between accuracy and precision. Here is an example to illustrate the difference. Let’s say Shana and Marius each buy digital thermometers for their homes. The thermometer Shana buys cost $10 and measures to the nearest 1°F. Marius pays $40 and gets one that reads to the nearest 0.1°F. Note that on a day when the actual temperature is 95.1°F, if the two thermometers are reading accurately Shana’s thermometer reads 95° and Marius’ reads 95.1°. Thus, Marius’ thermometer is more precise. Now suppose Shana reads the directions and properly in- Figure 2.1. The accuracy of these two scales is the same, but the stalls the sensor for her new ther- machinist’s rule on the top is more precise. 27

Chapter 2

mometer in the shade. Marius doesn’t read the directions and mounts his sensor in the direct sunlight, which causes a significant error in the measurement for much of the day. The result is that Shana has lower-precision, higher-accuracy measurements!

2.1.5 Significant Digits The precision in any measurement is indicated by the number of significant digits it contains. Thus, the number of digits we write in any measurement we deal with in science is very important. The number of digits is meaningful because it shows the precision present in the instrument used to make the measurement. Let’s say you are working a computational exercise in a science book. The problem tells you that a person drives a distance of 110 miles at an average speed of 55 miles per hour and wants you to calculate how long the trip takes. The correct answer to this problem will be different from the correct answer to a similar problem with given values of 110.0 miles and 55.0 miles per hour. And if the given values are 110.0 miles and 55.00 miles per hour, the correct answer is different yet again. Mathematically, of course, all three answers are the same. If you drive 110 miles at 55 miles per hour, the trip takes two hours. But scientifically, the correct answers to these three problems are different: 2.0 hours, 2.00 hours, and 2.000 hours, respectively. The difference between these cases is in the precision indicated by the given data, which are measurements. (Even though this is just a made-up problem in a book and not an actual measurement someone made in an experiment, the given data are still measurements. There is no way to talk about distances or speeds without talking about measurements, even if the measurements are only imaginary or hypothetical.) When you perform a calculation with physical quantities (measurements), you cannot simply write down all the digits shown by your calculator. The precision inherent in the measurements used in a computation governs the precision in any result you calculate from those measurements. And since the precision in a measurement is indicated by the number of significant digits, data and calculations must be written with the correct numbers of significant digits. To do this, you need to know how to count significant digits and you must use the correct number of significant digits in all your calculations and experimental data. Correctly counting significant digits involves four different cases: 1.

Rules for determining how many significant digits there are in a given measurement.

Rules for writing down the correct number of significant digits in a measurement you are making and recording.

Rules for computations you perform with measurements—multiplication and division.

Rules for computations you perform with measurements—addition and subtraction.

In this course, we do not use the rules for addition and subtraction, so we leave those for a future course (probably chemistry). We now address the first three cases, in order. Case 1

We begin with the rule for determining how many significant digits there are in a given measurement value. The rule is as follows: The number of significant digits (or figures) in a number is found by counting all the digits from left to right beginning with the first nonzero digit on the left. When no decimal is present, trailing zeros are not considered significant.

Motion

Let’s apply this rule to several example values to see how it works: 15,679

This value has five significant digits.

21.0005

This value has six significant digits.

37,000

This value has only two significant digits because when there is no decimal trailing zeros are not significant. Notice that the word significant here is a reference to the precision of the measurement, which in this case is rounded to the nearest thousand. The zeros in this value are certainly important, but they are not significant in the context of precision.

0.0105

This value has three significant digits because we start counting with the first nonzero digit on the left.

0.001350 This value has four significant digits. Trailing zeros count when there is a decimal. The significant digit rules enable us to tell the difference between two measurements such as 13.05 m and 13.0500 m. Mathematically, of course, these values are equivalent. But they are different in what they tell us about the process of how the measurements were made. The first measurement has four significant digits. The second measurement is more precise. It has six significant digits and would come from a more precise instrument. Now, just in case you are bothered by the zeros at the end of 37,000 that are not significant, here is one more way to think about significant digits that may help. The precision in a measurement depends on the instrument used to make the measurement. If we express the measurement in different units, this should not change the precision. A measurement of 37,000 grams is equivalent to 37 kilograms. Whether we express this value in grams or kilograms, it still has two significant digits. Case 2

The second case addresses the rules that apply when you record a measurement

yourself, rather than reading a measurement someone else has made. When you take measurements yourself, as you do in laboratory experiments, you need to know the rules for which digits are significant in the reading you are taking on the measurement instrument. The rule for taking measurements depends on whether the instrument you are using is a digital instrument or an analog instrument. Here are the rules for these two possibilities: Rule 1 for digital instruments For the digital instruments commonly found in high school or undergraduate science labs, assume all the digits in the reading are significant, except leading zeros.

Chapter 2

The first of these rules is illustrated in Figure 2.2. The reading on the left has leadFigure 2.2. With digital instruments, all digits are significant except ing zeros, which do not count leading zeros. Thus, the numbers of significant digits in these readings as significant. Thus, the first are, from left to right, three, three, five, and five. reading has three significant digits. The second reading also has three significant digits. The third reading has five significant digits. The fourth reading also has five significant digits because with a digital display, the only zeros that don’t count are the leading zeros. Trailing zeros are significant with a digital instrument. However, when you write this measurement down, you must write it in a way that shows those zeros to be significant. The way to do this is by using scientific notation. Thus, the right-hand value in Figure 2.2 must be written as 4.2000 × 104. Dealing with digital instruments is actually more involved than the simple rule above implies, but the issues involved go beyond what we typically deal with in introductory or intermediate science classes. So, simply take your readings and assume that all the digits in the reading except leading zeros are significant. Now let’s look at some examples illustrating the rule for analog instruments. Figure 2.3 shows a machinist’s rule being used to measure the length in millimeters (mm) of a brass block. We know the first two digits of the length with certainty; the block is clearly between 31 mm and 32 mm long. We have to estimate the third significant digit. The scale on the rule is marked in increments of 0.5 mm. Comparing the edge of the block with these marks, I would estimate the next digit to be a 6, giving a measurement of 31.6 mm. Others might estimate the last digit to be 5 or 7; these small differences in the last digit are unavoidable because the last digit is estimated. Whatever you estimate the last digit to be, two digits of this measurement are known with certainty, the third digit is estimated, and the measurement has three significant digits. The photograph in Figure 2.4 shows a measurement in milliliters (mL) being taken with a piece of apparatus called a buret—a long glass tube used for measuring liquid volumes. Notice in this figure that when measuring liquid volume the surface of the liquid curls up at the edge of the cylinder. This curved surface is called a meniscus. The liquid measurement must be made at the bottom of the meniscus for most liquids, including water. The scale on the buret shown is marked in increments of 0.1 mL. This means we estimate to the nearest 0.01 mL. To one person, the bottom of the meniscus (the black curve) may appear to be just below 2.2 mL, so that person would call this measurement 2.21 mL. To someone else, it may seem that the bottom of the meniscus is right on 2.2, in which case that person would call the reading 2.20 mL. Either way, the reading has three significant digits and the last digit is estimated to be either 1 or 0. As a third example, Figure 2.5 shows a liquid volume measurement being taken with a piece of apparatus called Figure 2.3. Reading the significant digits a graduated cylinder. (We use graduated cylinders in an experiment we perform later on in this course.) The scale with a machinist’s rule.

0042.0

42.0

42.000

42,000

Motion

on the graduated cylinder shown is marked in increments of 1 mL. In the photo, the entire meniscus appears silvery in color with a black curve at the bottom. For the liquid shown in the figure, we know the first two digits of the volume measurement with certainty because the reading at the bottom of the meniscus is clearly between 82 mL and 83 mL. We have to estimate the third digit, and I would estimate the black line to be at 40% of the distance between 82 and 83, giving a reading of 82.4 mL. Someone else might read 82.5 mL, or even 82.6 mL. It is important for you to keep the significant digits rules in mind when you are taking measurements and entering data for your lab reports. The data in your lab journal and the values you use in your calculations and report must correctly reflect the use of the significant digits rules as they apply to the actual instruments you use to take your measurements. Note also the helpful fact Figure 2.4. Reading the significant digits that when a measurement is written in scientific nota- on a buret. tion, the digits written in the stem (the numerals in front of the power of 10) are the significant digits. Case 3

The third case of rules for significant digits ap-

plies to the calculations (multiplication and division) you perform with measurements. The main idea behind the rule for multiplying and dividing is that the precision you report in your result cannot be higher than the precision you have in the measurements to start with. The precision in a measurement depends on the instrument used to make the measurement, nothing else. Multiplying and dividing things cannot improve that precision, and thus your results can be no more precise than the measurements that go into the calculations. In fact, your result can be no more precise than the least precise value used in the calculation. The least precise value is, so to speak, the “weak link” in the chain, and a chain is no stronger Figure 2.5. Reading the significant digits on a graduated cylinder. than its weakest link. There are two rules for combining the measured values into calculated values, including any unit conversions that must be performed. Here are the two rules for using significant digits in our calculations in this course: Rule 1 Count the significant digits in each of the values you use in a calculation, including the conversion factors you use. (Exact conversion factors are not considered.) Determine how many significant digits there are in the least precise of these values. The result of your calculation must have this same number of significant digits.

Chapter 2

Rule 1 is the rule for multiplying and dividing, which is what most of our calculations entail. (As I mentioned previously, there is another rule for adding and subtracting that you will learn when you take chemistry.) Rule 2 When performing a multi-step calculation, you must keep at least one extra digit during intermediate calculations and round off to the final number of significant digits you need at the very end. This practice ensures that small round-off errors don’t add up during the calculation. This extra digit rule also applies to unit conversions performed as part of the computation. As I present example problems in the coming chapters, I frequently refer to these rules and show how they apply to the example at hand. As you take your quizzes, your instructor might give you a few weeks to practice and master the correct use of significant digits without penalizing you for mistakes. But get this skill down as soon as you can because soon you must use significant digits correctly in your computations to obtain the highest scores on your quizzes.

2.1.6 Scientific Notation You have probably studied scientific notation before. However, in this course you must master it, including the use of the special key found on scientific calculators for working with values in scientific notation. Mastery of scientific notation is important because working with values in scientific notation is a basic and common occurrence in scientific work. We review the basic principles next. Mathematical Principles

Scientific notation is a way of expressing very large or very small numbers without all the zeros, unless the zeroes are significant. This is of enormous benefit when one is dealing with a value such as 0.0000000000001 cm (the approximate diameter of an atomic nucleus). The basic idea will be clear from a few examples. Let’s say we have the value 3,750,000. This number is the same as 3.75 million, which can be written as 3.75 × 1,000,000. Now, 1,000,000 itself can be written as 106 (which means one followed by six zeros), so our original number can be expressed equivalently as 3.75 × 106. This expression is in scientific notation. The numerals in front, the stem, are always written as one digit followed by a decimal and the other digits. The multiplied 10 raised to a power has the effect of moving the decimal over as many places as necessary to recreate our original number. As a second example, the current population of earth is about 7,290,000,000, or 7.29 billion. One billion has nine zeros, so it can be written as 109. So we can express the population of earth in scientific notation as 7.29 × 109. When dealing with extremely small numbers such as 0.000000016, the process is the same, except the power on the 10 is negative. The easiest way to think of it is to count how many places the decimal in the value must be moved over to get 1.6. To get 1.6, the decimal has to be moved to the right eight places, so we write our original value in scientific notation as 1.6 × 10−8. Using Scientific Notation with a Scientific Calculator

All scientific calculators have a key for entering values in scientific notation. This key is labeled EE or EXP on most 32

Motion

calculators, but others use a different label.3 It is very common for those new to scientific calculators to use this key incorrectly and obtain incorrect results. So read carefully as I outline the general procedure. The whole point of using the EE key is to make keying in the value as quick and errorfree as possible. When using the scientific notation key to enter a value, you do not press the × key, nor do you enter the 10. The scientific calculator is designed to reduce all this key entry, and the potential for error, by use of the scientific notation key. You only enter the stem of the value and the power on the ten and let the calculator do the rest. Here’s how. To enter a value, simply enter the digits and decimal in the stem of the number, then hit the EE key, then enter the power on the ten. The value is now entered and you may do with it as you wish. As an example, to multiply the value 7.29 × 109 by 25 using a standard scientific calculator, the sequence of key strokes is as follows: 7.29 EE 9 × 25 = Notice that between the stem and the power the only key pushed is the EE key. When entering values in scientific notation with negative powers on the 10, the +/− key is used before the power to make the power negative. Thus, to divide 1.6 × 10−8 by 36.17, the sequence of key strokes is: 1.6 EE +/− 8 ÷ 36.17 = Again, neither the “10” nor the “×” sign that comes before it is keyed in. The EE key has these built in. Students sometimes wonder why it is incorrect to use the 10x key for scientific notation. To execute 7.29 × 109 times 25, they are tempted to enter the following: 7.29 × 10x 9 × 25 = The answer is that sometimes this works, and sometimes it doesn’t, and calculator users must use key entries that always work. The scientific notation key ( EE ) keeps a value in scientific notation all together as one number. That is, when the EE key is used, then to the calculator 7.29 × 109 is not two numbers, it is a single numerical value. But when the × key is manually inserted, the calculator treats the numbers separated by the × key as two separate values. This causes the calculator to render an incorrect answer for a calculation such as 3.0 ×106 1.5 ×106 The denominator of this expression is exactly half the numerator, so the value of this fraction is obviously 2. But when using the 10x key, the 1.5 and the 106 in the denominator are separated and treated as separate values. The calculator then performs the following calculation: 3.0 ×106 ×106 1.5 3

One infuriating model uses the extremely unfortunate label x10x which looks a lot like 10x , a different key with a completely different function.

Chapter 2

This comes out to 2,000,000,000,000 (2 × 1012), which is not the same as 2! The bottom line is that the EE key, however it may be labeled, is the correct key to use for scientific notation.

2.1.7 Problem Solving Methods Organizing problems on your paper in a reliable and orderly fashion is an essential practice. Physics problems can get very complex, and proper solution practices can often make the difference between getting most or all of the points for a problem and getting few or none. Each time you start a new problem, you must set it up and follow the steps according to the outline presented in the box on pages 36 and 37, entitled Universal Problem Solving Method. It is very important that you always show all your work. Do not give in to the temptation to skips steps or take shortcuts. Develop correct habits for problem solving and stick with them!

2.2

Motion

In this course, we address two types of motion: motion at a constant velocity, when an object is not accelerating, and motion with a uniform acceleration. Defining these terms is a lot simpler if we stick to motion in one dimension, that is, motion in a straight line. So in this course, this is what we will do.

2.2.1 Velocity When thinking about motion, one of the first things we must consider is how fast an object is moving. The common word for how fast an object is moving is speed. A similar term is the word velocity. For the purposes of this course, you may treat these two terms as synonyms. The difference is technical. Technically, the term velocity means not only how fast an object is moving, but also in what direction. The term speed refers only to how fast an object is moving. But since we are only going to consider motion in one direction at Figure 2.6. A car traveling with the cruise control a time, we can use the terms speed and velocity inon is an example of an object moving with terchangeably. constant velocity. An important type of motion is motion at a constant velocity, like a car with the cruise control on (Figure 2.6). At a constant velocity, the velocity of an object is defined as the distance the object travels in a certain period of time. Expressed mathematically, the velocity, v, of an object is calculated as v=

d t

Motion

in a certain length of time. When performing calculations using the SI System of units, distances are measured in meters and times are measured in seconds. This means the units for a velocity are meters per second, or m/s. The relationship between velocity, distance, and time for motion at a constant velocity is shown graphically in Figure 2.7. Travel time is shown on the horizontal axis and distance traveled is shown on the vertical axis. The steeper curve4 shows distances and times for an object moving at 2 m/s. At a time of one second, the distance traveled is two meters because the object is moving at two meters per second (2 m/s). After two seconds at this speed, the object has moved four meters: (4 m)/(2 s) = 2 m/s. And after three seconds, the object has moved six meters: (6 m)/(3 s) = 2 m/s. The right-hand curve in Figure 2.7 represents an object traveling at the much slower velocity of 0.5 m/s. At this speed, the graph shows that an object travels two meters in four seconds, four meters in eight seconds, and so on. To see this algebraically, look again at the velocity equation above. If we multiply both sides of this equation by the time, t, and cancel, we have d = vt This is the same equation, just written in a different form. It still applies to objects moving at a constant velocity. Written this way, t is the independent variable, d is the dependent variable, and v serves as the slope of the line relating d to t. With this form of the velocity equation, we can calculate how far an object travels in a given amount of time, assuming the object is moving at a constant velocity. Now we work a couple of example problems, following the problem-solving method described on pages 36–37. And remember, all the unit conversion factors you need are listed in Appendix A.

v = 2 m/s

v = 0.5 m/s

distance (m)

6 5 4 3 2 1 1

6 7 8 9 10 11 12 time (s) Figure 2.7. A plot of distance versus time for an object moving at constant velocity. Two different velocity cases are shown. 4

Note that when discussing graphs, the lines or curves on the graph are all referred to as curves, whether they are curved or straight.

Chapter 2

Universal Problem Solving Method Solid Steps to Reliable Problem Solving In Introductory Physics, you learn how to use math to solve scientific problems. Developing a sound and reliable method for approaching problems is very important. The problem solving method shown below is used in scientific work everywhere. You must follow every step closely and show all your work. 1.

3. 4. 5. 6. 7. 8. 9.

Write down the given quantities at the left side of your paper. Include the variable quantities given in the problem statement and the variable you must solve for. Make a mental note of the precision in each given quantity. For each given quantity that is not already in MKS units, work immediately to the right of it to convert the units of measure into MKS units. To help prevent mistakes, always use horizontal fraction bars in your units and unit conversion factors. Write the results of these unit conversions with one extra digit of precision over what you need in your final result. Write the standard form of the equation you will use to solve the problem. If necessary, use algebra to get the variable you are solving for alone on the left side of the equation. Never put values into the equation until this step is done. Write the equation again with the values in it, using only MKS units, and compute the result. If you are asked to state the answer in non-MKS units, perform the final unit conversion now. Write the result with the correct number of significant digits and the correct units of measure. Check your work. Make sure your result is reasonable.

Example Problem If you want a complete and happy life, do ’em just like this! A car is traveling at 35.0 mph. The driver then accelerates uniformly at a rate of 0.15 m/s2 for 2 minutes and 10.0 seconds. Determine the final velocity of the car in mph. Step 1 Write down the given information in a column down the left side of your page, using horizontal lines for the fraction bars in the units of measure. mi hr m a = 0.15 2 s t = 2 min 10.0 s vf = ? vi = 35.0

Motion

Step 2 Perform the needed unit conversions, writing the conversion factors to the right of the given quantities you wrote in the previous step. mi 1609 m 1 hr m ⋅ ⋅ = 15.6 hr mi 3600 s s m a = 0.15 2 s t = 2 min 10.0 s = 130.0 s vf = ? vi = 35.0

Step 3 Write the equation you will use in its standard form. a=

v f − vi t

Step 4 Perform the algebra necessary to get the unknown you are solving for alone on the left side of the equation. a=

v f − vi

t at = v f − vi v f = vi + at Step 5 Using only values in MKS units, insert the values and compute the result. v f = vi + at = 15.6

m m m + 0.15 2 ⋅130.0 s = 35.1 s s s

Step 6 Convert to non-MKS units, if required in the problem. v f = 35.1

m 1 mi 3600 s mi ⋅ ⋅ = 78.5 s 1609 m 1 hr hr

Step 7 Write the result with correct significant digits and units of measure. v f = 79 mph Step 8 Check over your work, looking for errors. Step 9 Make sure your result is reasonable. First, check to see if your result makes sense. The example above is about an accelerating car, so the final velocity we calculate should be a velocity a car can have. A result like 14,000 mph is obviously incorrect. (And remember that nothing can travel faster than the speed of light, so make sure your results are reasonable in this way as well.) Second, if possible, estimate the answer from the given information and compare your estimate to your result. In step 6 above, we see that 3600/1609 is about 2, and 2∙35.1 is about 70. Thus our result of 79 mph makes sense. (Optional Step 10: Revel in the satisfaction of knowing that once you get this down you can work physics problems perfectly nearly every time!)

Chapter 2

Example 2.6 Sound travels 1,120 ft/s in air. How much time does it take to hear the crack of a gun fired 1,695.5 m away? First, write down the given information and perform the required unit conversions so that all given values are in MKS units. Check to see how many significant digits your result must have and do the unit conversions with one extra significant digit. The given speed of sound has three significant digits, so we perform our unit conversions with four digits. ft 0.3048 m m ⋅ = 341.4 s ft s d = 1695.5 m t =? v = 1120

Next, write the appropriate equation to use. v=

d t

Perform any necessary algebra, insert the values in MKS units, and compute the result. v= t=

d t d 1695.5 m = = 4.966 s v 341.4 m s

Next, round the result so that it has the correct number of significant digits. In the velocity unit conversion and in the calculated result, I used four significant digits. The given velocity has three significant digits and the given distance has five significant digits. Thus, our result must be reported with three significant digits, but all intermediate calculations must use one extra digit. This is why I used four digits. But now we are finished and our result must be rounded to three significant digits because the least precise measurement in the problem has three significant digits. Rounding our result accordingly, we have t = 4.97 s The final step is to check the result for reasonableness. The result should be roughly the same as 1500/300 or 2000/400, both of which equal 5. Thus, our result makes sense.

2.2.2 Acceleration An object’s velocity is a measure of how fast it is going; it is not a measure of whether its velocity is changing. The quantity we use to measure if a velocity is changing, and if so, how fast it is changing, is the acceleration. If an object’s velocity is changing, the object is accelerating, and the value of the acceleration is the rate at which the velocity is changing. 38

Motion

The equation we use to calculate uniform acceleration in terms of an initial velocity vi and a final velocity vf is a=

v f − vi t

where a is the acceleration (m/s2), t is the time spent accelerating (s), and vi and vf are the initial and final velocities, respectively, (m/s). Did you notice that the MKS units for acceleration are meters per second squared (m/s2)? These units often drive students crazy, and we need to pause here and discuss what this means so you can sleep peacefully tonight. I wrote just above that the acceleration is the rate at which the velocity is changing. The acceleration simply means that the velocity is increasing by so many meters per second, per second. Now, “per” indicates a fraction, and if a velocity is changing so many meters per second, per second, we write these units in a fraction this way and simplify the expression: m m s = s = m ⋅1 = m s s s s2 s 1 Because the acceleration equation results in negative accelerations when the initial velocity is greater than the final velocity, you can see that a negative value for acceleration means the object is slowing down. In future physics courses, you may learn more sophisticated interpretations for what a negative acceleration means, but in this course you are safe associating negative accelerations with decreasing velocity. In common speech, people sometimes use the term “deceleration” when an object is slowing down, but mathematically we just say the acceleration is negative. Before we work through some examples, let’s look at a graphical depiction of uniform acceleration the same way we did with velocity. Figure 2.8 shows two different acceleration curves, representing two different acceleration values. For the curve on the right, after 1 s a = 4 m/s2

a = 1 m/s2

velocity (m/s)

12 10 8 6 4 2 1

6 7 8 9 10 11 12 time (s) Figure 2.8. A plot of velocity versus time for an object accelerating uniformly. Two different acceleration cases are shown.

Chapter 2

ball falling

ball rising

the object is going 1 m/s. After 2 s, the object is going 2 m/s. After 12 s, the object is going 12 m/s. You can take the velocity that corresponds to any length of time (by finding where their lines intersect on the curve) and calculate the acceleration by dividing the velocity by the time to get a = 1 m/s2. The other curve has a higher acceleration, 4 m/s2. An acceleration of 4 m/s2 means the velocity is increasing by 4 m/s every second. Accordingly, after 2 s the velocity is 8 m/s, and after 3 s, the velocity is 12 m/s. No matter what point you select on that curve, v/t = 4 m/s2. We must be very careful to distinguish between velocity (m/s) and acceleration (m/s2). Acceleration is a measure of how fast an object’s velocity is changing. To see the difference, note that an object can be at rest (v = 0) and accelerating at the same instant. Now, although you may not see this at first, it is important for you to think this through and understand how this counter-intuitive situation can come about. Here are two examples. The instant an object starts from rest, such as when the driver hits the gas while sitting at a traffic light, the object is simultaneously at rest and accelerating. This is because if an object at rest is to ever begin moving, its velocity must Right here at change from zero to something else. In other words, the top the ball the object must accelerate. Of course, this situation is at rest for an only holds for an instant; the velocity instantly begins instant, but still changing and does not stay zero. accelerating Perhaps my point will be easier to see with this because of the pull of earth’s gravity. second example. As depicted in Figure 2.9, when a ball is thrown straight up and reaches its highest point, it stops for an instant as it starts to come back down. At its highest point, the ball is simultaneously at rest and accelerating due to the force of gravity pulling it down. As before, this situation only holds for a single instant. The point of these two examples is to help you understand the difference between the two variables we are discussing, velocity and acceleration. If an object is moving at all, then it has a velocity that is not zero. The object may or may not be accelerating. Figure 2.9. A rising and falling ball helps illustrate the difference between velocity But acceleration is about whether the velocity itself is and acceleration. changing. If the velocity is constant, then the acceleration is zero. If the object is speeding up or slowing down, then the acceleration is not zero. And now for another example problem, this time using the acceleration equation. Example 2.7 A truck is moving with a velocity of 42 mph (miles per hour) when the driver hits the brakes and brings the truck to a stop. The total time required to stop the truck is 8.75 s. Determine the acceleration of the truck, assuming the acceleration is uniform. Begin by writing the givens and performing the unit conversions.

Motion

vi = 42 vf = 0

mi 1609 m 1 hr m ⋅ ⋅ = 18.8 hr mi 3600 s s

t = 8.75 s a=? Now write the equation and complete the problem. a=

v f − vi t

m s = −2.15 m 8.75 s s2

0 −18.8

The initial velocity has two significant digits, so I did the calculations with three significant digits until the end. Now we round off to two digits giving a = −2.2

m s2

If you keep all the digits in your calculator throughout the calculation and round to two digits at the end, you have −2.1 m/s2. This answer is fine, too. Remember, the last digit of a measurement or computation always contains some uncertainty, so it is reasonable to expect small variations in the last significant digit. A check of our work shows the result should be about –20/10, which is –2. Thus the result makes sense. One more point on this example: Notice that the calculated acceleration value came out negative. This was because the final velocity was lower than the initial velocity. Thus we see that a negative acceleration means the vehicle is slowing down.

If you haven’t yet read the example problem in the yellow Universal Problem Solving Method box, you should read it now to see a slightly more difficult example using this same equation.

2.3

Planetary Motion and the Copernican Revolution

2.3.1 Science History and the Science of Motion People have been fascinated with the heavens since ancient times. In fact, astronomy is one of the oldest sciences, studied widely in Egypt, China, Central America, Persia, and other parts of the world. Ancient astronomers worked on developing maps showing the positions of stars at different times of the year. Eventually, these early maps led to the development of the first mathematical model of the stars that could be used for making predictions of celestial events. This model was developed by the Alexandrian astronomy Ptolemy, and we will study it is some detail shortly. The study of motion has always been associated with the motion of the heavenly bodies we see in the sky, so it is particularly fitting in this chapter on motion for us to review the history of views about the solar system and the rest of the universe, referred to as “the 41

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heavens” by those in ancient times. As we will see, the particular episode known as the Copernican Revolution was a pivotal moment in that history and was the setting for the emergence of our contemporary understanding of scientific epistemology—what knowledge is and how we know what we know. As you recall, Chapter 1 addresses the Cycle of Scientific Enterprise and examines the way science works. From that discussion you know that science is an ongoing process of modeling nature—at least that is the way we understand science now. We now understand that scientists use theories as models of the way nature works, and over time theories change and evolve as scientists learn more. Sometimes scientists find that a theory is so far off the mark that they have to toss it out completely and replace it with a different one. The present general understanding among scientists that science is a process of modelling nature took hold around the beginning of the 20th century. The ideas that led to this understanding began to emerge at the time of the Copernican Revolution in the 16th and 17th centuries. But since natural philosophy was then entering new territory, there was a period of difficult struggle that involved both theologians and philosophers. There are a lot of misconceptions about what happened at that time. The conflict in Galileo’s day is often regarded as a fight between faith and science, which it wasn’t. There were plenty of people in the Roman Catholic Church at the time who were big supporters of science and who followed all the latest developments. The real issue with Galileo was about epistemology. The so-called “faith versus science” debate rages today as much as ever, so it is worth spending some time to understand that crucial period in scientific history.

2.3.2 Aristotle As mentioned above, the study of astronomy and astrology has origins in many ancient cultures, but we pick up the story with the ancient Greeks and the Greek philosopher Aristotle in the 4th-century BCE (Figure 2.10). Aristotle was a highly influential philosopher who wrote a lot about philosophy, physics, biology, and other fields of learning. Back then, science was called natural philosophy and there was really no distinction between scientists and philosophers. That time was also many centuries before experiments became part of scientific research. Natural philosophy did involve making observations about the world, but the conclusions reached by ancient philosophers like Aristotle were based simply on observation and philosophical thought. It was still about 2,000 years before natural philosophers realized that the way things appear to our ordinary senses might not be the way they actually are and that to understand more about the world required scientific experiments. For example, if you just walk outside and quietly look around you notice that the earth does not appear to be in motion; it feels solid and at rest. The sun, planets, and stars appear to move across the sky each day. In fact, watching a sunrise gives the distinct impression that the sun is moving up and then across the sky. Today, we understand things differently, but that is the result of the revolution we are about to explore and the experimental science that emerged at that time. Aristotle’s ideas were grounded in the concept of telos—a Greek term meaning purpose, goal, or end. Aristotle believed Figure 2.10. Greek philosopher Aristotle (384–322 BCE). that each thing that exists has its own telos. He argued that 42

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objects acted in a fashioned determined by their telos, and this way of reasoning had a major affect on his ideas about physics. Aristotle observed the serene beauty of the stars, the planets, the sun, and moon as they appear majestically to rotate around the earth day after day. He also noticed that nothing in the heavens ever seems to change. Other than the motions of the heavenly bodies, everything in the heavens seems to be pure and eternal. On earth, of course, Aristotle was surrounded by change: decay, corruption, birth, and death are all around. Animals and plants live and die, forests grow and burn, rivers flow and flood, storms come and go. These observations led Aristotle to conclude that change and corruption occur only on the earth. He wrote that imperfection and change of any kind occur only on the earth, while the heavens are pure and unchanging. Aristotle taught that the heavenly bodies—planets, stars, sun, and moon—are eternal and perfect. Further, he said that their motions must be in perfect circles since the circle is the purest and most perfect geometric shape. He conceived of the sun, moon, and planets as inhabiting celestial spheres, centered on the earth, one inside the other—an exquisite geocentric (earth-centered) system. Aristotle was a tremendous moral philosopher whose ideas still have a profound influence on us today. Back in ancient times, he was regarded so highly that questioning his ideas was virtually unthinkable. Thus, his views about the heavenly motions became the basis for all further work on understanding the motions of the heavenly bodies.

2.3.3 Ptolemy In the second century CE, the famous Alexandrian astronomer Ptolemy (Figure 2.11) worked out a detailed mathematical system based on Aristotle’s ideas. (By the way, the “P” in Ptolemy is silent.) As with all ancient astronomers, Ptolemy’s goal was to be able to make predictions about the movements of the planets and stars, along with other astronomical events such as eclipses, because these events were widely used as omens signifying important events on earth. Ptolemy started with Aristotle’s basic ideas and developed a complex mathematical system—a model—that was quite effective in making the desired predictions. There were other astronomers around that time who developed different systems, but Ptolemy’s system became the most Figure 2.11. Alexandrian astronomer widely accepted understanding of the heavens for over a Ptolemy (c. 100–170 CE). thousand years.

2.3.4 The Ptolemaic Model The basic structure of Ptolemy’s geocentric model of the heavens is depicted in Figure 2.12. As with Aristotle, there are seven heavenly bodies, each inhabiting a sphere centered on the earth. Each of the heavenly bodies is also itself a perfect sphere. The contents of the spheres are summarized in Table 2.3. The first seven spheres contain the five planets (not including the earth), the sun, and the moon. Sphere 8 contains the so-called Firmament, the fixed layer of stars. The stars do not move relative to each other; their positions are fixed and they rotate as a body in the eighth sphere each day. Within the firmament, the stars are arranged according to the zodiac, a belt of twelve constellations 43

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around the earth. The term zodiac derives from the Latin and Greek terms meaning “circle Saturn of animals,” and is so named Jupiter because many of the constelMars lations in the zodiac represent Sun animals. Venus The ninth sphere contains Mercury the Primum Mobile, which is Moon Latin for “prime mover” (or “first mover”). The Primum Earth Mobile is the sphere set into motion by God or the gods. As the Primum Mobile turns, it pulls all the other spheres with it, making them rotate as well. Outside the ninth sphere is the so-called Empyrean, the dwelling place of God or the gods. Figure 2.12 shows the basic structure of Ptolemy’s modFigure 2.12. The Ptolemaic model of the heavens. el, but there is a great deal more to the model than shown there. This is because all seven of the heavenly bodies appear to move around in the nighttime sky against the background of the fixed stars. If all the heavenly bodies simply moved in their spheres around the earth together once each day, there would be no way to account for why the planets’ positions change relative to the stars. Ptolemy accounted for the changes by a system of epicycles. An epicycle is a circular planetary orbit with its center moving in a separate circular path, as

Empyrean

Primum Mobile Firmament

Sphere 1

Moon

Sphere 2

Mercury

Sphere 3

Venus

Sphere 4

Sun

Sphere 5

Mars

Sphere 6

Jupiter

Sphere 7

Saturn

Sphere 8

The Firmament. This region consists of the stars arranged in their constellations according to the zodiac.

Sphere 9

The Primum Mobile. This Latin name means “first mover.” This sphere rotates around the earth every 24 hours and drags all the other spheres with it, making them all move.

Beyond

The Empyrean. This is the region beyond the spheres. The Empyrean is the abode of God, or the gods.

Table 2.3. Contents of the spheres in the Ptolemaic model.

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planet

epicycle

earth

plan

etary sphere

depicted in Figure 2.13. As the center of an epicycle moves along its path in the sphere, the planet in the epicycle rotates about the center of the epicycle, as if the epicycle were a wheel rolling around a path centered on the earth. To help you understand why epicycles are necessary in Ptolemy’s model, we discuss them in more detail in the next section. A planet moving in an epicycle moves in a path similar to a person riding in a “tea cup ride” at an amusement park, like the one picture in Figure 2.14. To account for the complex motions of the heavenly bodies, Ptolemy’s model contained some 80 different epicycles. Some of the planets were located in a epicycle riding on the rim of another epicycle, which in turn moved in

Figure 2.13. A planet moving in a path defined by an epicycle around the earth.

the sphere around the earth. Ptolemy’s system was mathematically very complex, but its genius was that it worked pretty well! The main features of Ptolemy’s model are summarized in the box below. Among the different astronomers of the ancient world there were those who held to variations on this basic model. For example, some astronomers reckoned that Mercury and Venus orbited the sun while the other heavenly Figure 2.14. The people in the cups spin in a circle while the bodies orbited the earth. But the basic cup moves in a larger circle, motion like that of a planet Ptolemaic model is as described in the moving on an epicycle. box.

The Main Principles in Ptolemy’s Celestial Model 1. 2. 3. 4. 5. 6. 7.

There are seven heavenly bodies. All the heavenly bodies move in circular orbital regions called spheres. In the model, there are nine spheres plus the region beyond the spheres, with contents as listed in Table 2.3. All the heavenly bodies are perfectly spherical. All the spheres are centered on the earth, so this system is a geocentric system. Corruption and change only exist on earth. All other places in the universe, including all the heavenly bodies and stars, are perfect and unchanging. All the spheres containing the heavenly bodies and all the stars in the Firmament rotate completely around the earth every 24 hours. Epicycles are used to explain the motion of the planets relative to the stars. 45

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2.3.5 The Ancient Understanding of the Heavens We soon address the new ideas that began unfolding when Nicolaus Copernicus introduced his new heliocentric (sun-centered) model of the heavens. But before pressing on, let’s pause to consider a couple of things about the way the motion of the planets in the night sky appears to observers on earth. This will make it easier to understand why Ptolemy’s system became so widely accepted. Stationary Earth First, as I mention above, the earth does not seem to be moving. To you and I, who grew up in a time when everyone knows that the earth and other planets orbit the sun, it seems obvious that day and night are caused by the earth’s rotation on its axis. We have heard about this all our lives. But stop and consider that if all we had to go on was our simple observations, it does appear that everything is orbiting around the earth while the earth sits still: the sun and moon rise each day, track across the sky, and set, and the planets and stars all do the same thing. Also, it doesn’t feel at all like earth is rotating. We all know that anytime we spin in a circle, like people on a merry-go-round, we have to hold on to keep from falling off. We also feel the wind in our hair. Again, if we have something with us on the merry-go-round that is tall and flexible, such as a sapling, it does not remain vertical when it is moving in a circular fashion like this. Instead, it bends over because of the acceleration pulling it in its circular motion. Now, the ancients knew about the large size of the earth—the Greek mathematician and geographer Eratosthenes (Figure 2.15) made a very accurate estimate of the earth’s circumference—a bit under 25,000 miles—as far back as 240 BCE. If a sphere that size spins in a circle once a day, the people on its surface move very fast (over 1,000 miles per hour on the equator). For this to be the case, it seemed that we would be hanging on for dear life! The trees would be laying down and we would constantly feel winds that make a hurricane seem like a calm summer day! For all these reasons, it did not seem reasonable to believe that the apparent motion of the heavenly bodies across the sky every day was due to the earth’s rotation. These arguments seemed obvious to nearly everyone before 1500, and to everyone except a few cutting-edge asFigure 2.15. Greek mathematician and geographer Eratosthenes (c. 276–194 tronomers right up to the end of the 17th century. Only BCE). a crazy person imagined that the earth spins, and people used these arguments all the way up to the time of Galileo to prove that the earth was not orbiting the sun and spinning around once a day. Back then, these were persuasive arguments. Forward and Retrograde Motion

The second item to consider here has to do with the

apparent motion of the planets in the sky against the background of the stars. If you go out and look at, say, Mars each night and make a note of its location against the stars, you see that it is in a slightly different place each night. The planet gradually works its way along in a pathway against the starry background night after night. If you track the planet for sev46

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eral months or a year, it Firmament moves quite far. As mentioned above, Ptolsequence of a observations emy used epicycles to b account for this forretrograde motion ward motion of planets c against the background a b of fixed stars. c Going back to 3 3 watching Mars, if you follow the planet’s sequence of Earth 2 observations progress long enough, 2 you see that there are forward 1 motion Mars periods of time lasting 1 Sphere 5 several weeks when the (Mars) nightly progress of the planet reverses course. Mars appears to be backing up! This apparent backing up is called Figure 2.16. Using epicycles to explain the forward and retrograde motion of retrograde motion. heavenly bodies against the background of fixed stars. Ptolemy used epicycles to account for this, too. Figure 2.16 is a diagram showing how epicycles are used in the geocentric system to account for the planetary motions—both forward and retrograde—against the background of the fixed stars. Mars is shown in red moving on an epicycle, while the center of the epicycle moves around the earth. The dashed lines are the lines of sight from earth to Mars, and the letters 3 and numbers outside sequence of observations the firmament show 3 2 1 2 the locations where forward 1 2 1 motion c Mars appears among the stars at different 3 times. c b a b The lower right sequence of Sun a b observations a part of the diagram Mars c Earth retrograde shows Mars in three motion locations (labeled 1, 2, and 3) over the course of a few weeks. Compared to the back-

Figure 2.17. Explanation of forward and retrograde motion in the Copernican system.

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ground of fixed stars, Mars exhibits forward motion during a sequence of nighttime observations. The upper right part of the figure shows Mars’ locations during a different sequence of observations (a, b, and c) some months later. Mars is now on the other side of its epicycle. The center of the epicycle continues to move in the same direction in its sphere around earth. But since Mars is on the near side of its epicycle, the sequence of observations of its location against the starry background—a, b, and c—maps along the starry background in the opposite direction. This apparent motion of Mars in the opposite direction is retrograde motion. While we are on the subject, we may as well look at how forward and retrograde motion are explained in a heliocentric system—a system in which the planets orbit the sun. The system introduced by Copernicus is a heliocentric system. Assuming that the earth moves faster in its orbit than Mars (which is correct), the explanation is straightforward. As shown in the upper part of Figure 2.17, when the earth and Mars are on opposite side of their orbits, the observations of Mars’ location against the stars exhibit forward motion. But when the earth and Mars are on the same side of the sun, as in the center-right part of the figure, the earth’s greater velocity makes Mars’ position against the stars exhibit retrograde motion. To summarize, none of the planets actually reverses course in its orbit, and neither the geocentric nor heliocentric models depict planets as reversing direction. But depending on the system, the presence of epicycles and the relative locations of earth and a planet can combine to produce the appearance of forward or retrograde motion of the planet against the fixed background of the stars.

2.3.6 The Ptolemaic Model and Theology We soon continue our history of the science of planetary motion by reviewing the momentous events of the 16th and 17th centuries. Between Ptolemy and Copernicus were 1,300 years of Christian theology and philosophy. During this long period of history, a strong tradition emerged among many Christian theologians that the Ptolemaic model of the heavens aligned very well with certain passages in the Bible. This circumstance led theologians in this tradition to assume that such passages were to be interpreted as literal descriptions of the motions of the heavenly bodies. Here are a few examples of passages that seem to describe the earth as motionless, with the sun and stars going around the earth: He set the earth on its foundations, so that it should never be moved (Psalm 104:5). He made the moon to mark the seasons; the sun knows its time for setting (Psalm 104:19). [The sun’s] rising is from the end of the heavens, and its circuit to the end of them (Psalm 19:6). The sun rises and the sun goes down, and hastens to the place where it rises (Ecclesiastes 1:5). Additionally, other features in the Ptolemaic model (derived from Aristotle) seemed to line up with biblical symbolism. For example: •

Seven was regarded as the biblical number symbolizing perfection, so it made sense that the creation contains seven heavenly bodies (those in the first seven spheres).

•

Circles are the most perfect shape, regarded as divine from the times of the ancient Greeks, so the spherical bodies inhabiting spheres in which they move seemed to reflect the perfection of the creator.

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•

Corruption was thought to exist only on earth, and it seemed this was obviously because of the curse after the Fall of Adam and Eve in the Garden of Eden, described in the biblical book of Genesis.

The result of such teaching was that many theologians assumed that the biblical passages and doctrines described above, along with the Ptolemaic model of the heavens, were literal descriptions of the true nature of reality. To these theologians, anyone who had different ideas about the heavens—such as, for example, the idea that the earth moved and orbited the sun—should be censored and prevented from spreading teachings they felt were unbiblical. Although widespread, this tradition of associating the Ptolemaic model with the Bible was by no means universal. Many theologians took a completely different position, including the great theologian and philosopher Augustine, a bishop in northern Africa in the 4th and 5th centuries CE. An insightful and relevant passage from Augustine is found in his book On the Literal Meaning of Genesis: Usually, even a non-Christian knows something about the earth, the heavens, and the other elements of this world, about the motion and orbit of the stars and even their sizes and relative positions, about the predictable eclipses of the sun and moon, the cycles of the years and the seasons, about the kinds of animals, shrubs, stones, and so forth, and this knowledge he holds to as being certain from reason and experience. Now it is a disgraceful and dangerous thing for an infidel to hear a Christian, presumably giving the meaning of Holy Scripture, talking nonsense on these topics, and we should take all means to prevent such an embarrassing situation, in which people show up vast ignorance in a Christian and laugh it to scorn. To summarize, many Christian theologians of the day were convinced that the Ptolemaic model lined up with a literal reading of certain passages in the Bible and with Aristotle’s ideas about the perfection of the heavens. Other theologians were more open minded on such questions. As we open the curtain now on the rest of our story, it is key to remember that many church theologians were strong supporters of those engaged in natural philosophy (science). The Roman Catholic Church—which figures prominently in these events—had a long tradition of supporting intellectual inquiry, including natural philosophy, and many of the individual theologians in the church were admirers of the scientists involved in these events.

2.3.7 Copernicus and Tycho Nicolaus Copernicus (Figure 2.18), a Polish astronomer, first proposed a detailed, mathematical, heliocentric model of the heavens, with the earth rotating on its axis, all the planets moving in circular orbits around the sun, and the moon orbiting the earth. Copernicus’ system was about as accurate—and about as complex—as the Ptolemaic system. Copernicus’ model still used circular orbits and because of this he still had to use epicycles to make the model accurate. Still, the model is an arrangement that is a lot closer to today’s understanding than the Ptolemaic model is. Copernicus dedicated his famous work On the Revolutions of the Heavenly Spheres to Pope Paul III. This dedication indicates that the Roman Catholic Church itself was not opposed to Copernicus’ ideas. Nevertheless, Copernicus knew there were scholars in the Church who were strongly opposed to the suggestion that the earth moved. Being sen49

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sitive to this, he didn’t want to cause trouble so he published his work privately to his close friends in 1514. Just before Copernicus’ death in 1543, his student and admirer, mathematician and astronomer Georg Joachim Rheticus, persuaded Copernicus to publish the work. Rheticus delivered the manuscript to the printer and brought proofs back to Copernicus to review. RheFigure 2.18. Polish astronomer Nicolaus Copernicus (1473–1543). ticus was not continuously present with the printer, and during his absence a theologian named Andreas Osiander added an unsigned “note to the reader” to the front of Copernicus’ book stating that the heliocentric ideas were hypotheses (although theory is the better term, since we are talking about a model) that were useful for the purpose of performing computations and not descriptions of actual reality. Because of this note, people generally thought that it expressed Copernicus’ own viewpoint. However, Rheticus was outraged by the addition and marked it out with a red crayon in the copies he sent to people. Copernicus did not live to see the final printed version of his book, but Rheticus’ reaction to Osiander’s note suggests that Copernicus regarded his model as more than merely an imaginary convenience that made computations easier. Tycho Brahe (Figure 2.19), was a Danish nobleman and astronomer. Tycho5 built a magnificent observatory called the Uraniborg on an island Denmark ruled at the time. This observatory is depicted in Figure 2.20. Tycho was a passionate and hotheaded guy, as evidenced by the fact he had the bridge of his nose cut off in a duel. (You can see his prosthesis in Figure 2.19 if you look closely.) Even though Tycho’s Uraniborg must have been the most palatial observatory in the world, he had a falling out with the new King of Denmark and decided to leave. In 1597, Tycho moved to Prague in Bohemia (the modern-day Czech Republic) and became Imperial Mathematician for Figure 2.19. Danish astronomer Rudolph II, King of Bohemia and Holy Roman Emperor Tycho Brahe (1546–1601). 5

I know it is appropriate to refer to historical figures by their last names, but most references in the literature refer to Tycho; historians rarely call him Brahe. I love the name Tycho, so I also call him that.

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there. Tycho spent his life cataloging astronomical data for over 1,000 stars (with cleverly contrived instruments, but only a primitive telescope). His work was published much later (1627) by Johannes Kepler in a new star catalog that identified the positions of these stars with unprecedented accuracy. Tycho witnessed and recorded two astronomical events that became historically very important. First, in 1563 he observed a conjunction between Jupiter and Saturn. A conjunction, Figure 2.20. Tycho’s Danish observatory, the Uraniborg. illustrated in Figure 2.21, occurs when two planets are in a straight line with the earth so that from earth they appear to be in the same place in the sky. Tycho predicted the date for this conjunction using Copernicus’ new heliocentric model. The prediction was close (this is good) but was still off by a few days Figure 2.21. The alignment of three planets, called a conjunction.

(not so good). The error indicated that there was still something lacking in Copernicus’ model. (There was: the orbits are not circular as Copernicus assumed.) Second, in 1572 Tycho observed what he called a “nova” (which is Latin for new; today we would call it a supernova) and proved that it was a new star. This discovery rocked the Renaissance world because it was strong evidence that the stars are not perfect and unchanging as Aristotle had thought and as the Ptolemaic model of the heavens declared.

Saturn Jupiter Mars Venus Mercury Sun

Earth Moon

Figure 2.22. The Tychonic System of the world.

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Although familiar with Copernicus’ model, Tycho was a proud advocate of his own model, depicted in Figure 2.22. Tycho’s model is called the Tychonic System of the world. In the model, the sun and moon orbit the earth and the other planets orbit the sun, which in turn orbits the earth. His model did have the advantage of maintaining a stationary earth, which allowed Tycho to avoid controversy with those who insisted that the Bible taught that the earth did not move. Tycho also had a good technical reason for rejecting Copernicus’ model. If the earth moves in an orbit, then earth’s location is different in the summer from its location in the winter. This means the relative positions of the stars should be slightly different at these different times of the year, an effect called stellar parallax. (As an analogy, imagine yourself looking at the trees in a forest. If you take a few steps to one side, the positions of the trees relative to each other in your new location are different.) At that time, no stellar parallax had been observed, and Tycho knew that this meant that either the earth was stationary or the stars were incredibly far away. Copernicus had accepted the great distance of the stars but Tycho did not, and famously wondered, “What purpose would all that emptiness serve?” In fact, stellar parallax was not observed until 1838, when telescopes were finally up to the task. The discovery of stellar parallax in 1838 was the first actual evidence that Copernicus was right. It helps to keep this in mind when we get to the controversy surrounding Galileo.

2.3.8 Kepler and the Laws of Planetary Motion German mathematician Johannes Kepler (Figure 2.23), was perhaps the greatest mathematician of his day and one of the greatest astronomers. In 1600, Tycho Brahe invited him to move to Prague to work with him at his observatory. This invitation was good timing from Kepler’s point of view. Kepler was a devout man and took his faith very seriously, but he was caught in the middle during the Counter-Reformation, a time of serious disagreement between Roman Catholics and Protestants in Germany. Kepler was not interested in identifying with either the Roman Catholic or Protestant factions in the conflict, and every city in Germany was devoted to one of the two sides. Thus, Kepler was driven out of one city after another and was glad finally to have a place to settle down with his family. Tycho died in 1601 and Kepler became the Imperial Mathematician in Tycho’s place. Kepler had access to Tycho’s massive body of research data and used it to develop his famous three laws of planetary motion, the first two of which were published in 1609. He discovered the third law a few years later and published it in 1619. Today, Kepler’s laws Figure 2.23. German astronomer of planetary motion remain the currently accepted model and mathematician Johannes describing our solar system. Kepler (1571–1630). In addition to his astronomical discoveries, he made important discoveries in geometry and optics, he figured out some of the major principles of gravity later synthesized by Isaac Newton, and he was the first to hypothesize that the sun exerted a force on the earth. I want to show you the three beautiful laws of planetary motion Kepler discovered. For your memory work, you may focus on remembering only the first one. But I want you to 52

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see some amazing things about how the solar system works and how we can model it mathematically, so I am going to describe all three of Kepler’s Laws. Kepler’s first law of planetary motion is as follows: First Law

Each of the planetary orbits is an ellipse, with the sun at one focus.

A planet in an elliptical orbit is depicted in Figure 2.24. You may not have studied ellipses yet in planet math, so I will describe them. An ellipse is a geometric figure shaped like this: . An ellipse is similar to a circle, except that instead of having a single sun point locating the center, an ellipse has two points on either side of the center called foci that define its shape. (The term foci is plural, and pronounced elliptical orbit FOH-sigh; the singular is focus.) Out in space, each planet travels on a path defined by a geometrical el- Figure 2.24. A planet in an elliptical orbit lipse. The planetary orbits all have one focus located around the sun (Kepler’s First Law). at the same place in space and this is where the sun is. Think how incredible it is that Kepler figured this out! He was a monster mathematician (no calculator!) and an extremely careful scientist, and the fact that scientists had understood the orbits to be circular for two thousand years did not get in his way. To me, this is simply amazing. Kepler’s second law is not hard to understand. It is in the next box. Second Law A line drawn from the sun to any planet sweeps across the same area in space in any given period of time. The second law is planet travels a depicted in Figure 2.25. specific length specific area swept The idea is that for a of time out in space given period of time, say, a month or a week or whatever, the shaded region in the figure will have the same area, regardless of where the planet is in its orbit. equal area swept out Now, since the sun is in space off-center, this law im- planet plies that the planets travels an travel faster when they equal length of time are closer to the sun Figure 2.25. Equal areas are swept in space for equal periods of time (Kepler’s and slower when they Second Law). are farther away. Keep thinking about how stunning it is that a guy without a calculator or any modern computer could figure this out, all from the observational data that Tycho had assembled. 53

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Kepler’s third law is definitely more mathematically complex than the first two. This law is shown in the next box. Third Law 2

The orbits of any two planets are related as follows:

⎛ T1 ⎞ ⎛ R1 ⎞ ⎜⎝ T ⎟⎠ = ⎜⎝ R ⎟⎠ 2 2

where T1 and T2 are the planets’ orbital periods, and R1 and R2 are their mean distances from the sun. You might be relieved to learn that you will not be doing any computations with this law. However, Kepler’s third law of planetary motion is a stunning example of the mathematical modeling that physicists do all the time, so I want to comment on it here for a bit. The third law is quite accurate. The equation can be expressed in a way that shows that the orbital period, T, for any planet depends simply on the planet’s mean distance from the sun, R. This expression of the third law is written as T = kR

In this equation, k is simply a constant that depends on the units used for T and R. In applying this equation, these units are often based on earth’s orbit, and the units are chosen in a way that make both T and R equal to 1 for earth. The unit for the period is the earth year, since for earth the orbital period is one year. For the distance R, the unit is the astronomical unit (au), equal to the mean earth-sun distance, 149,597,870,700 metres (93,000,000 miles). Using these units, T and R for earth are 1 yr and 1 au, respectively. This gives us a value for the constant k of k =1

yr 3 ( au ) 2

As a quick application, we can use this equation to compute the orbital period of Mars in earth years. Mars is an average of 1.5237 au from the sun. This means the orbital period of Mars is T =1

( au )

⋅(1.5237 au ) 2 = 1.8808 yr 3

I am not planning to keeping going crazy with the math here, and I know you may be freaking out wondering what it means to raise a variable like R to the 3/2 power. Right now it doesn’t matter. You will learn all that when you get to Algebra 2. I just want to show how simple Kepler’s third law really is. Think about it. This simple equation accurately relates the period of any planet’s orbit to that planet’s mean distance from the sun. Now I don’t know about you, but when I see an equation that is as amazing and as simple as this, it sets me thinking. First, Kepler’s work as a scientist is first class. He figured this out from data collected in the era before calculators and before computers. This was only three years after Shakespeare died! 54

Motion

Second, this equation says something deep about the universe we live in. The universe can be modeled with simple mathematics that can be understood by high school kids! We do indeed live in a fascinating world.

2.3.9 Galileo Galileo Galilei (Figure 2.26) worked at the university at Padua, Italy, and later as chief mathematician and philosopher for the ruling Medici family in Florence, Italy. Galileo’s work in astronomy represents the climax of the Copernican Revolution. He made significant improvements to the telescope and used the telescope to see the craters on the moon and sunspots, which provided additional evidence that the heavens were not perfect and unchanging as Aristotle and Ptolemy had maintained. In 1610, he used the telescope to discover four of the moons around Jupiter, which was clearly in conflict with the idea that there had to be exactly seven heavenly bodies. He was fully on board with all the new science of the Copernican model, but, oddly, he never did accept Kepler’s discovery that the planets’ orbits were elliptical rather than circular. Galileo published his early astronomical discoveries in 1610 in Figure 2.26. Florentine scientist Galileo Galilei (1564–1642). a book called The Starry Messenger. Most people know that Galileo was put on trial in 1633 by the Holy Office of the Inquisition established by Roman Catholic Church. However, the reasons for that trial are widely and seriously misunderstood. The real story is rather illuminating, and I will explain it here as briefly as possible. Galileo is famous for this remark: “Philosophy is written in this grand book—I mean the universe—which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language in which it is written. It is written in the language of mathematics.” This beautiful statement calls attention to the mathematical structure of nature. However, Galileo erred in taking his statement too far, claiming that the mathematics he used in his astronomical discoveries was the truth about nature. Galileo felt that his work proved beyond question that Copernicus was right. Galileo, along with other scientists over the next three centuries, still had to learn the limits associated with human modeling of nature. In contrast to Galileo’s attitude toward his work is the attitude of his friend Cardinal Roberto Bellarmine, an important Church official. Bellarmine was a great admirer of Galileo’s work, but Bellarmine’s thinking was very much along the lines of our discussion in the previous chapter: scientific inquiry leads to theories which are models and cannot be regarded as truths; models are provisional and subject to change. In this attitude of an important Church official, we recognize a remarkable early statement of the attitude toward scientific knowledge that today is held as the correct way to think about scientific theories. Bellarmine cautioned Galileo that no natural science could make claims as to truth and urged him to present his ideas as everyone thought Copernicus had done—as hypotheses rather than as truths. 55

Chapter 2

There were definitely mistakes on both sides of the conflict that led eventually to Galileo’s trial by the Holy Office. On Galileo’s part, the mistake was in pushing his ideas too forcefully with the claim that they were true. Galileo’s position implied that the theologians who claimed that the Bible lined up with the Ptolemaic system were wrong in their interpretation of the Bible. Galileo wrote an important letter at that time explaining that the theologians needed to reconsider their interpretations of the Bible in light of what the scientific evidence was showing. Galileo was correct in his views about interpreting the Bible, but his claims pertaining to the truth of his discoveries went too far. To the theologians and church officials who held to the Ptolemaic view, having their views called into question was equivalent to calling the Bible itself into question. This was their mistake: they did not yet understand that the Bible has to be interpreted just as observations of nature have to be interpreted, and they were not yet ready to reconsider their views about the heavens and their interpretations of the Bible. It is interesting to note that in 1992, Pope John Paul II gave an address in which he commended Galileo’s comments on the necessity of interpreting the Bible! Recall that at this time, there was as yet no physical evidence that the earth was moving and rotating on an axis. As I mention in Section 2.3.7, the first actual evidence for earth’s motion around the sun came in 1838 with the discovery of stellar parallax. Evidence for the rotation of the earth came a bit later in 1851 with the invention of Foucault’s pendulum, like the one shown in Figure 2.27. The rotation of the earth causes a small change of direction in each swing of a very long, massive pendulum. If the earth were not rotating, the pendulum would swing steadily back and forth in the same direction. Now, briefly, here is the sequence of events that led to Galileo’s trial. Rumors got around that Galileo had been secretly examined by the Holy Office and forced to abjure (renounce) his views about Copernicus. To help Galileo fight these annoying rumors, Cardinal Bellar-

Figure 2.27. A Foucault Pendulum in the Panthéon in Paris.

Motion

mine wrote Galileo a letter in 1616 stating that the rumors were false. The letter went on to say that though Galileo had not been taken before the Holy Office, he had been told not to defend or teach as true the system of Copernicus. Then in 1632, Galileo published another major work on astronomy in which he did in fact uphold the system of Copernicus against the system of Ptolemy. The Pope at the time, Urban VIII, was also a friend and admirer of Galileo, but when he heard that Galileo had published such a book after having specifically been told not to, he was extremely upset and had no choice but to have Galileo examined by the Holy Office. This was Galileo’s famous 1633 trial. The details leading to Galileo’s trial are very complex, but the controversy boils down to the two issues I have emphasized. First, Galileo pushed his scientific claims too far, claiming truth for a scientific theory which could not be regarded as more than a model of nature. Second, he published a book in defiance of an injunction against doing so. Galileo was a devout man and there is good evidence that he never did actually intend to fall afoul of the injunction. But when the Holy Office persuaded him that he had, he was immediately ready to confess his actions and abjure them. This he did. Galileo was never tortured, but it was necessary that he be punished in some way. His friend Pope Urban VIII made it as easy on Galileo as he could by confining him to “house arrest” and prohibited him from further publishing. He lived for a few months in Rome in the palace of one of the cardinals, and then was allowed to return to his home in Florence were he lived in house arrest for the last eight years of his life. In addition to his work in astronomy, Galileo developed ground-breaking ideas in physics over the course of 30 years of work. These ideas were published after his trial in what would be his final book.6 Before Galileo, scientists had always accepted Aristotle’s physics, which held that a force was needed to keep an object moving. Galileo broke with this 2,000-year-old idea and hypothesized that force was needed to change motion but not to sustain motion as Aristotle had taught. Galileo was the first to formulate the idea of a friction force that caused objects to slow down. By conducting his own experiments, Galileo also discovered that all falling objects accelerate at the same rate (the acceleration of gravity, 9.80 m/s2), which is mathematically very close to Isaac Newton’s second law of motion (our topic in the next chapter). Galileo’s studies in physics thrust forward the Scientific Revolution and set the stage for the work of Isaac Newton, where the Scientific Revolution reached its climax. The saga of the Copernican Revolution ends more or less with Galileo. Within 50 years of Galileo’s death, the heliocentric model of the planetary orbits was becoming widely accepted. But while we are studying the planets and gravity, the whole story just isn’t complete unless we mention two more key figures in the history of science.

2.3.10 Newton, Einstein, and Gravitational Theory Sir Isaac Newton (Figure 2.28) is perhaps the most celebrated mathematician and scientist of all time. He was English, as his title implies, and he was truly phenomenal. He held a famous professorship in mathematics at Cambridge University. He developed calculus. He developed the famous laws of motion, which we will examine. He developed an entire theory of optics and light. He formulated the first quantitative law of gravity called the law of universal gravitation. His massive work on motion, gravity, and the planets, Principia

Since he was forbidden to publish through the Catholic Church, the book was published by a Protestant publisher in Holland.

Chapter 2

Mathematica, was published in 1687. This work is one of the most important publications in the history of science. In this course, we do not perform computations with Newton’s law of universal gravitation, and you do not need to memorize the equation for it. But let’s look at it here briefly. The law is usually written as F =G

m1m2 d2

Do You Know ...

Who built the first monster telescope?

William Herschel was a German astronomer who moved to England when he was a young man. He was a major contributor to pushing the technology of the reflecting telescope to new limits, and spent vast amounts of time casting and polishing his own mirrors. He constructed the largest telescopes ever built at the time. In 1781, Herschel discovered the planet Uranus. Herschel’s sister Caroline was an important astronomer herself. She worked closely with her brother. Herschel gave her a telescope of her own and with it she discovered many new comets, for which she became recognized. Herschel’s monster 40-foot telescope, shown to the left, had a primary mirror over four feet in diameter. In 1789, on the first night of using the new telescope, Herschel discovered a new moon of Saturn. He discovered another new moon about a month later.

Motion

the force of gravity has both masses in it multiplied together. Newton’s model implies that if either mass is zero, the force of gravitational attraction is zero. While we are here looking at Isaac Newton, we should pause and consider the relationship between his physical theories (including law of universal gravitation and his laws of motion) and Kepler’s mathematical theory of planetary motion. It turns out that Kepler’s discovery about the elliptical orbits and the relationship between the period and mean radius of the orbit can be directly derived from Newton’s theories, and Newton does derive them in Principia Mathematica. But Newton’s equations apply much more generally than Kepler’s do. As we see in the next chapter, Newton’s laws apply to all objects in motion— planets, baseballs, rockets—while Kepler’s laws apply to the special case of the planets’ orbits. If we consider this in light of my comments in Chapter 1 about the way theories work, we see that Newton’s laws explain everything Kepler’s laws explain, and more. This places Newton’s theory about motion and gravity above Kepler’s, so Newton’s theories took over as the most widely-accepted theoretical model explaining gravity and motion in general. However, even though Newton’s laws ruled the scientific world for nearly 230 years, they do not tell the whole story. This is where the German physicist Albert Einstein (Figure 2.29) comes in with his general theory of relativity, published in 1915. Einstein’s theory explains gravity in terms of the curvature of space (or more accurately, spacetime) around a massive object, such as the sun or a planet. This spacetime curvature is represented visually in Figure 2.30. Fascinatingly, since Einstein’s theory is about curving space, the theory predicts that even phenomena without mass, such as rays of light, are affected by gravity. Einstein noticed this and made the stunning prediction in 1917 that starlight bends as it travels through space when it passes near a massive object such as the sun. He formed this hypothesis, including the amount light bends, based on his general theory of relativity, which was based completely on mathematics. What do you think about that? It practically Figure 2.29. German physicist Albert Einstein (1879–1955). leaves me speechless. Einstein became instantly world famous in 1919 when his prediction was confirmed. To test this hypothesis, Einstein proposed photographing the stars we see near the sun during a solar eclipse. This has to be done during an eclipse because looking at the sky while the sun is nearby means it is broad daylight and we aren’t able to see the stars. Einstein predicted that the apparent position of the stars shifts a tiny amount relative to where they appear when the sun is not near the path of the starlight. British scientist Sir Arthur Eddington commissioned two teams of photographers to photograph the stars during the solar eclipse of 1919. After analyzing their photographic plates (one of which is shown on the opening page of Chapter 1), they found the starlight shifted by exactly the amount Ein- Figure 2.30. A visual representation of the curvature of stein said it would. Talk about sudden spacetime around the earth. 59

Chapter 2

fame—Einstein became the instant global rock star of physics when this happened! (And his puppy dog eyes contributed even more to his popularity!) Just as Kepler’s laws were superseded by Newton’s laws and can be derived from Newton’s laws, Newton’s law of universal gravitation was superseded by Einstein’s general theory of relativity and can be derived from general relativity. Einstein believed that his own theories would some day be superseded by an even more all-encompassing theory, but so far (after 102 years) that has not happened. The general theory of relativity remains today the reigning champion theory of gravity, our best understanding of how gravity works, and one of the most important theories in 20th- and 21st-century physics.

Chapter 2 Exercises Unit Conversions It is time for you to get busy learning the metric prefixes and unit conversion factors in Appendix A. Perform the following unit conversions, showing all your work in detail. (Showing just the answers is not adequate; show all the conversion factors involved in the conversion for each problem.) Check your work against the answer key on the next page. Where possible, express your results in both standard notation and scientific notation, using the correct number of significant digits. For the first 20 problems, use the standard method of multiplying conversion factors. The last problem requires an extra step that I think you can figure out. Convert this Quantity

Into these Units

1,750 meters (m)

feet (ft)

3.54 grams (g)

kilograms (kg)

41.11 milliliters (mL)

liters (L)

7 × 10 m (radius of the sun)

miles (mi)

1.5499 × 10–12 millimeters (mm)

750 cubic centimeters (cm or cc) (size of the engine in my old motorcycle)

2.9979 × 108 meters/second (m/s) (speed of light)

ft/s

168 hours (hr) (one week)

5,570 kilograms/cubic meter (kg/m3) (average density of the earth)

g/cm3

45 gallons per second (gps) (flow rate of Mississippi River at the source)

m3/minute

600,000 cubic feet/second (ft3/s) (flow rate of Mississippi River at New Orleans)

liters/hour (L/hr)

5,200 mL (volume of blood in a typical man’s body)

(m3/min)

Motion

Convert this Quantity

Into these Units

5.65 × 10 mm (area of a postage stamp)

square inches (in2)

32.16 ft/s2 (acceleration of gravity, or one “g”)

m/s2

5,001 μg/s

kg/min

4.771 g/mL

kg/m3

13.6 g/cm3 (density of mercury)

mg/m3

93,000,000 mi (distance from earth to the sun)

65 miles per hour (mph)

m/s

633 nanometers (nm) (wavelength of light from a red laser)

5.015% of the speed of light

mph

Answers (A dash indicates that it is either silly or incorrect to write the answer that way, so I didn’t: silly because there are simply too many zeros, or no zeros at all; incorrect because we are unable to express the result that way and still show the correct number of significant digits.) Standard Notation

Scientific Notation

5,740 ft

5.74 × 103 ft

0.00354 kg

3.54 × 10–3 kg

0.04111 L

4.111 × 10–2 L

400,000 mi

4 × 105 mi

–

1.5499 × 10–15 m

0.00075 m

7.5 × 10–4 m3

983,560,000 ft/s

9.8356 × 108 ft/s

605,000 s

6.05 × 105 s

5.57 g/cm3

–

1.0 × 101 m3/min

60,000,000,000 L/hr

6 × 1010 L/hr

0.0052 m3

5.2 × 10–3 m3

0.876 in2

8.76 × 10–1 in2

9.802 m/s

–

0.0003001 kg/min

3.001 × 10–4 kg/min

4,771 kg/m3

4.771 × 103 kg/m3

Chapter 2

Standard Notation

Scientific Notation

13,600,000,000 mg/m

1.36 × 1010 mg/m3

–

1.5 × 1013 cm

29 m/s

2.9 × 101 m/s

0.0000249 in

2.49 × 10–5 in

33,700,000 mph

3.37 × 107 mph

Motion Exercises 1.

A train travels 25.1 miles in 0.50 hr. Calculate the velocity of the train.

Convert your answer from the previous problem to km/hr.

How far can you walk in 4.25 hours if you keep up a steady pace of 5.0000 km/hr? State your answer in km.

For the previous problem, how far is this in miles?

On the German autobahn there is no speed limit and in good weather many cars travel at velocities exceeding 150.0 mi/hr. How fast is this in km/hr?

Referring again to the previous question, how long does it take a car at this velocity to travel 10.0 miles? State your answer in minutes.

An object travels 3.0 km at a constant velocity in 1 hr 20.0 min. Calculate the object’s velocity and state your answer in m/s.

A car starts from rest and accelerates to 45 mi/hr in 36 s. Calculate the car’s acceleration and state your answer in m/s2.

A rocket traveling at 31 m/s fires its retro-rockets, generating a negative acceleration (it is slowing down). The rockets are fired for 17 s and afterwards the rocket is traveling at 22 m/s. What is the rocket’s acceleration?

10. A person is sitting in a car watching a traffic light. The light is 14.5 m away. When the light changes color, how long does it take the new color of light to travel to the driver so that he can see it? State your answer in nanoseconds. (The speed of light in a vacuum or air, c, is one of the physical constants listed in Appendix A that you need to know.) 11. A proton is uniformly accelerated from rest to 80.0% of the speed of light in 18 hours, 6 minutes, 45 seconds. What is the acceleration of the proton? 12. A space ship travels 8.96 × 109 km at 3.45 × 105 m/s. How long does this trip take? Convert your answer from seconds to days. 13. An electron experiences an acceleration of 5.556 × 106 cm/s2 for a period of 45 ms. If the electron is initially at rest, what is its final velocity? 14. A space ship is traveling at a velocity of 4.005 × 103 m/s when it switches on its rockets. The rockets accelerate the ship at 23.1 m/s2 for a period of 13.5 s. What is the final velocity of the rocket? 15. A more precise value for c (the speed of light) than the value given in Appendix A is 2.9979 × 108 m/s. Use this value for this problem. On a particular day the earth 62

Motion

is 1.4965 × 108 km from the sun. If on this day a solar flare suddenly occurs on the sun, how long does it take an observer on the earth to see it? State your answer in minutes.

Answers 1.

22 m/s

79 km/hr

21.3 km

241.4 km/hr

4.00 min

0.63 m/s

–0.53 m/s

10. 48.3 ns

13. 2.5 × 10 m/s 3

11. 3,680 m/s

14. 4.32 × 10 m/s 3

13.2 mi

0.56 m/s2

12. 301 days

15. 8.3197 min

Ptolemaic Model and Copernican Revolution Study Questions 1.

Make a list of all the regions in the Ptolemaic Model in their correct order. (There are 10 of them and the first nine are called spheres.) For each of the last three regions write a brief description of the meaning of the name.

Describe why some theologians in the 16th century were strongly opposed to Copernicus’ heliocentric theory.

State six features of the Ptolemaic model other than the spheres.

Describe Copernicus’ model of the heavens.

What are some of the “proofs” people used in arguing that there is no way that the earth rotates on an axis?

For what reason did Copernicus decide to keep his theory private?

Write a description of the two key observations Tycho made (including dates) that challenged the Ptolemaic system.

Briefly describe the cosmological model put forward by Tycho.

State Kepler’s first law of planetary motion.

10. This is a bit difficult, but explain retrograde motion and epicycles as well as you can in a few sentences. 11. Explain the two main mistakes individuals made that led to Galileo’s trial. 12. Explain the actual cause of Galileo’s trial and the results of that trial. 13. Describe why Pope John Paul II commended Galileo in 1992. 14. Distinguish between Newton’s and Einstein’s theories of gravitation. According to each of these two geniuses, what is the cause of gravity and what are the effects of gravity? 15. The theories reviewed in this chapter suggest that the universe possesses a very deep mathematical structure. Even so, scientists do not claim that equations such as Kepler’s third law are true. Write a few sentences explaining why. 16. Describe some of Kepler’s scientific achievements, aside from his laws of planetary motion. 63

CHAPTER 4

Energy

Large Hadron Collider In the Large Hadron Collider (LHC) at CERN in Switzerland (see box on page 12), protons are accelerated and collided at extremely high energies. The purpose of these collisions is to help scientists discover more about the fundamental structure of matter. Theory predicts the existence of a particle called the Higgs Boson. The image above is a computer simulation of a Higgs detection event inside the CMS detector at the LHC. The CMS website states, “The lines represent the possible paths of particles produced by the proton-proton collision in the detector while the energy these particles deposit is shown in blue.” As you know, the speed of light is 300,000,000 m/s. To generate the energy needed to observe the Higgs Boson, protons are accelerated to a speed that is only 3 m/s slower than the speed of light! At this speed, the protons only require 90 μs (0.000090 s) to travel 17 miles around the main underground tunnel of the LHC. This huge kinetic energy is way beyond the energy produced by any other particle accelerator yet constructed. 84

OBJECTIVES Memorize and learn how to use these equations: EG = mgh

EK = 12 mv 2

2EK m

W = Fd

After studying this chapter and completing the exercises, students will be able to do each of the following tasks, using supporting terms and principles as necessary: 1. 2.

State the law of conservation of energy. Describe how energy can be changed from one form to another, including: a. different forms of mechanical energy (kinetic, gravitational potential, elastic potential) b. chemical potential energy c. electrical energy d. elastic potential energy e. thermal energy f. electromagnetic radiation g. nuclear energy h. acoustic energy 3. Briefly define each of the types of energy listed above. 4. Describe two processes by which energy can be transferred from one object to another (work and heat), and the conditions that must be present for the energy transfer to occur. 5. Describe in detail how energy from the sun is converted through various forms to end up as energy in our bodies, as energy used to run appliances in our homes, or as energy used to power machines in industry. 6. Explain why the efficiency of any energy conversion process is less than 100%. 7. Calculate kinetic energy, gravitational potential energy, work, heights, velocities, and masses from given information using correct units of measure. 8. Define friction. 9. Using the pendulum as a case in point, explain the behavior of ideal and actual systems in terms of mechanical energy. 10. Explain how friction affects the total energy present in a mechanical system.

4.1

What is Energy?

4.1.1 Defining Energy Defining energy is tricky. Dictionaries usually say, “the capacity to do mechanical work,” which is not particularly helpful. Actually, there is no definition for energy that gets at what it actually is, so I will not try to define it. We are just going to accept that energy exists in the universe, it was there from the moment the universe came into existence, and it exists in many different forms. It is fairly obvious that a bullet traveling at 2,000 ft/s has more energy than a bullet at rest. This is why the high speed bullet can kill but the bullet at rest cannot. This study is mainly about tracking energy as it changes from one form to another, and calculating the quantities of three particular forms of energy. 85

Chapter 4

4.1.2 The Law of Conservation of Energy The law of conservation of energy is as follows: Energy can be neither created nor destroyed, only changed in form. Energy can be in many different forms in different types of substances, such as in the molecules of gasoline, in the waves of a beam of light, in heat radiating through space, in moving objects, in compressed springs, or in objects lifted vertically on earth. As different physical processes occur—such as digesting food, throwing a ball, operating a machine, heating due to friction, or accelerating a race car—energy in one form is being converted into some other form. Energy might be in one form in one place, such as in the chemical potential energy in the muscles of your arm, and be converted through a process such as throwing a ball to become energy in another form in another place, such as kinetic energy in the ball.

4.1.3 Mass-Energy Equivalence In 1905, Albert Einstein published his now-famous equation, E = mc2. The E and m in this equation represent energy and mass; c represents the speed of light. With this equation, Einstein theorized that mass and energy are really just different forms of the same thing. That is, all mass has associated with it an equivalent amount of energy (given by E = mc2), and vice versa. This theory of mass-energy equivalence is now considered to be a fundamental property of the universe. The reason I mention mass-energy equivalence here is that since mass is a form of energy, matter must be taken into consideration for a completely accurate statement of the law of conservation of energy. In nuclear reactions, such as take place in the sun (fusion) or in nuclear power plants (fission), quantities of matter are converted completely into energy. Einstein’s equation E = mc2 also gives the amount of energy that appears when a quantity of matter in converted to energy in one of these nuclear processes. Thus, to be completely accurate, we need to state that the law of conservation of energy includes all mass as well, as one of the forms energy can take. Let’s restate the conservation law with this in mind: “mass-energy can be neither created nor destroyed, only changed in form.” Most of the problems we encounter in physics and chemistry don’t involve nuclear reactions (thankfully). This means that for most purposes, we can consider the common forms of energy listed below without worrying about the complicated issue of mass-energy equivalence.

4.2

Energy Transformations

4.2.1 Forms of Energy Here are some common forms energy can take: Gravitational Potential Energy

This is the energy an object possesses because it has been lifted up in a gravitational field. If such an object is released and allowed to fall, the gravitational potential energy converts into kinetic energy. The term potential in the name of this form of energy indicates that the energy is stored and converts into another form of

Energy

energy when released. There are other forms of energy listed below that use this term for the same reason. Kinetic Energy

This is the energy an object possesses because it is in motion. The faster an object is moving, the more kinetic energy the object has. Electromagnetic Radiation

This is the energy in electromagnetic waves traveling through space, or through media such as air or glass. This type of energy includes all forms of light, as well as radio waves, microwaves, and a number of other kinds of radiation. We study electromagnetic waves in some detail in a later chapter. Chemical Potential Energy

This energy is in the chemical bonds of molecules. In the case of substances that burn, the chemical potential energy in the molecules is released in large quantities as heat and light when the substance is burned, making these substances useful as fuel. Electrical Energy

This is energy flowing in electrical conductors, such as from a power station to your house to power your appliances. Thermal Energy This is the energy a substance possesses due to being heated. We examine thermal energy more closely in the next chapter. Elastic Potential Energy

This is the energy contained in any object that has been stretched (such as a bungee cord or a hunter’s bow) or compressed (such as a spring). Nuclear Energy

This is energy released from the nuclear processes of fission (when the nuclei of atoms are split apart) or fusion (when atomic nuclei are fused together). As I mention in the previous section, these processes convert mass into energy. Acoustic (Sound) Energy

This is the energy carried in sound waves, such as from a person’s voice, the speakers in a sound system, or the noise of an explosion. Since sound waves are carried by moving air molecules, this is really a special form of kinetic energy.

4.2.2 Energy Transfer Two more important energy-related terms are those associated with the process of energy being transferred from one place, substance, or object to another. These two terms are: Work

Work is a mechanical process by which energy is transferred from one object to another. Objects do not possess work like they do other forms of energy. Instead, one object “does work” on another object by applying a force to it and moving it a certain distance. When one object does work on another, energy is transferred from the first object to the second. We study work in more detail later in the chapter. Heat

The term heat is used as a general description of energy in transit, flowing by various means from a hot substance to a cooler substance when a difference in temperature is present. We study heat and the three ways it flows in more detail in Chapter 7. As with work, substances do not possess heat. What substances do possess is kinetic energy in their moving atoms, and we refer to this energy as the internal energy of the substance. 87

Chapter 4 acoustic (sound) energy in the sound wave of the bang

powder in the firecracker contains chemical potential energy

BANG!

kinetic energy in the flying debris energy in the electromagnetic radiation (light and heat) of the flash

Figure 4.1. Chemical potential energy in a firecracker is converted into other forms of energy when the firecracker explodes.

Let’s look at a common example of energy changing from one form into others. We all know what happens when a person lights a firecracker. (It explodes!) What forms of energy are present during the explosion, and where did all this energy come from? As illustrated in Figure 4.1, the energy released in the explosion is the chemical potential energy in the molecular bonds of the chemicals inside the firecracker. When these chemicals burn, they release a lot of energy. And as you already know, an exploding firecracker gives off a flash of light and heat (both are forms of electromagnetic radiation), a loud bang, and the fragments of the firecracker are blown all over the place. Thus, the chemical potential energy in the powder inside the firecracker is converted into several different kinds of energy during the explosion. Now consider how the law of conservation of energy applies to this explosion. All the energy present in the chemicals before the explosion is still present in various forms after the firecracker explodes. This is what “conservation” of energy means. We can represent the conservation of energy in a sort of equation like this: chemical potential energy in the firecracker

acoustic energy in the bang

kinetic energy of flying debris

energy in light and heat

We do not perform any calculations this complex in this course. But later in this chapter, we begin using the principle of conservation of energy to solve problems involving three of the forms of energy we have seen so far.

4.2.3 The “Energy Trail” Much of the energy we depend on here on earth comes to us from the sun. As we track the forms this energy takes in its journey from the sun to, say, the energy in our bodies, we might call this the “energy trail.” (This way we can have fun yelling Yee-Haw! while we are studying this. Ask your teacher for a demonstration.) We now follow the energy trail beginning with the sun, through different processes of conversion, and arriving at different places where energy is commonly used. 88

Energy

Do You Know ...

What is dark energy?

When Einstein first developed his equations for the general theory of relativity (published in 1915), the equations implied that the universe must be either expanding or contracting. At the time, the prevailing view was that the universe was doing neither, so Einstein put a fudge factor in the equations to keep the universe static. Just a few years later, astronomer Edwin Hubble made observations that enabled Belgian Georges Lemaître, a physicist and Catholic priest, to conclude that the universe was expanding, just like Einstein’s original equations implied! So Einstein took the fudge factor out of the equations and said that putting it in was “the biggest blunder of his life.” In the 1990s, astronomers discovered that the expansion of the universe is speeding up—the expansion of the universe is accelerating. This seems impossible, because the gravitational attraction of the galaxies pulling on each other should be slowing the expansion rate down. In a classic illustration of how the Cycle of Scientific Enterprise works, no known theory was able to account for the cause of the acceleration, so scientists had to get busy theorizing about this mystery. At present, the most accepted hypothesis for the cause of the acceleration is the presence of an unknown form of energy called dark energy that permeates all of space. Calculations indicate that on the basis of mass-energy equivalence, 68% of the energy in the universe is dark energy, 27% of the energy is in the form of dark matter (see the box on page 8), and only 5% of the energy is in the form of ordinary matter. The sun’s energy is produced by fusion reactions as the nuclei of hydrogen atoms “fuse” or stick together to form helium. This is so hard to do that we have not yet succeeded in doing it here on earth in a controlled way. However, we have succeeded in doing it in an uncontrolled way. Fusion is the same nuclear reaction as the main reaction in a thermonuclear bomb. A thermonuclear explosion is an uncontrolled nuclear fusion reaction. When referring to this energy being produced in the sun, we can simply call it nuclear energy. The energy leaves the sun as electromagnetic radiation, a different form of energy, and travels through space to us. When this energy arrives at earth, most of it warms the ground, oceans, and atmosphere. This is very important for stabilizing the earth’s climate and making earth habitable, but unless we collect the energy in a solar collector of some kind we are not able to use this energy directly. However, some of the electromagnetic radiation streaming from the sun is captured by plants and converted into chemical potential energy in the molecules in the plants through the process of photosynthesis. These plants eventually become the foods we eat or the fuels we burn. Current theory holds that in ancient eras in the earth’s history, many vast forests were buried and the plant matter was converted underground into what we now call “fossil fuels” (petroleum, coal, and natural gas). Some fuels come from living plants too, such as firewood from trees and automotive alcohol (ethanol) from corn. The energy in the molecules of these fuels is chemical potential energy that is converted into heat energy when the fuels are burned. Your task is to describe the energy conversions each step of the way from the sun all the way to your breakfast cereal or your computer. Tables 4.1 through 4.4 illustrate a few examples of following the energy from the sun to different places it can end up here on earth. When asked to outline one of these pathways in the “energy trail,” always list two things for each step of the way: (1) Where the energy is, and (2) what form the energy is in. 89

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Where is the energy?

The Sun

Electromagnetic waves in space

Plants on earth

Breakfast cereal

Muscles in the human body

Stretched bow

Flying arrow

What form is the energy in?

Nuclear energy

Electromagnetic radiation

Chemical potential energy

Elastic potential energy

Kinetic energy

Table 4.1. Energy transformations from the sun to a flying arrow, assuming the archer was on a vegetarian diet. Where is the energy?

The Sun

Electromagnetic waves in space

Plants on earth

Chicken feed

Muscles in the bodies of chickens

Muscles in the human body

Moving kid on skateboard

What form is the energy in?

Nuclear energy

Electromagnetic radiation

Chemical potential energy

Kinetic energy

Table 4.2. Energy transformations from the sun to a kid on a skateboard, assuming the kid was eating chicken.

Where is the energy?

The Sun

Electromagnetic waves in space

Plants on earth

Fossil fuel (crude oil, coal, natural gas)

Spinning gas turbine generator at power station

Wires from the power station to your house

Heat from the coils in the toaster

What form is the energy in?

Nuclear energy

Electromagnetic radiation

Chemical potential energy

Kinetic energy

Electrical energy

Electromagnetic radiation

Table 4.3. Energy transformations from the sun to the heat from a toaster in your house. Where is the energy?

The Sun

Electromagnetic waves in space

Plants on earth

Fossil fuel (coal)

Heat from burning coal

Steam in the boiler

Moving train

What form is the energy in?

Nuclear energy

Electromagnetic radiation

Chemical potential energy

Heat

Thermal energy

Kinetic energy

Table 4.4. Energy transformations from the sun to a moving steam locomotive.

4.2.4 The Effect of Friction on a Mechanical System You probably already have a feel for what people mean by the term friction. Friction is a force present any time one object or material comes in contact with another object or material. The effect of friction is to oppose any relative motion between the two objects in contact. The cause of friction is rather complicated, but down at the atomic level friction has to do with the electrical attractions and repulsions between the charged particles in the atoms of the objects.

Energy

Friction makes it harder for one object to slide on top of another, which is good if you are talking about the friction between the tires of a car and the pavement. If there were no friction, cars could not start or stop or steer. (You may experience this physically if you are ever in the undesirable position of trying to drive a car on ice.) Friction is also very nice to have around any time one is attempting to grab something or clamp something. Without friction, things would just slip through our fingers. But friction is undesirable when the goal is to design the parts of a machine so they slide smoothly against one another without wear or damage to the machine. And, of course, there is friction when an object moves through the air. This friction is usually called air resistance or drag. In this course, we are not considering friction in the calculations we do. However, in all real mechanical systems friction plays a significant role. Friction is caused when parts of the system rub against each other or when parts of the system move through a fluid such as air or water. Just as when you rub your hands together on a cold day, friction always results in heating. When the parts of a mechanical system such as a machine get warm from friction, heat flows from the warm parts into the cooler surrounding environment. (We will look more at how this happens in Chapter 7.) This heat energy flowing out of the system is energy that used to be in the system. When heat energy flows out of a system due to friction, the law of conservation of energy still applies: no energy is created or destroyed. However, the energy remaining in the system is reduced by the amount of energy that flows out of the system due to heating from friction. A scientist or engineer may refer to energy “lost” due to friction. This does not mean the energy is destroyed or ceases to exist, only that it flows out of the system as heat and is no longer available as energy in the system. The net effect, of course, is that things slow down as energy gradually leaves the system as heat due to friction.

4.2.5 Energy “Losses” and Efficiency For all the different forms of energy we have considered, there are many different kinds of processes that might be involved in converting energy from one form to another. Combustion is a process that converts chemical potential energy into heat. Photosynthesis converts electromagnetic energy from the sun into chemical potential energy in the cells of plants. The Industrial Revolution began when humans began learning how to design machines and systems to convert energy from various forms found in nature into forms that can be harnessed to do useful work for us. Let’s consider some process like this, such as an engine in a car converting the chemical potential energy in the gasoline into kinetic energy in the moving car. One of the facts of life on earth is that it is theoretically impossible for a conversion process to capture all the energy involved and convert it to a form that can do useful work. Whether we want it to or not, some of the energy always converts to heat, which radiates out into the environment. The laws of thermodynamics state that this is always the case. This situation is represented in Figure 4.2. It is common to speak of the energy converted into heat as “lost.” Keeping the law of conservation of energy in mind, it should be clear that what we mean by this is not that the energy ceases to exist, only that the energy escapes into the environment where it is no longer available to us in a usable form. It is lost from the system, not from the universe. The efficiency of an energy conversion process is the ratio of the usable energy coming out of the process to the energy that goes into the process:

Energy converted to heat

Chapter 4

Energy in

Energy conversion process

Useful energy out

Figure 4.2. In any energy conversion process, some energy is converted to heat that is not available to do useful work.

Efficiency =

useful energy out ×100% energy in

Since some energy is always lost as heat, the efficiencies of our machines are always less than 100%. As physical examples, the efficiency of typical automobile engines is only around 15%, which means that 85% of the energy in the fuel is not used to propel the car. (That’s a lot of lost energy.) Solar cells convert electromagnetic radiation from the sun into electricity. At present, the highest efficiency realized with these technologies is around 25%. The overall efficiency of the new electric cars is around 20–25%. This figure may seem low, but there are a lot of losses in generating the electrical power at a power station and transporting the power to the homes where people charge up the batteries in their electric cars.

4.3

Calculations with Energy

4.3.1 Gravitational Potential Energy and Kinetic Energy Two important forms energy can take in mechanical systems are gravitational potential energy, EG , and kinetic energy, EK . The gravitational potential energy an object possesses depends on how high up it is and the kinetic energy of an object depends on how fast it is moving. Both also depend on the object’s mass. Gravitational potential energy is calculated as EG = mgh where EG is energy in joules (J), m is the mass (kg), g = 9.80 m/s2, and h is the height (m). Notice that if you know how much gravitational potential energy an object has and its mass, you can solve this equation for h to find out how high the object is above the ground. Simply divide both sides of the equation by mg and you have h=

EG mg

Energy

Notice also that the gravitational potential energy of an object is directly proportional to its height. If the height of an object increases by 50%, the gravitational potential energy of the object also increases by 50%. When calculating gravitational potential energy, the energy you calculate always depends on the location you choose to use as your zero reference for the height. This zero reference might be sea level, or the ground, or the floor of your classroom, or a table top. It doesn’t matter, because the EG an object has is always relative to where h = 0 is. Usually, the most logical and convenient location for h = 0 is clear from the context. The equation for gravitational potential energy gives us another example of a derived MKS unit, the joule, for quantities of energy. Multiplying the units together for the terms on the right side of the EG equation, we can see that a joule is made up of primary units as follows: 1 J = 1 kg ⋅

m kg ⋅m 2 ⋅m = 1 s2 s2

You might compare this to the units described at the bottom of page 22. Example 4.1 A golf ball has a mass of 45.9 g. While climbing a tree near a golf course, little Janie finds a golf ball stuck in a branch 9.5 ft above the ground. What is the gravitational potential energy of the golf ball at that height? Start by writing the givens and doing the unit conversions to get all quantities into MKS units, keeping one extra significant digit in your intermediate calculations. 1 kg = 0.0459 kg 1000 g 0.3048 m h = 9.5 ft ⋅ = 2.90 m ft EG = ?

m = 45.9 g ⋅

Now write the equation and complete the problem. EG = mgh = 0.0459 kg ⋅9.80

m ⋅2.90 m = 1.30 J s2

These calculations all contain one extra significant digit. The given height only has two significant digits, so now we round our final result to two digits. EG = 1.3 J

Example 4.2 An ant carries a grain of sugar up the side of a building to its nest on the roof. The mass of the grain of sugar is 0.0356 μg. After it has been carried to the roof, the EG in the grain of sugar is 1.91 nJ. How high is the ant nest? 93

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Write the givens and do the unit conversions. 1g 1 kg ⋅ = 3.56 ×10−11 kg 6 10 µg 1000 g 1J EG = 1.91 nJ⋅ 9 = 1.91×10−9 J 10 nJ h=?

m = 0.0356 µg ⋅

Now write the equation, solve for h, and compute the result. EG = mgh h=

EG 1.91×10−9 J = = 5.47 m mg 3.56 ×10−11 kg ⋅9.80 m s2

Every value in this computation has three significant digits, as does this result, so the problem is complete.

Now we look at another important form of energy, kinetic energy. Kinetic energy is one of the most important concepts in physics because it relates to many other concepts. Kinetic energy is calculated as EK = 12 mv 2 where EK is the kinetic energy in joules (J), m is the mass (kg), and v is the velocity (m/s). The units for kinetic energy are joules, just as with all other forms of energy. Kinetic energy is proportional to the mass of an object and to the square of the object’s velocity. Notice that if you know how much kinetic energy an object has and its mass, you can solve this equation for v to find out how fast the object is moving. Since the algebra to do this may be unfamiliar to students in this course, you may want to just go ahead and memorize the equation for velocity as a function of kinetic energy. This equation is v=

2EK m

Example 4.3 An electron with a mass of 9.11 × 10–28 g is traveling at 1.066% the speed of light. Determine the amount of kinetic energy the electron has and state your result in nJ. Start by writing the givens and doing the unit conversions. To obtain the electron’s velocity, we must multiply the speed of light (from Appendix A) by 0.01066.

Energy

1 kg = 9.11×10−31 kg 1000 g m m v = 0.01066⋅3.00 ×108 = 3.198 ×106 s s EK = ?

m = 9.11×10−28 g ⋅

Now compute the kinetic energy. 2

m⎞ ⎛ EK = 12 mv 2 = 0.5⋅9.11×10−31 kg ⋅ ⎜ 3.198 ×106 ⎟ = 4.658 ×10−18 J ⎝ s⎠ The problem statement requires the result to be in units of nanojoules (nJ), so perform this conversion. 4.658 ×10−18 J⋅

109 nJ = 4.658 ×10−9 nJ J

Both the mass and the speed of light values have three significant digits, so rounding this result to three significant digits gives EK = 4.66 ×10−9 nJ

Example 4.4 A kid fires a plastic dart from a dart gun. The mass of the dart is 21.15 g and its kinetic energy is 0.3688 J when it flies out the dart gun. Determine the velocity of the dart. Write the givens and do the unit conversions. 1 kg = 0.02115 kg 1000 g EK = 0.3688 J v=?

m = 21.15 g ⋅

Now complete the problem using the memorized velocity equation. v=

2EK 2⋅0.3688 J m = = 5.905 0.02115 kg s m

Both the mass and the kinetic energy values have four significant digits, so this result is rounded to four significant digits.

Chapter 4

4.3.2 Work The way an object acquires kinetic energy or gravitational potential energy is that another object or person or machine does work on it. Work is the way mechanical energy is transferred from one machine or object to another. Work is a form of energy, but objects don’t possess work. Work is the process by which energy is transferred from one mechanical system to another. Work is defined as the energy it takes to push an object with a certain (constant) force over a certain distance. Work is calculated as W = Fd where W is the work done on the object in joules (J), F is the force on the object (N), and d is the distance the object moves (m). Notice from this equation that since work is energy, the units here come out to be 1 J = 1 N⋅m Let’s take a moment to convince ourselves that the units here are the same as the units described above right after the EG equation. The work equation says that joules are equal to newtons times meters. A newton is a force, and we know from Newton’s second law of motion that force equals mass times acceleration, or F = ma. If we multiply all these units together we have 1 J = 1 N⋅m = 1 kg ⋅

m kg ⋅m 2 ⋅m = 1 s2 s2

These units are indeed the same as the units we worked out for gravitational potential energy a few pages back. The concept of work is the basic principle governing how energy is transferred from one device to another in a mechanical system. For example, as depicted in Figure 4.3, if an electric motor is used to lift a piece of equipment, the motor must reel in a certain length, L, of steel cable, and it must pull on the cable with a certain force, F, while doing so. The pulling force times the length of cable is the amount of work done by the motor. And where does this work energy supplied by the motor go? Assuming 100% efficiency in the lifting motor and cables (and electric motors have very high efficiencies, so this is not a bad approximation), the energy all goes into the gravitational potential energy acquired by the piece of equipment being lifted. In actuality, since the efficiency of all systems is less than 100%, some of the energy leaves L the system as heat. In the end, the gravitational potential energy of the lifted piece F of equipment does not quite represent all L F of the energy the electric motor has to supply. electric equipment to We see here that work and conservamotor be lifted tion of energy are very closely related. As another example, if a man pushes a kid on Figure 4.3. An electric motor raising an object to a height a bicycle over a short distance to get the L by means of force F pulling on the cable.

Energy

kid going, the man delivers energy to the kid on the bicycle equal to the pushing force times the distance pushed. Ignoring friction for now, that work energy from the man is now in the kinetic energy of the kid on the bicycle. There are two more important details to note about work. First, the equation for work, W = Fd, requires that the force applied to an object and the distance the object travels must lie in the same direction. As depicted in Figure 4.4, if a person lifts a bucket of water, then work is done on the bucket of water. The force is applied vertically and the bucket moves vertically, so the work done to lift a bucket of water is the force required to lift it, its weight, times the distance it is lifted. But a person carrying a bucket of water down the road is not doing any work on the bucket. This is because the force on the bucket to hold it up is vertical, but the distance W = Fd F the bucket is moving is horizontal. These F two forces do not point in the same did W=0 rection. In fact, they are at right angles to d one another and no work is done on the bucket of water. Figure 4.4. In raising the bucket (left) work is done on the The second detail is foreshadowed bucket. In moving the bucket horizontally, no work is done in the previous paragraph. People often on the bucket. say that the force required to lift a bucket is just a little larger than the weight, but this is not correct. If the upward force is at all greater than the weight, then we have a net upward force on the bucket. Recall from Chapter 3 that Newton’s second law of motion says that a net force does not just raise the bucket, it accelerates the bucket, giving it kinetic energy. But for a bucket to move up with a constant speed requires no net force at all, according to the first law of motion. So after a little bump of force to get the bucket moving (which does briefly require a larger force and some energy), the bucket can be lifted to any height with a force equal to the bucket’s weight. In physics problems of this kind, we normally just neglect the little bump of force necessary to get the bucket started and assume that the force required to lift the bucket is equal to the weight of the bucket. So, a handy problem solving tip to keep in mind is this: The force required to lift an object is equal to its weight. Recall that you can always calculate the weight of an object from its mass as Fw = mg

Do You Know ...

What is alpha radiation?

Nuclear radiation is emitted from radioactive substances during the process of nuclear decay. One form of radiation, called alpha radiation, consists of alpha particles streaming out of the radioactive substance. These incredibly small and fast moving alpha particles each consist of two protons and two neutrons. They receive their kinetic energy from radioactive atoms as matter in the atom converts into kinetic energy during the process of nuclear decay. In Chapter 6, we encounter the story of Ernest Rutherford, who used the alpha particles from a compound called radium bromide to explore the structure of the atom. The alpha particles emitted from radium bromide are travelling at 15,000,000 m/s! This is 9,300 miles per second, a speed that is 5% the speed of light! 97

Chapter 4

Example 4.5 An elevator in a skyscraper has a mass of 904.9 kg. Inside the elevator are three people whose masses are 67.8 kg, 55.9 kg, and 75.1 kg. Determine how much work the elevator motor does in lifting this elevator and the people inside it from the ground floor up to the 47th floor, 564 ft above the ground floor. Assume there is no friction and state the result in kJ. Write the givens and do the unit conversions. m = 904.9 kg + 67.8 kg + 59.9 kg + 75.1 kg = 1103.7 kg 0.3048 m d = 564 ft ⋅ = 171.9 m ft W =? As I write just above, the force required to lift an object is equal to its weight. So next we need to compute the weight of the elevator and the people. Fw = mg = 1103.7 kg ⋅9.80

m = 10,820 N s2

My calculator has a lot more digits in it than this, but I see that several of the pieces of given information have three significant digits, and I only need one extra digit for intermediate calculations, so I round to four digits. Now complete the problem. W = Fd = Fw d = 10,820 N⋅171.9 m = 1,860,000 J This result is rounded to three significant digits. As a last step, we convert this value to kilojoules, as required by the problem statement. W = 1,860,000 J⋅

1 kJ = 1860 kJ 1000 J

4.3.3 Applying Conservation of Energy When an object is thrown or fired straight up from the ground, it leaves the ground with a certain velocity, and thus a certain amount of EK . As it goes up, what happens to this EK? It is converted to EG , of course, as the object goes higher and higher and goes slower and slower. At the top of its flight, all the energy the object has at the ground in EK has been converted into EG. We can use the law of conservation of energy, along with the equations for EG and EK, to determine how high the object goes. The same thing works in reverse. An object at a certain height has EG . If the object is then released, as it falls the EG is gradually and continuously converted into EK . Just before it hits the ground, all the EG it has at the top has been converted into EK . We can use the law of conservation of energy, along with the equations for EG and EK , to find out how fast the object is going just before it strikes the ground. 98

Energy

In all the problems we do in this course involving conservation of energy, we are ignoring friction. In reality, friction is always present in any so-called mechanical system, such as moving objects or machines. In Section 4.2.4, we considered the effect friction has on mechanical systems, but in our computations we ignore it. Many physical systems can be approximated pretty well even if friction is ignored. In the conservation of energy experiment (the Hot Wheels Experiment), friction is low enough that the experimental velocity you measure should agree fairly well with the prediction. Let us now look at a simple example of the conservation of energy in action. Figure 4.5 illustrates the application of conservation of energy to a person lifting a bucket and letting it drop. When a person lifts a bucket vertically, the person does work on the bucket. To compute this work, the force to lift the bucket is the weight of the bucket and the distance involved is the height it is lifted, so the work done on the bucket by the person is W = Fw h Since the weight, Fw , is equal to mg, this equation can be written as W = Fw h = mgh Energy is transferred from the person (the chemical potential energy in the person’s muscles) to the bucket, and the bucket now has gravitational potential energy equal to

v A bucket on the ground. EG = 0 because h = 0, and EK = 0 because v = 0.

A person doing work on the bucket. The total work done is the weight of the bucket (Fw = mg) times the distance lifted (height h) so W = mgh.

The bucket held at height h. Work done lifting the bucket is W = mgh. This is now the EG the bucket has, so EG = mgh. The bucket is at rest, so EK = 0.

The bucket is released. As the bucket falls, its total energy all the way down equals the original energy it had due to the work done on it, mgh. Some of this energy is EG at a certain height, and some is EK at a corresponding velocity.

Just before striking the ground, the height is essentially 0 and the bucket has velocity v. All the original energy is now EK , and EK = ½mv2.

Figure 4.5. Conservation of energy applied to a lifted and falling bucket.

Chapter 4

EG = mgh Right here we can see the conservation of energy at work. The work done by the person to lift the bucket is mgh. Where did that energy go? It went into the EG the bucket has at the top, which is mgh, the same amount of energy. If the person releases the bucket, then as the bucket falls the gravitational potential energy begins to convert to kinetic energy. At any point as the bucket is falling, energy is conserved, which means the total energy the bucket has is still the same as the energy it has at the top, but some of the energy is in kinetic energy and some of it is in gravitational potential energy. At the instant before the bucket hits the ground, there is no more gravitational potential energy because the height then is zero, so all the energy originally given to the bucket by the work done on it is in the kinetic energy of the bucket.

4.3.4 Conservation of Energy Problems Now we look at a couple of example problems with objects falling down or flying up. In these kinds of problems, we use the basic principle of conservation of energy to find out how high an object goes or how fast an object is going just before it lands. A helpful problem solving technique for these kinds of problems is to draw a little diagram for yourself to indicate whether the EG is converting to EK ( EG → EK ), or vice versa ( EK → EG ). This helps you keep track of what you are doing so you don’t become confused. I demonstrate this in the example problems we do below. One more thing before we do those examples. To help you continue to remember how to calculate an object’s mass from its weight, I like to design these problems by giving you the weight instead of the mass that you need. So just first do a separate little problem to obtain the object’s mass. Then proceed with the energy calculations. The examples that follow make all these things clear. Example 4.6 A certain bucket of paint weighs 8.55 lb and is carried up a ladder until it is 4.750 ft above the ground. Sadly, the bucket then falls off the ladder. How fast is the bucket of paint moving just before it hits the ground and makes a colossal mess? To start, use the weight equation to obtain the mass of the bucket. As always, we first convert the given weight into MKS units. Fw = 8.55 lb⋅ m=? Fw = mg m=

4.45 N = 38.05 N lb

Fw 38.05 N = = 3.883 kg g 9.80 m 2 s

Notice that I do these calculations with four significant digits. This is because the value for g has three significant digits and my intermediate results always have an extra digit before I round off at the end. 100

Energy

Just as a reminder, the beauty of working in the MKS unit system is that when we use MKS units in a calculation, the result always has MKS units. So when I divide newtons (N) by meters per second squared (m/s2), I don’t have to worry about puzzling out any unit issues. I know this calculation gives me a mass, and the MKS unit for mass is the kilogram (kg). Now that we have the mass, let’s write down everything and begin the energy calculation. m = 3.883 kg 0.3048 m h = 4.750 ft ⋅ = 1.448 m ft v=? Now here is our energy diagram for this problem: EG → EK This tells me that in this problem, all the EG we have to begin with converts into EK as the bucket falls. So I need to calculate the EG first and then use this amount of energy as the EK for calculating the velocity. EG = mgh = 3.883 kg ⋅9.80

m ⋅1.448 m = 55.10 J s2

Since this gravitational potential energy converts to kinetic energy, we now have EK = 55.10 J Finally, we use this value in the velocity equation to obtain our final result. v=

2EK 2⋅55.10 J m = = 5.33 3.883 kg s m

This result is rounded to three significant figures as required.

Example 4.7 A baseball batter hits a baseball straight up with a velocity of 180 ft/s. A regulation baseball has a mass of 144.3 g. Ignoring air friction, how high does the baseball go before it comes to a stop? I work this out using the same method as in the previous problem. First, write down the givens and do the unit conversions.

101

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m = 144.3 g ⋅ v = 180 h=?

1 kg = 0.1443 kg 1000 g

ft 0.3048 m m ⋅ = 54.9 s 1 ft s

Now draw the energy diagram that indicates what is happening in this problem. The kinetic energy the ball has as it leaves the bat converts to gravitational potential energy as it rises, so EK → EG Since we are starting with kinetic energy this time, compute that first. 2

m⎞ ⎛ EK = 12 mv 2 = 0.5⋅0.1443 kg ⋅ ⎜ 54.9 2 ⎟ = 217 J ⎝ s ⎠ Since this energy is converting to gravitational potential energy, at the top of the ball’s flight we have EG = 217 J Now use this value to solve for the height. EG = mgh E 217 J h= G = = 153 m mg 0.1443 kg ⋅9.80 m s2 Finally, recall that the given velocity has only two significant digits, so we must round this result to two digits, giving h = 150 m

4.3.5 Energy in the Pendulum A swinging pendulum provides us with one final example of the conservation of energy in action. To begin, note that because of friction between the swinging pendulum and the air and friction in the pivot at the top, any actual pendulum loses energy to the environment as heat. This is why any actual free-swinging pendulum always comes to a stop. But let’s imagine a perfect pendulum, one that loses no energy due to friction. We call this an ideal pendulum. In an ideal pendulum, no energy leaves the “system” (the swinging pendulum) as heat and the pendulum just keeps on swinging without slowing down. (Actually, it’s a bit more complicated because of the rotation of the earth, so even in a vacuum with a magnetic bearing at the pivot the pendulum still slows down, so don’t start getting visions of a perpetual motion machine! But our imaginary ideal pendulum is also free from such influences.) 102

Energy

From what you know about the forms of energy and energy conservation, you can probably already see how energy transformation will work in this ideal pendulum. As shown in Figure 4.6, we let the height of the pendulum when it is at rest (that is, not swinging) be our reference for height measurements. This means when the pendulum is straight down its height is zero and its gravitational potential energy is also zero. When the pendulum swings up to its highest point, it momentarily comes to rest. At this moment, its velocity is zero and so its kinetic energy is zero. Put these facts together and let the pendulum start swinging. Because of the conservation of energy, the pendulum always has the same total amount of energy, no matter where it is. When someone lifts the pendulum to get it started, the person does work on the pendulum equal to the force it takes to lift it times the height (just as we h = 0.400 m saw with the bucket example). v=0 Since the pendulum is ideal, h = 0.200 m no energy leaves the system as v = 1.98 m/s it swings. This means that no h=0 matter where the pendulum is, v = 2.80 m/s the total energy in the system is always the same and is equal to Figure 4.6. Conservation of energy in a swinging pendulum. the amount of energy put into the system in the first place by the work done on it to get it up to where it is released. As the pendulum swings down, EG converts into EK , and as the pendulum swings up, EK converts back into EG. At all times, the total energy the pendulum possesses always equals the sum of the EG and the EK , and this sum always adds up to the same value no matter where the swinging pendulum is. Just to run through a quick calculation, let’s say the mass at the end of the pendulum is 2.00 kg and we lift it up 0.400 m above its lowest point to release it. The total EG at this starting point is mgh, which gives EG = 7.84 J. The kinetic energy here is zero, so now we know that the pendulum has a total of 7.84 J of energy no matter where it is. How fast is it going when it is halfway down, 0.200 m high? Well, the EG at that position is 3.92 J, so the EK is 7.84 J – 3.92 J = 3.92 J. Using this kinetic energy value in the velocity equation gives a velocity of 1.98 m/s. At the bottom, all the energy (7.84 J) is kinetic energy, so the velocity is 2.80 m/s. As you see, if you know how high the pendulum is at any point, you can determine how fast it is moving, and vice versa.

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Chapter 4 Exercises Energy Study Questions 1.

Write out the stages in the “energy trail” for the following sequences. For each stage, list where the energy is and what form it is in. a. From the sun to a ball thrown straight up at the top of its flight. b. From the sun to a galloping horse. c. From the sun to water stored in a water tower, pumped up there by the city’s electric pumps. d. From the sun to a moving motorcycle. e. From the sun to an electric blow dryer for drying hair. f. From the sun to a diesel truck parked at the top of a steep hill.

Write down the law of conservation of energy from memory. Then write a paragraph explaining the law in your own words. Include some examples in your explanation.

From an energy standpoint, what is an “ideal system”? (Think of the ideal pendulum discussed at the end of the chapter.)

If you run out of gas, your car soon comes to a stop. What happens to all the kinetic energy the car has before running out of gas? Where does it go?

Classroom Energy Computation Examples These examples are written here so that as your teacher works examples in class you can focus on the solutions rather than on worrying about getting the problems written down. 1.

Water is pumped into a high water tower. If the total mass of the water is 1.00 × 105 kg and the tower is 240 feet high, what is the gravitational potential energy (EG) of the water in the tower?

A bullet of mass 25 g is fired at a velocity of 556 ft/s. How much kinetic energy (EK) does the bullet have?

A man lifts a bucket of sand 75 cm above the ground. If the bucket of sand has a mass of 12,500 g, how much work does the man do on the bucket?

Referring again to the previous problem, after the bucket of sand is lifted, how much EG does the sand have?

If the man releases the bucket of sand and lets it drop, what is its velocity the instant before it strikes the ground?

A boy carries a water balloon up to the top of a ladder to let it drop. The mass of the water balloon is 255.8 g and the ladder is 10.4 ft tall. a. How much EG does the balloon have at the top of the ladder? b. If the balloon is released, how fast is it going just before it splats on the ground?

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Answers 1. 2. 3. 4. 5. 6.

72,000,000 J 360 J 92 J 92 J 3.8 m/s a) 7.95 J; b) 7.88 m/s

Energy Calculations Set 1 1.

A load of building materials is hoisted to the top of a building under construction. If the total mass of the material is 1.31 × 103 kg and the building is 177.44 feet high, what is the gravitational potential energy (EG) of the material at the top of the building?

A car of mass 2,345 kg is traveling at a speed of 31 mph. How much kinetic energy (EK) does the car have?

A woman lifts a bucket of water 61.7 cm above the ground. If the bucket of water has a mass of 17.5 kg, how much work does the woman do?

How much EG does the woman’s bucket have after being lifted?

If the woman releases the bucket of water and lets it drop, what is its velocity the instant before it strikes the ground?

A kid shoots an arrow straight up. The arrow has a mass of 122 g and leaves the bow with a velocity of 13.75 m/s. How high does it go above the point where it is shot from the bow?

A girl drops a stone into the water from a bridge. The stone has a mass of 325 g and the bridge is 36.1 m above the water. How fast is the stone moving just before it hits the water?

A worker slides a carton across the floor. The force of friction between the carton and the floor is 735 N. If the worker pushes the carton 26 m, how much work does he do?

Answers 1. 2. 3. 4. 5. 6. 7. 8.

694,000 J 230,000 J 106 J (This answer is top secret!) 3.48 m/s 9.65 m 26.6 m/s 19,000 J

Energy Questions Set 2 1.

A carpenter hauls 20 bundles of shingles up onto a roof, each bundle weighing 80.0 lb. The roof is 8.5 m above the ground. 105

Chapter 4

a. b. c. d.

e. 2.

Compute the total mass of the shingles using the conversion factor for pounds to newtons found in Appendix A, and then the weight equation to get the mass. Compute the work the carpenter has to do to get the shingles up onto the roof. State your answer in joules (J). Using the equation for gravitational potential energy (EG), compute the EG of the shingles while they are on the roof. If the entire stack of shingles slides off the roof, how much kinetic energy does the stack have at the following times: i. At the moment it first begins to slide. ii. Just before it hits the ground. Compute how fast the stack is falling just before it hits the ground.

A car weighing 3,193 lb rests at the top of a hill, then begins to roll down the hill. (The engine is off.) Assume it rolls to the bottom of the hill with negligible friction. a. If the hill is 16 m high compared to the flat road at the bottom, compute the EG of the car while it is at the top of the hill. b. After the car rolls to the bottom of the hill, where is the energy that is in the EG of the car while it is at the top of the hill? c. Compute the EK of the car when it reaches the bottom of the hill. d. Compute the velocity of the car when it reaches the bottom of the hill. e. Explain how conservation of energy relates to this problem. f. Explain specifically how friction would change the results of the problem in a more realistic example.

Answers 1.

a) 727 kg; b) 6.1 × 104 J; c) 6.1 × 104 J; d) 0 J, 6.1 × 104 J; e) 13 m/s

a) 230,000 J; c) 230,000 J; d) 18 m/s

Energy Questions Set 3 1.

Consider a large, ideal (that is, frictionless) pendulum with a steel ball weighing 27.05 lb on the end. Keeping the pendulum cable tight, this ball is lifted to a height of 185 cm and released. Since the pendulum is frictionless, the ball swings back and forth forever. a. b. c. d. e. f.

106

How much work is done to lift the ball? What is the EG of the ball after it is lifted, but before it starts falling? At the bottom of the ball’s pathway, what is its EG? At the bottom of the ball’s pathway, what is its EK? How fast is the ball going at the bottom? Use friction and energy considerations to explain the difference between this ideal pendulum and an actual pendulum. Which one has the highest velocity at the bottom of its swing?

A group of city water pumps pushes water up into a water tower 197 feet high. The water tower holds 6.016 × 106 kg of water. Determine the amount of mechanical work the pumps must do to fill the water tower and state your answer in GJ.

Energy

Imagine a new frictionless roller coaster that uses magnetic levitation so that the cars float above the rails without actually touching them. Imagine also that the aerodynamic design of the cars is so brilliant that there is essentially no air friction. The car has a mass of 5,122 kg. From the top of a 25.0 m-hill, the car rolls down to a valley where the track is right on the ground. Assuming the roller coaster begins at rest at the top of the hill, determine how fast it is traveling when it reaches the bottom of the valley.

A boy weighing 104.6 lb runs up the steps two at a time. There are 13 steps, each one 16.5 cm high. a. How much work does he do to get to the top of the steps? b. If he steps over the hand rail and drops back down to the ground, how fast is he moving just before he lands?

An object with mass m = 351 g is sliding on a frictionless surface at 500.00 cm/s when it begins going up a ramp. What is its height, h, when it stops?

Answers 1. 2. 3. 4. 5.

a) 223 J; b) 223 J; c) 0 J; d) 223 J; e) 6.02 m/s 3.54 GJ 22.1 m/s a) 998 J; b) 6.48 m/s 1.28 m

Do You Know ...

Why are there pendulums in clocks?

Galileo first discovered that the period of a swinging pendulum depends only on its length (a fact you confirm in the Pendulum Experiment). Because of this, pendulums are used to regulate the motion of the mechanical systems in clocks. The weight on the end of the pendulum is supported by a nut on a threaded rod, and the vertical position of the nut is adjusted to give the pendulum the precise length needed for the clock to run at the correct speed. The pendulum in a grandfather clock is kept in motion by the gravitational potential energy in the weights hanging inside the cabinet. As the weights slowly descend, their gravitational potential energy is transferred to the energy in the swinging pendulum. The gravitational potential energy in the weights is replenished when a person does work on them, raising them back to their highest position to begin descending again. Because of friction, the pendulum would eventually stop swinging without receiving energy from the continuous action of the descending weights. As the pendulum swings, it continuously converts its energy from kinetic energy to gravitational potential energy and back again. 107

APPENDIX C

Laboratory Experiments

C.1

Important Notes

Important Note: Please refer to pages xii–xiii in the Preface for Teachers for important information pertaining to the terms “experimental error” and “percent difference” as used in this text. The following pages contain your guidelines for the five laboratory experiments you will conduct in Introductory Principles in Physics during the year. For each of these experiments, you will submit an individual written report. It is your responsibility to study The Student Lab Report Handbook thoroughly so that you can meet the expectations for lab reports in this course. The instructions written here are given to help you complete your experiment successfully. However, your report must be written in your own words. This applies to all sections of the report. Do not copy the descriptions in this appendix into your report in place of writing your report for yourself in your own words.

C.2

Lab Journals

The pages in the journal are quadrille ruled (graph paper) and the journal entries are in ink.

The journal is neatly maintained and free of sloppy marks, doodling, and messiness.

Each entry includes the date and the names of the team members present.

Every experiment and every demonstration that involves taking data or making observations is documented in the journal.

Entries for each experiment or demonstration include: •

the date 345

Appendix C

C.3

•

the team members’ names

•

the team’s hypothesis

•

an accurate list of materials and equipment, including make and model of any electronic equipment or test equipment used

•

tables documenting all the data taken during the experiment, including the units of measure and identifying labels for all data

•

all support calculations used during the experiment or in preparation of the lab report

•

special notes documenting any unusual events or circumstances, such as bad data that require doing any part of the experiment over, unexpected occurrences or failures, or changes to your experimental approach

•

little details about the experiment that need to be written in the report that you may forget about later

•

important observations or discoveries made during the experiment.

Experiments

Experiment 1 The Pendulum Experiment Variables and experimental methods Essential equipment: •

string

•

meter stick

•

paper clip

•

large steel washers

•

clock with second hand

This investigation involves a simple pendulum. The experiment is an opportunity for you to learn about conducting an effective experiment. In this investigation, you learn about controlling variables, collecting careful data, and organizing data in tables in your lab journal. To make your pendulum, bend a large paper clip into a hook. Then connect the hook to a string, and connect the string to the end of a meter stick. Then lay the meter stick on a table with the pendulum hanging over the edge and tape the meter stick down. Finally, hang one or more large metal washers on the hook for the weight. Your goal in this experiment is to identify the explanatory variables that affect the period of a simple pendulum. A pendulum is an example of a mechanical system that oscillates, that is, repeatedly “goes back and forth” in some regular fashion. In the study of any oscillating system, an important parameter is the period of the oscillation. The period is the length of time (in seconds) required for the system to complete one full cycle of its oscillation. In this experiment, the period of the pendulum is the response variable you monitor. (Actually, for convenience you monitor a slightly different variable, closely related to the period. This is explained on the next page.) After thinking about the possibilities and forming your team hypothesis, construct your own simple pendulum from string and some 346

Laboratory Experiments

weights and conduct tests on it to determine which variables affect its period and which variables do not. In class, explore the possibilities for variables that may affect the pendulum’s period. Within the pendulum system itself there are three candidates, and your instructor will lead the discussion until the class has identified them. (We ignore factors such as air friction and the earth’s rotation in this experiment. Just stick to the obvious variables that clearly apply to the problem at hand.) Then, as a team, continue the work by discussing the problem for a few minutes with your teammates. In this team discussion, form your own team hypothesis stating which variables you think affect the period. To form this hypothesis, you need not actually do any new research or tests. Just use what you know from your own experience to make your best guess. The central challenge for this experiment is to devise an experimental method that tests only one explanatory variable at a time. Your instructor will help you work this out, but the basic idea is to set up the pendulum so that two variables are held constant while you test the system with large and small values of the third variable to see if this change affects the period. You must test all combinations of holding two variables constant while manipulating the third one. All experimental results must be entered in tables in your lab journal. Recording the data for the different trials requires several separate tables. For each experimental setup, time the pendulum during three separate trials and record the results in your lab journal. Repeating the trials this way enables you to verify that you have valid, consistent data. To make sure you can tell definitively that a given variable is affecting the period, make the large value of the variable at least three times the small value in your trials. Here is bit of advice about how to measure the period of your pendulum. The period of your pendulum is likely to be quite short, only one or two seconds, so measuring it directly with accuracy is difficult. Here is an easy solution: assign one team member to hold the pendulum and release it on a signal. Assign another team member to count the number of swings the pendulum completes, and another member as a timer to watch the second hand on a clock. When the timer announces “GO” the person holding the pendulum releases it, and the swing counter starts counting. After exactly 10.0 seconds, the timer announces “STOP” and the swing counter states the number of swings completed by the pendulum during the trial. Record this value in a table in your lab journal. If you have four team members, the fourth person can be responsible for recording the data during the experiment. After the experiment, the data recorder reads off the data to the other team members as they enter the data in their journals. This method of counting the number of swings in 10 seconds does not give a direct measurement of the period, but you can see that your swing count works just as well for solving the problem posed by this experiment, and is a lot easier to measure than the period itself. (The actual period is equal to 10 seconds divided by the number of swings that occur in 10 seconds.) 347

Appendix C

One more thing on measuring your swing count: your swing counter should state the number of swings completed to the nearest 1/4 swing. When the pendulum is straight down, it has either completed 1/4 swing or 3/4 swing. When it stops to reverse course on the side opposite from where it is released, it has completed 1/2 swing. When you have finished taking data, review the data together as a team. If you did the experiment carefully, your data should clearly indicate which potential explanatory variables affect the period of the pendulum and which ones do not. If your swing counts for different trials of the same setup are not consistent, then something is wrong with your method. Your team must repeat the experiment with greater care so that your swing counts for each different experimental setup are consistent. Discuss your results with your team members and reach a consensus about the meaning of your data. Expect to spend at least four hours writing, editing, and formatting your report. Lab reports count a significant percentage of your science course grades throughout high school, so you should invest the time now to learn how to prepare a quality report. Your goal for this report is to begin learning how to write lab reports that meet all the requirements described in The Student Lab Report Handbook. One of our major goals for this year is to learn what these requirements are and become proficient at generating solid reports. Nearly all scientific reports involve reporting data, and a key part of this first report is your data tables, which should all be properly labeled and titled. After completing the experiment, all the information you need to write the report should be in your lab journal. If you properly journal the lab exercise, you will have all the data, your hypothesis, the materials list, your team members’ names, the procedural details, and everything else you need to write the report. Your report must be typed and will probably be around two or three pages long. You should format the report as shown in the examples in The Student Lab Report Handbook, including major section headings and section content. Here are a few guidelines to help you get started with your report: 1.

There is only a small bit of theory to cover in the Background section, namely, to describe what a pendulum and its period are. You should also explain the experimental method, that is, why we are using the number of swings completed in 10 seconds in our work in place of the actual period. As stated in The Student Lab Report Handbook, the Background section must include a brief overview of your experimental method and your team’s hypothesis.

Begin your Discussion section by describing your data and considering how they relate to your hypothesis. In this experiment, we are not making quantitative predictions, so there are no calculations to perform for the discussion. We are simply seeking to discover which variables affect the period of a pendulum and which do not. Your goal in the Discussion section is to identify what your data say and relate that to your reader.

Consider the following questions as you write your discussion. What variables did you manipulate to determine whether they had any effect on the period of the pendulum? What did you find? According to your data, which variables do affect the period? How do the data show this? Refer to specific data tables to explain specifically how the data support your conclusion. Are your findings consistent with your hypothesis? If not, then what conclusion do you reach about the question this experiment seeks to answer?

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Experiment 2 The Soul of Motion Experiment Newton’s second law of motion Essential equipment: •

vehicle

•

duct tape

•

stop watch

•

bathroom weigh scale (2)

Note: The report for this experiment requires you to set up a graph showing predicted and experimental curves on the same set of axes. Procedures for creating such a graph on a PC or Mac are described in detail in The Student Lab Report Handbook. You will have a great time with this experiment. You meet out in the parking lot as a class. The idea is to push a vehicle from the rear, using scales that measure the force the pushers are applying to the vehicle. You time the vehicle as it accelerates from rest through a ten-meter timing zone and use the time data to calculate the experimental values of the vehicle’s acceleration. Using the mass of the vehicle and Newton’s second law, you calculate a predicted acceleration for each amount of pushing force used. Your goal is to compare your predicted accelerations to the experimental values of acceleration for four different force values. You then graph the results and calculate the percent difference to help you see how they compare. This experiment is an excellent example of how experiments in physics actually work. The scientists have a theory that enables them to predict, in quantitative terms, the outcome of an experiment. Then the scientists carefully design the experiment to measure the values of these variables and compare them to the predictions, seeking to account for all factors that affect the results. If the theory is sound and the experiment is well done, the results should agree well with the theoretical predictions and the percent difference should be low. In our case, when a force is applied to a vehicle at rest, we expect the vehicle to accelerate in accordance with Newton’s second law of motion: a=

F m

Appendix C

you must use the actual net force on the vehicle in your predictions. Details are discussed below. For the data collection, you must have a way to measure the actual vehicle’s acceleration so that you can compare it to your predictions. You already know an equation that gives the acceleration based on velocities and time. However, you have no convenient way of measuring the vehicle’s velocity. (The vehicle moves too slowly for the speedometer to be of any use.) Fortunately, there is another equation you can use if you time the vehicle with a stop watch as it starts from rest and moves through a known distance. If you know the distance and the time, and the acceleration is uniform, you can calculate the vehicle’s acceleration as follows: a=

2d t2

Always have two students pushing on the vehicle. Thus, for each force value the pushers use, the total applied force is twice that amount. (You use four different force values in the experiment.)

Measure the friction on the vehicle so you can subtract it from the force the pushers are applying to get the net force applied for your predictions. To measure the friction, use one pusher and estimate the absolute minimum amount of force needed to keep the vehicle barely moving at a constant speed. As you know from our studies of the laws of motion, vehicles move at a constant speed when there is no net force. So if the vehicle is moving at a constant speed, it means that the friction and the applied force are exactly balanced. This allows you to infer what the friction force is.

Use four different values of pushing force. For each force value, time the vehicle over the ten-meter timing zone at least three times. The forces the pushers apply to the vehicle always vary quite a bit, so if you get three valid trials at each force you have three reliable data points for the time. You then calculate the average of these times and use it to calculate the experimental value of the acceleration of the vehicle for that force.

The major factor introducing error into this experiment is the forces applied by the pushers. Pushing at a constant force while the vehicle is accelerating is basically impossible. (The dial on the force scale jumps all over the place.) But if the pushers are careful, they can push with an average force that is pretty consistent. You need a standard to judge whether you have had a successful run with consistent pushing. Here is the criterion to use: when you obtain three trials with time measurements all within a range of one second from highest to lowest, accept those values as valid. If your times are not this close together, assume that the pushing forces are not consistent enough and keep running new trials until you get more consistent data.

The instructor will take the vehicle, with a full gas tank, to get it weighed and report this weight to the class. Make sure to measure the weight of the driver and the weight of the scale support rack (if there is one). Add these weights to the weight of the vehicle and determine the mass for this total weight. (Of course, the instructor must also make

350

Laboratory Experiments

Variable

Equation

Comments

force

net force = (2 × force for each pusher) – friction force estimate

There are four values of net force, one for each set of trials.

predicted acceleration

predicted accel = (net force)/(total mass)

Net force is as calculated above. Mass is determined from the total weight. There is a predicted acceleration for each value of net force.

experimental experimental accel = acceleration (2 × distance)/(avg time)2

Distance is the length of the timing zone. Average time is the average of the three valid times for a given trial. There is an experimental acceleration for each value of net force.

Table C.1. Summary of equations for the calculations.

351

Appendix C

For your predicted values of acceleration, use the total mass of the vehicle, driver, and support rack. The instructor will tell you the weight of the vehicle, which you record in your lab journal. Also record the weights of the driver and support rack determined during the experiment. Convert the total weight from pounds to newtons, then determine the mass in kilograms by using the weight equation, Fw = mg. For the force values in your predictions, use the nominal amount of force applied (the two pushers’ forces combined) less the amount of force necessary to overcome the friction (which is determined during the experiment). Table C.1 summarizes the calculations you need to perform for each set of trials. The heart of your discussion is a comparison of the two curves representing acceleration vs. force (displayed on the same graph), and a discussion of how well the actual values of acceleration match up with the predicted values. In addition to this graphical comparison, compare the four predictions to the four experimental acceleration values by calculating the percent difference for each one, presenting these values in a table and discussing them. To compare the curves, think about the questions below. Do not write your discussion section by simply going down this list and answering each question. (Please spare your instructor the pain of reading such a report!) Instead, use the questions as a guide to the kinds of things you should discuss and then write your own discussion section in your own words. Remember—this is an exercise in learning how to write a well-constructed lab report, not a boring fill-in-the-blank activity. Thought Questions and Considerations for Discussion 1.

Are both the curves linear? What does that mean?

Do they both look like direct proportions? What does that imply?

Do the curves have similar slopes? What does that imply?

How successful are the results? A percent difference of less than 5% for an experiment as crude as this can be considered a definite success. If the difference is greater than 5%, identify and discuss the factors that may have contributed to the difference between prediction and result. In this experiment, there are several such factors, including wind that may have been blowing on the vehicle.

Do not make the mistake of merely assuming that the fluctuation in the pushers’ forces explains everything without taking into account the precautions you took to eliminate this factor from being a problem (the time data validity requirement).

Also do not make the mistake of assuming that friction explains the difference between prediction and result. Friction can only affect the data one way (slowing the vehicle down). So if friction is a factor, the data have to make sense in light of how friction affects the data. But further, since measuring friction and taking it into account in your predictions is part of your procedure, a generic appeal to friction will not do.

Finally, do not make the mistake of asserting that errors in the timing or the timing zone distance measurement explain the difference between prediction and result. Consider just how large the percentage error could realistically be in these measurements, and whether that kind of percentage helps at all in explaining the difference you have between prediction and result. For example, the timing zone is 10 m long. If it is carefully laid out on the pavement, it is unlikely that the distance measurement is in error

352

Laboratory Experiments

by more than a centimeter or so. Even including the slight misalignments of the vehicle that crop up, the distance could probably not be off by more than, say, 10 or 20 cm. But this is only 1–2% of 10 m, and if you are trying to explain a percent difference of 5–10% or more this won’t do it. Similar considerations apply to the time values. Given the slow speed the vehicle moves, how far off can the timing be? What kind of percentage error would this produce? Alternate Experimental Method If your class is using digital devices such as the PASCO Xplorer GLX to read forces, you can use a slightly different experimental method that improves results and lowers the difference between prediction and result. One of the major sources of error in this experiment is the difficulty the pushers have in accurately applying the correct amount of force to the vehicle. If you use bathroom scales to measure the force, there is nothing that can be done about this problem and the pushers simply have to do the best they can. However, with the digital devices you can eliminate the problem of force accuracy by using the actual average values of the forces applied by the two pushers to calculate the predicted values. The Xplorer GLX can record a data file of the applied force during a given trial, and when reviewing the data file back at your computer you can view the mean value of the force during the trial. You can use this mean value to calculate the predicted acceleration from Newton’s second law. Using this method to form your predictions eliminates much of the uncertainty surrounding the forces applied to the car. Here are a few details to consider if you use this alternative approach to collecting data: 1.

You do not need to select four different force values in advance and push the vehicle repeatedly at each force value. Instead, only a single trial is needed for each force.

Select 10 or 12 different target force values and run a single trial with each. The force targets should range from low values that barely get the vehicle to accelerate, all the way up to the highest values the pushers can deliver. For each trial, tell the pushers the target force and tell them to do their best to stay on it during the trial. But it doesn’t matter nearly so much how accurate the pushers are because you are using the average of the actual data from the digital file to make the predictions, rather than relying on the pushers to maintain the target force accurately.

The method for determining values of net force for the predictions is similar to that shown in Table C.1. The difference is that instead of doubling the target force for each pusher, you add together the actual mean forces obtained from the data files for each pusher and subtract out the friction force.

Use the time of each trial to determine the experimental value of the acceleration for that trial.

Calculate the percent difference for each trial and report these values in the report. Also calculate the average of the percent difference values and use this figure in your discussion of the results.

353

Accelerated Studies in Physics and Chemistry PUBLISHED BY CENTRIPETAL PRESS Grade Level: 9th Grade Yearlong Course ACCELERATED TRACK RECOMMENDATION*

*Review the Overview section on pg. 5 for more information

CLICK TO VIEW IN OUR ONLINE CATALOG

ACCELERATED STUDIES IN PHYSICS AND CHEMISTRY Accelerated Studies in Physics and Chemistry (ASPC) is an introductory physics and chemistry text designed for accelerated students in 9th grade. The text contains approximately the same physics content as our grade-level text for 9th grade (Introductory Principles in Physics) but moves along more briskly and contains more advanced mathematical content. The physics content is completed in late February of a typical school year, at which time the subject matter switches to chemistry. Here we highlight a few distinctive features of this text. •

ASPC supports our signature philosophy of science education based on wonder, integration, and mastery. We always seek to build on and stimulate the student’s innate sense of wonder at the marvels found in nature. Integration refers to the epistemology, mathematics, and history embedded in the text, and a curriculum designed to help develop the student’s facility with using language to communicate scientific concepts. ASPC is designed to be used in a mastery-learning environment, using the mastery teaching model John developed and describes in From Wonder to Mastery: A Transformative Model for Science Education. When teachers use this mastery-learning model, the result is high student achievement and exceptional long-term retention for all students.

•

The text includes six unique experiments. These experiments are a departure from the norm. Students make pendulums with washers and string; make velocity predictions with Hot WheelsTM cars going down ramps; push real cars in the parking lot to measure acceleration; measure the density of aluminum and PVC; build test circuits using precision resistors, a voltmeter, and an electronic breadboard; and measure the solubility of salt and sugar.

•

The mathematical content is grade-appropriate and thorough. We leave the vector analysis for a later course. The math level is designed for students simultaneously enrolled in geometry and focuses strongly on basic skills such as unit conversions, scientific notation, metric units and prefixes, and isolating variables. In addition to learning these essential basic skills, students will practice using significant digits. The result of the numerous practice problems, combined with the mastery-learning teaching model referred to above, is that every student completely masters the basic math skills—a real game-changer for students’ preparedness for future science studies.

ACCELERATED STUDIES IN PHYSICS AND CHEMISTRY Support Resources A number of support resources are available to accompany ASPC. They include: The Student Lab Report Handbook Students should begin writing their lab reports from scratch in 9th grade. This popular manual gives them everything they need. We recommend supplying this handbook to every freshman so they can refer to it throughout high school. Read more about this resource on pg. 6. Solutions Manual to Accompany Accelerated Studies in Physics and Chemistry This book contains complete written solutions for all the computations in the chapter exercises. Favorite Experiments for Physics and Physical Science This reference book contains complete background information, procedures, and material lists for the experiments and demonstrations used in all three of our high school physics courses. Experiments for Introductory Physics and ASPC This reference book contains the material for the experiments in ASPC, taken from Favorite Experiments listed above. Digital Resources The following materials are accessible exclusively through Centripetal Press: • a full year of weekly, cumulative quizzes • two semester exams • a document containing answer keys for all the quizzes and tests and sample answers for all the verbal questions in the text • a full year of Weekly Review Guides • a document with recommendations for teaching the course • a lesson list and example calendar • a sample graded lab report Tips and Tools There are a variety of tips and tools available at www.centripetalpress.com. Here, teachers will find scores of images and short videos organized chapter by chapter. Teachers are free to use any of this content as they wish to assemble vivid and fascinating lessons.

ACCELERATED STUDIES IN PHYSICS AND CHEMISTRY Contact Us Our team is here to assist as you choose curricula that will be the best fit for your classroom. We encourage you to take advantage of the following resources: Questions & Guidance We look forward to answering any questions you might have regarding our curriculum, placement, or pricing. Please contact us! Email: info@classicalsubjects.com Phone: (717) 730-0711 Quotes and Vendor Applications We are able to provide formal quotes for all of our materials, and can also submit applications required to become an approved vendor for your school district. Inquire at info@classicalsubjects.com. Get the Latest Information from Centripetal Press and Classical Academic Press For more information on Centripetal Press products, visit www.centripetalpress.com. Sample Chapters The following pages contain samples from the text. The Table of Contents is shown, as well as Chapters 1, 2, and 4, and the first two experiments.

Accelerated Studies in Physics and Chemistry A Mastery-Oriented Curriculum

John D. Mays

Third Edition

CENTRIPETAL PRESS

Austin, Texas 2018

© 2012, 2015, 2018 Novare Science & Math LLC All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by information storage and retrieval systems, without the written permission of the publisher, except by a reviewer who may quote brief passages in a review. All images attributed to others under any of the Wikimedia Commons licenses, such as CC-BY-SA 3.0 and others, may be freely reproduced and distributed under the terms of those licenses. Published by Centripetal Press centripetalpress.com

CENTRIPETAL PRESS

Printed in the United States of America. Centripetal Press is an imprint of Novare Science & Math LLC. ISBN: 978-0-9989833-6-3 For a catalog of titles published by Centripetal Press, visit centripetalpress.com. Cover design by Scarlett Rugers, scarlettrugers.com.

Preface for Teachers

xiv

Preface for Students

xxii xxiv

A Solid Study Strategy

Chapter 1 The Nature of Scientific Knowledge 1.1

Modeling Knowledge 1.1.1 Kinds of Knowledge 1.1.2 What is Truth and How Do We Know It? 1.1.3 Propositions and Truth Claims 1.1.4 Truth and Scientific Claims 1.1.5 Truth vs. Facts Do you know ... Dark matter 1.1.6 Relating Scientific Knowledge and Truth 1.2 The Cycle of Scientific Enterprise 1.2.1 Science 1.2.2 Theories 1.2.3 Hypotheses 1.2.4 Experiments 1.2.5 Analysis 1.2.6 Review 1.3 The Scientific Method 1.3.1 Conducting Reliable Experiments 1.3.2 Experimental Variables 1.3.3 Experimental Controls Do you know ... Double-blind experiments Chapter 1 Exercises Do you know ... Hero Sir Humphry Davy

Chapter 2 Motion 2.1

Computations in Physics 2.1.1 The Metric System 2.1.2 MKS Units Do you know ... Defining base units 2.1.3 Dimensional Analysis 2.1.4 Accuracy and Precision 2.1.5 Significant Digits 2.1.6 Scientific Notation 2.1.7 Problem Solving Methods 2.2 Motion 2.2.1 Velocity Universal Problem Solving Method 2.2.2 Acceleration 2.2.3 Graphical Analysis of Motion 2.3 Planetary Motion and the Copernican Revolution 2.3.1 Science History and the Science of Motion 2.3.2 Aristotle

2 3 3 4 5 7 7 8 9 9 9 10 12 13 13 14 14 14 15 16 17 18 19 20 21 22 23 23 24 24 25 29 30 30 31 32 35 38 41 41 42 vii

Contents

2.3.3 Ptolemy 2.3.4 The Ptolemaic Model 2.3.5 The Ancient Understanding of the Heavens 2.3.6 The Ptolemaic Model and Theology 2.3.7 Copernicus and Tycho 2.3.8 Kepler and the Laws of Planetary Motion 2.3.9 Galileo Do you know ... The first monster telescope 2.3.10 Newton, Einstein, and Gravitational Theory Chapter 2 Exercises

Chapter 3 Newton’s Laws of Motion 3.1 3.2

Matter, Inertia, and Mass Newton’s Laws of Motion 3.2.1 The Three Laws of Motion 3.2.2 Actions and Reactions 3.2.3 Showing Units of Measure in Computations 3.2.4 Weight 3.2.5 Applying Newton’s Laws of Motion Thinking About Newton’s Laws of Motion 3.2.6 How a Rocket Works 3.3 Newton’s Laws of Motion and Momentum Do you know ... Gravity and elliptical orbits Chapter 3 Exercises Do you know ... Isaac Newton’s tomb

Chapter 4 Variation and Proportion 4.1 4.2

The Language of Nature The Mathematics of Variation 4.2.1 Independent and Dependent Variables 4.2.2 Common Types of Variation 4.2.3 Normalizing Equations Chapter 4 Exercises

Chapter 5 Energy 5.1

What is Energy? 5.1.1 Defining Energy 5.1.2 The Law of Conservation of Energy 5.1.3 Mass-Energy Equivalence 5.2 Energy Transformations 5.2.1 Forms of Energy 5.2.2 Energy Transfer 5.2.3 The “Energy Trail” Do you know ... Caloric theory 5.2.4 The Effect of Friction on a Mechanical System 5.2.5 Energy “Losses” and Efficiency

viii

43 43 45 47 49 51 53 54 56 58 64 65 66 66 71 72 73 74 76 77 78 79 80 83 84 85 86 86 86 88 90 102 103 103 104 104 104 104 105 106 108 109 110

Contents

5.3

Calculations With Energy 5.3.1 Gravitational Potential Energy and Kinetic Energy 5.3.2 Work Do you know ... Nuclear energy calculations (just for fun!) 5.3.3 Applying Conservation of Energy 5.3.4 Conservation of Energy Problems Do you know ... The expanding universe and dark energy 5.3.5 Energy in the Pendulum Chapter 5 Exercises

Chapter 6 Heat and Temperature 6.1

Measuring Temperature 6.1.1 Temperature Scales 6.1.2 Temperature Unit Conversions 6.2 Energy In Substances 6.2.1 How Atoms Possess Energy Do you know ... The velocity of air molecules 6.2.2 Internal Energy, Thermal Energy, and Temperature 6.2.3 Absolute Zero 6.2.4 Thermal Equilibrium 6.3 Heat Transfer Processes 6.3.1 Heat Conduction In Nonmetal Solids 6.3.2 Heat Conduction in Metals 6.3.3 Convection 6.3.4 Radiation 6.4 The Kinetic Theory of Gases 6.5 Thermal Properties of Substances 6.5.1 Specific Heat Capacity 6.5.2 Thermal Conductivity 6.5.3 Heat Capacity vs. Thermal Conductivity Chapter 6 Exercises Do you know ... The temperature in outer space

Chapter 7 Waves, Sound, and Light 7.1

Modeling Waves 7.1.1 Describing Waves 7.1.2 Categorizing Waves 7.1.3 Modeling Waves Mathematically 7.2 Wave Interactions 7.2.1 Reflection 7.2.2 Refraction 7.2.3 Dispersion 7.2.4 Diffraction 7.2.5 Resonance 7.2.6 Interference Do you know ... Resonant frequencies in skyscrapers 7.3 Sound Waves

111 111 114 116 118 119 123 124 125 130 131 131 132 133 133 134 134 135 135 135 135 136 138 138 140 141 141 141 142 144 147 148 149 150 150 152 155 155 155 156 156 157 160 161 163 ix

Contents

7.3.1 Pressure Variations in Air 7.3.2 Frequencies of Sound Waves 7.3.3 Loudness of Sound 7.3.4 Connections Between Scientific and Musical Terms Do you know ... Sonic booms 7.4 The Electromagnetic Spectrum and Light Chapter 7 Exercises

Chapter 8 Electricity and DC Circuits 8.1

The Amazing History of Electricity 8.1.1 Greeks to Gilbert 8.1.2 18th-Century Discoveries Intriguing Similarities between Gravity and Electricity 8.1.3 19th-Century Breakthroughs 8.2 Charge and Static Electricity 8.2.1 Electric Charge 8.2.2 How Static Electricity Forms Do you know ... Plasmas Do you know ... The first color photograph 8.3 Electric Current 8.3.1 Flowing Charge 8.3.2 Why Electricity Flows So Easily in Metals 8.3.3 The Water Analogy 8.4 DC Circuit Basics 8.4.1 AC and DC Currents 8.4.2 DC Circuits and Schematic Diagrams 8.4.3 Two Secrets 8.4.4 Electrical Variables and Units 8.4.5 Ohm’s Law 8.4.6 What Exactly Are Resistors and Why Do We Have Them? 8.4.7 Through? Across? In? 8.4.8 Voltages Are Relative 8.4.9 Power in Electrical Circuits 8.4.10 Tips on Using Metric Prefixes in Circuit Problems 8.5 Multi-Resistor Circuits 8.5.1 Two-Resistor Networks 8.5.2 Equivalent Resistance 8.5.3 Significant Digits in Circuit Calculations 8.5.4 Larger Resistor Networks 8.6 Solving DC Circuits 8.6.1 Kirchhoff ’s Laws Do you know ... Rectifiers and inverters 8.6.2 Putting it All Together to Solve DC Circuits Do you know ... The war of currents Chapter 8 Exercises

163 163 165 165 166 166 169 172 173 174 174 175 176 178 178 180 181 184 184 184 184 185 187 187 187 189 190 191 193 194 194 195 197 198 198 201 203 203 207 207 210 210 217 218

Contents

Chapter 9 Fields and Magnetism 9.1 9.2

Three Types of Fields Laws of Magnetism 9.2.1 Ampère’s Law 9.2.2 Faraday’s Law of Magnetic Induction 9.2.3 The Right-Hand Rule 9.3 Magnetic Devices 9.3.1 Solenoids 9.3.2 Motors and Generators 9.3.3 Transformers Do you know ... The first transformers Chapter 9 Exercises

Chapter 10 Substances

10.1 Review of Some Basics 10.2 Types of Substances 10.2.1 Major Types of Substances 10.2.2 Elements 10.2.3 Compounds 10.2.4 Heterogeneous Mixtures 10.2.5 Homogeneous Mixtures 10.3 Solubility Do you know ... Carbon structures 10.4 Phases and Phase Transitions 10.4.1 The Phases of Matter 10.4.2 Evaporation Do you know ... The triple point 10.4.3 Sublimation Do you know ... Crystalline beauty 10.5 Physical and Chemical Properties and Changes 10.5.1 Physical Properties 10.5.2 Chemical Properties Do you know ... The crystal structure of ice 10.5.3 Physical and Chemical Changes Chapter 10 Exercises

Chapter 11 Historical Atomic Models and Density 11.1 The History of Atomic Models 11.1.1 Ancient Greece 11.1.2 The Scientific Revolution 11.1.3 Dalton’s Model 11.1.4 New Discoveries 11.2 Density 11.2.1 Volume 11.2.2 Density Calculations Chapter 11 Exercises

228 229 231 231 231 233 234 234 235 237 240 242 244 245 246 246 246 251 252 253 255 256 257 257 261 261 263 263 264 264 265 266 266 268 270 271 271 272 274 275 280 280 280 284 xi

Contents

Chapter 12 The Bohr and Quantum Atomic Models

12.1 The Bohr Model 12.1.1 Bohr’s Planetary Model 12.1.2 Atomic Spectra 12.2 The Quantum Model 12.2.1 Quantum Numbers and Orbitals Do you know ... Energy transitions in the hydrogen atom 12.2.2 Electron Configurations 12.3 Atomic Masses and Isotopes 12.3.1 Atomic Number, Atomic Mass, and Mass Number 12.3.2 Isotopes Chapter 12 Exercises

Chapter 13 Atomic Bonding

13.1 Bonds and Valence Electrons 13.1.1 The Science of Atoms 13.1.2 The Three Types of Atomic Bonds 13.1.3 Valence Electrons and Atomic Shells 13.1.4 Goals Atoms Seek to Fulfill 13.1.5 Why Some Elements Are More Reactive than Others 13.2 Metallic Bonding 13.3 Ionic Bonds 13.3.1 Bonding by Electron Transfer 13.3.2 Valence Numbers and Ionic Compound Binary Formulas 13.4 Covalent Bonds 13.4.1 Bonding by Electron Sharing 13.4.2 Electron Dot Diagrams and Octets 13.4.3 Diatomic Gases 13.5 Hydrogen 13.6 Polyatomic Ions Do you know... X-ray crystallography Chapter 13 Exercises

Chapter 14 Chemical Reactions

14.1 Chemical Equations 14.2 Four Types of Chemical Reactions 14.2.1 Synthesis Reactions 14.2.2 Decomposition Reactions 14.2.3 Single Replacement Reactions 14.2.4 Double Replacement Reactions 14.3 Other Reactions 14.3.1 Salts 14.3.2 Combustion Reactions 14.3.3 Oxidation Reactions 14.3.4 Redox Reactions 14.3.5 Precipitation Reactions

xii

288 289 289 290 295 295 296 300 300 300 301 302 304 305 305 306 307 308 309 310 311 311 313 314 314 315 317 317 318 319 320 324 325 326 326 327 327 329 329 330 330 331 331 332

Contents

14.3.6 Acid-Base Reactions 14.4 Balancing Chemical Equations 14.4.1 The Law of Conservation of Mass 14.4.2 Balancing Chemical Equations Do you know... Metal pickling 14.5 Energy in Chemical Reactions 14.5.1 Activation Energy 14.5.2 Exothermic and Endothermic Reactions 14.6 Reaction Rates and Collision Theory Chapter 14 Exercises

333 336 336 336 337 340 340 341 342 344

Glossary

346

Appendix A Reference Data

368

Appendix B Chapter Equations and Objectives Lists

370

Appendix C Laboratory Experiments

378 378 378 379 379 382 387 389 396 400

Appendix D Scientists to Know About

403

Appendix E Unit Conversions Tutorial

404

Appendix F Making Accurate Measurements F.1 F.2 F.3 F.4 F.5

Parallax Error Measurements with a Meter Stick or Rule Liquid Measurements Measurements with a Triple-Beam Balance Measurements with an Analog Thermometer

408 408 409 409 410 410

Appendix G References

411

Image Credits

413

Index

417

xiii

CHAPTER 1

The Nature of Scientific Knowledge

OBJECTIVES After studying this chapter and completing the exercises, students will be able to do each of the following tasks, using supporting terms and principles as necessary: 1. 2.

1.1

Modeling Knowledge

1.1.1 Kinds of Knowledge There are many different kinds of knowledge. One kind of knowledge is truth. Each of us is concerned about truth and how to know the truth. But the facts and theories of science constitute a different kind of knowledge, and as students of the natural sciences we are also concerned about these. Some people handle the distinction between the truth and scientific knowledge by referring to religious teachings as one kind of truth and scientific teaching as a different kind of truth. The problem here is that there are not different kinds of truth. There is only one truth about reality, but there are different kinds of knowledge about reality. Knowing truth is one kind of knowledge, and scientific knowledge is a different kind of knowledge. We are going to unpack this further over the next few pages, but here is a taste of where we are going. Scientific knowledge is not static. It is always changing as new discoveries are made. On the other hand, the truth about reality does not change. It is always true. For example, you are reading this sentence right now; that is true and nothing can make it false. This difference between scientific knowledge and knowledge from other sources indicates to us that the knowledge we have of truth is a different kind of knowledge than what we learn from scientific investigations. I have developed a model of knowledge that emphasizes the differences between truths we can know and what scientific investigations teach us. This model is not perfect (no mod3

Chapter 1

el is), nor is it exhaustive, but it is very useful, as all good models are. Our main goal in the next few sections is to develop this model of knowledge. The material in this chapter is crucial if you wish to have a proper understanding of what science is all about. To understand science correctly, we need to understand what we mean by scientific knowledge. Unfortunately, there is much confusion among non-scientists about the nature of scientific knowledge and this confusion often leads to misunderstandings when we talk about scientific findings and scientific claims. This is nothing new. Misconceptions about scientific claims have plagued public discourse for thousands of years and continue to do so to this day. This confusion is a severe problem, one much written about within the scientific community in recent years. To clear the air on this issue, it is necessary to examine what we mean by the term truth, as well as the different ways we discover truth. Then we must discuss the specific characteristics of scientific knowledge, including the key scientific terms fact, theory, and hypothesis.

1.1.2 What is Truth and How Do We Know It? Epistemology, one of the major branches of philosophy, is the study of what we can know and how we know it. Both philosophers and theologians claim to have important insights on the issue of knowing truth, and because of the roles science and religion have played in our culture over the centuries, we need to look at what both philosophers and theologians have to say. The issue we need to treat briefly here is captured in this question: what is truth and how do we know it? In other words, what do we mean when we say something is true? And if we can agree on a definition for truth, how can we know whether something is true? These are really complex questions, and philosophers and theologians have been working on them for thousands of years. But a few simple principles will be adequate for our purpose. As for what truth is, my simple but practical definition is this: Truth is the way things really are. Whatever reality is like, that is the truth. If there really is life on other planets, then it is true to say, “There is life on other planets.” If you live in Poughkeepsie, then when you say “I live in Poughkeepsie” you are speaking the truth. The harder question is: how do we know the truth? According to most philosophers, there are two ways that we can know truth, and these involve either our senses or our use of reason. First, truths that are obvious to us by direct observation of the world around us are said to be evident. It is evident that birds can fly. No proof is needed; we all observe this for ourselves. So the proposition, “Birds can fly,” conveys truth. Similarly, it is evident that humans can read books and birds cannot. Of course, when we speak of people knowing truth this way we are referring to people whose perceptive faculties are functioning normally. The second way philosophers say we can know truth is through the valid use of logic. Logical conclusions are typically derived from a sequence of logical statements called a syllogism, in which two or more statements (called premises) lead to a conclusion. For example, if we begin with the premises, “All men are mortal,” and, “Socrates was a man,” then it is a valid conclusion to state, “Socrates was mortal.” The truth of the conclusion of a logical syllogism definitely depends on the truth of the premises. The truth of the conclusion also depends on the syllogism having a valid structure. Some logical structures are not logically 4

The Nature of Scientific Knowledge

valid. (These invalid structures are called logical fallacies.) If the premises are true and the structure is valid, then the conclusion must be true. So the philosophers provide us with two ways of knowing truth that most people agree upon—truths can be evident (according to our senses) or they can be proven with reason (by valid use of logic, starting from true premises). Believers in some faith traditions argue for a third possibility for knowing truth, which is by revelation from supernatural agents such as God or angels. As examples, Christians believe that God has revealed truth to humanity in the Bible, through various appearances of angels, and through Jesus; Jews believe in God’s revelation through the Torah; and Muslims regard the Koran as God’s revelation. Obviously, not everyone accepts the possibility of knowing truth by revelation. Specifically, those who do not believe in God do not accept the possibility of revelations from God. Additionally, there are some who accept the existence of a transcendent power or being, but do not accept the possibility of revelations of truth from that power. So this third way of knowing truth is embraced by many people, but certainly not by everyone. Few people would deny that knowing truth is important. This is why we started our study by briefly exploring what truth is. But this is a book about science, and we need now to move to addressing a different question: what does science have to do with truth? The question is not as simple as it seems, as evidenced by the continuous disputes between religious and scientific communities stretching back over the past 700 years. To get at the relationship between science and truth, we first look at the relationship between propositions and truth claims.

1.1.3 Propositions and Truth Claims Not all that passes as valid knowledge can be regarded as truth, which I defined in the previous section as “the way things really are.” In many circumstances—maybe most—we do not actually know the way things really are. People do, of course, often use propositions or statements with the intention of conveying truth. But with other kinds of statements, people intend to convey something else. Let’s unpack this with a few example statements. Consider the following propositions: 1. I have two arms. 2. My wife and I have three children. 3. I worked out at the gym last week. 4. My car is at the repair shop. 5. Texas gained its independence from Mexico in 1836. 6. Atoms are composed of three fundamental particles—protons, neutrons, and electrons. 7. God made the world. Among these seven statements are actually three different types of claims. From the discussion in the previous section you may already be able to spot two of them. But some of these statements do not fit into any of the categories we explored in our discussion of truth. We can discover some important aspects about these claims be examining them one by one. So suppose for a moment that I, the writer, am the person asserting each of these statements as we examine the nature of the claim in each case.

Chapter 1

I have two arms. This is true. I do have two arms, as is evident to everyone who sees me. My wife and I have three children. This is true. To me it is just as evident as my two arms. I might also point out that it is true regardless of whether other people believe me when I say it. (Of course, someone could claim that I am delusional, but let’s just keep it simple here and assume I am in normal possession of my faculties.) This bit about the statement being true regardless of others’ acceptance of it comes up because of a slight difference here between the statement about children and the statement about arms. Anyone who looks at me will accept the truth that I have two arms. It will be evident, that is, obvious, to them. But the truth about my children is only really evident to a few people (my wife and I, and perhaps a few doctors and close family members). Nevertheless, the statement is true. I worked out at the gym last week. This is also true; I did work out last week. The statement is evident to me because I clearly remember going there. Of course, people besides myself must depend on me to know it because they cannot know it directly for themselves unless they saw me there. Note that I cannot prove it is true. I can produce evidence, if needed, but the statement cannot be proven without appealing to premises that may or may not be true. Still, the statement is true. My car is at the repair shop. Here is a statement that we cannot regard as a truth claim. It is merely a proposition about where I understand my car to be at present, based on where I left it this morning and what the people at the shop told me they were going to do with it. For all I know, they may have taken my car joy riding and presently it may be flying along the back roads of the Texas hill country. I can say that the statement is correct so far as I know. Texas gained its independence from Mexico in 1836. We Texans were all taught this in school and we believe it to be correct, but as with the previous statement we must stop short of calling this a truth claim. It is certainly a historical fact, based on a lot of historical evidence. The statement is correct so far as we know. But it is possible there is more to that story than we know at present (or will ever know) and none of those now living were there. Atoms are composed of three fundamental particles—protons, neutrons, and electrons. This statement is, of course, a scientific fact. But like the previous two statements, this statement is not—surprise!—a truth claim. We simply do not know the truth about atoms. The truth about atoms is clearly not evident to our senses. We cannot guarantee the truth of any premises we might use to construct a logical proof about the insides of atoms, so proof is not able to lead us to the truth. And so far as I know, there are no supernatural agents who have revealed to us anything about atoms. So we have no access to knowing how atoms really are. What we do have are the data from many experiments, which may or may not tell the whole story. Atoms may have other components we don’t know about yet. The best we can say about this statement is that it is correct so far as we know (that is, so far as the scientific community knows). God made the world. This statement clearly is a truth claim, and those who believe in God accept it as the truth. But other people disagree on whether the statement is true. I include this example here because we soon see what happens when scientific claims and religious truth claims get confused. Regardless of whether a person is religious or not, the issue is important. We all need to learn to speak correctly about the different claims people make. To summarize this section, some statements we make are evidently or obviously true. But for many statements, we must recognize that we don’t know if they actually are true. The 6

The Nature of Scientific Knowledge

best we can say about these kinds of statements—and scientific facts are like this—is that they are correct so far as we know. Finally, there are metaphysical or religious statements about which people disagree; some claim they are true, some deny the same, and some say there is no way to know.

1.1.4 Truth and Scientific Claims Let’s think a bit further about the truth of reality, both natural and supernatural. Most people agree that regardless of what different people think about God and nature, there is some actual truth or reality about nature and the supernatural. Regarding nature, there is some full reality about the way, say, atoms are structured, regardless of whether we currently understand that structure correctly. So far as we know, this reality does not shift or change from day to day, at least not since the early history of the universe. So the reality about atoms—the truth about atoms—does not change. And regarding the supernatural, there is some reality about the supernatural realm, regardless of whether anyone knows what that is. Whatever these realities are, they are truths, and these truths do not change either. Now, I have observed over the years that since (roughly) the beginning of the 20th century, careful scientists do not refer to scientific claims as truth claims. They do not profess to knowing the ultimate truth about how nature really is. For example, Niels Bohr, one of the great physicists of the 20th century, said, “It is wrong to think that the task of physics is to find out how nature is. Physics concerns what we can say about nature.” Scientific claims are understood to be statements about our best understanding of the way things are. Most scientists believe that over time our scientific theories get closer and closer to the truth of the way things really are. But when they are speaking carefully, scientists do not claim that our present understanding of this or that is the truth about this or that.

1.1.5 Truth vs. Facts Whatever the truth is about the way things are, that truth is presumably absolute and unchanging. If there is a God, then that’s the way it is, period. And if matter is made of atoms as we think it is, then that is the truth about matter and it is always the truth. But what we call scientific facts, by their very nature, are not like this. Facts are subject to change, and sometimes do, as new information comes becomes known through ongoing scientific research. Our definitions for truth and for scientific facts need to take this difference into account. As we have seen, truth is the way things really are. By contrast, here is a definition for scientific facts: A scientific fact is a proposition supported by a great deal of evidence. Scientific facts are discovered by observation and experiment, and by making inferences from what we observe or from the results of our experiments. A scientific fact is correct so far as we know, but can change as new information becomes known. So facts can change. Scientists do not put them forward as truth claims, but as propositions that are correct so far as we know. In other words, scientific facts are provisional. They are always subject to revision in the future. As scientists make new scientific discoveries,

Chapter 1

Do you know ...

Dark matter

Many astrophysicists now hypothesize that most of the matter in the universe is dark matter—about 80% of it, in fact. Dark matter does not interact at all with light the way ordinary matter does (this is why it is called dark matter), and this makes dark matter extremely difficult to detect using ordinary astronomical instruments such as telescopes. The primary evidence we have for the existence of dark matter is in calculations associated with the gravitational attraction in spiral galaxies, such as the Whirlpool Galaxy, shown below. Spiral galaxies are rotating, which means they would fly apart if there weren’t strong enough forces of gravitational attraction to hold them together. The thing is, calculations show that the gravitational attraction between the stars in any galaxy is not even close to what is required to hold a rotating galaxy together. As we see in Chapters 2 and 9, gravitational fields are caused by mass, in this case, the mass of the stars. Calculating the mass of all the stars in a galaxy is actually quite simple, and the calculations show that the amount of mass required is simply not there. But obviously there is something holding the spiral galaxies together, and this is what has led the scientists to think that there may be another kind of mass that we have so far not been able to detect. 8

The Nature of Scientific Knowledge

The distinction between truth and scientific facts is crucial for a correct understanding of the nature of scientific knowledge. Facts can change; truth does not.

1.1.6 Relating Scientific Knowledge and Truth There are two ways in which our discussion of scientific knowledge relates to our discussion of truth. First, we have seen that one way to know truth is by direct observation. We have also seen that scientists use observation as a way of discovering scientific facts. How do these two uses of observation relate to each other? Imagine a scientist studying tigers who observes a tiger eating the meat of another animal. The scientist can say, “It is true that this tiger eats meat.” The scientist might observe 25 other tigers exhibiting this same behavior. She can then say, “These 25 tigers all eat meat.” So far, all the scientist has done is to say things that she has found to be true by direct observation. But now suppose the scientist takes this information and makes a general claim about tigers: “It is a scientific fact that tigers are carnivores.” The scientist has now made a leap from tigers she has directly observed to many other tigers she has not directly observed. Who knows whether there might be a species of vegetarian tiger somewhere out there? We have no way of knowing the eating habits of every single tiger. This is why we cannot say that meat eating is a truth about all tigers. We can only say that it is a scientific fact about tigers. The scientific fact about tigers is a statement based on a lot of evidence that is correct so far as we know, but it may need to be changed if further research shows that there are species of tigers that do not eat meat. Second, we have studied tigers for a long time and are pretty sure that the statement, “all tigers are carnivores” is true. We are so sure that most of us probably do regard this statement as true. This is fine, but we must keep in mind that it is always possible that a scientific claim may turn out to be false.

1.2

The Cycle of Scientific Enterprise

1.2.1 Science Having established some basic principles about the distinction between scientific facts and truth, we are now ready to define science itself and examine what science is and how it works. Here is a definition: Science is the process of using experiment, observation, and logical thinking to build “mental models” of the natural world. These mental models are called theories. We do not and cannot know the natural world perfectly or completely, so we construct models of how it works. We explain these models to one another with descriptions, diagrams, and mathematics. These models are our scientific theories. Theories never explain the world to us perfectly. To know the world perfectly, we would have to know the absolute truth about reality, which we certainly do not and never will. So theories always have their limits, but we hope they become more accurate and more complete over time, accounting for more and more physical phenomena (data, facts), and helping us to understand the natural world as a coherent whole.

Chapter 1

Hypothesis An informed prediction, based on a theory

Scientific Fact

Theory

Experiment

Our best explanation at present

Putting the hypothesis to the test

Scientific Fact Scientific Fact

New Scientific Fact

Yes

Analysis Are the experimental results consistent with the theory we started with?

No Review

Reconsider experimental methods, appropriateness of hypothesis, adequacy of theory

Figure 1.1. The Cycle of Scientific Enterprise.

Scientific knowledge is continuously changing and advancing through a cyclic process that I call the Cycle of Scientific Enterprise, represented in Figure 1.1. In the next few sections, we examine the individual parts of this cycle in detail.

1.2.2 Theories Theories are the grandest thing in science. In fact, it is fair to say that theories are the glory of science, and developing good Theory Scientific theories is what science is all about. Electromagnetic field theoOur best explanation Fact at present ry, atomic theory, quantum theory, the general theory of relativScientific ity—these are all theories in physics that have had a profound Fact Scientific effect on scientific progress and on the way we all live.1 Fact Now, even though many people do not realize it, all scientific knowledge is theoretically based. Let me explain. A theory is a mental model or explanatory system that explains and relates together most or all of the facts (the data) in a certain sphere of knowledge. A theory is not a hunch or a guess or a wild idea. Theories are the mental structures we use to make sense of the data we have. We cannot understand any scientific data without a theory to organize it and explain it. This is why I write that all scientific knowledge is theoretically based. And for this reason, it is inappropriate and scientifically incorrect to scorn these explanatory systems as “merely a theory” or “just a theory.” Theories are explanations that account for a lot of different facts. If a theory has stood the test of time, that means it has wide support within the scientific community.

The term law is just a historical (and obsolete) term for what we now call a theory.

The Nature of Scientific Knowledge

It is popular in some circles to speak dismissively of certain scientific theories, as if they represent some kind of untested speculation. It is simply incorrect—and very unhelpful—to speak this way. As students in high-school science, one of the important things you need to understand is the nature of scientific knowledge, the purpose of theories, and the way scientific knowledge progresses. These are the issues this chapter is about. All useful scientific theories possess several characteristics. The two most important ones are:

Examples of Famous Theories In the next chapter, we encounter Einstein’s general theory of relativity, one of the most important theories in modern physics. Einstein’s theory represents our best current understanding of how gravity works. Another famous theory we address later is the kinetic theory of gases, our present understanding of how molecules of gas too small to see are able to create pressure inside a container.

•

The theory accounts for and explains most or all of the related facts.

•

The theory enables new hypotheses to be formed and tested.

Theories typically take decades or even centuries to gain credibility. If a theory gets replaced by a new, better theory, this also usually takes decades or even centuries to happen. No theory is ever “proven” or “disproven” and we should not speak of them in this way. We also should not speak of them as being “true” because, as we have seen, we do not use the word “truth” when speaking of scientific knowledge. Instead, we speak of facts being correct so far as we know, or of current theories as representing our best understanding, or of theories being successful and useful models that lead to accurate predictions. An experiment in which the hypothesis is confirmed is said to support the theory. After such an experiment, the theory is stronger but it is not proven. If a hypothesis is not confirmed by an experiment, the theory might be weakened but it is not disproven. Scien-

Key Points About Theories 1.

A theory is a way of modeling nature, enabling us to explain why things happen in the natural world from a scientific point of view.

A theory tries to account for and explain the known facts that relate to it.

Theories must enable us to make new predictions about the natural world so we can learn new facts.

Strong, successful theories are the glory and goal of scientific research.

It is incorrect to speak dismissively of successful theories because theories are not just guesses.

We don’t speak of theories as being proven or disproven. Instead, we speak of them in terms such as how successful they have been at making predictions and how accurate the predictions have been.

Figure 1.2. Key points about theories.

Chapter 1

tists require a great deal of experimental evidence before a new theory can be established as the best explanation for a body of data. This is why it takes so long for theories to become widely accepted. And since no theory ever explains everything perfectly, there are always phenomena we know about that our best theories do not adequately explain. Of course, scientists continue their work in a certain field hoping eventually to have a theory that does explain all the facts. But since no theory explains everything perfectly, it is impossible for one experimental failure to bring down a theory. Just as it takes a lot of evidence to establish a theory, so it takes a large and growing body of conflicting evidence before scientists abandon an established theory. At the beginning of this section, I state that theories are mental models. This statement needs a bit more explanation. A model is a representation of something, and models are designed for a purpose. You have probably seen a model of the organs in the human body in a science classroom or textbook. A model like this is a physical model and its purpose is to help people understand how the human body is put together. A mental model is not physical; it is an intellectual understanding, although we often use illustrations or physical models to help communicate to one another our mental ideas. But as in the example of the model of the human body, a theory is also a model. That is, a theory is a representation of how part of the world works. In physics and chemistry, scientific models generally take the form of mathematical equations that allow scientists to make numerical predictions and calculate the results of experiments. The more accurately a theory represents the way the world works, which we judge by forming new hypotheses and testing them with experiments, the better and more successful the theory is. To summarize, a successful theory represents the natural world accurately. This means the model (theory) is useful because if a theory is an accurate representation, then it leads to accurate predictions about nature. When a theory repeatedly leads to predictions that are confirmed in scientific experiments, it is a strong, useful theory. The key points about theories are summarized in Figure 1.2.

1.2.3 Hypotheses

Hypothesis

A hypothesis is a positively stated, informed prediction An informed prediction, about what will happen in certain circumstances. We say a based on a theory hypothesis is an informed prediction because when we form hypotheses we are not just speculating out of the blue. We are applying a certain theoretical understanding of the subject to the new situation before us and predicting what will happen or what we expect to find in the new situation based on Key Points About Hypotheses the theory the hypothesis is coming from. Ev1. A hypothesis is an informed preery scientific hypothesis is based on a particular diction about what will happen in theory,and competing theories can lead to differcertain circumstances. ent hypotheses. 2. Every hypothesis is based on a parOften hypotheses are worded as if-then stateticular theory. ments, such as, “If various forces are applied to a 3. Well-formed scientific hypotheses vehicle, then the vehicle accelerates at a rate that must be testable, which is what is in direct proportion to the net force.” Every sciscientific experiments are de- entific hypothesis is based on a theory and it is the hypothesis that is directly tested by an experiment. signed to do. If the experiment turns out the way the hypothesis Figure 1.3. Key points about hypotheses. predicts, the hypothesis is confirmed and the the12

The Nature of Scientific Knowledge

ory it came from is strengthened. Of course, the hypothesis may not An informed prediction, be confirmed by the experiment. based on a theory We see how scientists respond to this situation in Section 1.2.6. The terms theory and hypothesis are often used interchangeably in common speech, but in science Theory Scientific Our best explanation Fact they mean different things. For this at present reason, you should make note of Scientific Fact the distinction. Scientific One more point about hypothFact eses. A hypothesis that cannot be tested is not a scientific hypothesis. For example, horoscopes purport to predict the future with statements like, “You will meet someone important to your career in the coming weeks.” Statements like this are so vague they are untestable and do not qualify as scientific hypotheses. The key points about hypotheses are summarized in Figure 1.3.

Hypothesis

Examples of Famous Hypotheses Einstein used his general theory of relativity to make an incredible prediction in 1917: that gravity causes light to bend as it travels through space. In the next chapter, you read about the stunning result that occurred when this hypothesis was put to the test. The year 2012 was a very important year for the standard theory in the world of subatomic particles, called the Standard Model. This theory led in the 1960s to the prediction that there are weird particles in nature, now called Higgs bosons, which no one had ever detected. Until 2012, that is! An enormous machine that could detect these particles, called the Large Hadron Collider, was built in Switzerland and completed in 2008. In 2012, scientists announced that the Higgs boson had been detected at last, a major victory for the Standard Model, and for Peter Higgs, the physicist who first proposed the particle that now bears his name.

1.2.5 Analysis In the Analysis phase of the Cycle of Scientific Enterprise, researchers must interpret the experimental results. The results of an experiment are essentially data, and data must always be interpreted.

Yes

Experiment

Putting the hypothesis to the test

Analysis Are the experimental results consistent with the theory we started with?

No 13

Chapter 1

The main goal of this analysis is to determine whether the original hypothesis is confirmed by the experiment. If it is, then the result of the experiment is new facts that are consistent with the original theory because the hypothesis is based on that theory. As a result, the support for the theory is increased—the theory was successful in generating a hypothesis that was confirmed by experiment. As a result of the experiment, our confidence in the theory as a useful model is increased and the theory is even more strongly supported than before.

No Review

Reconsider experimental methods, appropriateness of hypothesis, adequacy of theory

•

The values used in the calculation of the hypothesis’ predictions may not have been accurate or precise enough, throwing off the hypothesis’ predictions.

•

1.3

The Scientific Method

1.3.1 Conducting Reliable Experiments The so-called scientific method that you have been studying ever since about 4th grade is simply a way of conducting reliable experiments. Experiments are an important part 14

The Nature of Scientific Knowledge

of the Cycle of Scientific Enter- The Scientific Method prise, so the scientific method is 5. Collect data. important to know. You probably 1. State the problem. 6. Analyze the data. remember studying the steps in 2. Research the problem. the scientific method from prior 3. Form a hypothesis. 7. Form a conclusion. courses, so they are listed in Table 4. Conduct an experiment. 8. Repeat the work. 1.1 without further comment. We discuss variables and Table 1.1. Steps in the scientific method. measurements a lot in this course, so we take the opportunity here to identify some of the language researchers use during the experimental process. In a scientific experiment, the researchers have a question they are trying to answer (from the State the Problem step in the scientific method), and typically it is some kind of question about the way one physical quantity affects another one. So the researchers design an experiment in which one quantity can be manipulated (that is, deliberately varied in a controlled fashion) while the value of another quantity is monitored. A simple example of this in everyday life that you can easily relate to is varying the amount of time you spend each week studying for your math class in order to see what effect the time spent has on the grades you earn. If you reduce the time you spend, will your grades go down? If you increase the time, will they go up? A precise answer depends on a lot of things, of course, including the person involved, but in general we would expect that if a student varies the study time enough we will see the grades vary as well. And in particular, we expect more study time to result in higher grades. The way your study time and math grades relate together can be represented in a diagram such as Figure 1.4. Now let us consider this same concept in the context of scientific experiments. An experiment typically involves some kind of complex system that the scientists are modeling. The system could be virtually anything in the natural world—a galaxy, a system of atoms, a This is the quantity you adjust. mixture of chemicals, a protein, or a badger. The variables in the scientists’ mathematiStudy Time cal models of the system correspond to the Grades in Math Experimental System physical quantities that can be manipulated or measured in the system. As I describe the different kinds of variables, refer to Figure This is where you look to see the effect. 1.5. Figure 1.4. Study time and math grades in a simple

1.3.2 Experimental Variables

experimental system.

When performing an experiment, the variable that is deliberately manipulated by the researchers is called the explanatory variable. As the explanatory variable is manipulated, the researchers monitor the effect this variation has on the response variable. In the example of study time versus math grade, the study time is the explanatory variable and the grade earned is the response variable. Usually, a good experimental design allows only one explanatory variable to be manipulated at a time so that the researchers can tell definitively what its effect is on the response variable. If more than one explanatory variable is changing during the course of the experiment, researchers may not be able to tell which one is causing the effect on the response variable. A third kind of variable that plays a role in experiments is the lurking variable. A lurking variable is a variable that affects the response variable without the researchers being 15

Chapter 1

aware of it. This is undesirable, of course, because with unknown influences present the Explanatory Variable researchers may not Response Variable be able to make a Experimental System correct conclusion Researchers measure about the effect of the known explanaoutcome here Figure 1.5. The variables in an experimental system. tory variables on the response variable under study. So researchers study their experimental projects very carefully to minimize the possibility of lurking variables affecting their results. In our example about study time and math grades, there could be a number of lurking variables affecting the results of the experiment. Possible lurking variables include changes in the difficulty of the material from one chapter to the next and variations in the student’s ability to concentrate due to fatigue from seasonal sports activities. Researchers manipulate input here

Lurking Variable

The Nature of Scientific Knowledge

Do you know ...

Double-blind experiments

The human mind is so powerful that if a person believes a new medication might help, the person’s condition can sometimes improve even if the medication itself isn’t doing a thing! This is amazing, but in medical research it means that the researchers can have a hard time determining whether a person is helped by the new medication, or by feeling positively about the medication, or even by the attention given to him or her by the doctor. Pictured below is Lauren Wood, a clinician involved in vaccine research at the Center for Cancer Research, which is part of the National Cancer Institute. Just as with every other scientific researcher, Dr. Wood’s research is conducted according to methods that have been developed to ensure that people’s beliefs about the research don’t influence the outcome of the research. The approach is to divide the patients who will participate in testing a new medication into two groups, control and experimental. The experimental group is given the new medication. The control group is given a placebo—a fake medication such as a sugar pill—that has no effect on the person’s medical condition. Further, none of the patients know whether they are given the placebo or the real medication. This technique, called a blind experiment, allows the researchers to determine whether a new medication actually helps, as they compare the results of the control and experimental groups. But there’s more. It turns out that the researchers themselves can affect the results of the experiment if they know which patients are receiving a placebo and which ones are receiving the medication under study. How can this happen? Well, if the researchers know who is getting the real medication, they might subconsciously act more positively with them than with other patients. This might be because the researchers expect those getting the new medication to improve, and this expectation gets subconsciously communicated to the patients. The positive attitude might be perceived as more encouraging and patients might improve just because of the encouragement! The way around this dilemma is to use a double-blind experiment. In a double-blind experiment, neither the patients nor the researchers know which patients are getting the placebo and which are getting the real treatment. A team of technicians is in the middle, administering the medication and keeping records of who receives what. The researchers are not allowed to see the lists until the research results are finalized. The double-blind experiment is the standard protocol followed today for new medical research. The response variable is the quality of the plant’s fruit. Researchers expect that under drought conditions, the fruit of the new variety will be better than the fruit of the other varieties. The explanatory variable is the unique feature of the new variety that relates to the plant’s use of water. The trees are exposed to drought conditions in the experiment. If the new variety produces higher quality fruit than the control group, then the hypothesis is confirmed, and the theory that led to the hypothesis has gained credibility through this success. One can imagine many different lurking variables that could affect the outcome of this experiment without the scientists’ awareness. For example, the new variety trees could be planted in locations that receive different amounts of moisture or sun than the locations 17

Chapter 1

where the control group trees are, or, the nutrients in the soil in different locations might vary. In a good experimental design, researchers seek to identify such factors and take measures to ensure that they do not affect the outcome of the experiment. They do this by making sure there are trees from both the experimental group and the control group in all the different conditions the trees will experience. This way, variations in sunlight, soil type, soil water content, elevation, exposure to wind, and other factors are experienced equally by trees in both groups.

Chapter 1 Exercises As you go through the chapters in this book, always answer the questions in complete sentences, using correct grammar and spelling. Here is a tip to help improve the quality of your written responses: avoid pronouns! Pronouns almost always make your responses vague or ambiguous. If you want to receive full credit for written responses, avoid them. (Oops. I mean, avoid pronouns!)

Study Questions Answer the following questions with a few complete sentences. 1.

Distinguish between theories and hypotheses.

Explain why a single experiment can never prove or disprove a theory.

Explain how an experiment can still provide valuable data even if the hypothesis under test is not confirmed.

Explain the difference between truth and facts and describe the sources of each.

State the two primary characteristics of a theory.

Does a theory need to account for all known facts? Why or why not?

It is common to hear people say, “I don’t accept that; it’s just a theory.” What is the error in a comment like this?

Distinguish between facts and theories.

Distinguish between explanatory variables, response variables, and lurking variables.

10. Why do good experiments that seek to test some kind of new treatment or therapy include a control group? 11. Explain specifically how the procedure students follow in the Pendulum Experiment satisfies every step of the “scientific method.” 12. This chapter argues that scientific facts should not be regarded as true. Someone might question this and ask, If they aren’t true, then what are they good for? Develop a response to this question. 13. Explain what a model is and why theories are often described as models. 18

The Nature of Scientific Knowledge

14. Consider an experiment that does not deliver the result the experimenters expect. In other words, the result is negative because the hypothesis is not confirmed. There are many reasons why this might happen. Consider each of the following elements of the Cycle of Scientific Enterprise. For each one, describe how it might be the driving factor that results in the experiment’s failure to confirm the hypothesis. a. the experiment b. the hypothesis c. the theory 15. Identify the explanatory and response variables in the Pendulum Experiment, and identify two realistic possibilities for ways the results may be influenced by lurking variables.

Do you know ...

Hero Sir Humphry Davy Sir Humphry Davy (1778–1829) was one of the leading experimenters and inventors in England in the early nineteenth century. He conducted many early experiments with gases; discovered sodium, potassium, and numerous other elements; and produced the first electric light from a carbon arc. In the early nineteenth century, explosions in coal mines were frequent, resulting in much tragic loss of life. The explosions were caused by the miners’ lamps igniting the methane gas found in the mines.

Davy became a national hero when he invented the Davy Safety Lamp (below). This lamp incorporated an iron mesh screen around the flame. The cooling from the iron reduces the flame temperature so the flame does not pass through the mesh, and thus cannot cause an explosion. The Davy Lamp was produced in 1816 and was soon in wide use. Davy’s experimental work proceeded by reasoning from first principles (theory) to hypothesis and experiment. Davy stated, “The gratification of the love of knowledge is delightful to every refined mind; but a much higher motive is offered in indulging it, when that knowledge is felt to be practical power, and when that power may be applied to lessen the miseries or increase the comfort of our fellow-creatures.” 19

CHAPTER 2

Motion

OBJECTIVES Memorize and learn how to use these equations: v=

d t

v f − vi t

After studying this chapter and completing the exercises, students will be able to do each of the following tasks, using supporting terms and principles as necessary: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 13.

2.1

Define and distinguish between velocity and acceleration. Use scientific notation correctly with a scientific calculator. Calculate distance, velocity, and acceleration using the correct equations, MKS units, and correct dimensional analysis. Use from memory the conversion factors, metric prefixes, and physical constants listed in Appendix A. Explain the difference between accuracy and precision and apply these terms to questions about measurement. Demonstrate correct understanding of precision by using the correct number of significant digits in calculations and rounding. Draw and interpret graphs of distance, velocity, and acceleration vs. time and describe an object’s motion from the graphs. Describe the key features of the Ptolemaic model of the heavens, including all the spheres and regions in the model. State several additional features of the medieval model of the heavens and relate them to the theological views of the Christian authorities opposing Copernicanism. Briefly describe the roles and major scientific models or discoveries of Copernicus, Tycho, Kepler, and Galileo in the Copernican Revolution. Also, describe the significant later contributions of Isaac Newton and Albert Einstein to our theories of motion and gravity. Describe the theoretical shift that occurred in the Copernican Revolution and how Christian officials (both supporters and opponents) were involved. State Kepler’s three laws of planetary motion. Describe how the gravitational theories of Kepler, Newton, and Einstein illustrate the way the Cycle of Scientific Enterprise works.

Computations in Physics

In this chapter. you begin mastering the skill of applying mathematics to the study of physics. To do this well, you must know a number of things about the way measurements are handled in scientific work. You must also have a solid problem-solving strategy that you can depend on to help you solve problems correctly without becoming confused. These are the topics of the next few sections.

Chapter 2

2.1.1 The Metric System Units of measure are crucial in science. Science is about making measurements and a measurement without its units of measure is a meaningless number. For this reason, your answers to computations in scientific calculations must always show the units of measure. The two major unit systems you must know about are the SI (from the French Système international d’unités), typically known in the United States as the metric system, and the USCS (U.S. Customary System). You have probably studied these systems before and should already be familiar with some of the SI units and prefixes, so our treatment here is brief. If you think about it, you would probably agree that the USCS is cumbersome. One problem is that there are many different units of measure for every kind of physical quantity. For example, just for measuring length or distance we have the inch, foot, yard, and mile. The USCS is also full of random numbers like 3, 12, and 5,280, and there is no inherent connection between units for different types of quantities. By contrast, the SI system is simple and has many advantages. There is only one basic unit for each kind of quantity, such as the meter for measuring length. Instead of having many unrelated units of measure for measuring quantities of different sizes, fractional and multiple prefixes based on powers of ten are used with the units to accommodate various sizes of measurements. A second advantage is that since quantities with different prefixes are related by some power of ten, unit conversions can often be performed mentally. To convert 4,555 ounces into gallons, we first have to look up the conversion from ounces to gallons (which is hard to remember), and then use a calculator to perform the conversion. But to convert 40,555 cubic centimeters into cubic meters is simple—simply divide by 1,000,000 and you have 0.040555 m3. (If this doesn’t seem to click for you, take time to study the unit conversions tutorial in Appendix E. Another SI advantage is that the units for different types of quantities relate to one another in some way. Unlike the gallon and the foot, which have nothing to do with each other, the liter (a volume) relates to the centimeter (a length): 1 liter = 1,000 cubic centimeters.1 For all these reasons, the USCS is not used much in scientific work. The SI system is the international standard and it is important to know it well. In the SI unit system, there are seven base units, listed in Table 2.1. (In this text, Unit Symbol Quantity we use only the first five of them.) There meter m length are also many additional units of measure, known as derived units. All the derived units kilogram kg mass are formed by various combinations of the second s time seven base units. To illustrate, below are a ampere A electric current few examples of derived units that we diskelvin K temperature cuss and use in this book. Note, however, candela Cd luminous intensity that we won’t be working much with the messy fractions; they are simply shown to mole mol amount of substance illustrate how base units are combined to Table 2.1. The seven base units in the SI unit system. form derived units.

The liter is not actually an official SI unit of measure, but it is used all the time anyway in scientific work.

Motion

Do You Know ...

How are the base units defined?

the newton (N) is the SI unit for measuring force: 1 N = 1

•

the joule (J) is the SI unit for measuring energy: 1 J = 1

•

the watt (W) is the SI unit for measuring power: 1 W = 1

kg ⋅m s2

kg ⋅m 2 s2 kg ⋅m 2 s3

Using the SI system requires knowing the units of measure—base and derived—and the prefixes that are applied to the units to form fractional units (such as the centimeter) and multiple units (such as the kilometer). The complete list of metric prefixes is shown in Appendix A in Table A.1. The short list of prefixes you must know by memory for use in this course is in Table A.2. Note that even though the kilogram is a base unit, prefixes are not added to the kilogram. Instead, prefixes are added to the gram to form units such as the milligram and microgram.

2.1.2 MKS Units A subset of the SI system is the MKS system. The MKS system, summarized in Table 2.2, uses the meter, the kilogram, and the second (hence, “MKS”) as primary Variable Variable Unit Unit units. Dealing with different systems of Symbol Symbol units can become very confusing. But length d (distance) meter m the wonderful thing about sticking to L (length) the MKS system is that any calculation h (height) r (radius), etc. performed with MKS units gives a result in MKS units. This is why the MKS mass m kilogram kg system is so handy and why we use it time t second s almost exclusively in physics. Table 2.2. The three base units in the MKS system.

Chapter 2

To convert the units of measure given in problems into MKS units, you must know the conversion factors listed in Appendix A (Tables A.2 and A.3). Appendix A also lists several physical constants you must know (Table A.4) and some common unit conversion factors that you are not required to memorize, but should have handy when working problem assignments.

2.1.3 Dimensional Analysis Dimensional analysis is a term that refers to all the work of dealing with units of measure in computations. This work includes converting units from one set of units to another and using units consistently in equations. You are probably already familiar with methods for performing unit conversions. I have a lot of practice problems cued up for you (coming up soon!), but if you need a refresher on using unit conversion factors to convert from one set of dimensions to another, please refer to the tutorial in Appendix E.

2.1.4 Accuracy and Precision The terms accuracy and precision refer to the limitations inherent in making measurements. Science is all about investigating nature and to do that we must make measurements. Accuracy relates to error, which is the difference between a measured value and the true value of a given quantity. The lower the error is in a measurement, the better the accuracy. Error can be caused by a number of different factors, including human mistakes, malfunctioning equipment, incorrectly calibrated instruments, or unknown factors that influence a measurement without the knowledge of the experimenter. All measurements contain error because (alas!) perfection is simply not a thing we have access to in this world. Precision refers to the resolution or degree of “fine-ness” in a measurement. The limit to the precision obtained in a measurement is ultimately dependent on the instrument used to make the measurement. If you want greater precision, you must use a more precise instrument. The precision of a measurement is indicated by the number of significant digits (or significant figures) included when the measurement is written down (see next section). Figure 2.1 is a photograph of a machinist’s rule and an architect’s scale placed side by side. Since the marks on the two scales line up consistently, these two scales are equally accurate. But the machinist’s rule (on top) is more precise. The architect’s scale is marked in 1/16-inch increments, but the machinist’s rule is marked in 1/64-inch increments. The machinist’s rule has higher resolution, and thus greater precision. It is important that you are able to distinguish between accuracy and precision. Here is an example to illustrate the difference. Let’s say Shana and Marius each buy digital thermometers for their homes. The thermometer Shana buys cost $10 and measures to the nearest 1°F. Marius pays $40 and gets one that reads to the nearest 0.1°F. Note that on a day when the actual temperature is 95.1°F, if the two thermometers are reading accurately Shana’s thermometer reads 95° and Marius’ reads 95.1°. Thus, Marius’ thermometer is more precise.

Figure 2.1. The accuracy of these two scales is the same, but the machinist’s rule on the top is more precise.

Motion

Now suppose Shana reads the directions and properly installs the sensor for her new thermometer in the shade. Marius doesn’t read the directions and mounts his sensor in the direct sunlight, which causes a significant error in the measurement for much of the day. The result is that Shana has lower-precision, higher-accuracy measurements!

Chapter 2

Let’s apply this rule to several example values to see how it works: 15,679

This value has five significant digits.

21.0005

This value has six significant digits.

37,000

This value has only two significant digits because when there is no decimal, trailing zeros are not significant. Notice that the word significant here is a reference to the precision of the measurement, which in this case is rounded to the nearest thousand. The zeros in this value are certainly important, but they are not significant in the context of precision.

0.0105

This value has three significant digits because we start counting with the first nonzero digit on the left.

0.001350 This value has four significant digits. Trailing zeros count when there is a decimal. The significant digit rules enable us to tell the difference between two measurements such as 13.05 m and 13.0500 m. Mathematically, of course, these values are equivalent. But they are different in what they tell us about the process of how the measurements were made. The first measurement has four significant digits. The second measurement is more precise. It has six significant digits and would come from a more precise instrument. Now, just in case you are bothered by the zeros at the end of 37,000 that are not significant, here is one more way to think about significant digits that may help. The precision in a measurement depends on the instrument used to make the measurement. If we express the measurement in different units, this does not change the precision. A measurement of 37,000 grams is equivalent to 37 kilograms. Whether we express this value in grams or kilograms, it still has two significant digits. Case 2

The second case addresses the rules that apply when you record a measurement yourself, rather than reading a measurement someone else has made. When you take measurements yourself, as you do in laboratory experiments, you must know the rules for which digits are significant in the reading you are taking on the measurement instrument. The rule for taking measurements depends on whether the instrument you are using is a digital instrument or an analog instrument. Here are the rules for these two possibilities: Rule 1 for digital instruments For the digital instruments commonly found in high school or undergraduate science labs, assume all the digits in the reading are significant, except leading zeros.

Rule 2 for analog instruments The significant digits in a measurement include all the digits known with certainty, plus one digit at the end that must be estimated between the finest marks on the scale of your instrument. The first of these rules is illustrated in Figure 2.2. The reading on the left has leading zeros, which do not count as significant. Thus, the first reading has three significant digits. 26

Motion

The second reading also has 42.0 42.000 42,000 three significant digits. The 0042.0 third reading has five signifiFigure 2.2. With digital instruments, all digits are significant except cant digits. leading zeros. Thus, the numbers of significant digits in these readings The fourth reading also are, from left to right, three, three, five, and five. has five significant digits because with a digital display, the only zeros that don’t count are the leading zeros. Trailing zeros are significant with a digital instrument. However, when you write this measurement down, you must write it in a way that shows those zeros to be significant. The way to do this is by using scientific notation. Thus, the right-hand value in Figure 2.2 must be written as 4.2000 × 104. Dealing with digital instruments is actually more involved than the simple rule above implies, but the issues involved go beyond what we typically deal with in introductory or intermediate science classes. So, simply take your readings and assume that all the digits in the reading except leading zeros are significant. Now let’s look at some examples illustrating the rule for analog instruments. Figure 2.3 shows a machinist’s rule being used to measure the length in millimeters (mm) of a brass block. We know the first two digits of the length with certainty; the block is clearly between 31 mm and 32 mm long. We have to estimate the third significant digit. The scale on the rule is marked in increments of 0.5 mm. Comparing the edge of the block with these Figure 2.3. Reading the significant digits marks, I would estimate the next digit to be a 6, giving a with a machinist’s rule. measurement of 31.6 mm. Others might estimate the last digit to be 5 or 7; these small differences in the last digit are unavoidable because the last digit is estimated. Whatever you estimate the last digit to be, two digits of this measurement are known with certainty, the third digit is estimated, and the measurement has three significant digits. The photograph in Figure 2.4 shows a measurement in milliliters (mL) being taken with a piece of apparatus called a buret—a long glass tube used for measuring liquid volumes. Notice in this figure that when measuring liquid volume, the surface of the liquid curls up at the edge of the cylinder. This curved surface is called a meniscus. The liquid measurement must be made at the bottom of the meniscus for most liquids, including water. The scale on the buret shown is marked in increments of 0.1 mL. This means we estimate to the nearest 0.01 mL. To one person, the bottom of the meniscus (the black Figure 2.4. Reading the significant digits curve) may appear to be just below 2.2 mL, so that per- on a buret. 27

Chapter 2

son would call this measurement 2.21 mL. To someone else, it may seem that the bottom of the meniscus is right on 2.2, in which case that person would call the reading 2.20 mL. Either way, the reading has three significant digits and the last digit is estimated to be either 1 or 0. As a third example, Figure 2.5 shows a liquid volume measurement being taken with a piece of apparatus called a graduated cylinder. (We use graduated cylinders in an experiment we perform later on in this course.) The scale on the graduated cylinder shown is marked in increments of 1 mL. In the photo, the entire meniscus appears silvery in color with a black curve at the bottom. For the liquid shown in the figure, we know the first two digits of the volume measurement with certainty because the reading at the bottom of the meniscus is clearly between 82 mL and 83 mL. We have to estimate the third digit, and I Figure 2.5. Reading the significant digits would estimate the black line to be at 40% of the distance between 82 and 83, giving a reading of 82.4 mL. Someone on a graduated cylinder. else might read 82.5 mL, or even 82.6 mL. It is important for you to keep the significant digits rules in mind when you are taking measurements and entering data for your lab reports. The data in your lab journal and the values you use in your calculations and report must correctly reflect the use of the significant digits rules as they apply to the actual instruments you use to take your measurements. Note also the helpful fact that when a measurement is written in scientific notation, the digits written in the stem (the numerals in front of the power of 10) are the significant digits. Case 3 The third case of rules for significant digits applies to the calculations (multiplication and division) you perform with measurements. The main idea behind the rule for multiplying and dividing is that the precision you report in your result cannot be higher than the precision you have in the measurements to start with. The precision in a measurement depends on the instrument used to make the measurement, nothing else. Multiplying and dividing things cannot improve that precision, and thus your results can be no more precise than the measurements that go into the calculations. In fact, your result can be no more precise than the least precise value used in the calculation. The least precise value is, so to speak, the “weak link” in the chain, and a chain is no stronger than its weakest link. There are two rules for combining the measured values into calculated values, including any unit conversions that must be performed. Here are the two rules for using significant digits in our calculations in this course: Rule 1 Count the significant digits in each of the values you use in a calculation, including the conversion factors you use. (Exact conversion factors are not considered.) Determine how many significant digits there are in the least precise of these values. The result of your calculation must have this same number of significant digits.

Motion

Rule 1 is the rule for multiplying and dividing, which is what most of our calculations entail. (As I mentioned previously, there is another rule for adding and subtracting that you learn later in chemistry.) Rule 2 When performing a multi-step calculation, you must keep at least one extra digit during intermediate calculations and round off to the final number of significant digits you need at the very end. This practice ensures that small round-off errors don’t add up during the calculation. This extra digit rule also applies to unit conversions performed as part of the computation. As I present example problems in the coming chapters, I frequently refer to these rules and show how they apply to the example at hand. Get this skill down as soon as you can because soon you must use significant digits correctly in your computations to obtain the highest scores on your quizzes.

2.1.6 Scientific Notation In this course, we are assuming you already know how to use scientific notation in computations. However, you must also make sure you are correctly using the EE or EXP feature on your scientific calculator for executing computations that involve values in scientific notation. All scientific calculators have a key for entering values in scientific notation. This key is labeled EE or EXP on most calculators, but others use a different label.2 It is very common for those new to scientific calculators to use this key incorrectly, sometimes obtaining incorrect results. So read carefully as I outline the general procedure. The whole point of using the EE key is to make keying in the value as quick and errorfree as possible. When using the scientific notation key to enter a value, you do not press the × key, nor do you enter the 10. The scientific calculator is designed to reduce all this key entry, and the potential for error, by use of the scientific notation key. You only enter the stem of the value and the power on the ten and let the calculator do the rest. Here’s how. To enter a value, simply enter the digits and decimal in the stem of the number, then hit the EE key, then enter the power on the ten. The value is now entered and you may do with it as you wish. As an example, to multiply the value 7.29 × 109 by 25 using a standard scientific calculator, the sequence of key strokes is as follows: 7.29 EE 9 × 25 = Notice that between the stem and the power, the only key pushed is the EE key. When entering values in scientific notation with negative powers on the 10, the +/− key is used before the power to make the power negative. Thus, to divide 1.6 × 10−8 by 36.17, the sequence of key strokes is: 1.6 EE +/− 8 ÷ 36.17 =

One infuriating model uses the extremely unfortunate label x10x which looks a lot like 10x , a different key with a completely different function.

Chapter 2

Again, neither the “10” nor the “×” sign that comes before it is keyed in. The EE key has these built in. Students sometimes wonder why it is incorrect to use the 10x key for scientific notation. To execute 7.29 × 109 times 25, they are tempted to enter the following: 7.29 × 10x 9 × 25 = The answer is that sometimes this works, and sometimes it doesn’t, and calculator users must use key entries that always work. The scientific notation key ( EE ) keeps a value in scientific notation all together as one number. That is, when the EE key is used, the calculator regards 7.29 × 109 not as two numbers but as a single numerical value. But when the × key is manually inserted, the calculator treats the numbers separated by the × key as two separate values. This causes the calculator to render an incorrect answer for a calculation such as 3.0 ×106 1.5 ×106 The denominator of this expression is exactly half the numerator, so the value of this fraction is obviously 2. But when using the 10x key, the 1.5 and the 106 in the denominator are separated and treated as separate values. The calculator then performs the following calculation: 3.0 ×106 ×106 1.5 This comes out to 2,000,000,000,000 (2 × 1012), which is not the same as 2! The bottom line is that the EE key, however it may be labeled, is the correct key to use for scientific notation.

2.1.7 Problem Solving Methods Organizing problems on your paper in a reliable and orderly fashion is an essential practice. Physics problems can get very complex and proper solution practices can often make the difference between getting most or all the points for a problem and getting few or none. Each time you start a new problem, you must set it up and follow the steps according to the outline presented in the box on pages 32 and 33, entitled Universal Problem Solving Method. It is important that you always show all your work. Do not give in to the temptation to skips steps or take shortcuts. Develop correct habits for problem solving and stick with them!

2.2

Motion

In this course, we address two types of motion: motion at a constant velocity, when an object is not accelerating, and motion with uniform acceleration. Defining these terms is a lot simpler if we stick to motion in one dimension, that is, motion in a straight line. So in this course, this is what we do.

Motion

2.2.1 Velocity When thinking about motion, one of the first things we must consider is how fast an object is moving. The common word for how fast an object is moving is speed. A similar term is the word velocity. For the purposes of this course, you may treat these two terms as synonyms. The difference is technical. Technically, the term velocity means not only how fast an object is moving, but also in what direction. The term speed refers only to how fast an object is moving. But since we only consider motion in one direction at a time, we can use the terms speed and velocity interchangeably. An important type of motion is motion at a constant velocity, as with a car with the cruise control on. At a constant velocity, the velocity of an object is defined as the distance the object travels in a certain period of time. Expressed mathematically, the velocity, v, of an object is calculated as v=

d t

The velocity is calculated by dividing the distance the object travels, d, by the amount of time, t, it takes to travel that distance. So, if you walk 5.0 miles in 2.0 hours, your velocity is v = (5.0 miles)/(2.0 hours), or 2.5 miles per hour. Notice that for a given length of time, if an object covers a greater distance it is moving with a higher velocity. In other words, the velocity is proportional to the distance traveled in a certain length of time. When performing calculations using the SI System of units, distances are measured in meters and times are measured in seconds. This means the units for a velocity are meters per second, or m/s. The relationship between velocity, distance, and time for motion at a constant velocity is shown graphically in Figure 2.6. Travel time is shown on the horizontal axis and distance traveled is shown on the vertical axis. The steeper curve3 shows distances and times for an v = 2 m/s

v = 0.5 m/s

distance (m)

6 5 4 3 2 1 1

6 7 8 9 10 11 12 time (s) Figure 2.6. A plot of distance versus time for an object moving at constant velocity. Two different velocity cases are shown.

The lines or curves on a graph are all referred to as curves, whether they are curved or straight.

Chapter 2

Universal Problem Solving Method Solid Steps to Reliable Problem Solving In ASPC, you learn how to use math to solve scientific problems. Developing a sound and reliable method for approaching problems is crucial. The problem solving method shown below is used in scientific work everywhere. Always follow every step closely and show all your work. 1.

3. 4. 5. 6. 7. 8. 9.

Write down the given quantities at the left side of your paper. Include the variable quantities given in the problem statement and the variable you must solve for. Make a mental note of the precision in each given quantity. For each given quantity that is not already in MKS units, work immediately to the right of it to convert the units of measure into MKS units. To help prevent mistakes, always use horizontal fraction bars in your units and unit conversion factors. Write the results of these unit conversions with one extra digit of precision over what is required in your final result. Write the standard form of the equation to be used in solving the problem. If necessary, use algebra to isolate the variable you are solving for on the left side of the equation. Never put values into the equation until this step is done. Write the equation again with the values in it, using only MKS units, and compute the result. If you are asked to state the answer in non-MKS units, perform the final unit conversion now. Write the result, with the correct number of significant digits and the correct units of measure. Check your work. Make sure your result is reasonable.

Motion

Step 2 Perform the needed unit conversions, writing the conversion factors to the right of the given quantities you wrote in the previous step. Use only horizontal bars in unit fractions. mi 1609 m 1 hr m ⋅ ⋅ = 15.6 hr mi 3600 s s m a = 0.15 2 s t = 2 min 10.0 s = 130.0 s vf = ? vi = 35.0

Step 3 Write the equation to be used in its standard form. a=

v f − vi t

Step 4 Perform the algebra necessary to isolate the unknown you are solving for on the left side of the equation. a=

v f − vi

t at = v f − vi v f = vi + at Step 5 Using only values in MKS units, insert the values and compute the result. v f = vi + at = 15.6

m m m + 0.15 2 ⋅130.0 s = 35.1 s s s

Step 6 Convert to non-MKS units, if required in the problem. v f = 35.1

m 1 mi 3600 s mi ⋅ ⋅ = 78.5 s 1609 m 1 hr hr

Chapter 2

object moving at 2 m/s. At a time of one second, the distance traveled is two meters because the object is moving at two meters per second (2 m/s). After two seconds at this speed, the object has moved four meters: (4 m)/(2 s) = 2 m/s. And after three seconds, the object has moved six meters: (6 m)/(3 s) = 2 m/s. The right-hand curve in Figure 2.6 represents an object traveling at the much slower velocity of 0.5 m/s. At this speed, the graph shows that an object travels two meters in four seconds, four meters in eight seconds, and so on. To see this algebraically, look again at the velocity equation above. This equation can be written as d = vt Written this way, t is the independent variable, d is the dependent variable, and v serves as the slope of the line relating d to t. With this form of the velocity equation, we can calculate how far an object travels in a given amount of time, assuming the object is moving at a constant velocity. Now we work a couple of example problems, following the problem-solving method described on pages 32–33. And remember, all the unit conversion factors you need are listed in Appendix A. Example 2.1 Sound travels 1,120 ft/s in air. How much time does it take to hear the crack of a gun fired 1,695.5 m away? First, write down the given information and perform the required unit conversions so that all given values are in MKS units. Check to see how many significant digits your result must have and do the unit conversions with one extra significant digit. The given speed of sound has three significant digits, so we perform our unit conversions with four digits. ft 0.3048 m m ⋅ = 341.4 s ft s d = 1695.5 m t =? v = 1120

Next, write the appropriate equation to use. v=

d t

Perform any necessary algebra, insert the values in MKS units, and compute the result. t=

d 1695.5 m = = 4.966 s v 341.4 m s

Motion

result must be reported with three significant digits, but all intermediate calculations must use one extra digit. This is why I use four digits. But now we have he result, and it must be rounded to three significant digits because the least precise measurement in the problem has three significant digits. Rounding our result accordingly, we have t = 4.97 s The final step is to check the result for reasonableness. The result should be roughly the same as 1500/300 or 2000/400, both of which equal 5. Thus, our result makes sense.

v f − vi t

where a is the acceleration (m/s2), t is the time spent accelerating (s), and vi and vf are the initial and final velocities, respectively, (m/s). Notice that the MKS units for acceleration are meters per second squared (m/s2). These units sometimes drive students crazy, so we pause here to discuss what this means so you can sleep peacefully tonight. I mention just above that the acceleration is the rate at which the velocity is changing. The acceleration simply means that the velocity is increasing by so many meters per second, every (per) second. Now, “per” indicates a fraction, and if a velocity is changing so many meters per second, per second, we write these units in a fraction this way and simplify the expression: m m s = s = m ⋅1 = m s s s s2 s 1 Because the acceleration equation results in negative accelerations when the initial velocity is greater than the final velocity, you can see that a negative value for acceleration means the object is slowing down. In future physics courses, you may learn more sophisticated interpretations for what a negative acceleration means, but in this course you are safe associating negative accelerations with decreasing velocity. In common speech, people sometimes use the term “deceleration” when an object is slowing down, but mathematically we just say the acceleration is negative. Before we work through some examples, let’s look at a graphical depiction of uniform acceleration the same way we did with velocity. Figure 2.7 shows two different acceleration curves, representing two different acceleration values. For the curve on the right, after 1 s 35

Chapter 2 a = 4 m/s2

a = 1 m/s2

velocity (m/s)

12 10 8 6 4 2 1

6 7 8 9 10 11 12 time (s) Figure 2.7. A plot of velocity versus time for an object accelerating uniformly. Two different acceleration cases are shown.

ball falling

ball rising

the object is going 1 m/s. After 2 s, the object is going 2 m/s. After 12 s, the object is going 12 m/s. You can take the velocity that corresponds to any length of time (by finding where their lines intersect on the curve) and calculate the acceleration by dividing the velocity by the time to get a = 1 m/s2. The other curve has a higher acceleration, 4 m/s2. An acceleration of 4 m/s2 means the velocity is increasing by 4 m/s every second. Accordingly, after 2 s the velocity is 8 m/s, and after 3 s, the velocity is 12 m/s. No matter what point you select on that curve, v/t = 4 m/s2. We must be careful to distinguish between velocity (m/s) and acceleration (m/s2). Acceleration is a measure of how fast an object’s velocity is changing. To see the difference, note that an object can be at rest (v = 0) and accelerating at the same instant. Now, although you may not see this at first, it is important for you to think this through and understand how this counter-intuitive situation Right here at can come about. Here are two examples. The instant the top the ball an object starts from rest, such as when the driver hits is at rest for an the gas while sitting at a traffic light, the object is siinstant, but still multaneously at rest and accelerating. This is because accelerating if an object at rest is to ever begin moving, its velocbecause of the pull ity must change from zero to something else. In other of earth’s gravity. words, the object must accelerate. Of course, this situation only holds for an instant; the velocity instantly begins changing and does not stay zero. Perhaps my point will be easier to see with this second example. As depicted in Figure 2.8, when a ball is thrown straight up and reaches its highest point, it stops for an instant as it starts to come back down. At its highest point, the ball is simultaneously at rest and accelerating due to the force of gravity pulling it down. As before, this situation only holds for a single instant. Figure 2.8. A rising and falling ball The point of these two examples is to help you illustrates the difference between velocity understand the difference between the two variables and acceleration. 36

Motion

we are discussing, velocity and acceleration. If an object is moving at all, then it has a velocity that is not zero. The object may or may not be accelerating. But acceleration is about whether the velocity itself is changing. If the velocity is constant, then the acceleration is zero. If the object is speeding up or slowing down, then the acceleration is not zero. And now for another example problem, this time using the acceleration equation. Example 2.2 A truck is moving with a velocity of 42 mph (miles per hour) when the driver hits the brakes and brings the truck to a stop. The total time required to stop the truck is 8.75 s. Determine the acceleration of the truck, assuming the acceleration is uniform. Begin by writing the givens and performing the unit conversions. vi = 42 vf = 0

mi 1609 m 1 hr m ⋅ ⋅ = 18.8 hr mi 3600 s s

t = 8.75 s a=? Now write the equation and complete the problem. a=

v f − vi t

m s = −2.15 m 8.75 s s2

0 −18.8

The initial velocity has two significant digits, so I perform the calculations with three significant digits until the end. Now we round off to two digits giving a = −2.2

m s2

If you keep all the digits in your calculator throughout the calculation and round to two digits at the end, you have −2.1 m/s2. This answer is fine, too. Remember, the last digit of a measurement or computation always contains some uncertainty, so it is reasonable to expect small variations in the last significant digit. A check of our work shows the result should be about –20/10, which is –2. Thus the result makes sense. One more point on this example: Notice that the calculated acceleration value is negative. This is because the final velocity is lower than the initial velocity. Thus we see that a negative acceleration means the vehicle is slowing down.

If you haven’t yet read the example problem in the yellow Universal Problem Solving Method box on page 32, you should read it now to see a slightly more difficult example using this same equation.

Chapter 2

2.2.3 Graphical Analysis of Motion Analyzing motion graphically is a powerful tool. When you understand the graphical analysis in this section, you will be off to a solid start in being able to think conceptually and quantitatively about motion the way a student of physics should be able to do. The graphs in Figure 2.9 show representative curves for three different motion states an object can be in: at rest (no motion), moving at a constant velocity, and accelerating uniformly. Each vertical group of curves depicts distance, velocity, and acceleration as functions of time. In the first group, the object is at rest, which means the distance from the object to the “starting line” (from which we measure how far it has gone) is a constant. The only way this can happen is for the object to be at rest. (Well, yes, technically the object could be moving in a circle, but we are not going to consider that in this course!) v = constant (constant velocity)

time

Figure 2.9. Graphical depictions of states of motion.

time

acceleration

time

acceleration

time

velocity

time

a = constant (uniform acceleration)

distance

d = constant (at rest)

time

Motion

distance

In the second group, the velocity is a constant (and not zero). This means the distance to the starting line is changing at a constant rate, and the object is not accelerating. In the third group, the object is accelerating uniformly, which means its velocity is changing at a constant rate. Notice that I always start the distance graphs at an arbitrary point. This is just to make the curves as generic as possible. When drawing your own curves if you wish to start the distance curves at the origin that is fine. However, the origin on a velocity graph represents a velocity of zero, meaning the object is at rest. Thus, you can only start a velocity graph at the origin if you are depicting an object that is starting from rest. Interestingly, when an object is accelerating the distance graph is no longer linear. Instead, it is a type of curve called a quadratic. We address this type of curve in Chapter 4, but essentially it is the type of curve that occurs when the relationship between y and x, that is, between the distance and time in this case, can be modeled by an equation such as y = kx 2, or, in our specific case, d = kt 2. In an equation like this, k is simply a constant that depends on the circumstances. In these time diagrams, the distance graph is the only one that is ever curved. All the others are linear. Also make note that in this course if an object is accelerating the acceleration is always uniform. On graphs like these, this means the acceleration is always a horizontal line. This horizontal line is at a positive value when the object is speeding up and at a negative value when the object is slowing down. As I mention above, when an object is accelerating the graph of the object’s distance vs. time has a quadratic curvature. This curvature makes this graph more complex than the others, so we now look more closely at this type of graph. There are four ways this graph can curve, depending on what the object is doing, shown in Figure 2.10. The first thing to notice is that if the object is going forward the distance is increasing, so the curve slopes upward. If the slope is getting steeper, the object is speeding up. If it is getting less steep, the object is slowing down. The only way the curve can slope downward is if the object is going backwards, so that the distance to the starting line is decreasing. Just as before, if the curve is getting steeper, the object is going faster. If the curve is getting less steep (more horizontal), the object is slowing down. Figure 2.11 highlights additional details of distance and velocity graphs. When a distance graph curves all the way over to horizontal it means the object stops. If it stays stopped, then the distance graph becomes a horizontal line. A downward sloping velocity graph means the object is slowing, but if the velocity curve actually goes below the horizontal axis that means the velocity is negative and the object is going in reverse. When given a description of an object’s motion for a graphing exercise, your task is to piece together segments from different representative curves to represent motion in different time intervals. For example, a vehicle could be traveling at one velocity, accelerate for a

time

going forward and speeding up

going forward and slowing down

going backward and speeding up

going backward and slowing down

Figure 2.10. Curvature possibilities for distance vs. time graphs.

velocity

distance

Chapter 2

time

object slows but does not stop

object slows and comes to a complete stop

object slows to a complete stop and remains at rest

time object slows to a stop and reverses direction

Figure 2.11. Details for distance and velocity graphs.

while, and then travel at a new velocity. In such a case, there are three distinct time intervals associated with this motion—one for the constant speed at the beginning, one for the acceleration in the middle, and one for the new constant speed motion at the end. Example 2.3

velocity

There are three distinct time intervals in this scenario. First, there is a period of time at the initial constant velocity. Then there is an interval when the car is accelerating. Finally, the car continues at the new velocity.

distance

Consider a car driving down the road at a constant velocity. The driver then accelerates uniformly to a higher velocity and continues at this new, higher velocity. Draw diagrams of d vs. t, v vs. t, and a vs. t decar moving at car car moving at constant, picting this scenario. Show the constant velocity speeding up higher velocity time intervals distinctly in your diagrams and align your time intervals vertically.

acceleration

The graphical depiction of this sequence of events is shown in Figure 2.12. The three graphs (distance, velocity and acceleration vs. time) are drawn above one another so the time intervals can be aligned in each graph. The three time intervals are separated by the dashed lines. Notice some key details. First, on the distance graph, the slope is higher in the third intime terval than in the first because the car’s velocity is higher. Sec- Figure 2.12. Combining time intervals to make a complete graph of ond, the two linear sections on an object’s motion. 40

Motion

the distance graph are smoothly connected to the curved (quadratic) section in the middle. There should never be kinks (sharp corners) in a distance graph. The quadratic curvature only occurs in the distance graph, and only when the car is accelerating. Finally, the acceleration graph is zero everywhere except in the middle when the car is accelerating, and there the curve is a horizontal line representing positive, uniform acceleration.

2.3

Planetary Motion and the Copernican Revolution

2.3.1 Science History and the Science of Motion People have been fascinated with the heavens since ancient times. Major cultures all over the globe and throughout history have engaged in mapping the stars, predicting celestial events such as solar and lunar eclipses, observing the difference between the unified movement of the stars and the individual movements of the planets, observing unusual events such as the appearance of comets and supernovae, and speculating on the cause of the movements of the planets and stars in the night-time sky. Babylonians, Egyptians, Persians, Chinese, Maya, Greeks, Indians, and other indigenous people of North and South America are some of the major cultures with records of astronomical study. The study of the heavens is one of the oldest natural sciences. The study of motion has always been associated with the motion of the heavenly bodies we see in the sky, so it is particularly fitting in this chapter on motion for us to review the history of views about the solar system and the rest of the universe, referred to as “the heavens” by those in ancient times. As we will see, the particular episode known as the Copernican Revolution was a pivotal moment in that history and was the setting for the emergence of our contemporary understanding of scientific epistemology—what knowledge is and how we know what we know. As you recall, Chapter 1 addresses the Cycle of Scientific Enterprise and examines the way science works. From that discussion you know that science is an ongoing process of modeling nature—at least that is the way we understand science now. We now understand that scientists use theories as models of the way nature works, and over time theories change and evolve as scientists learn more. Sometimes scientists find that a theory is so far off the mark that they have to toss it out completely and replace it with a different one. The present general understanding among scientists that science is a process of modelling nature took hold around the beginning of the 20th century. The ideas that led to this understanding began to emerge at the time of the Copernican Revolution in the 16th and 17th centuries. But since natural philosophy was then entering new territory, there was a period of difficult struggle that involved both theologians and philosophers. There are a lot of misconceptions about what happened at that time. The conflict in Galileo’s day is often regarded as a fight between faith and science, and these misconceptions have led many people in today’s world to the position that faith is dead and only science gives us real knowledge. But that depiction is not even close to what really happened, and that belief about science is not even close to the truth. The real issue with Galileo was about epistemology. The so-called “faith versus science” debate rages today as much as ever, so it is worth spending some time to understand that crucial period in scientific history.

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2.3.2 Aristotle The study of astronomy and astrology dates back to the ancient Babylonians, but we pick up the story with the ancient Greeks and the Greek philosopher Aristotle in the 4th century BCE (Figure 2.13). Aristotle was a highly influential philosopher who wrote a lot about philosophy, physics, biology, and other fields of learning. Back then, science was called natural philosophy and there was really no distinction between scientists and philosophers. That time was also many centuries before experiments became part of scientific research. Natural philosophy did involve making observations about the world, but the conclusions reached by ancient philosophers like Aristotle were based simply on observation and philosophical thought. It was still about 2,000 years before natural philosophers realized that the way things appear to our ordinary senses might not be the way they actually are and that to understand more about the world required scientific experiments. For example, if you just walk outside and quietly look around you notice that the earth does not appear to be in motion; it feels solid and at rest. The sun, planets, and stars appear to move across the sky each day. In fact, watching a sunrise gives the distinct Figure 2.13. Greek philosopher impression that the sun is moving up and then across the sky. Aristotle (384–322 BCE). Today, we understand things differently, but that is the result of the revolution we are about to explore and the experimental science that emerged at that time. Aristotle’s ideas were grounded in the concept of telos—a Greek term meaning purpose, goal, or end. Aristotle believed that each thing that exists has its own telos, and that an object’s behavior was to be understood in terms of its particular telos. He applied this idea to everything from falling objects to the rising flames from a fire. Aristotle observed the serene beauty of the stars, the planets, the sun, and moon as they appear majestically to rotate around the earth day after day. He also noticed that nothing in the heavens ever seems to change. Other than the motions of the heavenly bodies, everything in the heavens seems to be pure and eternal. On earth, of course, Aristotle was surrounded by change: decay, corruption, birth, and death are all around. Animals and plants live and die, forests grow and burn, rivers flow and flood, storms come and go. These observations led Aristotle to conclude that change and corruption occur only on the earth. He wrote that imperfection and change of any kind occur only on the earth, while the heavens are pure and unchanging. Aristotle taught that the heavenly bodies—planets, stars, sun, and moon—are eternal and perfect. Further, he said that their motions must be in perfect circles since the circle is the purest and most perfect geometric shape. He conceived of the sun, moon, and planets as inhabiting celestial spheres, centered on the earth, one inside the other—an exquisite geocentric (earth-centered) system. Aristotle was a tremendous moral philosopher whose ideas still have a profound influence on us today. Back in ancient times, he was regarded so highly that questioning his ideas was virtually unthinkable. Thus, his views about the heavenly motions became the basis for all further work on understanding the motions of the heavenly bodies.

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2.3.3 Ptolemy In the second century CE, the famous Alexandrian astronomer Claudius Ptolemy (Figure 2.14) worked out a detailed mathematical system based on Aristotle’s ideas. (By the way, the “P” in Ptolemy is silent.) Many Greek philosophers, including Ptolemy, believed the cosmos to be pervaded by a moral order. Contemplating the heavens and studying the mathematics were stages toward contemplation of the divine. Ptolemy’s practical goal was to be able to make predictions about the movements of the planets and stars, along with other astronomical events such as eclipses, because these events were widely believed to be omens signifying important events on earth. Ptolemy started with Aristotle’s basic ideas and de- Figure 2.14. Alexandrian astronomer veloped a complex mathematical system—a model—that Claudius Ptolemy (c. 100–170 CE). was quite effective in making the desired predictions. There were other astronomers around that time who developed different systems, but Ptolemy’s system became the most widely accepted understanding of the heavens for over a thousand years.

2.3.4 The Ptolemaic Model The basic structure of Ptolemy’s geocentric model of the heavens is depicted in Figure 2.15. As with Aristotle, there are seven heavenly bodies, each inhabiting a sphere centered on the earth. Each of the heavenly bodies is also itself a perfect sphere. The contents of the spheres are summarized in Table 2.3. The first seven spheres contain the five planets (not includEmpyrean ing the earth), the sun, and the Primum Mobile Firmament moon. Sphere 8 contains the soSaturn called Firmament, the fixed layer Jupiter of stars. The stars do not move Mars relative to each other; their positions are fixed and they rotate Sun as a body in the 8th sphere each Venus day. Within the firmament, the Mercury stars are arranged according to Moon the zodiac, a belt of twelve constellations around the earth. The Earth term zodiac derives from the Latin and Greek terms meaning “circle of animals,” and is so named because many of the constellations in the zodiac represent animals. The ninth sphere contains the Primum Mobile, which is Latin for “prime mover” (or “first mover”). The Primum Figure 2.15. The Ptolemaic model of the heavens. 43

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Mobile is the sphere set into motion by God or the gods. As the Sphere 2 Mercury Primum Mobile turns, it pulls all Sphere 3 Venus the other spheres with it, making them rotate as well. Outside the Sphere 4 Sun ninth sphere is the so-called EmSphere 5 Mars pyrean, the dwelling place of God Sphere 6 Jupiter or the gods. There is a great deal more Sphere 7 Saturn to the model than shown in the Sphere 8 The Firmament. This region consists of the diagram of Figure 2.15. This is stars arranged in their constellations acbecause all seven of the heavenly cording to the zodiac. bodies appear to move around Sphere 9 The Primum Mobile. This Latin name means in the nighttime sky against the “first mover.” This sphere rotates around the background of the fixed stars. If earth every 24 hours and drags all the oth- all the heavenly bodies simply er spheres with it, making them all move. moved in their spheres around Beyond The Empyrean. This is the region beyond the earth together once each day, the spheres. The Empyrean is the abode of there would be no way to account God, or the gods. for why the planets’ positions change relative to the stars. PtolTable 2.3. Contents of the spheres in the Ptolemaic model. emy accounted for the changes by a system of epicycles. An epicycle is a circular planetary orbit with its center moving in planet epicycle a separate circular path, as depicted in Figure 2.16. As the center of an epicycle moves along its path in the sphere, the planet in the epicycle rotates about the center of the epicycle, as if the epicycle were a wheel rolling around a path centered on the earth. earth To help you understand why epicycles are necessary in Ptolemy’s model, we discuss them in more detail in the next section. A planet moving in an epiSphere 1

Moon

plan

etary sphere

Figure 2.16. A planet moving in a path defined by an epicycle around the earth.

cycle moves in a path similar to a person riding in a “tea cup ride” at an amusement park, like the one pictured in Figure 2.17. To account for the complex motions of Figure 2.17. The people in the cups spin in a circle while the heavenly bodies, Ptolemy’s model con- the cup moves in a larger circle, motion like that of a tained some 80 different epicycles. Ptol- planet moving on an epicycle. 44

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The Main Principles in Ptolemy’s Celestial Model 1. 2. 3. 4. 5. 6. 7.

There are seven heavenly bodies. All the heavenly bodies move in circular orbital regions called spheres. In the model, there are nine spheres plus the region beyond the spheres, with contents as listed in Table 2.3. All the heavenly bodies are perfectly spherical. All the spheres are centered on the earth, so this system is a geocentric system. Corruption and change exist only on earth. All other places in the universe, including all the heavenly bodies and stars, are perfect and unchanging. All the spheres containing the heavenly bodies and all the stars in the Firmament rotate completely around the earth every 24 hours. Epicycles are used to explain the motion of the planets relative to the stars.

Figure 2.18. The main principles in the Ptolemaic model.

emy’s model has some of the planets in an epicycle riding on the rim of another epicycle, which in turn moved in the sphere around the earth. Ptolemy’s system was mathematically very complex, but its genius was that it worked pretty well! The main features of Ptolemy’s model are summarized in the box in Figure 2.18. Among the different astronomers of the ancient world, there were those who held to variations on this basic model. For example, some astronomers reckoned that Mercury and Venus orbited the sun while the other heavenly bodies orbited the earth. But the basic Ptolemaic model is as described in Figure 2.15.

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on the equator). For this to be the case, it seemed that we would be hanging on for dear life! The trees would be laying down and we would constantly feel winds that make a hurricane seem like a calm summer day! For all these reasons, it did not seem reasonable to believe that the apparent motion of the heavenly bodies across the sky every day was due to the earth’s rotation. These arguments seemed obvious to nearly everyone before 1500, and to everyone except a few cutting-edge astronomers right up to the end of the 17th century. Only a crazy person imagined that the earth spins, and people used these arguments all the way up to the time of Galileo to prove that the earth was not orbiting the sun and spinning around once a day. Back then, these were persuasive arguments. Forward and Retrograde Motion

The second item to Figure 2.19. Greek mathematician

consider here has to do with the apparent motion of the and geographer Eratosthenes (c. planets in the sky against the background of the stars. If 276–194 BCE). you go out and look at, say, Mars each night and make a note of its location against the stars, you see that it is in a slightly different place each night. The planet gradually works its way along in a pathway against the starry background night after night. If you track the planet for several months or a year, it moves quite far. As mentioned above, Ptolemy used epicycles to account for this forward motion of planets against the background of fixed stars. Going back to watching Mars, if you follow the planet’s progress long enough, you see that there are periods of time lasting several weeks when the nightly progress of the planet reverses course. Mars appears to be backing Firmament up! This apparent backsequence of ing up is called retroa observations b grade motion. Ptolemy retrograde used epicycles to acmotion c count for this, too. a Figure 2.20 illusb c trates how epicycles 3 are used in the geo3 centric system to acsequence of Earth 2 observations count for the planetary 2 motions—both forforward 1 ward and retrograde— motion Mars 1 against the background Sphere 5 of the fixed stars. Mars (Mars) is shown in red moving on an epicycle, while the center of the epicycle moves around the Figure 2.20. Using epicycles to explain the forward and retrograde motion of earth. The dashed lines heavenly bodies against the background of fixed stars.

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are the lines of sight from earth to Mars, and the letters and numbers outside the 3 firmament show the sequence of observations locations where Mars 3 2 1 2 appears among the forward 1 2 1 motion c stars at different times. The lower right 3 c part of the diagram b a b shows Mars in three sequence of Sun a b observations a Mars locations (labeled 1, 2, c Earth retrograde and 3) over the course motion of a few weeks. Compared to the background of fixed stars, Mars exhibits forward motion during a sequence of nighttime observations. Figure 2.21. Explanation of forward and retrograde motion in the Copernican The upper right system. part of the figure shows Mars’ locations during a different sequence of observations (a, b, and c) some months later. Mars is now on the other side of its epicycle. The center of the epicycle continues to move in the same direction in its sphere around earth. But since Mars is on the near side of its epicycle, the sequence of observations of its location against the starry background—a, b, and c—maps along the starry background in the opposite direction. This apparent motion of Mars in the opposite direction is retrograde motion. While we are on the subject, we may as well look at how forward and retrograde motion are explained in a heliocentric system—a system in which the planets orbit the sun. The system introduced by Copernicus is a heliocentric system. Assuming that the earth moves faster in its orbit than Mars (which is correct), the explanation is straightforward. As shown in the upper part of Figure 2.21, when the earth and Mars are on opposite sides of their orbits, the observations of Mars’ location against the stars exhibit forward motion. But when the earth and Mars are on the same side of the sun, as in the center-right part of the figure, the earth’s greater velocity makes Mars’ position against the stars exhibit retrograde motion. To summarize, none of the planets actually reverses course in its orbit, and neither the geocentric nor heliocentric models depict planets as reversing direction. But depending on the system, the presence of epicycles and the relative locations of earth and a planet can combine to account for the appearance of forward or retrograde motion of the planet against the fixed background of the stars.

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very well with certain passages in the Bible. This circumstance led theologians in this tradition to assume that such passages were to be interpreted as literal descriptions of the motions of the heavenly bodies. Here are a few examples of passages that seem to describe the earth as motionless, with the sun and stars going around the earth: He set the earth on its foundations, so that it should never be moved (Psalm 104:5). He made the moon to mark the seasons; the sun knows its time for setting (Psalm 104:19). [The sun’s] rising is from the end of the heavens, and its circuit to the end of them (Psalm 19:6). The sun rises and the sun goes down, and hastens to the place where it rises (Ecclesiastes 1:5). Additionally, other features in the Ptolemaic model (derived from Aristotle) seemed to line up with biblical symbolism. For example: •

Seven was understood to be the biblical number symbolizing perfection, so it made sense that God’s creation contains seven heavenly bodies.

•

Circles were considered to be the most perfect shape, regarded as divine from the times of the ancient Greeks, so the spherical bodies inhabiting spheres in which they move seemed to reflect the perfection of their creator.

•

Corruption was thought to exist only on earth, and it seemed this was obviously because of the curse that resulted from the Fall of man described in the book of Genesis.

The result of such teaching was that many theologians assumed that the biblical passages and doctrines described above, along with the Ptolemaic model of the heavens, were literal descriptions of the true nature of reality. To these theologians, anyone who had different ideas about the heavens—such as, for example, the idea that the earth moved and orbited the sun—should be censored and prevented from spreading teachings they felt were unbiblical. Although widespread, this tradition of associating the Ptolemaic model with the Bible was by no means universal. Many theologians took a completely different position, including the great theologian and philosopher Augustine, a bishop in northern Africa in the 4th and 5th centuries in the common era. An insightful and relevant passage from Augustine is found in his book On the Literal Meaning of Genesis: Usually, even a non-Christian knows something about the earth, the heavens, and the other elements of this world, about the motion and orbit of the stars and even their sizes and relative positions, about the predictable eclipses of the sun and moon, the cycles of the years and the seasons, about the kinds of animals, shrubs, stones, and so forth, and this knowledge he holds to as being certain from reason and experience. Now it is a disgraceful and dangerous thing for an infidel to hear a Christian, presumably giving the meaning of Holy Scripture, talking nonsense on these topics, and we should take all means to prevent such an embarrassing situation, in which people show up vast ignorance in a Christian and laugh it to scorn. The shame is not so much that an ignorant individual is derided, but that people outside the household of faith think our sacred writers held such opinions, and, to the great loss of those for whose salvation we toil, the writers of our Scripture are criticized and rejected as unlearned men.

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As we open the curtain now on the rest of our story, it is key to remember that many church theologians were strong supporters of those engaged in natural philosophy. The Roman Catholic Church—which figures prominently in these events—had a long tradition of supporting intellectual inquiry, including natural philosophy, and many of the individual theologians in the church were admirers of the scientists involved in these events.

2.3.7 Copernicus and Tycho Nicolaus Copernicus (Figure 2.22), a Polish astronomer, first proposed a detailed, mathematical, heliocentric model of the heavens, with the earth rotating on its axis, all the planets moving in circular orbits around the sun, and the moon orbiting the earth. Copernicus’ system was about as accurate—and about as complex—as the Ptolemaic system. The Copernican model still used circular orbits and because of this he still had to use epicycles to make the model accurate. Still, the model is an arrangement that is a lot closer to today’s understanding than the Ptolemaic model is. Copernicus dedicated his famous work On the Revolutions of the Heavenly Spheres to Pope Paul III. This dedication indicates that the Roman Catholic Church itself was not opposed to Copernicus’ ideas. Figure 2.22. Polish astronomer Nicolaus Nevertheless, Copernicus knew there were scholars in Copernicus (1473–1543). the Church who were strongly opposed to the suggestion that the earth moved. Being a sensitive man, he didn’t want to cause trouble so he published his work privately to his close friends in 1514. Just before Copernicus’ death in 1543, his student and admirer, mathematician and astronomer Georg Joachim Rheticus, persuaded Copernicus to publish the work. Rheticus delivered the manuscript to the printer and brought proofs back to Copernicus to review. Rheticus was not continuously present with the printer, and during his absence a theologian named Andreas Osiander added an unsigned “note to the reader” to the front of Copernicus’ book stating that the heliocentric ideas were hypotheses (although theory is the better term, since we are talking about a model) that were useful for the purpose of performing computations and not descriptions of actual reality. Because of this note, people generally thought that it expressed Copernicus’ own viewpoint. However, Rheticus was outraged by the addition and marked it out with a red crayon in the copies he sent to people. Copernicus did not live to see the final printed version of his book, but Rheticus’ reaction to Osiander’s note suggests that Copernicus regarded his model as more than merely an imaginary convenience that made computations easier. Tycho Brahe (Figure 2.23), was a Danish nobleman and astronomer. Tycho4 built a magnificent observatory called the Uraniborg on an island Denmark ruled at the time. This observatory is depicted in Figure 2.24.

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Tycho was a passionate and hotheaded guy, as evidenced by the fact he had the bridge of his nose cut off in a duel. (You can see his prosthesis in Figure 2.23 if you look closely.) Even though Tycho’s Uraniborg must have been the most palatial observatory in the world, he had a falling out with the new King of Denmark and decided to leave. In 1597, Tycho moved to Prague in Bohemia (the modern-day Czech Republic) and became Imperial Mathematician for Rudolph II, King of Bohemia and Holy Roman Emperor there. Tycho spent his life cataloging astronomical data for over 1,000 stars (with cleverly contrived instruments, but only a primitive telescope). His work was published much later (1627) by Johannes Kepler in a new star catalog that identified the positions of these stars with unprecedented accuracy. Tycho witnessed and recorded two astronomical events that became historically very important. First, in 1563 he observed a conjunction between Jupiter and Figure 2.23. Danish astronomer Tycho Saturn. A conjunction, illustrated in Figure 2.25, occurs Brahe (1546–1601). when two planets are in a straight line with the earth so that from earth they appear to be in the same place in the sky. Tycho predicted the date for this conjunction using Copernicus’ new heliocentric model. The prediction was close (this is good) but was still off by a few days (not so good). The error indicated there was still something lacking in Copernicus’ model. (There was: the orbits are not circular as Copernicus assumed.) Second, in 1572 Tycho observed what he called a “nova” (which is Latin for new; today we call this event a supernova) and proved that it was a new star. This discovery rocked the Renaissance world because it was strong evidence that the stars are not perfect and unchanging as Aristotle had thought and as the Ptolemaic model of the heavens declared. Although familiar with Copernicus’ model, Tycho was a proud advocate of his own model, in which the sun and moon orbit the earth and the other planets orbit the sun, which in turn orbits the earth. His model did have the advantage of maintaining a stationary earth, which allowed Tycho to avoid controversy with those who insisted that the Bible taught that the earth did not move. Tycho also had a good technical reason for rejecting Copernicus’ model. If the earth moves in an orbit, then earth’s location is different in the summer from its location in the winter. This means the relative positions of Figure 2.24. Tycho’s Danish observatory, the Uraniborg. the stars should be slightly differ50

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ent at these different times of the year, an effect called stellar parallax. (As an analogy, imagine yourself looking at the trees in a forest. If you take a few steps to one side, the positions of the trees relative to each other in your new location are different.) At that time, no stellar parallax had been observed, and Tycho knew this meant that either the earth was stationary or the stars were incredibly far away. Copernicus had accepted the great distance of the Figure 2.25. The alignment of three planets, called a stars but Tycho did not, and famously won- conjunction. dered, “What purpose would all that emptiness serve?” In fact, stellar parallax was not observed until 1838, when telescopes were finally up to the task. The discovery of stellar parallax in 1838 was one of the first pieces of actual evidence that Copernicus was right. It helps to keep this in mind when we get to the controversy surrounding Galileo.

2.3.8 Kepler and the Laws of Planetary Motion Johannes Kepler (Figure 2.26), a German astronomer and mathematician, was invited in 1600 to join the research staff at Tycho’s observatory in Prague and replaced Tycho as Imperial Mathematician there the following year, after Tycho’s death. Kepler had access to Tycho’s massive body of research data and used it to develop his famous three laws of planetary motion, the first two of which were published in 1609. He discovered the third law a few years later and published it in 1619. Today, Kepler’s laws of planetary motion remain the currently accepted model describing our solar system. Kepler was a man of faith, but he was caught in the middle during the Counter-Reformation, a time of serious conflict between Roman Catholics and Protestants. His unwillingness to align with either side in the conflict meant that he had to move around a lot, and this is why Figure 2.26. German astronomer and he eventually ended up in Prague with Tycho. Kepler was mathematician Johannes Kepler an amazing and creative scientist who was convinced (1571–1630). that there were mathematical relationships that could be applied to the motions and positions of the heavenly bodies. In addition to his astronomical discoveries, Kepler made important discoveries in geometry and optics, he figured out some of the major principles of gravity later synthesized by Isaac Newton, and he was the first to hypothesize that the sun exerted a force on the earth. Kepler’s first law of planetary motion is as follows: First Law

Each of the planetary orbits is an ellipse, with the sun at one focus.

A planet in an elliptical orbit is depicted in Figure 2.27. You may not have studied ellipses yet in math, so I will describe them. An ellipse is a geometric figure shaped like 51

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this: . An ellipse is similar to a circle, except that instead of having a single point locating the center, an ellipse has two points on either side of the center called foci that define its shape. (The term foci sun is plural, and pronounced FOH-sigh; the singular is focus.) Out in space, each planet travels on a path defined by a geometrical ellipse. The planetary orelliptical orbit bits all have one focus located at the same place in space and this is where the sun is. Think how inFigure 2.27. A planet in an elliptical orbit credible it is that Kepler figured this out! He was a around the sun (Kepler’s first law). monster mathematician (no calculator!) and an extremely careful scientist, and the fact that scientists had understood the orbits to be circular for two thousand years did not get in his way. To me, this is simply amazing. Kepler’s second law of planetary motion is: planet

Second Law A line drawn from the sun to any planet sweeps out a region in space that has equal area for any equivalent length of time. The second law is depicted in Figure specific area swept 2.28. The idea is that out in space for a given period of time—say, a month or a week—the shaded region in the figure has the same area, regardless of where the planet is in its orbit. Since the equal area swept out in space sun is off-center, this planet law implies that the travels an planets travel faster equal length of time when they are closer Figure 2.28. Equal areas are swept in space for equal periods of time (Kepler’s to the sun and slower second law). when they are farther away. Keep thinking about how stunning it is that a guy without a calculator or any modern computer could figure this out, all from the observational data that Tycho had assembled. Kepler’s third law is: planet travels a specific length of time

Third Law 2

⎛ T1 ⎞ ⎛ R1 ⎞ ⎜⎝ T ⎟⎠ = ⎜⎝ R ⎟⎠ 2 2

The orbits of any two planets are related as follows: 3

where T1 and T2 are the planets’ orbital periods, and R1 and R2 are their mean distances from the sun. 52

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This third law of Kepler is a stunning example of mathematical modeling and is quite accurate. The equation can be expressed in a way that shows that the ratio of the square of the period to the cube of the mean distance is a constant. That is, this ratio has the same value for every planet. It can also be expressed such that the orbital period, T, for any planet is a function of the planet’s mean distance from the sun, R. In equation form, this expression of the third law can be written as T = k R3 or simply T = kR

In these equations, k is just a constant that depends on the units used for T and R. I write the equation in two ways to show that taking a square root of a quantity is equivalent to raising the quantity to the 1/2 power, which multiplies by the power of 3 already there to give the 3/2 power on R. I am not trying to go crazy with the math here. I just want to show how simple Kepler’s third law really is. Think about it. This simple equation accurately relates the period of any planet’s orbit to that planet’s mean distance from the sun. Now I don’t know about you, but when I see an equation that is as amazing and as simple as this, it sets me thinking. First, Kepler’s work as a scientist is first class. He figured this out from data collected in the era before calculators and before computers. This was only three years after Shakespeare died! Second, this equation says something deep about the universe we live in. The universe can be modeled with simple mathematics that can be understood by high school kids! This is one of the most astonishing and wonderful things about the world we live in. Mathematical patterns are found everywhere in nature, from the arrangements of the leaves on plants to the numbers of spirals on pinecones and sunflowers. Moreover, mathematics can be (and is) used to model everything from the launch of a rocket to the motion of the planets to the gravitational fields of the cosmos. The existence of mathematical patterns is part of the deep order that pervades the natural world, and this pervading order is one of the characteristics of the world that make science possible. Many great scientists and mathematicians have called attention to the beautiful mathematical structure that appears everywhere in nature and have called it wonderful. And so it is.

2.3.9 Galileo Galileo Galilei (Figure 2.29) worked at the university at Padua, Italy, and later as chief mathematician and philosopher for the ruling Medici family in Florence, Italy. Galileo’s work in astronomy represents the climax of the Copernican Revolution. He made significant improvements to the telescope and used the telescope to see the craters on the moon and sunspots, which provided additional evidence that the Figure 2.29. Florentine scientist Galileo heavens were not perfect and unchanging as Aristotle Galilei (1564–1642). 53

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and Ptolemy had maintained. In 1610, he used the telescope to discover four of the moons around Jupiter, which was clearly in conflict with the idea that there had to be exactly seven heavenly bodies. He was fully on board with all the new science of the Copernican model, but, oddly, he never did accept Kepler’s discovery that the planets’ orbits are elliptical rather than circular. Galileo published his early astronomical discoveries in 1610 in a book called The Starry Messenger. Most people know that Galileo was put on trial in 1633 by the Holy Office of the Inquisition established by Roman Catholic Church. However, the reasons for that trial are widely and seriously misunderstood. The real story is rather illuminating, as I explain here as briefly as possible. Galileo is famous for this remark: “Philosophy is written in this grand book—I mean the universe—which stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language in which it is written. It is written in the language of mathematics.” This beautiful statement again calls attention to the mathematical structure in nature, which I mention at the end of the previous section. However, Galileo erred in taking his statement too far, claiming that the mathematics he used in his astronomical discoveries was the truth about nature. Galileo felt that his work proved beyond question that Copernicus was right. Galileo, along with other scientists over the next three centuries, still had to learn the limits associated with human modeling of nature. In contrast to Galileo’s attitude toward his work is the attitude of his friend Cardinal Roberto Bellarmine, an important Church official. Bellarmine was a great admirer of Galileo’s work, but Bellarmine’s thinking was very much along the lines of our discussion in the previous chapter: scientific inquiry leads to theories which are models and cannot be regarded as truths; models are provisional and subject to change. In this attitude of an

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The first monster telescope

Motion

important Church official, we recognize a remarkable early statement of the attitude toward scientific knowledge that today is held as the correct way to think about scientific theories. Bellarmine cautioned Galileo that no natural science could make claims as to truth and urged him to present his ideas as everyone thought Copernicus had done—as hypotheses rather than as truths. There were definitely mistakes on both sides of the conflict that led eventually to Galileo’s trial by the Holy Office. On Galileo’s part, the mistake was in pushing his ideas too forcefully with the claim that they were true. Galileo’s position implied that the theologians who claimed that the Bible lined up with the Ptolemaic system were wrong in their interpretation of the Bible. Galileo wrote an important letter at that time explaining that the theologians needed to reconsider their interpretations of Scripture in light of what the scientific evidence was showing. Galileo was correct in his views about interpreting Scripture, but his claims pertaining to the truth of his discoveries went too far. To the theologians and church officials who held to the Ptolemaic view, having their views called into question was equivalent to calling the Bible itself into question. This was their mistake: they did not yet understand that the Bible has to be interpreted just as observations of nature have to be interpreted, and they were not yet ready to reconsider their views about the heavens and their interpretations of the Bible. It is interesting to note that in 1992, Pope John Paul II gave an address in which he commended Galileo’s comments on the necessity of interpreting the Bible! Recall that at this time, there was as yet no physical evidence that the earth was moving and rotating on an axis. As I mention in Section 2.3.7, the first actual evidence for earth’s motion around the sun came in 1838 with the discovery of stellar parallax. Evidence for the rotation of the earth came a bit later in 1851 with the invention of Foucault’s pendulum, an example of which is shown in Figure 2.30. The rotation of the earth causes a small change of direction in each swing of a very long, massive pendulum. If the earth were not rotating, the pendulum would swing steadily back and forth in the same direction. Now, briefly, here is the sequence of events that led to Galileo’s trial. Rumors got around that Galileo had been secretly examined by the Holy Office and forced to abjure (renounce) his views about Copernicus. To help Galileo fight these annoying rumors, Cardinal Bellarmine wrote Galileo a letter in 1616 stating that the rumors were false. The letter went on to say that though Galileo had not been taken before the Holy Office, he had been told not to defend or teach as true the system of Copernicus. Then in 1632, Galileo published another major work on astronomy in which he did in fact uphold the system of Copernicus against the system of Ptolemy. The Pope at the time, Urban VIII, was also a friend and admirer of Galileo, but when he heard that Galileo had published such a book after having specifically been told not to, he was extremely upset and had no choice but to have Galileo examined by the Holy Office. This was Galileo’s famous 1633 trial. The details leading to Galileo’s trial are very complex, but the controversy boils down to the two issues I have emphasized. First, Galileo pushed his scientific claims Figure 2.30. A Foucault Pendulum in the Panthéon in too far, claiming truth for a scientific the- Paris. 55

Chapter 2

ory which could not be regarded as more than a model of nature. Second, he published a book in defiance of an injunction against doing so. Galileo was a pious and godly man. There is good evidence that he never did actually intend to fall afoul of the injunction. But when the Holy Office persuaded him that he had, he was immediately ready to confess his actions and abjure them. This he did. Galileo was never tortured, but it was necessary that he be punished in some way. His friend Pope Urban VIII made it as easy on Galileo as he could by confining him to “house arrest” and prohibited him from further publishing. He lived for a few months in Rome in the palace of one of the cardinals, and then was allowed to return to his home in Florence were he lived in house arrest for the last eight years of his life. In addition to his work in astronomy, Galileo developed ground-breaking ideas in physics over the course of 30 years of work. These ideas were published after his trial in what would be his final book.5 Before Galileo, scientists had always accepted Aristotle’s physics, which held that a force was needed to keep an object moving. Galileo broke with this 2,000-year-old idea and hypothesized that force was needed to change motion but not to sustain motion as Aristotle had taught. Galileo was the first to formulate the idea of a friction force that caused objects to slow down. By conducting his own experiments, Galileo also discovered that all falling objects accelerate at the same rate (the acceleration of gravity, 9.80 m/s2), which is mathematically very close to Isaac Newton’s second law of motion (our topic in the next chapter). Galileo’s studies in physics thrust forward the Scientific Revolution and set the stage for the work of Isaac Newton, where the Scientific Revolution reached its climax. The saga of the Copernican Revolution ends more or less with Galileo. Within 50 years of Galileo’s death, the heliocentric model of the planetary orbits was becoming widely accepted. But while we are studying the planets and gravity, the whole story just isn’t complete unless we mention two more key figures in the history of science.

2.3.10 Newton, Einstein, and Gravitational Theory

Figure 2.31. English scientist Isaac Newton (1643–1727). 5

Sir Isaac Newton (Figure 2.31) is perhaps the most celebrated mathematician and scientist of all time. He was English, as his title implies, and he was truly phenomenal. He held a famous professorship in mathematics at Cambridge University. He developed calculus. He developed the famous laws of motion, which we examine in the next chapter. He developed an entire theory of optics and light. He formulated the first quantitative law of gravity called the law of universal gravitation. His massive work on motion, gravity, and the planets, Principia Mathematica, was published in 1687. This work is one of the most important publications in the history of science. In this course, we do not perform computations with Newton’s law of universal gravitation, and you do not need to memorize the equation for it. But let’s look at it here briefly. The law is usually written as

Since he was forbidden to publish through the Catholic Church, the book was published by a Protestant publisher in Holland.

Motion

F =G

m1m2 d2

where G is a constant, m1 and m2 are the masses of any two objects (such as the sun and a planet), and d is the distance between the centers of the two objects. Newton theorized that every object in the universe pulls on every other object in the universe, which is why his law is called the law of universal gravitation. We now understand that he was correct. Everything in the universe pulls on everything else. I have no idea how Newton figured this out. The equation above gives the force of gravitational attraction between any two objects in the universe. Amazingly, this equation is quite accurate, too! Notice from the equation that Newton’s model depends on each object having mass because the force of gravity has both masses in it multiplied together. Newton’s model implies that if either mass is zero, the force of gravitational attraction is zero. While we are here looking at Isaac Newton, we should pause and consider the relationship between his physical theories (including law of universal gravitation and his laws of motion) and Kepler’s mathematical theory of planetary motion. It turns out that Kepler’s discovery about the elliptical orbits and the relationship between the period and mean radius of the orbit can be directly derived from Newton’s theories, and Newton does derive them in Principia Mathematica. But Newton’s equations apply much more generally than Kepler’s do. As we see in the next chapter, Newton’s laws apply to all objects in motion— planets, baseballs, rockets—while Kepler’s laws apply to the special case of the planets’ orbits. If we consider this in light of my comments in Chapter 1 about the way theories work, we see that Newton’s laws explain everything Kepler’s laws explain, and more. This places Newton’s theory about motion and gravity above Kepler’s, so Newton’s theories took over as the most widely-accepted theoretical model explaining gravity and motion in general. However, even though Newton’s laws ruled the scientific world for nearly 230 years, they do not tell the whole story. This is where the German physicist Albert Einstein (Figure 2.32) comes in with his general theory of relativity, published in 1915. Einstein’s theory explains gravity in terms of the curvature of space (or more accurately, spacetime) around a massive object, such as the sun or a planet. This spacetime curvature is represented visually in Figure 2.33. Fascinatingly, since Einstein’s theory is about curving space, the theory predicts that even phenomena without mass, such as rays of light, are affected by gravity. Einstein noticed this and made the stunning prediction in 1917 that starlight bends as it travels through space when it passes near a massive object such as the sun. He formed this hypothesis, including the amount light bends, based on his general theory of relativity, which is based completely on mathematics. What do you think about that? It practically leaves me speechless. Figure 2.32. German physicist Albert Einstein became instantly world famous in 1919 when Einstein (1879–1955). his prediction was confirmed. To test this hypothesis, Einstein proposed photographing the stars we see near the sun during a solar eclipse. This has to be done during an eclipse because looking at the sky while the sun is nearby means it is broad daylight and we aren’t able to see the stars. Einstein predicted that the apparent position of the stars shifts a tiny amount relative to where they appear when the sun is not near 57

Chapter 2

the path of the starlight. British scientist Sir Arthur Eddington commissioned two teams of photographers to photograph the stars during the solar eclipse of 1919. After analyzing their photographic plates (one of which is shown on the opening page of Chapter 1), they found the starlight shifted by exactly Figure 2.33. A visual representation of the curvature of the amount Einstein said it would. Talk spacetime around the earth. about sudden fame—Einstein became the instant global rock star of physics when this happened! (And his puppy dog eyes contributed even more to his popularity!) Just as Kepler’s laws were superseded by Newton’s laws and can be derived from Newton’s laws, Newton’s law of universal gravitation was superseded by Einstein’s general theory of relativity and can be derived from general relativity. Einstein believed that his own theories would some day be superseded by an even more all-encompassing theory, but so far (after 103 years) that has not happened. The general theory of relativity remains today the reigning champion theory of gravity, our best understanding of how gravity works, and one of the most important theories in 20th- and 21st-century physics.

Chapter 2 Exercises Unit Conversions Perform the following unit conversions. Express your results both in standard notation and in scientific notation using the correct number of significant digits. The unit conversion factors you need are all in Appendix A. On the first 20 problems, you use the standard method of multiplying conversion factors. The last four problems require somewhat different methods, which you can figure out. Convert this Quantity

Into these Units

1,750 meters (m)

feet (ft)

3.54 grams (g)

kilograms (kg)

41.11 milliliters (mL)

liters (L)

7 × 10 m (radius of the sun)

miles (mi)

1.5499 × 10–12 millimeters (mm)

750 cubic centimeters (cm or cc) (size of the engine in my old motorcycle)

2.9979 × 108 meters/second (m/s) (speed of light)

ft/s

168 hours (hr) (one week)

5,570 kilograms/cubic meter (kg/m3) (average density of the earth)

g/cm3

Motion

Convert this Quantity

Into these Units

45 gallons per second (gps) (flow rate of Mississippi River at the source)

m3/minute (m3/min)

600,000 cubic feet/second (ft3/s) (flow rate of Mississippi River at New Orleans)

liters/hour (L/hr)

5,200 mL (volume of blood in a typical man’s body)

5.65 × 102 mm2 (area of a postage stamp)

square inches (in2)

32.16 ft/s2 (acceleration of gravity, or one “g”)

m/s2

5,001 μg/s

kg/min

4.771 g/mL

kg/m3

13.6 g/cm3 (density of mercury)

mg/m3

93,000,000 mi (distance from earth to the sun)

65 miles per hour (mph)

m/s

633 nanometers (nm) (wavelength of light from a red laser)

5.015% of the speed of light

mph

98.6 degrees Fahrenheit (°F) (human body temperature)

degrees Celsius (°C)

50.0 °C (It gets this hot in northwest Australia.)

°F

4.3 lightyears (lt-yr) kilometers (km) (distance to the closest star, Proxima Centauri) (A lightyear is the distance light travels in one year. You will have to work this out first.)

Answers (A dash indicates that it would be either silly or incorrect to write the answer that way, so I didn’t. Silly is because there are simply too many zeros, or no zeros at all. Incorrect is because we are unable to express the result this way and still show the correct number of significant digits.) Standard Notation

Scientific Notation

5,740 ft

5.74 × 103 ft

0.00354 kg

3.54 × 10–3 kg

0.04111 L

4.111 × 10–2 L

400,000 mi

4 × 105 mi 59

Chapter 2

Standard Notation

Scientific Notation

–

1.5499 × 10–15 m

0.00075 m3

7.5 × 10–4 m3

983,560,000 ft/s

9.8356 × 108 ft/s

605,000 s

6.05 × 105 s

5.57 g/cm3

–

1.0 × 101 m3/min

60,000,000,000 L/hr

6 × 1010 L/hr

0.0052 m3

5.2 × 10–3 m3

0.876 in2

8.76 × 10–1 in2

9.802 m/s2

–

0.0003001 kg/min

3.001 × 10–4 kg/min

4,771 kg/m3

4.771 × 103 kg/m3

13,600,000,000 mg/m3

1.36 × 1010 mg/m3

–

1.5 × 1013 cm

29 m/s

2.9 × 101 m/s

0.0000249 in

2.49 × 10–5 in

33,700,000 mph

3.37 × 107 mph

37.0 °C

3.70 × 101 °C

122 °F

1.22 × 102 °F

–

4.1 × 1013 km

Motion Study Questions Set 1 1.

A train travels 25.1 miles in 0.50 hr. Calculate the velocity of the train.

Convert your answer to the previous problem to km/hr.

How far can you walk in 4.25 hours if you keep up a steady pace of 5.0000 km/hr? State your answer in km.

For the previous problem, how far is this in miles?

On the German autobahn, there is no speed limit and many cars travel at velocities exceeding 150.0 mi/hr. How fast is this in km/hr?

Referring again to the previous question, how long does it take a car at this velocity to travel 10.0 miles? State your answer in minutes.

An object travels 3.0 km at a constant velocity in 1 hr 20.0 min. Calculate the object’s velocity and state your answer in m/s.

A car starts from rest and accelerates to 45 mi/hr in 36 s. Calculate the car’s acceleration and state your answer in m/s2.

Motion

10. A person is sitting in a car watching a traffic light. The light is 14.5 m away. When the light changes color, how long does it take the new color of light to travel to the driver so that he can see it? State your answer in nanoseconds. (The speed of light in a vacuum or air, c, is one of the physical constants listed in Appendix A that you must know.) 11. A proton is uniformly accelerated from rest to 80.0% of the speed of light in 18 hours, 6 minutes, 45 seconds. What is the acceleration of the proton? 12. A space ship travels 8.96 × 109 km at 3.45 × 105 m/s. How long does this trip take? Convert your answer from seconds to days. 13. An electron experiences an acceleration of 5.556 × 106 cm/s2 for a period of 45 ms. If the electron is initially at rest, what is its final velocity? 14. A space ship is traveling at a velocity of 4.005 × 103 m/s when it switches on its rockets. The rockets accelerate the ship at 23.1 m/s2 for a period of 13.5 s. What is the final velocity of the rocket? 15. A more precise value for c (the speed of light) than the value given in Appendix A is 2.9979 × 108 m/s. Use this value for this problem. On a particular day, the earth is 1.4965 × 108 km from the sun. If on this day a solar flare suddenly occurs on the sun, how long does it take an observer on the earth to see it? State your answer in minutes.

Answers 1.

22 m/s

79 km/hr

21.3 km

13.2 mi

241.4 km/hr

4.00 min

0.63 m/s

0.56 m/s2

–0.53 m/s2

10. 48.3 ns

13. 2.5 × 10 m/s 3

11. 3,680 m/s2

14. 4.32 × 10 m/s 3

12. 301 days

15. 8.3197 min

Motion Study Questions Set 2 1.

Construct three simple graphs like those discussed in this chapter showing distance vs. time, velocity vs. time, and acceleration vs. time. Draw your graphs one beneath the other and line them up vertically so that the time intervals on each of them line up. Here is the situation: An SUV is at rest at a traffic signal. Then the car starts moving, accelerating uniformly to a certain velocity, and then continues at a constant velocity after that. Be sure to label the axes of your graphs.

Sketch the same three graphs for this situation: A pick-up truck is traveling at a constant velocity. Then the driver slams on the brakes. Before the car stops, the driver releases the brakes so that the car continues to roll along at a constant (but much slower) velocity. Be sure to label the axes of your graphs.

Sketch the same three graphs for this situation: An MG convertible is coasting out of gas and gradually slowing down when it arrives at the entrance to a gas 61

Chapter 2

station. Next, the car begins rolling up the entrance ramp to the gas station, but the ramp is steep and the car slows down rapidly to a stop. Next, the driver fails to step on the brake, so after stopping at the top of the ramp the car immediately begins to roll backwards down the ramp. Note that there are only two time intervals here: before the ramp and on the ramp. (Ignore the up and down motion due to the ramp and any turning the car would do to turn into the gas station ramp. Just think about how far the car has traveled from the point where it ran out of gas.) 4.

A space ship in space is cruising at a constant velocity. The captain switches on the retro-rockets, which begin slowing the vehicle. The captain never turns the rockets off, so after it comes to rest it immediately begins moving (still accelerating) in the opposite direction. Note that there are only two time intervals in this scenario.

The three graphs below represent three different scenarios involving an object in motion. For each one write a one-sentence description of the object’s motion. In your descriptions use terms like “speeding up,” “slowing down,” “at rest,” “backing up,” “constant velocity,” etc.

velocity

(b)

acceleration

(a)

time

Answers 1

4 d

t v

t t a

t t

a t

5. a) The object is speeding up (accelerating) uniformly, then it begins moving at a 62

Motion

constant velocity. b) The object slows rapidly to a complete stop, stays at rest for a while, then begins accelerating again, moving in the same direction as before. c) The object is speeding up (accelerating) uniformly, then it begins moving at a constant velocity.

Ptolemaic Model and Copernican Revolution Study Questions 1.

Describe why some theologians in the 16th century were strongly opposed to Copernicus’ heliocentric theory.

State six features of the Ptolemaic model other than the spheres.

Describe Copernicus’ model of the heavens.

What are some of the “proofs” people used in arguing that there is no way that the earth rotates on an axis?

For what reason did Copernicus decide to keep his theory private?

Write a description of the two key observations Tycho made (including dates) that challenged the Ptolemaic system.

Briefly describe the cosmological model put forward by Tycho.

State Kepler’s three laws of planetary motion.

10. This is a bit difficult, but explain retrograde motion and epicycles as well as you can in a few sentences. 11. Explain the two main mistakes individuals made that led to Galileo’s trial. 12. Explain the actual cause of Galileo’s trial and the results of that trial. 13. Describe why Pope John Paul II commended Galileo in 1992. 14. Distinguish between Newton’s and Einstein’s theories of gravitation. According to each of these two geniuses, what is the cause of gravity and what are the effects of gravity? 15. Describe some of Kepler’s scientific achievements, aside from his laws of planetary motion. 16. Write a paragraph or two explaining how the Copernican Revolution illustrates the Cycle of Scientific Enterprise.

CHAPTER 4

Variation and Proportion

The Planets with Their Sizes to Scale In Chapter 2, we briefly consider Newton’s law of universal gravitation, an equation that describes the force of attraction between any two objects, including the sun and planets in the solar system. The law of universal gravitation is an example of an inverse-square law. In an inverse-square law, the strength of one quantity depends on the inverse of the square of another quantity. The force of gravity between the sun and a planet is inversely proportional to the square of the distance between their centers. The masses of Uranus and Neptune are roughly equal, but Neptune is about three times farther away from the sun than Uranus is. Since the strength of the gravitational force is inversely proportional to the square of the distance, the force of the sun’s gravity on Neptune is only 1/9 the strength of the sun’s gravity on Uranus.

OBJECTIVES After studying this chapter and completing the exercises, students will be able to do each of the following tasks, using supporting terms and principles as necessary: 1. 2. 3. 4. 5. 6. 7. 8.

Identify and graph linear and nonlinear variation. Identify the constant of proportionality in a physical equation. Identify direct, inverse, square, inverse square, and cubic proportions when seen in equations. Identify direct, square, and inverse proportions when seen graphically. Identify dependent and independent variables in physical equations. Graph functional relationships, making proper use of dependent and independent variables. Normalize non-essential constants and variables in a given physical equation, and describe how the remaining two variables vary with respect to each other. Describe the way variations occur between variables in physical equations, including: a. area of a triangle b. area of a circle c. volume of a sphere d. gravitational potential energy e. kinetic energy f. pressure under water g. force of gravitational attraction h. Charles’ law i. Boyle’s law

Note: The purpose of this study is to learn about how physical quantities vary in equations, not to understand completely all the physical equations used as examples. The student should gain some understanding of the physical relationships involved, but detailed study of the physics involved occurs later in the course or in future courses.

4.1

The Language of Nature

The fact that nature can be modeled with mathematics is so surprising that famous scientists have repeatedly referred to it as some kind of miracle. As Galileo famously said, “The laws of nature are written in the language of mathematics.” One of the key ways scientists describe the mathematical properties in the laws of nature is in terms of the different proportionalities that are present. In other words, they describe how one variable in an equation varies with respect to another variable. This chapter is about learning how to see these relationships in equations and appreciate what they indicate about what one quantity does when another quantity is manipulated. The presentation in this chapter is short. This is because there is not that much content to learn. Most of the time spent in this chapter comes not from learning new laws and equations, but from working through a set of exercises that immerses you in the principles and language of variation and proportionality, with both analytic and graphical exercises. These exercises are in the Variation and Proportion Study Packet presented at the end of the chapter. 85

Chapter 4

4.2

The Mathematics of Variation

4.2.1 Independent and Dependent Variables As every algebra student knows, the generic equation for a line is y = mx + b This equation is called the slope-intercept form of an equation for a line because the constant m is the slope of the line and the constant b is the y-intercept of the line. In this equation, the variable x is called the independent variable and the variable y is called the dependent variable because the value of y depends on the value we choose for x. In general, we write our equations so the dependent variable is alone on the left side and all other variables and constants are on the right side. When graphing a line or curve relating two variables, say p and r, we often speak of “a graph of p vs. r.” The phrase “p vs. r” means that p is the dependent variable, represented by the scale on the vertical axis of the graph, and r is the independent variable, represented by the scale on the horizontal axis of the graph. Referring to “r vs. p” would call to mind a different graph, with the variables on the axes switched. The order matters in this expression. The graph of “p vs. r” has p on the vertical axis and r on the horizontal axis. It is important that you think this through and get this down. In this course, you are repeatedly asked to generate a graph representing the relationship between two variables, both in the exercises and in your lab reports. It is quite common for students to see a statement like “develop a graph of p vs. r” and fail to consider which of these variables should be placed on the vertical axis and which on the horizontal axis. Now, if a line passes through the origin, the y-intercept b is zero. The equation for the line then becomes y = mx Many of the equations in physics are linear functions of the form y = mx. We have already learned several of them, such as d = vt and Fw = mg. In many physical linear equations like these, the y-intercept is zero; a graph of the equation is a line that passes through the origin. These are examples of direct variation, as described in more detail below.

4.2.2 Common Types of Variation Quantities in nature often exhibit common types of variation with respect to other quantities. This is part of what Galileo was talking about, and if you think about it, it is quite wonderful that nature behaves this way. Just as an aside, I discuss this a bit back in Chapter 2 where we consider the work of Johannes Kepler. The regular, mathematical way nature behaves, from planets and stars down to quarks and molecules, is wonderful indeed. It is even more wonderful that we humans have the mental capacity to apply mathematics to nature in the models (theories) we develop. Now going back to the common types of variation found in physics, we desire to know how to describe these common types of variation using appropriate language. The graphs in Figure 4.1 show five common types of variation. Each of these is a graph of y vs. x, with k as a generic constant. The constant is usually called the constant of proportionality. There are

Variation and Proportion

y = kx 2

y = kx y

y = kx 3

y = k/x y

y = k/x 2 y

Figure 4.1. Five common types of variation.

many other common types of variation in addition to those shown in Figure 4.1. You will learn about those in future math or science classes. One of our objectives is to use correct scientific language to describe the variation going on between the variables in an equation. Here are the correct phrases we use to describe each of the types of variation represented in the figure. The prepositions are important. y = kx

y varies directly with x, or y varies in direct proportion to x

y = kx

y varies as the square of x, or y varies as x squared

y = kx

y varies as the cube of x, or y varies as x cubed

y = k/x y = k/x

y varies inversely with x 2

y varies inversely with the square of x

This last type of variation is often called an “inverse square law.” There are many quantities in nature that follow inverse square laws. Here are a couple of examples of how to apply this language using equations we have already learned. Example 4.1 In the weight equation, how does the weight of an object vary with respect to the acceleration of gravity? The weight equation is Fw = mg In this equation, the acceleration of gravity, g, is the independent variable, weight, Fw , is the dependent variable, and the mass, m, is the constant of proportionality. Thus, this equa-

Chapter 4

tion represents the same kind of variation as in the first graph of Figure 4.1, and Fw varies directly with g.

Example 4.2 In the standard equation for calculating uniform acceleration, how does acceleration vary with time? The equation is a=

v f − vi t

Time, t, is the independent variable and acceleration, a, is the dependent variable. Since the time is in the denominator without an exponent, this variation is like the fourth graph in Figure 4.1 and a varies inversely with t.

4.2.3 Normalizing Equations Often equations involve many variables and constants, such as Newton’s law of universal gravitation, considered in Chapters 2 and 3. We frequently want to consider only two of the variables to see how they vary with respect to one another, while holding everything else constant. To do this we normalize the equation and rewrite it as a proportion. To normalize an equation, take all the constants and all the extra variables and set them equal to “1.” Write what is left as a proportion. For example, let’s say we wish to consider the variables E and V in the equation E = 12 CV 2 Don’t worry what this equation is about; we do not address it in this course.1 All we are doing here is using the equation to learn about variation. We desire to determine how E, the dependent variable, varies with respect to V, the independent variable, assuming everything else is held constant. Normalizing the equation means we set the 1/2 and the variable C each equal to 1, and write what remains as a proportion. Doing so gives E ∝V 2 This expression is read “E is proportional to V squared.” When we write a normalized equation, we must replace the equals (=) sign with the mathematical symbol meaning “is proportional to” ( ∝ ). This is because although the proportionality is the same, the two sides

For those who are curious, this equation relates the energy, E, (measured in joules) stored in a capacitor to the capacitance of the capacitor, C, (measured in farads) and the voltage, V, across the capacitor (measured in volts).

Variation and Proportion

are not strictly equal any more. With the normalized expression, it is easier to see that this expression has the same form as y = x 2, which is graphed in the second graph of Figure 4.1. So we say that “E varies as the square of V.” Be sure to notice that any exponent that is present on the independent variable is a crucial part of the proportionality and appears in the normalized expression. Exponents are not variables, nor are they constants, so they must be retained. Only the variables and the constants are replaced with “1,” and only these vanish from the normalized expression. The graph of a normalized expression has the same basic shape as the original equation; it is only stretched horizontally or vertically. This means normalizing a complicated equation helps us to see how the variation between the two variables of interest works because the nature of the relationship between the two variables is preserved while making the equation (which becomes a proportion) simpler and easier to analyze. Example 4.3 Consider the following equation from the field of fluid dynamics: Q=

−kA ( Pb − Pa ) µ L

Assume that we are interested in L as the independent variable and Q as the dependent variable. Normalize this equation and describe how Q varies with respect to L, assuming all else is held constant. To normalize, we set all constants and variables other than the ones under consideration equal to 1. Treating the parenthetical sum as a single variable and normalizing, we have Q∝

1 L

From this expression, we see that Q varies inversely with L.

Chapter 4

Chapter 4 Exercises The Variation and Proportion Study Packet Completion of this packet counts as a major grade. For full credit, do all the activities and graphs with great care, thoroughness, and neatness. Packet Requirements: •

Use a separate sheet of graph paper for each activity.

•

Answer each question on a separate line on your graph paper.

•

Each graph must include a minimum of five plotted points, and must be accompanied by the table of values used to make the plot.

•

Make it neat.

•

Use a straightedge for all axes.

•

Choose scales to fit the data well so that the graph is well proportioned and large enough to read easily.

•

Label all axes and scales with the variable name and units of measure. Make sure your scales are linear.

•

Title each graph.

•

Label each curve as the normalized curve or the regular curve.

•

All graphs should be approximately the size of an index card (3” × 5”).

The packet consists of the following activities: •

Activity 1: How does the area of a triangle vary with its height?

•

Activity 2: How does the area of a circle vary with its radius?

•

Activity 3: How does the volume of a sphere vary with its radius?

•

Activity 4: How does the gravitational potential energy of an object vary with its height?

•

Activity 5: How does the kinetic energy of an object vary with its velocity?

•

Activity 6: How does pressure vary with depth under water?

•

Activity 7: How does the force of gravitational attraction between two objects vary with the distance between their centers?

•

Activity 8: At constant pressure, how does the volume of a gas vary with its temperature?

•

Activity 9: At constant temperature, how does the volume of a gas vary with its pressure?

Variation and Proportion

Activity 1. How does the area of a triangle vary with its height, if all else is held constant? 1. 2. 3. 4.

What is the relation for the area of a triangle? What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating area to height? 5. Select any value other than 2 to use for the base of a triangle. Using this value for the base, choose several values for the height of the triangle and calculate the area of the triangle for each height. Enter all these in a table of values. 6. Prepare a graph of area vs. height using the values you computed in the previous step. 7. Compute another table of values for the normalized expression from step 4. Treat the proportional sign as an equals sign for this. 8. Graph the normalized equation on the same set of coordinate axes you used for the graph in step 6. 9. Describe the similarities and the differences between the two “curves” on your graph. 10. Answer the main question (the title) for this activity.

Activity 2. How does the area of a circle vary with its radius, if all else is held constant? 1. 2. 3. 4.

What is the relation for area of a circle? What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating area to radius? 5. Choose several values for the radius of the circle and calculate the area of the circle for each radius. Enter all these in a table of values. 6. Prepare a graph of area vs. radius using the values you computed in the previous step. 7. Compute another table of values for the normalized expression from step 4. Treat the proportional sign as an equals sign for this. 8. Graph the normalized equation on the same set of coordinate axes you used for the graph in step 6. 9. Describe the similarities and the differences between the two curves on your graph. 10. Answer the main question for this activity.

Chapter 4

Activity 3. How does the volume of a sphere vary with its radius, if all else is held constant? 1. 2. 3. 4.

What is the relation for volume of a sphere? Look it up if you don’t know it. What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating volume to radius? 5. Choose several values for the radius of the sphere and calculate the volume of the sphere for each radius. Enter all these in a table of values. 6. Prepare a graph of volume vs. radius using the values you computed in the previous step. 7. Compute another table of values for the normalized expression from step 4. Treat the proportional sign as an equals sign for this. 8. Graph the normalized equation on the same set of coordinate axes you used for the graph in step 6. 9. Describe the similarities and the differences between the two curves on your graph. 10. Answer the main question for this activity.

Activity 4. How does the gravitational potential energy (EG ) of an object vary with its height, if all else is held constant? The equation for EG is EG = mgh where EG is the gravitational potential energy in joules (J), m is the mass of an object in kilograms (kg), g is the acceleration of gravity, 9.80 m/s2, and h is the height above the ground (or other reference) of the object in meters (m). 1. 2. 3. 4. 5. 6. 7. 8. 9. 92

What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating EG to height? Select a value to use for the mass of the object for this activity. Using this value, choose several values for the height of the object and calculate the EG of the object for each height. Enter all these in a table of values. Prepare a graph of EG vs. height using the values you computed in the previous step. Compute another table of values for the normalized expression from step 3. Treat the proportional sign as an equals sign for this. Graph the normalized equation on the same set of coordinate axes you used for the graph in step 5. Describe the similarities and the differences between the two “curves” on your graph. Answer the main question for this activity.

Variation and Proportion

Activity 5. How does the kinetic energy (EK ) of an object vary with its velocity, if all else is held constant? The equation for EK is EK = 12 mv 2 where EK is the kinetic energy in joules (J), m is the mass of an object in kilograms (kg), and v is the velocity of the object in meters per second (m/s). 1. 2. 3. 4. 5. 6. 7. 8. 9.

What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating EK to velocity? Select a value other than 2 kg to use for the mass of the object for this activity. Using this value, choose several values for the velocity of the object and calculate the EK of the object for each velocity. Enter all these in a table of values. Prepare a graph of EK vs. velocity using the values you computed in the previous step. Compute another table of values for the normalized expression from step 3. Treat the proportional sign as an equals sign for this. Graph the normalized equation on the same set of coordinate axes you used for the graph in step 5. Describe the similarities and the differences between the two curves on your graph. Answer the main question for this activity.

Chapter 4

Activity 6. How does pressure under water vary with depth, if all else is held constant? The equation for pressure under any liquid is P = ρ gh

The Pascal is the derived unit we use for pressure in the MKS system. If you multiply all the units together for the variables in this equation you get kg/(ms2). Thus, 1 Pa = 1 kg/(ms2).

where P is the pressure in pascals (Pa), ρ (that is, the Greek letter rho, the r in the Greek alphabet) is the density of the liquid in kg/m3, g is the acceleration of gravity, 9.80 m/s2, and h is the depth under the liquid in meters (m). For water, ρ = 998 kg/m3 (at room temperature). 1. 2. 3. 4. 5. 6. 7. 8. 9.

What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating pressure to depth? Using the value for the density of water given above, choose several values for the depth under water and calculate the pressure for each depth. Enter all these in a table of values. Prepare a graph of pressure vs. depth using the values you computed in the previous step. Compute another table of values for the normalized expression from step 3. Treat the proportional sign as an equals sign for this. Graph the normalized equation on a separate set of coordinate axes. Describe the similarities and the differences between the curves in the two graphs. Answer the main question for this activity.

Variation and Proportion

Activity 7. How does the force of gravitational attraction vary with the distance between the centers of two objects? The equation for the force of gravitational attraction between any two objects is given by Newton’s law of universal gravitation: F =G

m1m2 d2

The units of measure for this constant may look strange, but when G is placed in the equation, all the units cancel except for newtons, which are the appropriate units for force.

where F is the force in newtons (N), G is the gravitational constant, 6.67 x 10–11 Nm2/kg2, m1 and m2 are the masses of the two objects in kilograms (kg), and d is the distance between the centers of the two objects in meters (m). 1. 2. 3. 4. 5. 6. 7. 8. 9.

What are the two key variables to be compared in this activity? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables and constants, what is the expression relating the gravitational force between two objects to the distance between them (that is, between their centers)? Select values to use for the two masses of the objects for this activity. Using these values, choose several values for the distance between the objects and calculate the force of attraction for each distance. Enter all these in a table of values. Prepare a graph of force vs. distance using the values you computed in the previous step. (Hint: Convert all the force values in your table of values so that they have the same power of 10. Then scale your vertical axis using this power of 10.) Compute another table of values for the normalized expression from step 3. Treat the proportional sign as an equals sign for this. Graph the normalized equation on a separate set of coordinate axes. Describe the similarities and the differences between the curves in the two graphs. Answer the main question for this activity.

Chapter 4

Activity 8. How does the volume of a gas vary with its temperature? This activity is different from those you have completed so far. Instead of graphing an equation and the normalized version of it and comparing them, you prepare a graph of predicted and experimental values, similar to the one you made for the experiment on Newton’s second law. In this activity, we do a class demonstration in which we take data measuring the volume and temperature of a gas (air) as it changes temperature. Using the values from this demonstration and the equation for Charles’ law, you prepare a predicted curve of volume (V) vs. temperature (T). Then, on the same set of axes, you graph the data from the class demonstration. Charles’ law for gases can be written as: ⎛V ⎞ V = ⎜ i ⎟T ⎝T ⎠ i

The Charles’ law equation does not require us to use MKS units for the volume. Using cubic centimeters is more convenient for our demonstration.

where V is the volume of the gas in cubic centimeters (cm3), T is the temperature of the gas in kelvins (K), Vi is the initial gas volume in cubic centimeters (cm3), and Ti is the initial gas temperature in kelvins (K). If a certain quantity of a gas (a specific number of gas molecules) is placed in a container which holds the gas pressure constant, Charles’ law shows how volume varies with temperature. The values for Vi and Ti are the conditions of the gas at the start of the demonstration. Charles’ law applies to so-called ideal gases, a topic addressed in chemistry. Air behaves like an ideal gas in our demonstration, so long as we keep the temperature below about 40°C. 1. 2. 3. 4.

If one wishes to compare how the volume of a gas changes with temperature, what are the two key variables to be compared? Which of these is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables, what is the expression relating volume to temperature? How are these variables related to one another (i.e., what kind of proportion, etc.)? As the instructor demonstrates the change of volume of a gas with temperature, take volume and temperature data in your lab journal. Document the materials, experimental set-up and the procedure just as you do for an experiment. The Charles’ law demonstration notes below guide you in what data to take and what calculations must be performed. Follow the notes to get your data and computations set up in a table of experimental values. Use the Vi and Ti values from the demonstration (in units of cm3 and K) and the Charles’ law equation to calculate predicted volume values for each of the temperature values recorded during the demonstration. As shown in the example table below, these values can be entered in the same table with the experimental data. Prepare a graph of V vs. T. On the same set of axes, graph your predicted values and your experimental values. Be sure to label which curve is which. Reproduce

Variation and Proportion

your data table next to your graph. Use temperatures in degrees Celsius for the graph (even though for the calculations you must use temperatures in kelvins, as explained below). Also, to make your graph as accurate as possible, I encourage you to use software such as Microsoft Excel to prepare the graph for this activity.

Charles’ Law Demonstration: Notes for Activity 8 Use these notes to compute the values you need to prepare the graph for this activity. The volume calculations described below are a bit complicated. It is not necessary that you understand everything about where the equations came from. For those who are interested and want to learn where our equations come from, these notes explain the whole thing. If you find it confusing or would rather not get into it, that’s fine. Just use the equations from the notes to fill in the table and prepare your graphs. At a minimum, each student must collect data for the activity, use equations to calculate the values needed for the graphs of predicted and experimental volume vs. temperature, and prepare the graphs. A few words about the experimental set-up are necessary here. In the demonstration, a small volume of air is trapped in an upside down buret. A buret is a glass tube marked in increments of 0.1 mL (which equals 0.1 cm3), and normally used for measuring volumes of liquid. We use a buret that has been cut down to a much smaller size. It is placed upside down in a beaker of automotive antifreeze (ethylene glycol). The antifreeze is cold because it spends the night before the demo in a freezer. With the air trapped in the buret, which is immersed in the cold liquid, we insert a thermocouple wire up in the buret in the small air space at the top. A thermocouple is an electrical probe used to measure temperature. We then slowly heat up the antifreeze in the beaker, reading the temperature and volume of air on the buret as the air warms up. Because the buret is upside down, we are not able to measure the initial volume of the air (Vi) directly. This requires us to get mathematically creative to get an approximation for the initial air volume. Additionally, since the buret is upside down the readings of air volume we take are read with numbers on the buret that become smaller and smaller as the volume increases. We must take all this into account to figure out the volumes of air we have in the buret at different temperatures. Let’s begin with the data table for this activity. Make a table like the one below in your lab journal. You need about 10 or 15 rows in the table for all the data. The green columns are where you enter your data from the demonstration. The values that go in the other four columns must be calculated with the equations I present next.

Chapter 4

TC (°C) Demo Data

TK (K)

Vburet (mL=cm3) Demo Data

Ti =

ΔV (cm3)

Vexperimental (cm3)

Vi =

Vpredicted (cm3)

Vi =

First, enter the data from the demo in your table. Next, convert all the temperature values from degrees Celsius (TC ) to kelvins (TK ). Enter all the temperatures in kelvins in the table. To do this use this conversion equation: TK = TC + 273.2 Use this equation to convert all your temperature values into kelvins for the second column in the table. Next, we obtain the values for ΔV. The Greek letter delta (Δ) means change when used in math and science. I use the symbol ΔV to mean the total change in gas volume from when the demo started. For example, assume the buret reading is 49.8 mL when the demo begins. Assume you have a data value in your table with a buret reading, Vburet , of 46.1 mL. The value of ΔV for this data point is 49.8 cm3 – 46.1 cm3 = 3.7 cm3. So, the ΔV values are all calculated using this equation: ΔV = ( first reading of Vburet ) −Vburet Use this equation to calculate the values of ΔV for the fourth column in the table. The first value for ΔV is zero, as I show in the table above. Next, we must determine the value of the initial air volume in the buret, Vi . Now, note that the actual air volume in the buret, V, always equals the initial volume Vi , plus the total change in volume from Vi , which is ΔV. In other words, after we determine the value of Vi , the equation to use to compute all the experimental values of V is Vexperimental = Vi + ΔV Use this equation to compute all your experimental values for V after you determine Vi . These experimental values go in the fifth column in the table.

To get Vi , we substitute the expression from the previous equation for V into the Charles’ law equation to get ⎛V ⎞ Vi + ΔV = ⎜ i ⎟ T ⎝ Ti ⎠ 98

Variation and Proportion

Now we solve this equation for Vi , which gives us Vi =

ΔV ⎛T ⎞ ⎜⎝ T −1⎟⎠ i

Use this equation to compute Vi using the data from a temperature around 35 °C. I suggest that if you really want to follow what’s going on here (which you should if you intend to study science or engineering in college), you should work out the algebra for yourself to verify this new equation for Vi . (It won’t take long and it is good practice for you.) This equation allows us to get a good approximation for Vi . Simply pick one of the data points and insert the values of ΔV, T, and Ti into the equation and you’ve got a good estimate for Vi . Voila! Make sure you use temperature values in kelvins for this. I usually use one of the last data points for this calculation. Somewhere around 35°C (that is, 308 K) works well. Finally, now that we have Vi , we use it with Ti and the values of T to calculate the predicted value of the volume for each temperature. For this calculation, we use Charles’ law, which is what this investigation is all about. ⎛V ⎞ Vpredicted = ⎜ i ⎟ T ⎝ Ti ⎠ Use this equation to compute the predicted values for V. These go in the sixth column of the table. The result of our work is a graph of volume vs. temperature, showing both predicted and experimental values of volume for each value of temperature (two curves on the same pair of axes).

Chapter 4

Activity 9. How does the volume of a gas vary with its pressure? In Activity 8, we look at Charles’ law for gases, which relates the volume and temperature of an ideal gas. Another law for ideal gases is Boyle’s law, which relates together the volume and the pressure of a gas. Boyle’s law for gases can be written as: V=

Vi Pi P

where V is the volume of the gas in cubic meters (m3), P is the pressure of the gas in pascals (Pa), Vi is the initial gas volume in cubic meters (m3), and Pi is the initial gas pressure in pascals (Pa). If a certain quantity of a gas (a specific number of gas molecules) is placed in a container which holds the gas temperature constant, Boyle’s law shows how volume varies with pressure. The values for Vi and Pi are the conditions of the gas at the start of the event or experiment. 1. 2. 3. 4. 5.

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If one wishes to compare how volume changes with pressure, what are the two key variables to be compared? Which variable is the independent variable and which is the dependent variable? After combining and normalizing all non-essential variables, what is the expression relating volume to pressure? How are these variables related to one another (i.e., what kind of proportion, etc.)? To graph the variation of volume with pressure, let’s imagine that we are out in the middle of a deep lake with a large balloon and a kit of SCUBA diving gear. Let’s take our balloon and fill it with air so that it is the size of a basketball. This balloon is the size of a basketball without stretching. All we have to do is fill it; no extra pressure is required. Now let’s take our balloon down below the water. Your job is to use Boyle’s law to calculate the volume of the balloon at various pressures from atmospheric pressure at the surface of the water down to the pressure at a depth of 50.0 m. So your lowest pressure value is at the surface, the highest pressure value is the pressure at 50.0 m deep. You can pick several pressures in between and calculate the volume for them as well. Here are the values you need. A sphere the size of a basketball has a volume of 0.0071 m3 (a diameter of 9.39 inches for a regulation basketball, FYI.) This is Vi . Atmospheric pressure, Pi , is 101.3 kilopascals, or 101.3 kPa. Use the equation from Activity 6 to determine the pressure at a depth of 50.0 m, and convert your value from pascals to kilopascals. Prepare a table of values for pressures ranging from 101.3 kPa to the pressure 50.0 m down, and the corresponding volumes. Graph the changes in the volume as the pressure is increased from atmospheric pressure at the surface to pressure at final depth. Just for fun, let’s now find out how big this balloon is at this depth. To do this, we make the slightly silly assumption that the balloon can shrink right along with the air inside it and remain spherical. Take the final volume you calculate for a depth of 50.0 m and use the equation below for the radius of a sphere to calculate the

Variation and Proportion

radius of the balloon at the 50.0 m depth. Then double it to get the diameter. Give your final result in inches. 1

⎛ 3V ⎞ 3 r =⎜ ⎝ 4π ⎟⎠

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APPENDIX C

Laboratory Experiments

C.1

Important Notes

Please refer to pages xviii–xix in the Preface for Teachers for important information pertaining to the terms “experimental error” and “percent difference” as used in this text. The following pages contain your guidelines for six five laboratory experiments conducted in this course. For each of these experiments, students submit an individual written report. It is your responsibility to study The Student Lab Report Handbook thoroughly so that you can meet the expectations for lab reports in this course. The instructions written here are given to help you complete your experiment successfully. However, your report must be written in your own words. This applies to all sections of the report. Do not copy the descriptions in this appendix into your report in place of writing your report for yourself in your own words.

C.2

Lab Journals

You must maintain a proper lab journal throughout the year. Your lab journal contributes to your lab grade along with your lab reports. Chapter 1 of The Student Lab Report Handbook contains a detailed description of the kind of information you should carefully include in your lab journal entries. The following are highlights from that description. A good lab journal includes the following features: 1. The pages in the journal are quadrille ruled (graph paper) and the journal entries are in ink. 2. The journal is neatly maintained and free of sloppy marks, doodling, and messiness. 3. Each entry includes the date and the names of the team members present. 4. Every experiment and every demonstration that involves taking data or making observations is documented in the journal. 5. Entries for each experiment or demonstration include: 378

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C.3

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the date

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the team members’ names

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the team’s hypothesis

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an accurate list of materials and equipment, including make and model of any electronic equipment or test equipment used

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tables documenting all the data taken during the experiment, including the units of measure and identifying labels for all data

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all support calculations used during the experiment or in preparation of the lab report

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little details about the experiment that need to be written in the report that you may forget about later

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important observations or discoveries made during the experiment

Experiments

Experiment 1 The Pendulum Experiment Variables and experimental methods Essential equipment: •

string

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meter stick

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paper clip

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large steel washers

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clock with second hand

This investigation involves a simple pendulum. The experiment is an opportunity for you to learn about conducting an effective experiment. In this investigation, you learn about controlling variables, collecting careful data, and organizing data in tables in your lab journal. To make your pendulum, bend a large paper clip into a hook. Then connect the hook to a string, and connect the string to the end of a meter stick. Lay the meter stick on a table with the pendulum hanging over the edge and tape the meter stick down. Finally, hang one or more large metal washers on the hook for the weight. Your goal in this experiment is to identify the explanatory variables that affect the period of a simple pendulum. A pendulum is an example of a mechanical system that oscillates, that is, repeatedly “goes back and forth” in some regular fashion. In the study of any oscillating system, an important parameter is the period of the oscillation. The period is the length of time (in seconds) required for the system to complete one full cycle of its oscillation. In this experiment, the period of the pendulum is the response variable you monitor. (Actually, for convenience you monitor a slightly different variable, closely related to the period. This is explained on the next page.) After thinking about the possibilities and 379

Appendix C

forming your team hypothesis, construct your own simple pendulum from string and some weights and conduct tests on it to determine which variables affect its period and which variables do not. In class, explore the possibilities for variables that may affect the pendulum’s period. Within the pendulum system itself there are three candidates, and your instructor will lead the discussion until the class has identified them. (We ignore factors such as air friction and the earth’s rotation in this experiment. Just stick to the obvious variables that clearly apply to the problem at hand.) Then, as a team, continue the work by discussing the problem for a few minutes with your teammates. In this team discussion, form your own team hypothesis stating which variables you think affect the period. To form this hypothesis, you need not actually do any new research or tests. Just use what you know from your own experience to make your best guess. The central challenge for this experiment is to devise an experimental method that tests only one explanatory variable at a time. Your instructor will help you work this out, but the basic idea is to set up the pendulum so that two variables are held constant while you test the system with large and small values of the third variable to see if this change affects the period. You must test all combinations of holding two variables constant while manipulating the third one. All experimental results must be entered in tables in your lab journal. Recording the data for the different trials requires several separate tables. For each experimental setup, time the pendulum during three separate trials and record the results in your lab journal. Repeating the trials this way enables you to verify that you have valid, consistent data. To make sure you can tell definitively that a given variable is affecting the period, make the large value of the variable at least three times the small value in your trials. Here is bit of advice about how to measure the period of your pendulum. The period of your pendulum is likely to be quite short, only one or two seconds, so measuring it directly with accuracy is difficult. Here is an easy solution: assign one team member to hold the pendulum and release it on a signal. Assign another team member to count the number of swings the pendulum completes, and another member as a timer to watch the second hand on a clock. When the timer announces “GO” the person holding the pendulum releases it, and the swing counter starts counting. After exactly 10.0 seconds, the timer announces “STOP” and the swing counter states the number of swings completed by the pendulum during the trial. Record this value in a table in your lab journal. If you have four team members, the fourth person can be responsible for recording the data during the experiment. After the experiment, the data recorder reads off the data to the other team members as they enter the data in their journals. This method of counting the number of swings in 10 seconds does not give a direct measurement of the period, but you can see that your swing count works just as well for solving the problem posed by this experiment, and is a lot easier to measure than the period 380

Laboratory Experiments

itself. (The actual period is equal to 10 seconds divided by the number of swings that occur in 10 seconds.) One more thing on measuring your swing count: your swing counter should state the number of swings completed to the nearest 1/4 swing. When the pendulum is straight down, it has either completed 1/4 swing or 3/4 swing. When it stops to reverse course on the side opposite from where it is released, it has completed 1/2 swing. When you have finished taking data, review the data together as a team. If you did the experiment carefully, your data should clearly indicate which potential explanatory variables affect the period of the pendulum and which ones do not. If your swing counts for different trials of the same setup are not consistent, then something is wrong with your method. Your team must repeat the trials with greater care so that your swing counts for each different experimental setup are consistent. Discuss your results with your team members and reach a consensus about the meaning of your data. Expect to spend at least four hours writing, editing, and formatting your report. Lab reports count a significant percentage of your science course grades, so you should invest the time now to learn how to prepare a quality report. Your goal for this report is to begin learning how to write lab reports that meet all the requirements described in The Student Lab Report Handbook. One of our major goals for this year is to learn what these requirements are and become proficient at generating solid reports. Nearly all scientific reports involve reporting data, and a key part of this first report is your data tables, which should all be properly labeled and titled. After completing the experiment, all the information you need to write the report should be in your lab journal. If you properly journal the lab exercise, you will have all the data, your hypothesis, the materials list, your team members’ names, the procedural details, and everything else you need to write the report. Your report must be typed and will probably be around three pages long. You should format the report as shown in the examples in The Student Lab Report Handbook, including major section headings and section content. Here are a few guidelines to help you get started with your report: 1. There is only a small bit of theory to cover in the Background section, namely, to describe what a pendulum and its period are. You should also explain the experimental method, that is, why we are using the number of swings completed in 10 seconds in our work in place of the actual period. As stated in The Student Lab Report Handbook, the Background section must include a brief overview of your experimental method and your team’s hypothesis. 2. Begin your Discussion section by describing your data and considering how they relate to your hypothesis. In this experiment, we do not make quantitative predictions, so there are no calculations to perform for the discussion. We only seek to discover which variables affect the period of a pendulum and which do not. Your goal in the Discussion section is to identify what your data say and relate that to your reader. 3. Consider the following questions as you write your discussion. What variables did you manipulate to determine whether they had any effect on the period of the pendulum? What did you find? According to your data, which variables do affect the period? How do the data show this? Refer to specific data tables to explain specifically how the data support your conclusion. Are your findings consistent with your hypothesis? If not, then what conclusion do you reach about the question this experiment seeks to answer?

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Appendix C

Experiment 2 The Soul of Motion Experiment Newton’s second law of motion Essential equipment: •

vehicle

•

duct tape

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stop watch

•

bathroom weigh scale (2)

Note: The report for this experiment requires you to set up a graph showing predicted and experimental curves on the same set of axes. Procedures for creating such a graph on a PC or Mac are described in detail in The Student Lab Report Handbook. You will have a great time with this experiment, which is conducted in a flat parking lot as a class or other group. The idea is to push a vehicle from the rear, using scales that measure the force the pushers are applying to the vehicle. You time the vehicle as it accelerates from rest through a ten-meter timing zone and use the time data to calculate the experimental values of the vehicle’s acceleration. Using the mass of the vehicle and Newton’s second law, you calculate a predicted acceleration for each amount of pushing force used. Your goal is to compare your predicted accelerations to the experimental values of acceleration for four different force values. You then graph the results and calculate the percent difference to help you see how they compare. This experiment is an excellent example of how experiments in physics actually work. The scientists have a theory that enables them to predict, in quantitative terms, the outcome of an experiment. Then the scientists carefully design the experiment to measure the values of these variables and compare them to the predictions, seeking to account for all factors that affect the results. If the theory is sound and the experiment is well done, the results should agree well with the theoretical predictions and the percent difference should be low. In our case, when a force is applied to a vehicle at rest, we expect the vehicle to accelerate in accordance with Newton’s second law of motion: a=

F m

This equation predicts that the acceleration depends on the force applied. So Newton’s second law is our theoretical model for the motion of an accelerating object. Now, we know that a motor vehicle has a fair amount of friction in the brakes and wheel bearings, which means that not all the force applied by the pushers serves to accelerate the car. Some of it only overcomes the friction. Also, if the ground is not be perfectly level, this affects the acceleration as well. So to make the model as useful as possible, you must use the actual net force on the vehicle in your predictions. Details are discussed below. For the data collection, you must have a way to measure the actual vehicle’s acceleration so that you can compare it to your predictions. You already know an equation that gives the acceleration based on velocities and time. However, you have no convenient way of measuring the vehicle’s velocity. (The vehicle moves too slowly for the speedometer to be of any use.) Fortunately, there is another equation you can use if you time the vehicle with a stop watch as it starts from rest and moves through a known distance. If you know the distance 382

Laboratory Experiments

and the time, and the acceleration is uniform, you can calculate the vehicle’s acceleration as follows: a=

2d t2

Use this equation to determine the experimental acceleration value for each force, using the average time for each set of trials. Here are some crucial details to help make this experiment as successful as possible: 1. Always have two students pushing on the vehicle. Thus, for each force value the pushers use, the total applied force is twice that amount. (You use four different force values in the experiment.) 2. Measure the friction on the vehicle so you can subtract it from the force the pushers are applying to get the net force applied for your predictions. To measure the friction, use one pusher and estimate the absolute minimum amount of force needed to keep the vehicle barely moving at a constant speed. As you know from our studies of the laws of motion, vehicles move at a constant speed when there is no net force. So if the vehicle is moving at a constant speed, it means that the friction and the applied force are exactly balanced. This allows you to infer what the friction force is. 3. Use four different values of pushing force. For each force value, time the vehicle over the ten-meter timing zone at least three times. The forces the pushers apply to the vehicle always vary quite a bit, so if you get three valid trials at each force you have three reliable data points for the time. You then calculate the average of these times and use it to calculate the experimental value of the acceleration of the vehicle for that force. 4. The major factor introducing error into this experiment is the forces applied by the pushers. Pushing at a constant force while the vehicle is accelerating is basically impossible. (The dial on the force scale jumps all over the place.) But if the pushers are careful, they can push with an average force that is pretty consistent. You need a standard to judge whether you have had a successful run with consistent pushing. Here is the criterion to use: when you obtain three trials with time measurements all within a range of one second from highest to lowest, accept those values as valid. If your times are not this close together, assume that the pushing forces are not consistent enough and keep running new trials until you get more consistent data. 5. The instructor will take the vehicle, with a full gas tank, to get it weighed and report this weight to the class. Make sure to measure the weight of the driver and the weight of the scale support rack (if there is one). Add these weights to the weight of the vehicle and determine the mass for this total weight. (Of course, the instructor must also make 383

Appendix C

sure the gas tank is full on the day of the experiment, since the fuel in the tank typically amounts to 1–2% of the vehicle weight.) Considerations for Your Report In the Background section of your report, be sure to give adequate treatment to the theory you are using for this experiment. In the Newton’s second law equation, acceleration is directly proportional to force, so a graph of acceleration vs. force should be linear. In the Background, you should use this concept to explain why you expect your experimental acceleration values to vary in direct proportion to the force. Explain the equations you are using to get the predicted and experimental acceleration values. Since you are using two different equations, your Background section should include explanations for both of them and why they are needed. The force you are using to make your predictions takes friction into account. You need to explain how friction is taken into account, why you are doing so, and how this relates to the equations. In the Procedure section, don’t forget the important details, such as how you measured the friction force, weighed the driver, and judged the validity of your time data. In the Results section, present all time data in a single table, along with the average times for the trials at each force value applied by the pushers. Present all the predicted values, experimental values, and percent differences (see Preface, pages xviii–xix) in another table or two. Do not forget to state all the other values used in the experiment, such as the vehicle weight, the weights of the driver and support rack, the distance, the total mass you calculate, and the friction force you measure. As The Student Lab Report Handbook describes, in any report, all the data collected must be presented, and they all must be placed either in a table or in complete sentences. In the Discussion, the main feature is a graph of acceleration vs. force, showing both the predicted and experimental values on the same graph for all four force values. Carefully study Chapter 7 on graphs in The Student Lab Report Handbook and make sure your graph meets all the requirements listed. For your predicted values of acceleration, use the total mass of the vehicle, driver, and support rack. The instructor will tell you the weight of the vehicle, which you record in your lab journal. Also record the weights of the driver and support rack determined during the experiment. Convert the total weight from pounds to newtons, then determine the mass in kilograms by using the weight equation, Fw = mg. For the force values in your predictions, use the nominal amount of force applied (the two pushers’ forces combined) less the amount of force necessary to overcome the friction (which is determined during the experiment). Table C.1 summarizes the calculations you need to perform for each set of trials. The heart of your discussion is a comparison of the two curves representing acceleration vs. force (displayed on the same graph), and a discussion of how well the actual values of acceleration match up with the predicted values. In addition to this graphical comparison, compare the four predictions to the four experimental acceleration values by calculating the percent difference for each one, presenting these values in a table and discussing them. To compare the curves, think about the questions below. Do not write your discussion section by simply going down this list and answering each question. (Please spare your instructor the pain of reading such a report!) Instead, use the questions as a guide to the kinds of things you should discuss and then write your own discussion section in your 384

Laboratory Experiments

Variable

Equation

force

net force = There are four values of net force, one (2 × force for each pusher) – friction for each set of trials. force estimate

predicted acceleration

predicted accel = (net force)/(total mass)

experimental experimental accel = acceleration (2 × distance)/(avg time)2

Comments

Net force is as calculated above. Mass is determined from the total weight. There is a predicted acceleration for each value of net force. Distance is the length of the timing zone. Average time is the average of the three valid times for a given trial. There is an experimental acceleration for each value of net force.

Table C.1. Summary of equations for the calculations.

own words. Remember—this is an exercise in learning how to write a well-constructed lab report, not a boring fill-in-the-blank activity. Thought Questions and Considerations for Discussion 1. Are both the curves linear? What does that mean? 2. Do they both look like direct proportions? What does that imply? 3. Do the curves have similar slopes? What does that imply? 4. How successful are the results? A percent difference of less than about 5% for an experiment as crude as this can be considered a definite success. If the difference is greater than 5%, identify and discuss the factors that may have contributed to the difference between prediction and result. In this experiment, there are several such factors, including wind that may have been blowing on the vehicle. 5. Do not make the mistake of merely assuming that the fluctuation in the pushers’ forces explains everything without taking into account the precautions you took to eliminate this factor from being a problem (the time data validity requirement). 6. Also do not make the mistake of assuming that friction explains the difference between prediction and result. Friction can only affect the data one way (slowing the vehicle down). So if friction is a factor, the data have to make sense in light of how friction affects the data. But further, since measuring friction and taking it into account in your predictions is part of your procedure, a generic appeal to friction will not do. 7. Finally, do not make the mistake of asserting that errors in the timing or the timing zone distance measurement explain the difference between prediction and result. Consider just how large the percentage error could realistically be in these measurements, and whether that kind of percentage helps at all in explaining the difference you have between prediction and result. For example, the timing zone is 10 m long. If it is carefully laid out on the pavement, it is unlikely that the distance measurement is in error by more than a few centimeters or so. Even including the slight misalignments of the vehicle that crop up, the distance could probably not be off by more than, say, 10 or 20 cm. But this is only 1–2% of 10 m, and if you are trying to explain a percent difference of 5–10% or 385

Appendix C

more this won’t do it. Similar considerations apply to the time values. Given the slow speed the vehicle moves, how far off can the timing be? What kind of percentage error would this produce? Alternate Experimental Method If your class is using digital devices such as the PASCO Xplorer GLX to read forces, you can use a slightly different experimental method that improves results and lowers the difference between prediction and result. One of the major sources of error in this experiment is the difficulty the pushers have in accurately applying the correct amount of force to the vehicle. If you use bathroom scales to measure the force, there is nothing that can be done about this problem and the pushers simply have to do the best they can. However, with the digital devices you can eliminate the problem of force accuracy by using the actual average values of the forces applied by the two pushers to calculate the predicted values. The Xplorer GLX can record a data file of the applied force during a given trial, and when reviewing the data file back at your computer you can view the mean value of the force during the trial. You can use this mean value to calculate the predicted acceleration from Newton’s second law. Using this method to form your predictions eliminates much of the uncertainty surrounding the forces applied to the car. Here are a few details to consider if you use this alternative approach to collecting data: 1. You do not need to select four different force values in advance and push the vehicle repeatedly at each force value. Instead, only a single trial is needed for each force. 2. Select 10 or 12 different target force values and run a single trial with each. The force targets should range from low values that barely get the vehicle to accelerate, all the way up to the highest values the pushers can deliver. For each trial, tell the pushers the target force and tell them to do their best to stay on it during the trial. But it doesn’t matter nearly so much how accurate the pushers are because you are using the average of the actual data from the digital file to make the predictions, rather than relying on the pushers to maintain the target force accurately. 3. The method for determining values of net force for the predictions is similar to that shown in Table C.1. The difference is that instead of doubling the target force for each pusher, you add together the actual mean forces obtained from the data files for each pusher and subtract out the friction force. 4. Use the time of each trial to determine the experimental value of the acceleration for that trial. 5. Calculate the percent difference for each trial and report these values in the report. Also calculate the average of the percent difference values and use this figure in your discussion of the results.

386

Principles of Chemistry PUBLISHED BY CENTRIPETAL PRESS Grade Level: 11th Grade Yearlong Course

CLICK TO VIEW IN OUR ONLINE CATALOG

PRINCIPLES OF CHEMISTRY Principles of Chemistry is an introductory chemistry text designed for grade-level students in high school. This course is typically taught in 11th grade, while students are concurrently enrolled in algebra 2, and after students have taken Introductory Physics in 9th grade and General Biology in 10th grade (forthcoming). Here we highlight a few distinctive features of this text. •

Principles of Chemistry supports our signature philosophy of science education based on wonder, integration, and mastery. We always seek to build on and stimulate the student’s innate sense of wonder at the marvels found in nature. Integration refers to the epistemology, mathematics, and history embedded in the text, and a curriculum designed to help develop the student’s facility with using language to communicate scientific concepts. Principles of Chemistry is designed to be used in a mastery-learning environment, using the mastery teaching model John developed and describes in From Wonder to Mastery: A Transformative Model for Science Education. When teachers use this mastery-learning model, the result is high student achievement and exceptional long-term retention for all students.

•

The text includes a unique organizing introduction. The introduction, entitled “What Is Chemistry All About?,” sets forth five general organizing principles—chemistry is all about electrons, chemistry is all about electrical forces, chemistry is all about minimizing energy, chemistry is all about whole-number ratios of atoms, and chemistry is all about modeling. Throughout the balance of the text, students are repeatedly notified when a topic illustrates an instance of one of these five general principles.

•

The chapter exercises include plenty of content review. Beginning with Chapter 2, the chapter exercises following each chapter include 6–10 review problems to help students keep the content from old chapters fresh.

Support Resources A number of support resources are available to accompany Principles of Chemistry. They include: The Student Lab Report Handbook Students should begin writing their lab reports from scratch in 9th grade. This popular manual gives them everything they need. We recommend supplying this handbook to every freshman so they can refer to it throughout high school. Read more about this resource on pg. 6. Solutions Manual to Accompany Principles in Chemistry This book contains complete written solutions for all the computations in the chapter exercises. Chemistry Experiments for High School This student manual includes 20 excellent chemistry experiments that illustrate concepts and foster real-world lab skills. Using this book requires a standard high school laboratory facility.

PRINCIPLES OF CHEMISTRY Digital Resources The following materials are accessible exclusively through Centripetal Press: • a full year of chapter exams • two short quizzes for each chapter • two semester exams • a document containing answer keys for all the quizzes and tests and sample answers for all the verbal questions in the text • a document with recommendations for teaching the course • a lesson list and example calendar Tips and Tools A variety of tips and tools are available at www.centripetalpress.com. Here, teachers will find scores of images and short videos organized chapter by chapter. Teachers are free to use any of this content as they wish to assemble vivid and fascinating lessons.

PRINCIPLES OF CHEMISTRY Sample Chapters The following pages contain samples from the text. The Table of Contents is shown, as well as Chapters 1, 2, and 7. Note that the sample chapters shown are from the new edition planned for 2022. The new edition has been revised to accommodate the extensive changes to the metric system that went into effect in 2019. The edition currently in print looks exactly the same but uses the former definitions for the Avogadro constant, the mole, and other constants.

Principles of Chemistry A Mastery-Oriented Curriculum

Second Edition

John D. Mays

Camp Hill, Pennsylvania 2020

Principles of Chemistry © Classical Academic Press®, 2020 Edition 2.0 ISBN: 978-0-9972845-2-2 All rights reserved. Except as noted below, this publication may not be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior written permission of Classical Academic Press. All images attributed to others under any of the Wikimedia Commons licenses, such as CCBY-SA-3.0 and others, may be freely reproduced and distributed under the terms of those licenses. Classical Academic Press 515 S. 32nd Street Camp Hill, PA 17011 www.ClassicalAcademicPress.com/Novare/ Cover design by Nada Orlic, http://nadaorlic.info/ ISP.04.20

Contents Preface For Teachers

1. Student Audience and Preparedness 2. Our Emphasis on Mastery, Integration, and Wonder 3. Recommendations for Teaching With This Text 4. Laboratory Work and Lab Reports 5. Revisions in the Second Edition

Preface For Students Introduction What is Chemistry All About? I.1

A Few Major Themes I.1.1 Chemistry Is All About Electrons I.1.2 Chemistry Is All About Electrical Forces

Hmm... Interesting. Why water forms beads

I.1.3 Chemistry Is All About Minimizing Energy I.1.4 Chemistry Is All About Whole-Number Ratios of Atoms I.1.5 Chemistry Is All About Modeling I.2 Conclusion Introduction Study Questions

Chapter 1 Measurements 1.1

1.2

1.3 1.4

Science and Measurements 1.1.1 No Measurements, No Science 1.1.2 Matter, Volume, and Mass 1.1.3 The US Customary System 1.1.4 The SI Unit System 1.1.5 Metric Prefixes Converting Units of Measure 1.2.1 Basic Principles of Unit Conversion Factors 1.2.2 Tips for Converting Units of Measure 1.2.3 Converting Temperature Units Accuracy and Precision 1.3.1 Distinguishing Between Accuracy and Precision 1.3.2 Significant Digits Other Important Math Skills 1.4.1 Scientific Notation 1.4.2 Calculating Percent Difference

xii xii xiii xiv xvi xvii xviii 2 3 3 5 6 6 9 10 12 12 14 16 16 16 18 18 20 22 22 24 27 28 28 29 35 35 37

Contents Chapter 1 Exercises

Chapter 2 Atoms and Substances 2.1 2.2

Atoms and Molecules 2.1.1 Atomic Facts 2.1.2 The History of Atomic Models Types of Substances 2.2.1 Pure Substances: Elements and Compounds 2.2.2 Mixtures

Hmm... Interesting. Brownian motion

2.2.3 Physical and Chemical Properties 2.3 Isotopes and Atomic Masses 2.3.1 Isotopes 2.3.2 The Unified Atomic Mass Unit 2.3.3 Atomic Masses 2.4 Density and Quantity of Substances 2.4.1 Density 2.4.2 The Mole and the Avogadro Constant 2.4.3 Molar Mass and Formula Mass 2.4.4 Gram Masses of Atoms and Molecules Chapter 2 Exercises

Chapter 3 Atomic Structure 3.1

Atomic Spectra 3.1.1 The Electromagnetic Spectrum 3.1.2 Energy in Atoms

Hmm... Interesting. Neon signs and phonons

3.1.3 The Hydrogen Atom 3.2 The Bohr Model of the Atom 3.3 The Quantum Model of the Atom 3.3.1 Schrödinger and Pauli 3.3.2 Shells, Subshells, and Orbitals 3.3.3 The Aufbau Principle, the Madelung Rule, and Hund’s Rule 3.4 Electron Configurations 3.4.1 Electron Configurations 3.4.2 Condensed Electron Configurations 3.5 Empirical Formulas 3.5.1 Percent Composition and Empirical Formulas 3.5.2 Determining a Molecular Formula from an Empirical Formula Chapter 3 Exercises

Chapter 4 The Periodic Law 4.1 4.2 4.3

The Periodic Table of the Elements Periodic Table Nomenclature Periodic Physical Properties 4.3.1 Atomic Radius

37 42 44 44 45 49 49 53 56 57 59 59 60 60 62 62 64 65 69 71 74 76 76 78 79 81 82 83 83 83 88 90 90 92 93 93 95 96 100 102 104 105 105

Contents 4.3.2 Ionic Radius Periodic Chemical Properties 4.4.1 Core and Valence Electrons 4.4.2 Ionization Energy 4.4.3 Electron Affinity 4.4.4 Electronegativity 4.5 A Few Notes About Hydrogen Chapter 4 Exercises 4.4

Hmm... Interesting. Hydrogen in space

107 108 108 109 113 115 116 117 117

Chapter 5 Chemical Bonding 5.1

120 122 122 122 123 123 125 127 129 129 130 131 131 133 134 134 135 136 140 141 142 143 144 145 145 146 147 Hmm... Interesting. The molecular structure of glass and quartz 148

Preliminaries 5.1.1 Chemical Possibilities 5.1.2 The Octet Rule 5.2 Ionic Bonding 5.2.1 Ionic Bonds and Crystals 5.2.2 Naming Ionic Compounds 5.2.3 Energy in Ionic Bonds 5.2.4 Hydrates 5.2.5 Intensive and Extensive Properties 5.2.6 Physical Properties of Ionically Bonded Substances 5.3 Covalent Bonding 5.3.1 Covalent Bonds and Molecules 5.3.2 Polyatomic Ions 5.3.3 Ionic Compounds with Polyatomic Ions 5.3.4 Polyatomic Ion Names 5.3.5 Naming Acids 5.3.6 Lewis Structures 5.3.7 Exceptions to the Octet Rule 5.3.8 Resonance Structures 5.3.9 Naming Binary Covalent Compounds 5.3.10 Energy in Covalent Bonds 5.3.11 Physical Properties of Covalently Bonded Substances 5.4 Electronegativity, Polarity, and Bond Character 5.4.1 Polarity and Dipoles 5.4.2 The Nature of the Bond Chapter 5 Exercises

Chapter 6 Molecular Theory and Metallic Bonding 6.1

6.2

Molecular Structure 6.1.1 Covalent Bond Theory 6.1.2 Valence Shell Electron Pair Repulsion (VSEPR) Theory 6.1.3 The Effect of Nonbonding Domains on Bond Angle Metallic Bonding 6.2.1 Metallic Lattices 6.2.2 Physical Properties of Metals

150 152 152 152 156 157 157 158 vii

Contents 6.3

Intermolecular Forces 6.3.1 Bonding Forces 6.3.2 Intermolecular Forces

Hmm... Interesting. Tin pest

6.3.3 Hydrogen Bonding 6.3.4 Van der Waals Forces Chapter 6 Exercises

Chapter 7 Chemical Reactions and Stoichiometry 7.1

Introduction to Chemical Equations 7.1.1 Fascinating Chemistry 7.1.2 The Law of Conservation of Mass in Chemical Reactions 7.1.3 Reaction Notation 7.1.4 Balancing Chemical Equations 7.1.5 Oxidation States

7.2

General Types of Chemical Reactions 7.2.1 Synthesis Reactions 7.2.2 Decomposition Reactions 7.2.3 The Activity Series of Metals 7.2.4 Single Replacement Reactions 7.2.5 Double Replacement Reactions

Hmm... Interesting. Why nitrates and nitros blow up

Hmm... Interesting. A story about aqua regia

7.2.6 Combustion Reactions 7.2.7 Acid-Base Neutralization Reactions 7.2.8 Oxidation-Reduction Reactions 7.3 Stoichiometry 7.3.1 Stoichiometric Calculations 7.3.2 Limiting Reactant 7.3.3 Theoretical Yield and Percent Yield Chapter 7 Exercises

Chapter 8 Kinetic Theory and States of Matter 8.1

Temperature, Kinetic-Molecular Theory, and Pressure 8.1.1 Temperature and Molecular Energy 8.1.2 Velocity Distribution of Gases 8.1.3 The Kinetic-Molecular Theory of Gases 8.1.4 Gas Pressure

8.2

States of Matter 8.2.1 The Four Basic States of Matter 8.2.2 Solids 8.2.3 Liquids 8.2.4 Gases 8.2.5 Plasmas

Hmm... Interesting. How barometers work

Hmm... Interesting. Gas diffusion

8.2.6 Phase Transitions and Phase Diagrams viii

160 160 160 160 161 162 164 168 170 170 170 171 172 177 179 180 180 181 181 182 184 184 185 185 186 187 187 192 194 195 200 202 202 202 203 204 206 208 208 208 209 211 212 212 213

Contents 8.2.7 Heat Capacity, Heat of Fusion, and Heat of Vaporization 8.2.8 Evaporation 8.2.9 Vapor Pressure Chapter 8 Exercises

Chapter 9 The Gas Laws 9.1

9.2

Early Formulations of the Gas Laws 9.1.1 Boyle’s Law 9.1.2 Charles’ Law 9.1.3 Avogadro’s Law The Ideal Gas Law 9.2.1 Standard Temperature and Pressure 9.2.2 The Ideal Gas Law

Hmm... Interesting. The gas laws as models

9.3 9.4

9.2.3 Using the Ideal Gas Law to Find Molar Mass and Density The Law of Partial Pressures 9.3.1 Dalton’s Law of Partial Pressures 9.3.2 Collecting a Gas Over Water Stoichiometry of Gases and Effusion 9.4.1 Stoichiometry of Gases 9.4.2 Gas Diffusion and Effusion

Hmm... Interesting. Uranium enrichment

Chapter 9 Exercises

Chapter 10 Solution Chemistry

10.1 Dissolution 10.1.1 The Process of Dissolving 10.1.2 Heat of Solution 10.1.3 Entropy and Free Energy 10.1.4 Electrolytes 10.2 Solubility 10.2.1 Ionic Solids in Water 10.2.2 Ionic Solids in Nonpolar Solvents 10.2.3 Polar Liquids 10.2.4 Nonpolar Liquids 10.2.5 Solutions of Solids

Hmm... Interesting. How soap works

10.2.6 Gases in Liquid Solutions 10.2.7 The Effect of Temperature on Solubility 10.3 Quantifying Solution Concentration 10.3.1 Molarity 10.3.2 Molality 10.4 Compounds in Aqueous Solution 10.4.1 Ionic Equations and Precipitates 10.4.2 Net Ionic Equations and Spectator Ions 10.5 Colligative Properties of Solutions 10.5.1 Vapor Pressure Lowering

216 220 222 223 226 228 228 229 232 233 233 233 234 240 243 243 246 248 248 249 250 251 256 258 258 261 261 263 264 264 265 266 267 267 268 270 271 272 272 274 275 275 277 278 278 ix

Contents 10.5.2 Freezing Point Depression and Boiling Point Elevation Chapter 10 Exercises

Chapter 11 Acids and Bases

11.1 Properties and Nomenclature of Acids and Bases 11.1.1 Introduction 11.1.2 Properties of Acids and Bases 11.1.3 Acid Names and Formulas 11.2 Acid-Base Theories 11.2.1 Arrhenius Acids and Bases 11.2.2 Brønsted-Lowry Acids and Bases

Hmm... Interesting. What is an alkali?

11.2.3 Lewis Acids and Bases 11.2.4 Strength of Acids and Bases 11.3 Aqueous Solutions and pH 11.3.1 The Self-ionization of Water 11.3.2 Calculating [H3O+] and [OH–] 11.3.3 pH as a Measure of Ion Concentration and Acidity 11.3.4 pH Measurement, pH Indicators, and Titration 11.3.5 Titration Procedure 11.3.6 Determining [H3O+] or [OH–] from Titration Data Chapter 11 Exercises

Chapter 12 Redox Chemistry

290 292 292 292 294 295 296 298 299 302 303 306 306 306 307 312 316 318 319

12.3.4 Electrode Potentials 12.3.5 Electrochemical Applications Chapter 12 Exercises

324 326 326 326 330 332 332 336 342 342 343 344 348 350 353 358

Glossary

362

Answers to Selected Exercises

378

Appendix A Reference Data

390

12.1 Oxidation and Reduction 12.1.1 Introduction to Redox Reactions 12.1.2 Oxidation States 12.1.3 Strengths of Oxidizing and Reducing Agents 12.2 Redox Reaction Equations 12.2.1 Redox Half-Reactions 12.2.2 Balancing Redox Equations 12.3 Electrochemistry 12.3.1 Copper and Zinc Redox 12.3.2 Electricity Instead of Heat 12.3.3 Electrochemical Cells

Hmm... Interesting. How are salt bridges made?

280 285

Appendix B Scientists to Know About

394

Appendix C Memory Requirements

396

References and Citations

397

Image Credits

399

Index

401

Chapter 1 Measurements ΔνCs c

s m

mol K kB

cd Kcd

The SI unit system—or metric system—underwent a serious overhaul in 2019. Prior to the makeover, the kilogram was still defined by a physical object kept in a vault in France. Now, however, the seven base units in the SI system, represented by the circles in the inner ring in the graphic above, are defined in terms of physical constants (the outer ring) and each other. For example, the meter (m) is defined as the distance light travels in 1/299,792,458 seconds. That big number in the denominator is the speed of light in meters per second. So, the definition of the meter involves the speed of light (c) and the definition for the second (s). The arrows in the graphic indicate which units and constants affect others. As you see, the definition of the second affects every other unit definition except one.

Measurements

Objectives for Chapter 1 After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary. SECTION 1.1

1. Define matter and mass. 2. Describe the advantages the SI system has over the USCS system for scientific work. 3. State the SI base units for length, mass, and time. 4. For the SI system of units, define the terms base unit and derived unit. 5. State several examples of base units and derived units in the SI system of units. 6. Use the metric prefixes listed in Table 1.4 from memory with various units of measure to perform unit conversions and solve problems. SECTION 1.2

7. Given appropriate unit conversion factors, convert the units of measure for given quantities to different units of measure. 8. Given appropriate conversion factors, convert USCS units to SI units and vice versa. 9. Convert units of temperature measurements between °F, °C, and K. SECTION 1.3

10. Define accuracy and precision. 11. Name several possibilities for sources of error in experimental measurements. 12. Explain how the significant digits in a measurement relate to the precision of the measurement. 13. Use significant digits correctly to record measurements from digital and analog measurement instruments. 14. Use significant digits correctly to perform computations, including multiplication, division, addition, subtraction, and combinations of these operations. SECTION 1.4

15. Convert numerical values from standard notation into scientific notation and vice versa. 16. Use scientific notation when recording measurements and performing computations. 17. Calculate the percent difference between an experimental value and a theoretical value or accepted value.

Chapter 1

1.1

Science and Measurements

1.1.1 No Measurements, No Science One of the things that distinguishes scientific research from other fields of study is the central role played in science by measurement. In every branch of science, researchers study the natural world, and they do it by making measurements. The measurements we make in science are the data we use to quantify the facts we have and to test new hypotheses. These data—our measurements—answer questions such as What is its volume?, How fast is it moving?, What is its mass?, How much time does it take?, What is its diameter?, What is its frequency and wavelength?, When will it occur again?, and many others. Without measurements, modern science would not exist. Units of measure are crucial in science. Since science is so deeply involved in making measurements, you will work with measurements a lot in this course. The value of a measurement is always accompanied by the units of measure—a measurement without its units of measure is a meaningless number. For this reason, your answers to computations in scientific calculations must always show the units. In this chapter, we discuss units of measure at some length. The material in the chapter is very important—for the rest of this book, we will be engaged with calculations involving units of measure. However, before we launch into our study of units, there is a topic of great importance we need to nail down—mass. Students often do not fully understand what mass is, so we will start by reviewing the topic.

1.1.2 Matter, Volume, and Mass The best way to understand mass is to begin with matter and its properties. The term matter refers to anything composed of atoms or parts of atoms. Note that there are many things that are not material, that is, they are not matter, as illustrated in Figure 1.1. Your thoughts, concepts such as justice and equity, and your favorite song are not matter. You can write down your thoughts in ink, which is matter, and your song can be recorded onto a CD, which is matter. But ideas and songs are not material and are not made of what we call matter. Another part of this

A person’s thoughts are not matter.

I love science!

Ideas are not matter.

E = mc 2

A badger is matter. Potassium permanganate is matter.

Light is not matter. Figure 1.1. Some things are matter and some things are not.

Measurements world that is not matter is electromagnetic radiation—light, radio waves, X-rays, and all other forms of electromagnetic radiation. Light is pure energy; it is not matter and it has no mass. We get into electromagnetic radiation a bit in Chapter 3. For now, we are going to focus on matter. A lot of what we discuss in this text is about different properties of matter. There are many different properties to discuss, but here we focus on just two properties that all matter possesses: all matter takes up space and all matter has inertia. Describing and comparing these two properties helps make clear what we mean by the term mass. All matter takes up space. Even individual atoms and protons inside of atoms take up space. Now, how do we quantify how much space an object takes up? That is, how do we put a numerical measurement to it? The answer is, of course, by specifying its volume. Volume is the name of the variable we use to quantify how much space an object takes up. There are many different units of measure we use to specify an object’s volume. Examples are gallons, liters, cubic meters, and pints. When we say that the volume of an object is 338 cubic centimeters, what we mean is that if we could hollow the object out and fill it up with little cubes, each with a volume of one cubic centimeter, 338 of them are needed to fill up the hollowed object. All matter has inertia. The effect of this property is that objects resist being accelerated. The more inertia an object has, the more difficult it is to accelerate the object. For example, if the inertia of an object is small, as with, say, a golf ball, the object is easy to accelerate. Golf balls are easy to throw, and if you hit one with a golf club it accelerates at a high rate to a very high speed. But if the amount of inertia an object has is large, as with say, a grand piano, the object is difficult to accelerate. Just try throwing a grand piano or hitting one with a golf club and you will see that it doesn’t accelerate at all. This is because the piano has a great deal more inertia than a golf ball. As with the property of taking up space, we need to quantify the property of inertia. The way we do this is with the variable we call mass. The mass of an object is a numerical measurement specifying the amount of inertia the object has. Since inertia is a property of matter, and since all matter is composed of atoms, it should be pretty obvious that the more atoms there are packed into an object, the more mass it has. And since the different types of atoms themselves have different masses, an object made of more massive atoms has more mass than an object made of an equal number of less massive atoms. The main unit of measure we use to specify an object’s mass is the kilogram. There are other units such as the gram and the microgram. The kilogram (kg) is one of the base units in the metric system, our topic in Section 1.1.4. On the earth, an object weighing 2.2 pounds (lb) has a mass of one kilogram. To give you an idea of what a kilogram mass feels like in your hand on the earth, the lantern battery pictured in Figure 1.2 weighs 2.2 lb, and thus has a mass of 1 kg. We have established that the mass of an object is a measure of its inertia, which in turn depends on how many atoms it is composed of and how massive those atoms are. The implication of this is that an object’s mass does not depend on where it is. A golf ball on the earth has the same mass as a golf ball at the bottom of the ocean, on the moon, or in outer space. Even where there is no gravity, the Figure 1.2. The mass of this mass of the golf ball is the same. This is what distinguishes the mass battery is about one kilogram. of an object from its weight. Weight is caused by the force of gravity acting on an object composed of matter (which we often simply refer to as a mass). The weight of an object depends on where it is. An object—or mass—on the moon weighs only about 1/6 its weight on earth, and in outer space, where there is no gravity, a mass has no weight at all. But the mass of an object 17

Chapter 1 does not depend on where it is. This is because an object’s mass is based on the matter the object is made of. The lantern battery in the figure has a certain weight on the earth (2.2 lb). In outer space, it weighs nothing and floats right in front of you. But if you try to throw the battery, the force you feel on your hand is the same on the earth or in space. That’s because the force you feel depends on the object’s mass. Here is a summary using slightly different terminology that may help even more. Inertia is a quality of all matter; mass is the quantity of a specific portion of matter. Inertia is a quality or property all matter possesses. Mass is a quantitative variable, and it specifies an amount of matter, a quantity of matter.

1.1.3 The US Customary System The two major systems of units students should know about are the International System of Units, known as the SI system or the metric system, and the U.S. Customary System, or USCS. You have probably studied these systems before and should be already familiar with some of the SI units and prefixes. In this course, we do not make much use of the measurement system you are most familiar with—the USCS. For scientific work, the entire international scientific community uses the SI system. But here I address the USCS briefly before moving on. Americans are generally comfortable with measurements in feet, miles, gallons, inches, and degrees Fahrenheit because they grow up using this system and are very familiar with it. But in fact, the USCS is rather cumbersome. One problem is that there are many different units of measure for every kind of physical quantity. Just for measuring length or distance, for example, Unit Symbol Quantity we have the inch, foot, yard, and mile. The USCS meter m length is also full of random numbers such as 3, 12, and kilogram kg mass 5,280. A third problem is that there is no inherent connection between units for different types second s time of quantities. Gallons have nothing whatsoever ampere A electric current to do with feet, and quarts have nothing to do kelvin K temperature with miles. candela Cd luminous intensity The USCS may be familiar ground, and it may even feel patriotic to prefer it, but it is not mole mol amount of substance the system of measurement scientists use. ScienTable 1.1. The seven base units in the SI unit system. tists everywhere use the SI, and it is to that system we now turn.

1.1.4 The SI Unit System In contrast to the USCS, the SI system is simple and has many advantages. There is usually only one basic unit for each kind of quantity, such as the meter for measuring length. Instead of having many unrelated units of measure for measuring quantities of different sizes, prefixes based on powers of ten are used on all the units to accommodate various sizes of quantities. And units for different types of quantities relate to one another in some way. Unlike the gallon and the Unit Symbol Quantity foot, which have nothing to do with each other, joule J energy the cubic meter is 1,000,000 cubic centimeters. newton N force For all these reasons, the USCS is not used much volume cubic meter m3 at all in scientific work. The SI system is the international standard. watt W power There are seven base units in the SI System, pascal Pa pressure listed in Table 1.1. All other SI units of measure,

Table 1.2. Some SI System derived units.

Measurements such as the joule (J) for measuring quantities of energy and the newton (N) for measuring amounts of force, are based on these seven base units. Units based on combinations of the seven base units are called derived units. A few common derived units are listed in Table 1.2. You are already familiar with the SI unit for time: the second. You may or may not be familiar with some of the other base units, so here are some facts and photos to help familiarize you with these. A meter is just a few inches longer than a yard (3 feet). Figure 1.3 shows a wooden measuring rule one meter long, commonly called a meter stick, along with a metal yardstick for comparison. On earth, a mass of one kilogram weighs about 2.2 pounds. The six-volt lantern battery shown in Figure 1.2 weighs just under 2.2 pounds, so the mass of the battery is just about one kilogram, as I mentioned before. I can’t show you a picture of one ampere of electric current, but it may be helpful to know that a standard electrical receptacle (or “outlet”) such as the one shown in Figure 1.4 is rated to carry 15 amperes Figure 1.3. A meter stick of current. (However, the largest continuous current that the recep- (left), with a yardstick for comparison. tacle is allowed to supply is 80% of its rating, or 12 amperes. This is why vacuum cleaners are often advertised as having 12-amp motors. That’s the upper limit of the current available to run them.) Regarding temperature units, the Celsius scale is generally used for making scientific temperature measurements, but the Kelvin scale must be used for nearly all scientific calculations involving temperature. You must become familiar with both scales. On the Celsius scale, water freezes at 0°C and boils at 100°C. Since Celsius temperature measurements can take on negative values, the Celsius scale (like the Fahrenheit scale) is not an absolute temperature scale. The Kelvin scale is an absolute scale, with 0 K (0 kelvins) being equal to absolute zero, theoretically the lower limit of possible Figure 1.4. A standard temperatures. Note that the term “degrees” is not used when stating or receptacle in American writing values in kelvins. homes is rated for a current A temperature change of one kelvin is the same as a temperature of 15 amperes. change of one degree Celsius, and both are almost double the change that a change of one degree Fahrenheit is. For reference, room temperature on the three scales is 72°F, 22.2°C, and 295.4 K.

All seven of the SI base units are defined in terms of physical constants, each one of which is defined with its own exact value. For example, the speed of light in a vacuum is defined to be 299,792,458 meters per second. The meter is defined from this: the distance light travels in 1/299,792,458 seconds. (And the second has its own definition.) Formerly, units such as the kilogram and the meter were defined by man-made physical objects (artifacts), such as the one shown in Figure 1.5. The bar shown in the figure was the standard in the U.S. for the Figure 1.5. Standard meter bar number meter from 1893 to 1960. But this method of definition 27, owned by the U.S. and used as the was not at all convenient. standard meter from 1893 to 1960. 19

Chapter 1 Prefix Multiples

Symbol Factor Prefix

Fractions

Symbol Factor

deca– hecto– kilo– mega– giga– da 10

k 2

M 3

tera–

peta–

exa–

zetta– yotta– Z

1024

deci– centi– milli– micro– nano– pico– femto– atto– zetto– yocto– d 1/10

m 2

1/10

μ 3

1/10

n 6

1/10

p 9

1/10

f 12

1/10

a 15

1/10

z 18

1/10

y 21

1/10

1/1024

Table 1.3. The SI System prefixes.

In 1960, the definition of the meter was changed so that the meter was equal to a certain number of wavelengths of a certain color of light emitted by a certain isotope of the element krypton. In 1983, the definition of the meter was changed again to its present definition in terms of the speed of light and the second. The kilogram has a similar history, and its definition in terms of an artifact only ended in 2019. The SI system includes not only the base and derived units, but all the other units that can be formed by adding metric prefixes to these units. We address the prefixes in the next section. But first I will mention a particular subset of the SI system known as the MKS system. MKS stands for meter-kilogram-second. The MKS system uses only the base and derived units without the prefixes (except for the kilogram, the only base unit with a prefix). The nice thing about the MKS system is that any calculation performed with MKS units produces a result in MKS units. For this reason, the MKS system is used almost exclusively in physics. However, in chemistry, it is common to use SI units that are not MKS units. Some commonly used non-MKS units are the gram (g), the centimeter (cm), the cubic centimeter (cm3), the liter (L), and the milliliter (mL). The liter is not actually an official SI unit, but it is used all the time in chemistry anyway.

1.1.5 Metric Prefixes In the system of units commonly used in the U.S., different units are used for different sizes of objects. For example, for short distances we might use the inch or the foot, whereas for longer distances we switch to the mile. For the small volumes used in cooking, we use the fluid ounce (or pint, quart, teaspoon, tablespoon, etc.), but for larger volumes like the gasoline in the gas tank of a car, we switch to the gallon. (That’s six different volume units I just listed!) The SI System is much simpler. Each type of quantity—such as length or volume—has one main unit of measure. Instead of using several different units for different sizes of quantities, the SI System uses multipliers on the units to multiply them for large quantities, or to scale them down for smaller quantities. We call these multipliers the metric prefixes. The complete list of the 20 metric prefixes is in Table 1.3. You do not need to memorize all these; some are rarely used. But you do need to memorize some of them. I recommend that all science students commit to memory the prefixes listed in Table 1.4. Fractions Multiples Table 1.5 shows a few Prefix Symbol Factor Prefix Symbol Factor representative examples kilo– k 103 centi– c 1/102 of how to use the prefixes to represent mulmilli– m 1/103 mega– M 106 6 9 tiples (quantities larger micro– μ 1/10 giga– G 10 than the SI base unit) nano– n 1/109 tera– T 1012 pico–

1/1012

Table 1.4. SI System prefixes to commit to memory.

Measurements

Multiples

Prefix

Symbol

Meaning

Examples of usage

kilo–

1,000

One kilojoule is 1,000 joules. There are 1,000 joules in one kilojoule, so 1,000 J = 1 kJ.

mega–

1,000,000

One megawatt is 1,000,000 watts. There are 1,000,000 watts in one megawatt, so 1,000,000 W = 1 MW.

centi–

1/100

One centimeter is 1/100 of a meter. There are 100 centimeters in one meter, so 100 cm = 1 m.

milli–

1/1,000

One milligram is 1/1,000 of a gram. There are 1,000 milligrams in one gram, so 1,000 mg = 1 g.

micro–

1/1,000,000

One microliter is 1/1,000,000 of a liter. There are 1,000,000 microliters in one liter, so 1,000,000 μL = 1 L.

Fractions

Table 1.5. Examples of correct usage of metric prefixes.

and fractions (quantities smaller than the SI base unit). As illustrations, let’s look at a couple of these more closely. The prefix kilo– is a multiple, and it means 1,000. One kilogram is 1,000 grams, one kilometer is 1,000 meters, and so on. Figure 1.6 illustrates with my favorite dairy product—chocolate chip cookie dough ice cream. One gram is only a fraction of a taste (and contains only two chocolate chips and no cookie dough). A kilogram of ice cream is 1,000 grams of ice cream, equivalent to two large bowls of ice cream.

1,000 g 1g

Figure 1.6. The bowl in the photo above contains one gram (1 g) of chocolate chip cookie dough ice cream. Each of the two bowls shown to the right contains 500 g of chocolate chip cookie dough ice cream, so together they contain 1,000 g, or one kilogram of ice cream.

Chapter 1 The prefix milli– is a fraction, and it means one thousandth. One millimeter is one thousandth of a meter, and so on. The wooden rule in Figure 1.3 is one meter in length. A millimeter is one thousandth of this length, equal to the width of the line in Figure 1.7. We conclude this introduction to metric prefixes with a few brief notes. First, when using the prefixes for quantities of mass, prefixes are never added to the kilogram. Prefixes are only added to the gram, even though the kilogram is the base 1 mm unit in the SI system, not the gram. Second, note that when writing the symbols for metric prefixes, the case of the letter matters: kilo– always takes a lower-case k, mega– always takes an upper-case M, and so on. Third, one of the prefix symbols is not an English letter. The prefix μ for micro– is the lower-case Greek letter Figure 1.7. One millimeter, mu, the m in the Greek alphabet. Finally, pay close attention to the difference which is one between multiplier prefixes and fraction prefixes. Learning to use the fraction thousandth of a prefixes properly is the most challenging part of mastering the SI System of units, meter. and using them incorrectly in unit conversion factors (our next topic) is a common student error.

1.2

Converting Units of Measure

1.2.1 Basic Principles of Unit Conversion Factors For scientists and engineers, one of the most commonly used skills is re-expressing quantities into equivalent quantities with different units of measure. These calculations are called unit conversions. Mastery of this skill is essential for all students studying science. In this section, I describe what unit conversion factors are and how they are used. Let’s begin with the basic principles of how unit conversions work. First, we all know that multiplying any value by unity (one) leaves its value unchanged. Second, we also know that in any fraction, if the numerator and denominator are equivalent, the value of the fraction is unity (one). For example, the expression “12 bricks over 12 bricks” is equal to one: 12 bricks =1 12 bricks This is because the numerator and denominator are equivalent, and any time this is the case, the value of the fraction is one, or unity. A unit conversion factor is simply a fractional expression in which the numerator and denominator are equivalent ways of writing the same physical quantity with different units of measure. This means a conversion factor is just a special way of writing unity (one). The third basic principle is that when multiplying fractions, factors that appear in both the numerator and denominator may be “cancelled out.” So when performing ordinary unit conversions, what we are doing is repeatedly multiplying a given quantity by unity so that cancellations alter the units of measure until they are expressed the way we wish. Since all we are doing is multiplying by one, the value of our original quantity is unchanged; it simply looks different because it is expressed with different units of measure. There are many different units of measure and there are many different conversion factors used for performing unit conversions. Table A.3 in Appendix A lists a number of important ones. Let me elaborate a bit more on the idea of unity I mentioned above, using one common conversion factor as an example. American school kids all learn that there are 5,280 feet in one mile, which means 5,280 ft = 1 mi. One mile and 5,280 feet are equivalent ways of writing the same length. If we place these two expressions into a fraction, the numerator and denominator 22

Measurements are equivalent, so the value of the fraction is unity, regardless of the way we write it. The equation 5,280 ft = 1 mi can be written as a conversion factor two different ways, and the fraction equals unity either way: 5280 ft 1 mi = =1 1 mi 5280 ft

(1.1)

Now, I need to make an important clarification about the use of the equal sign in the expressions I just wrote. In mathematics, the equal sign means identity. An expression such as 3 miles = 3 miles is a mathematical identity; the first expression, “3 miles” is identical to the second expression, also “3 miles.” But when we are dealing with converting units of measure from one set of units to a different set of units, we don’t use the equal sign to mean identity. Obviously, the expression 5280 ft 1 mi is not identical to the expression 1 mi 5280 ft One of these has units of ft/mi and the other has units of mi/ft. But even though these expressions are not identical, they are equivalent. When we are dealing with converting units of measure, this is the sense in which we interpret the equal sign. We are using it to mean equivalent. This is why I can write Equation (1.1) using equal signs. The three terms in Equation (1.1) are not identical, but they are equivalent. Suppose you have a measurement such as 43,000 feet that you wish to re-express in miles. To convert the units from feet to miles, first write down the quantity you are given, with its units of measure: 43,000 ft Next, select a unit conversion factor containing the units you presently have and the ones you want to convert to. (This is not always possible. Sometimes more than one conversion factor is required, as Example 1.1 below illustrates.) In this case, those units are feet and miles, and the conversion factors containing these units are the two written in Equation (1.1). As I explain below, the one we need for converting 43,000 ft into miles is the second one. So to perform the conversion, you multiply your given quantify by the conversion factor. Then you cancel any units that appear both in the numerator and the denominator, as follows: 43,000 ft ⋅

1 mi = 8.1 mi 5280 ft

Chapter 1 the “feet” in the given quantity (which is in the numerator), so the conversion factor needs to be written with feet in the denominator and miles in the numerator. Second, if you perform the calculation above (43,000 ÷ 5,280), the result that appears on your calculator screen is 8.143939394. So why didn’t I write down all those digits in my result? Why did I round my answer off to simply 8.1 miles? The answer to that question has to do with the significant digits in the value 43,000 ft that we started with. We address the issue of significant digits later in this chapter, but in the examples that follow I always write the results with the correct number of significant digits for the values involved in the problem. The following example illustrates the use of conversion factors based on metric prefixes. This example also illustrates how to perform a conversion when more than one conversion factor is required. Example 1.1 Convert the value 2,953,000 μg into kilograms. Referring to Tables 1.4 and 1.5, you see that the symbol μ means micro–, which means one millionth. Thus, μg means millionths of a gram. We use this information to make conversion factors. Since it takes 1,000,000 millionths of a gram to make one gram, 1,000,000 μg = 1 g, and thus 1,000,000 µg 1g = =1 1g 1,000,000 µg Note that converting from μg to g only gets us part of the way toward the solution. We need another conversion factor to get from g to kg. Looking again at Tables 1.4 and 1.5, we see that there are 1,000 grams in one kilogram, or 1,000 g = 1 kg. From this we can make additional conversion factors: 1000 g 1 kg = =1 1 kg 1000 g Now to perform the conversion, first convert micrograms into grams. Then convert grams into kilograms. You can do this two-step conversion at the same time by simply multiplying both conversion factors at the same time as follows: 2,953,000 µg ⋅

1g 1 kg ⋅ = 0.002953 kg 1,000,000 µg 1000 g

The μg in the given quantity cancels with the μg in the denominator of the first conversion factor. The g in the first conversion factor cancels with the g in the denominator of the second conversion factor. The units we are left with are the kg in the numerator of the second conversion factor. The kg did not cancel out with anything, so these are the units of our result.

1.2.2 Tips for Converting Units of Measure There are several important points you must remember in order to perform unit conversions correctly. I illustrate them below with examples. You should rework each of the examples on your own paper as practice to make sure you can do them correctly. The conversion factors used in the examples below are all listed in Table A.3 in Appendix A. 24

Measurements Point 1 Never use slant bars in your unit fractions. Use only horizontal bars. In printed materials, we often sees values written with a slant fraction bar in the units, as in the value 35 m/s. Although writing the units this way is fine for a printed document, you should not write values this way when you are performing unit conversions. This is because it is easy to get confused and not notice that one of the units is in the denominator in such an expression (s, or seconds, in my example), and the conversion factors used must take this into account. Example 1.2 Convert 57.66 mi/hr into m/s. Writing the given quantity with a horizontal bar makes it clear that the “hours” is in the denominator. This helps you write the hours-to-seconds factor correctly. To perform this conversion, we must convert the miles in the given quantity into meters, and we must convert the hours into seconds. From Table A.3, we select the two conversion factors we need. Then we multiply the given quantity by them to convert the mi/hr into m/s. When doing the multiplying, we write all the unit fractions with horizontal bars. 57.66

mi 1609 m 1 hr m ⋅ ⋅ = 25.77 hr 1 mi 3600 s s

As you see, the miles cancel and the hours cancel, leaving meters in the numerator and seconds in the denominator. Now that you have your result, you may write it as 25.77 m/s if you wish, but do not use slant fraction bars in the units when you are working out the unit conversion.

Point 2 The term “per,” abbreviated p, implies a fraction. Some units of measure are commonly written with a “p” for “per,” such as mph for miles per hour, or gps for gallons per second. Change these expressions to fractions with horizontal bars when you work out the unit conversion. Example 1.3 Convert 472.2 gps to L/hr. When you write down the given quantity, change the gps to gal/s and write these units with a horizontal bar: 472.2

gal 3.785 L 3600 s L ⋅ ⋅ = 6,436,000 s 1 gal hr hr

Point 3 Use the × and ÷ keys correctly when entering values into your calculator. When dealing with several numerator terms and several denominator terms, multiply all the numerator terms together first, hitting the × key between each, then hit the ÷ key and enter 25

Chapter 1 all the denominator terms, hitting the ÷ key between each. This way you do not need to write down intermediate results, and you do not need to use any parentheses. Example 1.4 Convert 43.2 mm/hr into km/yr. The setup with all the conversion factors is as follows: 43.2

mm 1m 1 km 24 hr 365 dy km ⋅ ⋅ ⋅ ⋅ = 0.378 hr 1000 mm 1000 m 1 dy 1 yr yr

To execute this calculation in your calculator, you enter the values and operations in this sequence: 43.2 × 24 × 365 ÷1000 ÷1000 = If you do so, you get 0.37843200. (Again, significant digits rules require us to round to 0.378.)

Point 4

When converting units for area and volume such as cm2 or m3, you must use the appropriate length conversion factor twice for areas and three times for volumes. The units “cm2” for an area mean the same thing as “cm × cm.” Likewise, “m3” means “m × m × m.” So when you use a length conversion factor such as 100 cm = 1 m or 1 in = 2.54 cm, you must use it twice to get squared units (areas) or three times to get cubed units (volumes). Example 1.5 Convert 3,550 cm3 to m3. 3550 cm3 ⋅

1m 1m 1m ⋅ ⋅ ⋅ = 0.00355 m3 100 cm 100 cm 100 cm

Notice in this example that the unit cm occurs three times in the denominator, giving us cm3 when they are all multiplied together. This cm3 term in the denominator cancels with the cm3 term in the numerator. And since the m unit occurs three times in the numerator, they multiply together to give us m3 for the units in our result.

The issue of needing to repeat conversion factors only arises when you are using a unit raised to a power, such as a when a length unit is used to represent an area or a volume. When using a conversion factor such as 3.785 L = 1 gal, the units of measure are written using units that are strictly volumetric (liters and gallons), and are not obtained from lengths the way they are with in2, ft2, cm3, and m3. Another common unit that uses an exponent is acceleration, which has units of m/s2 in the MKS unit system. Example 1.6 Convert 5.85 mi/hr2 into MKS units.

Measurements The MKS unit for length is meters (m), so we must convert miles to meters. The MKS units for time is seconds (s), so we must convert hr2 into s2. 5.85

mi 1609 m 1 hr 1 hr m ⋅ ⋅ ⋅ = 0.000726 2 hr 2 1 mi 3600 s 3600 s s

With this example, you see that since the “hours” unit is squared in the given quantity, the conversion factor converting the hours to seconds must appear twice in the conversion calculation. The “miles” unit in the given quantity has no exponent, so the conversion factor used to convert miles to meters only appears once in the calculation.

1.2.3 Converting Temperature Units Converting temperature values from one scale to another requires the use of equations rather than conversion factors. This is due to the fact that the Fahrenheit and Celsius scales are not absolute temperature scales. If all temperature scales were absolute scales like the Kelvin scale is, temperature conversions could be performed with conversion factors just as other conversions are. To convert a temperature in degrees Fahrenheit (TF) into degrees Celsius (TC), use this equation: TC =

5 (TF − 32°) 9

Using a bit of algebra, we can work this around to give us an equation that can be used to convert Celsius temperatures to Fahrenheit values: 9 TF = TC + 32° 5 To convert a temperature in degrees Celsius into kelvins (TK), use this equation: TK = TC + 273.15 Again, some algebra gives us the equation the other way around. TC = TK − 273.15 All four of the temperature conversion equations above are exact. This is important to know later when we discuss significant digits. These equations are listed in Table A.3 in Appendix A. Example 1.7 The normal temperature of the human body is 98.6°F. Express this value in degrees Celsius and kelvins. Since the given value is in degrees Fahrenheit, write down the equation that converts values from °F to °C. TC =

5 (TF − 32°) 9 27

Chapter 1 Now insert the Fahrenheit value and calculate the Celsius value. TC =

5 ( 98.6° − 32°) = 37.0°C 9

Now we are able to use the Celsius value to compute the Kelvin value. TK = TC + 273.15 = 37.0 + 273.15 = 310.2 K The reason the answer is 310.2 K instead of 310.15 K is due to the significant digits rule for addition. Again, the topic of significant digits is coming up next.

Example 1.8 The melting point of aluminum is 933.5 K. Express this temperature in degrees Celsius and degrees Fahrenheit. Write down the equation that converts Kelvin values to Celsius values. TC = TK − 273.15 From this we calculate the Celsius value as TC = TK − 273.15 = 933.5 − 273.15 = 660.3°C Next, write down the equation for converting a Celsius value to a Fahrenheit value. 9 TF = TC + 32° 5 Insert the Celsius value into this equation and calculate the Fahrenheit value. 9 9 TF = TC + 32° = ⋅660.3°C + 32° = 1220.5°F 5 5

1.3

Accuracy and Precision

1.3.1 Distinguishing Between Accuracy and Precision The terms accuracy and precision refer to the practical limitations inherent in making measurements. Science is all about investigating nature, and to do that we must make measurements. Accuracy relates to error—that is, to the lack of it. Error is the difference between a measured value and the true value. The lower the error is in a measurement, the better the accuracy. Error arises from many different sources, including human mistakes, malfunctioning equipment, incorrectly calibrated instruments, vibrations, changes in temperature or humidity, or unknown causes that are influencing a measurement without the knowledge of the experimenter. All measurements contain error, because (alas!) perfection is simply not a thing we have access to in our world. 28

Measurements Precision refers to the resolution or degree of “fine-ness” in a measurement. The limit to the precision that can be obtained in a measurement is ultimately dependent on the instrument being used to make the measurement. If you want greater precision, you must use a more precise instrument. The degree of precision in every measurement is signified by the measurement value itself because the precision is a built-in part of the measurement. The precision of a measurement is indicated by the number of significant digits (or significant figures) included in the measurement value when the measurement is written down (see below). Here is an example that illustrates the idea of precision and also helps distinguish between precision and accuracy. The photograph in Figure 1.8 shows a machinist’s rule and an architect’s scale set one above the other. Since the marks on the two scales line up consistently, these two scales are equally accurate. But the machinist’s rule (on top) is more precise. The architect’s scale is marked in 1/16-inch increments, but the machinist’s rule is marked in 1/64-inch increments. Thus, the machinist’s rule is more precise. It is important that you are able to distinguish between accuracy and precision. Here is another example to help illustrate the difference. Let’s say Shana and Marius each buy digital thermometers for their homes. The thermometer Figure 1.8. The accuracy of these two scales is the same, but the machinist’s rule (above) is more precise than the architect’s scale Shana buys costs $10 and measures (below). to the nearest 1°F. Marius pays $40 and gets one that reads to the nearest 0.1°F. Shana reads the directions and properly installs the sensor for her new thermometer in the shade. Marius doesn’t read the directions and mounts his sensor in the direct sunlight, which causes a significant error in the thermometer reading when the sun is shining on it; thus Marius’ measurements are not accurate. The result is that Shana has lower-precision, higher-accuracy measurements!

1.3.2 Significant Digits The precision in any measurement is indicated by the number of significant digits it contains. Thus, the number of digits we write in any measurement we deal with in science is very important. The number of digits is meaningful because it shows the precision inherent in the instrument used to make the measurement. Let’s say you are working a computational exercise in a science book. The problem tells you that a person drives a distance of 110 miles at an average speed of 55 miles per hour and wants you to calculate how long the trip takes. The correct answer to this problem is different from the correct answer to a similar problem with given values of 110.0 miles and 55.0 miles per hour. And if the given values are 110.0 miles and 55.00 miles per hour, the correct answer is different yet again. Mathematically, of course, all three answers are the same. If you drive 110 miles at 55 miles per hour, the trip takes two hours. But scientifically, the correct answers to these three problems are different: 2.0 hours, 2.00 hours, and 2.000 hours, respectively. The difference between these cases is in the precision indicated by the given data, which are measurements. (Even though this is just a made-up problem in a book and not an actual measurement someone made in an experiment, the given data are still measurements. There is no way to talk about distances or speeds without talking about measurements, even if the measurements are only imaginary or hypothetical.) 29

Chapter 1 So when you perform a calculation with physical quantities (measurements), you can’t simply write down all the digits shown by your calculator. The precision inherent in the measurements used in a computation governs the precision in any result you might calculate from those measurements. And since the precision in a measurement is indicated by the number of significant digits, data and calculations must be written with the correct numbers of significant digits. To do this, you need to know how to count significant digits, and you must use the correct number of significant digits in all your calculations and experimental data. Correctly counting significant digits involves four different cases: 1. Rules for determining how many significant digits there are in a given measurement. 2. Rules for writing down the correct number of significant digits in a measurement you are making and recording. 3. Rules for computations you perform with measurements—multiplication and division. 4. Rules for computations you perform with measurements—addition and subtraction. We address each of these cases below, in order. Case 1

We begin with the rule for determining how many significant digits there are in a given measurement value. The rule is as follows: •

The number of significant digits in a number is found by counting all the digits from left to right beginning with the first nonzero digit on the left. When no decimal is present, trailing zeros are not considered significant.

Let’s apply this rule to the following values to see how it works. 15,679

This value has five significant digits.

21.0005

This value has six significant digits.

37,000

0.0105

This value has three significant digits because we start counting digits with the first nonzero digit on the left.

0.001350 This value has four significant digits. Trailing zeros count when there is a decimal. The significant digit rules enable us to tell the difference between two measurements such as 13.05 m and 13.0500 m. Again, these values are obviously equivalent mathematically. But they are different in what they tell us about the process of how the measurements were made—and science deals in measurements. The first measurement has four significant digits. The second measurement is more precise—it has six significant digits and was made with a more precise instrument. Now, just in case you are bothered by the zeros at the end of 37,000 that are not significant, here is one more way to think about significant digits that may help. The precision in a measurement depends on the instrument used to make the measurement. If we express the measure-

Measurements ment in different units, this cannot change the precision of the value. A measurement of 37,000 grams is equivalent to 37 kilograms, as shown in the following calculation: 37,000 g ⋅

1 kg = 37 kg 1000 g

Whether we express this value in grams or kilograms, it still has two significant digits. Case 2 The second case addresses the rules that apply when you are recording a measurement yourself, rather than reading a measurement someone else has made. When you make measurements yourself, as when conducting the laboratory experiments in this course, you must know the rules for which digits are significant in the reading you are making on the measurement instrument. The rule for making measurements depends on whether the instrument you are using is a digital instrument or an analog instrument. Here are the rules for these two possibilities: •

Rule 1 for digital instruments For the digital instruments commonly found in introductory science labs, assume all the digits in the reading are significant except leading zeros.

•

Rule 2 for analog instruments The significant digits in a measurement include all the digits known with certainty, plus one digit at the end that is estimated between the finest marks on the scale of your instrument.

The first of these rules is illustrated in Figure 1.9. The reading on the left has leading zeros, which do not count as significant. Thus, the first reading has three significant digits. The second reading also has three significant digits. The third reading has five significant digits. The fourth reading also has five significant digits because with a digital display the only zeros that don’t count are the leading zeros. Trailing zeros are significant with a digital instrument. However, when you write this measurement down, you must write it in a way that shows those zeros to be significant. The way to do this is by using scientific notation. When a value is written in scientific notation, the digits that are written down in front of the power of 10 (the stem, also called the mantissa) are the significant digits. Thus, the right-hand value in Figure 1.9 must be written as 4.2000 × 104. We address scientific notation in more 0042.0 42.0 42.000 42,000 detail in the next section. Dealing with digital instruFigure 1.9. With digital instruments, all digits are significant except ments is actually more involved leading zeros. Thus, the numbers of significant digits in these than the simple rule above im- readings are, from left to right, three, three, five, and five. plies, but the issues involved go way beyond what we can deal with in introductory science classes. So, simply make your readings and assume that all the digits in the reading except leading zeros are significant. Now let’s look at some examples illustrating the rule for analog instruments. Figure 1.10 shows a machinist’s rule being used to measure the length in millimeters (mm) of a brass block. We know the first two digits of the length with certainty; the block is clearly between 31 mm and 32 mm long. We have to estimate the third significant digit. The scale on the rule is marked in increments of 0.5 mm. Comparing the edge of the block with these marks, I estimate the next digit to be a 6, giving a measurement of 31.6 mm. Others might estimate the last digit to be a 5 or a 7; these small differences in the last digit are unavoidable because the last digit is estimated. Whatever you estimate the last digit to be, two digits of this measurement are known with certainty, the third digit is estimated, and the measurement has three significant digits. 31

Chapter 1 The photograph in Figure 1.11 shows a liquid volume measurement in milliliters (mL) being made with an article of apparatus called a buret. Notice in this figure that when measuring liquid volume the surface of the liquid curls up at the edge of the cylinder, forming a bowl-shaped surface on the liquid. This curved surface is called a meniscus. For most liquids, liquid measurement readings are taken at the bottom of the meniscus. Liquid mercury is the major exception, because the meniscus in liquid mercury is inverted—liquid mercury curves down at the edges. In that case, the measurement is read at the top of the meniscus. But that is an unusual case. For most liquids, the reading is made at the bottom of the meniscus. For the buret in the figure, you can see that the scale is marked in increments of 0.1 milliliters (mL). This means we are to estimate to the nearest 0.01 mL. To one person, it may look like the bottom of the meniscus (where the black curve touches the bottom of the silver bowl) is just above 2.2 mL, so that person would call this measurement 2.19 mL. To someone else, it may seem that the bottom of the meniscus is right on Figure 1.10. Reading the significant digits 2.2, in which case that person would call the reading with a machinist’s rule. 2.20 mL. Either way, the reading has three significant digits and the last digit is estimated to be either 9 or 0. The third example involves a liquid volume measurement with an article of apparatus called a graduated cylinder. The scales on small graduated cylinders like this one are marked in increments of 1 mL. In the photo of Figure 1.12, the entire meniscus appears silvery in color with a black curve at the bottom. For the liquid shown in the figure, we know the first two digits of the volume measurement with certainty because the reading at the bottom of the meniscus is clearly between 82 mL and 83 mL. We have to estimate the third digit, and I estimate the edge of the meniscus to be at 60% of the distance between 82 and 83, giving a reading of 82.6 mL. Others may prefer a different valFigure 1.11. Reading the significant digits ue for that third digit. Figure 1.12. Reading the significant digits on a buret.

on a graduated cylinder.

Measurements It is important for you to keep the significant digits rules in mind when you are making measurements and entering data for your lab reports. The data in your lab journal and the values you use in your calculations and report should correctly reflect the use of the significant digits rules as they apply to the actual instruments you use to make your measurements. Case 3

The third and fourth cases of rules for significant digits apply to the calculations you perform with measurements. In Case 3, we deal with multiplication and division. The main idea behind the rule for multiplying and dividing is that the precision you report in your result cannot be higher than the precision that is in the measurements you start with. The precision in a measurement depends on the instrument used to make the measurement, nothing else. Multiplying and dividing things cannot increase that precision, and thus your results can be no more precise than the measurements used in the calculations. In fact, your result can be no more precise than the least precise value used in the calculation. The least precise value is, so to speak, the “weak link” in the chain, and a chain is no stronger than its weakest link. Here are the two rules for using significant digits in calculations involving multiplication and division: •

Rule 1 When multiplying or dividing, count the significant digits in each of the values you are using in the calculation, including any conversion factors involved. (However, note: Conversion factors that are exact are not considered.) Determine how many significant digits there are in the least precise of these values. The result of your calculation must have this same number of significant digits.

•

Rule 2 When performing a multi-step calculation, keep at least one extra digit during intermediate calculations, and round off to the final number of significant digits you need at the very end. This practice ensures that small round-off errors don’t accumulate during a multi-step calculation. This extra digit rule also applies to unit conversions performed as part of the computation.

I illustrate the two rules above, along with some more unit conversions, in the following example problem and calculation. Example 1.9 At a chemical research lab, a stream of a reactant solution is flowing into a reaction vessel at a rate of 56.75 μL per second. A volume of 1.0 ft3 of this solution is required in the vessel for the reaction. Determine the amount of time needed for the required volume to be collected. State your result in hours. First note that the value of the flow rate has four significant digits, and the required volume has two significant digits. The two-digit value is the least precise of these, so our result must be rounded to two significant digits. But to avoid rounding error, we must work with values having at least three significant digits (one more than we need) until the very end. One of the volumes in this problem is in μL and the other is in ft3. I begin by converting the required volume from ft3 to μL so our volumes all have the same units. We have no conversion factor that goes directly from ft3 to μL, so we must use a chain of conversion factors that we know or have available. Since we are dealing with relatively small volumes based on length units, the main conversion from USCS units to SI units is the inch to centimeter factor of 1 in = 2.54 cm. This factor is exact and should be committed to memory. If we were starting with gallons instead of ft3, we might use the conversion 1 gal = 3.785 L, although this factor is not exact. We first convert ft3 to in3, then from in3 to cm3, then from cm3 to L, and finally from L to μL. 33

Chapter 1 1.0 ft 3 ⋅

12 in 12 in 12 in 2.54 cm 2.54 cm 2.54 cm 1L 1×106 µL ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 28,300,000 µL 3 1 ft 1 ft 1 ft 1 in 1 in 1 in 1000 cm 1L

This result has three significant digits. We are keeping one extra digit during the intermediate calculations. It is not incorrect to write down all the digits your calculator shows. But it is pointless. A person who writes all the digits regardless of whether they are needed simply shows that he or she doesn’t understand significant digits. Those extra digits are meaningless. Notice that all the conversion factors used in the calculation above are exact; none of them are approximations. Since they are all exact, they play no role in limiting the significant digits in our result. Note that if you must use a conversion factor that is approximate, you should make sure the precision of the value in the conversion factor is at least as high as the precision in your data (two significant digits, in this case). That way your conversion factor does not limit the precision of your result. If, for some reason, you do not have a conversion factor with as many significant digits as your least precise measurement, then the precision of your result must match the precision of the conversion factor. The least precise value in the entire calculation always governs the precision in your result. Now we compute the time required by dividing the required volume by the flow rate. t=

28,300,000 µL = 499,000 s µL 56.75 s

This value also has three significant digits—one more than we need. We now convert this value from seconds to hours as the problem statement requires. 499,000 s⋅

1 hr = 139 hr 3600 s

Finally, we need to round the result to the required two significant digits. The second non-zero digit from the left (3) is in the tens place, so we round to the nearest ten. t = 140 hr

Case 4

The fourth case of rules for significant digits also applies to the calculations you perform with measurements. In Case 4, we deal with addition and subtraction. The rule for addition and subtraction is completely different from the rule for multiplication and division. When performing addition, it is not the number of significant digits that governs the precision of the result. Instead, it is the place value of the last digit that is farthest to the left in the numbers being added that governs the precision of the result. This rule is quite wordy and is best illustrated by an example. Consider the following addition problem: 13.65 1.9017 + 1,387.069 1,402.62

Measurements Of the three values being added, 13.65 has digits out to the hundredths place, 1.9017 has digits out to the ten thousandths place, and 1,387.069 has digits out to the thousandths place. Looking at the final digits of these three, you can see that the final digit farthest to the left is the 5 in 13.65, which is in the hundredths place. This is the digit that governs the final digit of the result. There can be no digits to the right of the hundredths place in the result. The justification for this rule is that one of our measurements is precise only to the nearest hundredth, even though the other two are precise to the nearest thousandth or ten thousandth. We are going to add these values together, and one of them is precise only to the nearest hundredth. It makes no sense to have a result that is precise to a place more precise than that, so hundredths are the limit of the precision in the result. Correctly performing addition problems in science (where nearly everything is a measurement) requires that you determine the place value governing the precision of your result, perform the addition, then round the result. In the above example, the sum is 1,4602.6207. Rounding this value to the hundredths place gives 1,4602.62. Going back to Example 1.7, we saw the following equation for converting a temperature from kelvins to degrees Celsius: TK = TC + 273.15 = 37.0 + 273.15 = 310.2 K The two values we are adding are 37.0, which has digits out to the tenths place, and 273.15, which has digits out to the hundredths place. The final digit in 37.0 is the one farther to the left, so it governs the final digit we can have in the sum. The final digit in 37.0 is in the tenths place, so the final digit in the sum must also be in the tenths place. Adding the values gives 310.15 K. Rounding this to the tenths place gives us 310.2 K for our result.

1.4

Other Important Math Skills

1.4.1 Scientific Notation No doubt you have studied scientific notation in your math classes. However, beginning in high school, scientific notation is used all the time in scientific study. Knowing how to use scientific notation correctly—including the use of the special key found on scientific calculators for working with values in scientific notation—is very important. Mathematical Principles Scientific notation is a way of expressing very large or very small numbers without all the zeros, unless the zeros are significant. This is of enormous benefit when one is dealing with a value such as 0.0000000000001 cm (the approximate diameter of an atomic nucleus). The basic idea will be clear from a few examples. Let’s say we have the value 3,750,000. This number is the same as 3.75 million, which can be written as 3.75 × 1,000,000. Now, 1,000,000 itself can be written as 106 (which means one followed by six zeros), so our original number can be expressed equivalently as 3.75 × 106. This expression is in scientific notation. The number in front, the stem, is always written with one digit followed by the decimal and the other digits. The multiplied 10 raised to a power has the effect of moving the decimal over as many places as necessary to recreate our original number. As a second example, the current population of earth is about 7,200,000,000, or 7.2 billion. One billion has nine zeros, so it can be written as 109. So we can express the population of earth in scientific notation as 7.2 × 109.

Chapter 1 When dealing with extremely small numbers such as 0.000000016, the process is the same, except the power on the 10 is negative. The easiest way to think of it is to visually count how many places the decimal in the value has to be moved over to get 1.6. To get 1.6, the decimal has to be moved to the right 8 places, so we write our original value in scientific notation as 1.6 × 10−8. Using Scientific Notation with a Scientific Calculator All scientific calculators have a key for entering values in scientific notation. This key is labeled EE or EXP on most calculators, but others use a different label.1 It is very common for those new to scientific calculators to use this key incorrectly and obtain incorrect results. So read carefully as I outline the procedure. The whole point of using the EE key is to make keying in the value as quick and error free as possible. When using the scientific notation key to enter a value, you do not press the × key, nor do you enter the 10. The scientific calculator is designed to reduce all this key entry, and the potential for error, by use of the scientific notation key. You only enter the stem of the value and the power on the ten and let the calculator do the rest. Here’s how. To enter a value, simply enter the digits and decimal in the stem of the number. Then hit the EE key, and then enter the power on the ten. The value is then in the calculator and you may do with it whatever you need to. As an example, to multiply the value 7.2 × 109 by 25 using a standard scientific calculator, the sequence of key strokes is as follows: 7.2 EE 9 × 25 = Notice that between the stem and the power, the only key pushed is the EE key. When entering values in scientific notation with negative powers on the 10, the +/− key is used before the power to make the power negative. Thus, to divide 1.6 × 10−8 by 36.17, the sequence of key strokes is: 1.6 EE +/− 8 ÷ 36.17 = Again, neither the 10 nor the × sign that comes before it are keyed in. The EE key has these built in. x Students sometimes wonder why it is incorrect to use the 10 key for scientific notation. To 9 calculate 7.2 × 10 times 25, they are tempted to enter the following: x 7.2 × 10 9 × 25 =

The problem with this approach is that sometimes it works and sometimes it doesn’t, and calculator users need to use key entries that always work. The scientific notation key ( EE ) keeps all the parts of a value in scientific notation together as one number. That is, when the EE key is used, a value such as 7.2 × 109 is not two separate numbers to the calculator; it is a single numerical value. But when the × key is manually inserted, the calculator treats the numbers separated by the × key as two separate values, and this can caused the calculator to use a different order x of operations than you intend. For example, using the 10 key causes the calculator to render an incorrect answer for a calculation such as this: 3.0 ×106 1.5 ×106 x x 1 One infuriating model uses the extremely unfortunate label x10 which looks a lot like 10 , a different key with a completely different function.

Measurements The denominator of this expression is exactly half the numerator, so the value of this fraction is x obviously 2.0. But when using the 10 key, the 1.5 and the 106 in the denominator are separated and treated as separate values. The calculator then performs the following calculation: 3.0 ×106 ×106 1.5 This comes out to 2,000,000,000,000 (2 × 1012), which is not the same as 2.0! The bottom line is that the EE key, however it may be labeled, is the correct key to use for scientific notation. Finally, when writing a result in scientific notation, it is not acceptable to write it using the EE notation your calculator uses. For example, your calculator might display a result as 3.14EE8 or 3.14E8, but you must write this as 3.14 × 108.

1.4.2 Calculating Percent Difference One of the conventional calculations in science experiments is the so-called “experimental error.” Experimental error is typically defined as the difference between a predicted value and an experimental value, expressed as a percentage of the predicted value, or experimental error =

predicted or accepted value − experimental value ×100% predicted or accepted value

Although the term “experimental error” is widely used, it is a poor choice of words. When there is a mismatch between theory and experiment, the experiment may not be the source of the error. Often, it is the theory that is found wanting—this is how science advances. I now prefer to use the phrase percent difference to describe the value computed by the above equation. When quantitative results are compared to quantitative predictions or accepted values, students should compute the percent difference as percent difference =

predicted or accepted value − experimental value ×100% predicted or accepted value

Chapter 1 Exercises For all exercises, note that physical constants and unit conversion factors are found in Tables A.2 and A.3 of Appendix A. SECTION 1.1

1. Write a paragraph distinguishing between matter and mass. 2. Distinguish between base units and derived units in the SI system of units and give three examples of each. 3. Describe the advantages the SI system has over the USCS system for scientific work. 4. Why does the SI system use prefixes on the units of measure?

Chapter 1 5. Re-express the quantities in the following table using only a single numerical digit followed by an SI unit symbol, with a metric prefix if necessary. Example: 5 thousand liters = 5 kL a. 8 pascals

b. 5 hundredths of a meter

c. 3 million amperes

d. 2 thousand meters

e. 4 thousandths of a second

f. 6 thousand newtons

g. 8 thousand grams

h. 7 millionths of a liter

i. 1 thousandth of a joule

6. Re-write the quantities in the following table by writing out the unit names without symbols. Example: 5 km = 5 kilometers a. 14 m3

b. 164.1 kg

c. 250 MPa

d. 16.533 ms

e. 160 kA

f. 19.55 cL

g. 31.11 μJ

h. 2300 K

i. 13.0 mmol

SECTION 1.2

7. Why must equations be used instead of conversion factors for most temperature unit conversions? 8. Perform the USCS unit conversions required in the following table. (Note: The answers in the back of the book are given with the correct number of significant digits.) Convert This Quantity

Into These Units

Convert This Quantity

Into These Units

a. 12.55 ft

b. 0.44556 mi

c. 147.55 in

d. 55.08 gal

ft3

e. 934 ft

f. 739.22 ft /s

g. 12.4 yr

h. 51,083 in

gal/hr

i. 14,560.77 gal/hr

qt/s

j. 15.90 mi/dy

mi in/hr

9. Perform the SI/metric unit conversions required in the following table. Convert This Quantity a. 35.4 mm

Convert This Quantity

Into These Units

b. 76.991 mL

c. 34.44 cm

d. 6.33 g/cm

kg/m2

e. 9.35 m/s2

mm/ms2

f. 542.2 mJ/s

J/s

g. 56.6 μs

h. 44.19 mL

cm3

i. 532 nm

μm

j. 96,963,000 mL/ms

Into These Units

μL

m3/s

k. 295.6 cL

μL

l. 0.007873 m

m. 8,750 mm2

n. 87.1 cm/s2

m/s2

o. 15.75 kg/m3

g/cm3

p. 0.875 km

q. 16,056 MPa

kPa

r. 7,845 μA

Measurements 10. Reproduce the following table on your own paper and fill in the empty cells. °F a.

°C

°F

431.1

16.0

–77.0

0.0 (exact) 4,002

h. 1,958

–56.1

–32.0

°C

65.25

998.0

SECTION 1.3

11. Distinguish between accuracy and precision. 12. Describe the measurements you would obtain from an instrument that was very precise but not very accurate. 13. Which is more important on the speedometer of a car—accuracy or precision? 14. Explain why accuracy is important on a heart rate monitor but precision is not. 15. Sometimes we want high accuracy in a measurement, but are not too concerned about high precision. Sometimes we want both high accuracy and high precision. Explain why no one wants low accuracy and high precision. 16. On the package of a digital stopwatch I once purchased was the phrase: “1/100th second accuracy.” The stopwatch readings in seconds contained two decimal places, but the values the stopwatch actually displayed were spaced 0.03 seconds apart. Thus, it could read 12.31 s, 12.34 s, 12.37 s, etc. Comment on the accuracy and precision of this stopwatch with respect to the claim on the package. 17. Using the correct number of significant digits and the correct units of measure, record the measurements represented by the following instruments.

11 10

13 1 4

(c)

gallons

20 21 22

(d)

010 wind speed (mph)

Absolute Pressure (psi)

01320

(b) (a) SECTION 1.4

18. Using the correct number of significant digits, compute the percent difference for the experimental results in each of the following cases: 39

Chapter 1 a. A scientist measures the masses of three compounds resulting from a certain chemical reaction. Her measurements are 0.234 g, 1.678 g, and 4.446 g. Her calculations predict that the reaction results in masses of 0.239 g, 1.688 g, and 4.678 g, respectively. Determine the percent difference for each of the three compounds. b. A student measures the density of aluminum and finds it to be 2.81 g/cm3. The accepted density value for this alloy is 2.72 g/cm3. c. According to the Periodic Table of the Elements, the atomic mass of carbon is 12.011 g/mol. A calculation from experimental data results in a figure of 12.0117 g/mol. d. The predicted yields for the products of certain chemical reaction are 23.4 kg of compound A and 2.21 kg of compound B. Careful measurements of the masses of the compounds produced indicate masses of 21.610 kg for compound A and 1.995 kg of compound B. 19. Perform each of the unit conversions indicated in the table below. Express each result using the correct number of significant digits. Where possible and appropriate, express your result in both standard notation and scientific notation. (Note: By possible, I refer to the fact that sometimes a result can only be expressed with the correct number of significant digits if it is written in scientific notation, such as a value of 100 with two or three significant digits. By appropriate, I refer to the fact that it is silly to write a value with a very large number of zeros. Such values should always be expressed in scientific notation. It is also silly to use scientific notation to express a value such as 3 or 4.1. Such values should only be expressed in standard notation.) Convert This Quantity a. 1,737 km (radius of the earth’s moon)

b. 2.20 g (mass of a single peanut m&m)

c. 591 mL (volume of a typical water bottle)

μL

d. 7 × 108 m (radius of the sun)

e. 1.616 × 10

–35

m (Planck length, a fundamentally small length)

f. 750 cm (size of the engine in my old motorcycle) g. 2.9979 × 108 m/s (speed of light in a vacuum) h. 168 hr (one week) 3

i. 5,570 kg/m (average density of the earth)

ft m3 mi/hr s g/cm3

j. 45 gps (gal/sec, flow rate of Mississippi River at the source)

m3/min

k. 600,000 ft3/s (flow rate of Mississippi River at New Orleans)

L/hr

l. 5,200 mL (volume of blood in a typical man’s body)

m. 5.65 × 102 mm2 (area of a postage stamp)

in2

n. 32.16 ft/s2 (acceleration of gravity, or one “g”)

m/s2

o. 10.6 μm (wavelength of light from a CO2 laser)

p. 1.1056 g/mL (density of heavy water) 40

Into These Units

kg/m3

Measurements Convert This Quantity q. 13.6 g/cm3 (density of liquid mercury metal)

Into These Units mg/m3

r. 93,000,000 mi (distance from earth to the sun)

s. 65 mph (typical highway speed limit)

m/s

t. 633 nm (wavelength of light from a red laser)

u. 5.015% of the speed of light (see item g, or Table A.2)

mph

v. 6.01 kJ/mol (molar heat of fusion of water)

J/mol

w. 32.1 bar (pressure in saltwater at 318 m, free diving record depth)

psi

x. 0.116 nm (radius of a sodium atom)

y. 6.54 × 10–24 cm3 (volume of a sodium atom)

in3

z. 0.385 J/(g ∙K) (specific heat capacity of copper) aa. 370 mL (volume of a soft drink can)

J/(mg∙K) ft3

ab. 268,581 mi2 (land area of Texas)

mm2

ac. 50,200 mi2/yr (current rate of global deforestation)2

ft2/s

2 As a science educator who believes that we should take care of our planet, I note that items (ab) and (ac) above indicate that every 5.4 years, we lose an area of forest the size of Texas. Texas is a big place. Think about it. 41

Chapter 2 Atoms and Substances

Some substances pose a minimal threat to human health and safety. Others need to be labeled and classified so we know about the risks they pose. If you examine the packaging on a shipment of chemicals or look on the back of a tank truck on the highway, you may find the Fire Diamond, a symbolic representation of the various hazards associated with the substance inside. The numerical values in each of the three colored zones range from 0 to 4, with 4 representing the most extreme hazard. The blue region pertains to health, the red to flammability, the yellow to instability or reactivity, and the white to special notices. In the sample symbol above, the numerical codes specify the following: Blue: Health 3—Short exposure could cause serious temporary or moderate residual injury. (Example: chlorine) Red: Flammability 2—Must be moderately heated before ignition can occur (Example: diesel fuel) Yellow: Instability/reactivity 1—Normally stable, but can become unstable at elevated temperature and pressure (Example: alcohol) White: Special notices “W bar”—Reacts with water in an unusual or dangerous manner (Example: sodium)

Atoms and Substances

Objectives for Chapter 2 After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary. SECTION 2.1

1. Define and describe atom and molecule. 2. State the five points of John Dalton’s atomic model. 3. Write brief descriptions of J.J. Thomson’s cathode ray tube experiment, Robert Millikan’s oil drop experiment, and Ernest Rutherford’s gold foil experiment. In your descriptions, include definitions for the terms cathode ray and alpha particle. 4. Describe the atomic models proposed by J.J. Thomson and Ernest Rutherford. SECTION 2.2

5. Define pure substance, element, compound, mixture, heterogeneous mixture, and homogeneous mixture and give several examples of each. 6. Define suspension and colloidal dispersion and give several examples of each. 7. Explain the Tyndall effect and how it can be used to identify a mixture as colloidal. 8. Describe the two basic types of structures atoms form when bonding together. 9. Distinguish between compounds and mixtures. 10. Define and distinguish between physical properties, chemical properties, physical changes, and chemical changes. Give several examples of each. 11. Define and give examples of the terms malleable and ductile. SECTION 2.3

12. Define isotope, nuclide, and atomic mass. 13. Given isotope mass and abundance data, calculate the atomic mass of an element. 14. Given the periodic table, determine the number of protons, electrons, and neutrons in the atoms of a given nuclide. 15. Define the unified atomic mass unit, u. SECTION 2.4

16. Use the density equation to calculate the density, volume, or mass of a substance. 17. Define the mole. 18. State the Avogadro constant to four digits of precision. 19. Define molar mass, formula mass, and molecular mass. 20. Calculate the molar mass, formula mass, or molecular mass of a compound or molecule. 21. Calculate the mass in grams of a given mole quantity of a compound or molecule, or vice versa. 22. Calculate the number of atoms or molecules in a given quantity of substance. 23. Calculate the gram masses of an atom or molecule of a given pure substance.

Chapter 2

2.1

Atoms and Molecules

2.1.1 Atomic Facts We begin this chapter with a summary of the basic facts about atoms and molecules. Much of this information you probably already know. As we saw in the previous chapter, all matter is made of atoms, the smallest basic units matter is composed of. An atom of a given element is the smallest unit of matter that possesses all the properties of that element. Atoms are almost entirely empty space. Each atom has an incredibly tiny nucleus in the center containing all the atom’s protons and neutrons. Since the protons and the neutrons are in the nucleus, they are collectively called nucleons. The masses of protons and neutrons are very nearly the same, although the neutron mass is slightly greater. Each proton and neutron has nearly 2,000 times the mass of an electron, so the nucleus of an atom contains practically all the atom’s mass. Outside the nucleus is a weird sort of cloud surrounding the nucleus containing the atom’s electrons. We address the details about electrons in the next chapter, but here is a brief preview. The electron cloud consists of different orbitals where the electrons are contained. Electrons are sorted into the atomic orbitals according to the amount of energy they have. For an electron to be in a specific orbital means the electron has a certain amount of energy—no more, no less. I wrote above that atoms are almost entirely empty space because the nucleus is incredibly small compared to the overall size of the atom with its electron cloud. It’s quite easy for us to pass over that remark without pausing to consider what it means. To help visualize the meaning, consider the athletic stadium pictured in Figure 2.1. Using this stadium as an enlarged atomic model, the electrons in their orbitals would be zipping around in the region where the red seating sections are in the stadium. Each electron in this enormous atomic model is far smaller than the period at the end of this sentence. The atomic nucleus containing the protons and neutrons is located at the center of the playing field, and is the size of a pinhead. And what fills all the vast space inside the atom? Nothing, not even air, since air, of course, is also made of atoms. The inside of an atom is empty space. Returning to our discussion of atomic facts, one of the fundamental physical properties of the subatomic particles is electric charge. Neutrons have no electric charge. They are electrically neutral, hence their name. Protons and electrons each contain exactly the same amount of charge, but the charge on protons is positive and the charge on electrons is negative. If an atom or molecule has no net electric charge, it contains equal numbers of protons and electrons. Atoms are significantly smaller than the wavelengths of light, which means light does not reflect off atoms and there is no way to see them. The same is true of molecules. Molecules are clusters of atoms chemically bonded together. When atoms of different elements are bonded together in a molecule they form a compound, which we discuss later in this chapter. But sometimes atoms of the same element bond together in molecules, as illustrated in Figure 2.2. Oxygen and chlorine are two of the diatomic gases that form molecules consisting of a pair of atoms chemicalFigure 2.1. The head of a pin at center field in a stadium ly bound together. Hydrogen, nitrogen, and is analogous to the nucleus in the center of an atom. fluorine also exist naturally as diatomic gases. 44

Atoms and Substances

2.1.2 The History of Atomic Models The story of atomic theory starts back with the ancient Greeks. As we look at how the contemporary model of the atom developed, we also hit on some of the great milestones in the history of chemistry and physics along the way. In the 5th century BCE,1 the Greek philosopher Democritus proposed that everything was made of tiny, indivisible particles. Our word atom comes from the Greek word atomos, meaning “indivisible.” Democritus’ idea was that the properties of substances were due to characteristics of the atoms they are made from. So atoms of metals were supposedly hard and strong, atoms of water were assumed to be wet and slippery, and so Figure 2.2. Space-filling on. At this same time, there were various views about what the most basic models of the diatomic substances—that is, the elements—were. One of the most common views oxygen (top) and was that there are four elements—earth, air, water, and fire—and that ev- chlorine molecules. erything is composed of these. Not much real chemistry went on for a very long time. During the medieval period, of course, there were the alchemists, who sought to transform lead and other materials into gold. But this cannot be done by the methods available to them, so their efforts were not successful. But in the 17th century, things started changing as scientists became interested in experimental research. The goal of the scientists described here was to figure out what the fundamental constituents of matter are. This meant figuring out how atoms are put together, what the basic elements are, and understanding what is going on when various chemical reactions take place. The nature of earth, air, fire, and water was under intense scrutiny over the next 200 years. In 1803, English scientist John Dalton (Figure 2.3) produced the first scientific model of the atom. Dalton’s atomic model is based on five main points, listed in Table 2.1. The impressive thing about Dalton’s atomic theory is that even today the last three of these points are regarded as Figure 2.3. English scientist John Dalton correct, and the first two are at least partially correct. On the (1766–1844). first point, it is still scientifically factual that all substances are made of atoms, but we now know that atoms are not indivisible. This is now obvious, since atoms themselves are composed of protons, neutrons, and electrons. The second point is correct in every respect but one. Except for the number of neutrons in the nucleus, every atom of a given element is identi- 1. All substances are composed of tiny, indivisible substances called atoms. cal. However, we 2. All atoms of the same substance are identical. now know that 3. Atoms of different elements have different weights. atoms of the same 4. Atoms combine in whole-number ratios to form compounds. element can vary in the number of 5. Atoms are neither created nor destroyed in chemical reactions. neutrons they have Table 2.1. The five tenets of Dalton’s 1803 atomic model. 1 BCE stands for before common era. 45

Chapter 2

Figure 2.4. English scientist Joseph John (J.J.) Thomson (1856–1940).

in the nucleus. These varieties of nuclei are called isotopes, a topic we return to soon. After Dalton, the next breakthrough in our understanding of atomic structure came from English scientist J.J. Thomson (Figure 2.4). Thomson worked at the Cavendish Laboratory in Cambridge, England. In 1897, he conducted a series of landmark experiments that revealed the existence of electrons. Because of his work, he won the Nobel Prize in Physics in 1906. A photograph of the cathode ray tube Thomson used for his work is shown in Figure 2.5. Thomson placed electrodes from a high-voltage electrical source inside a very elegantly made, sealed-glass vacuum tube. This apparatus can generate a so-called cathode ray from the negative electrode (1), called the cathode, to the positive one, called the anode (2). A cathode ray is simply a beam of electrons, but this was not known at the time. The anode inside Thomson’s vacuum tube had a hole in it for some of the electrons to escape through, which created a beam of cathode rays heading toward the other end of the tube (5). Thomson placed the electrodes of another voltage source inside the tube (3), above and below the cathode ray, and discov5

Figure 2.5. J.J. Thomson’s cathode ray tube.

ered that the beam of electrons deflected when this voltage was turned on. He also placed magnetic coils on the sides of the tube (4) and discovered that the electrons also deflected as they passed through the magnetic field produced by the coils. The deflection of the beam toward the positive electrode led Thomson to theorize that the beam was composed of negatively charged particles, which he called “corpuscles.” (The name electron was first used a few years later by a different scientist.) By trying out many different arrangements of cathode ray tubes, Thomson confirmed that the ray was negatively charged. Then using the scale on the end of the tube to measure the deflection 46

positive material electrons

Figure 2.6. Thomson’s plumb pudding model.

Atoms and Substances angle (5), he was able to determine the charge-to-mass ratio of the individual electrons he had discovered. This value is 1.8 × 1011 C/kg, where C stands for coulomb, the SI unit of electric charge. Thomson went on to theorize that electrons came from inside atoms. He developed a new atomic model that envisions atoms as tiny clouds of massless, positive charge sprinkled with thousands of the negatively charged electrons, as illustrated in Figure 2.6. Thomson’s model is usually called the plum pudding model. In 1911, American scientist Robert Millikan (Figure 2.7) devised his famous oil drop experiment, an extraordinary procedure that allowed him to determine the charge on individual electrons, 1.6 × 10–19 C. Once this value was known, Millikan used Thomson’s charge-to-mass ratio and calculated the mass of the electron, 9.1 × 10–31 kg. Millikan’s apparatus is pictured Figure 2.7. American scientist Robert in Figure 2.8, and his Millikan (1868–1953). sketch of the system is 1 3 shown in Figure 2.9. Inside a heavy metal drum (1) about the size of a 5-gallon bucket, Millikan placed a pair of horizontal 4 metal plates (2) connected to an adjustable high-voltage source. The upper plate had a hole in the center and was connected to the positive voltage, the lower plate to the negative. He used an atomizer spray pump (3) to spray in a fine mist of watchmaker’s oil above the positive plate. Some of the oil droplets would fall through the hole in the upper plate and move into the region between the plates. Connected through the side of the drum between Figure 2.8. Apparatus for the oil drop experiment. the two plates was a telescope eyepiece (4) and lamp (5) so that Millikan could see the oil droplets between the plates. The process of squirting in the oil droplets with the atomizer sprayer caused some of the droplets to acquire a charge of static electricity. This means the droplets had excess electrons on them 5 and carried a net nega2 tive charge. They picked up these extra electrons by friction as the droplets squirted through the rubber sprayer tube. As Millikan looked at an oil droplet through the eyeFigure 2.9. Millikan’s schematic of his oil drop apparatus.

Chapter 2 piece and adjusted the voltage between the plates, he could make the charged oil droplet hover when the voltage was just right. Millikan took into account the weight of the droplets and the viscosity of the air as the droplets fell and was able to determine that every droplet had a charge on it that was a multiple of 1.6 × 10–19 C. From this he deduced that this must be the charge on a single electron, which it is. Millikan won the Nobel Prize in Physics in 1923 for this work. The last famous experiment in this basic history of atomic models was initiated in 1909 by one of Thomson’s students at Cambridge, New Zealander Lord Ernest Rutherford (Figure 2.10). Rutherford was already famous when this experiment occurred, having just won the Nobel Prize in Chemistry the previous year. Rutherford’s gold foil experiment resulted in the discovery of the atomic nucleus. To understand this experiment you need to know that an alpha particle, or α-particle (using the Greek letter alpha, α) is a particle composed of two Figure 2.10. New Zealander and protons and two neutrons. Alpha particles are naturally emitphysicist Ernest Rutherford (1871– ted by some radioactive materials in a process called nuclear 1937). decay. Rutherford created a beam of α-particles by placing some radioactive material (radium bromide) inside a lead box with a hole in one end. The α-particles from the decaying radium atoms streamed out the hole at very high speed (15,000,000 m/s!). Rutherford aimed the α-particles at an extremely thin sheet of gold foil only a few hundred atoms thick. Surrounding the gold foil was a ring-shaped screen coated with a material that glows when hit by α-particles. Rutherford could then determine where the α-particles went after encountering the gold foil. – When Rutherford began taking data, he had Thomson’s – – plum pudding model in mind and was expecting results – – – – – consistent with that atomic model. This scenario is depicted – – – – – – in the upper part of Figure 2.11. The atom’s positive charge, – – – – – shown in the light orange color, is spread throughout the – – – – – atom and the negative electrons are embedded in the posi– – – tive material. Rutherford expected the massive and positively – – – – – charged α-particles to blow right through the gold foil. – – What Rutherford found was astonishing. Most of the α-particles passed straight through the foil and struck the screen on the other side, just as Rutherford expected. However, occasionally an α-particle (one particle out of every several thousand) deflected with a small angle. And sometimes the deflected particles bounced almost straight back. This situation is depicted in the lower part of Figure 2.11. The astonished Rutherford commented that it was like firing a huge artillery shell into a piece of tissue paper and having it bounce back and hit you! Rutherford’s work led to his new proposal in 1911 for a model of the atom. Rutherford’s Figure 2.11. Alpha-particle pathways through the gold atoms expected model included the key points listed in Table 2.2. from Thomson’s plum pudding model In 1917, Rutherford became the first to “split the atom.” (top) and the pathways Rutherford In this experiment he used α-particles again, this time strikobserved (bottom).

Atoms and Substances ing nitrogen atoms. His work 1. The positive charge in atoms is concentrated in a tiny region led to the discovery of the in the center of the atom, which Rutherford called the nucleus. positively-charged particles in the atomic nucleus, which he 2. Atoms are mostly empty space. 3. The electrons, which contain the atoms’ negative charge, are named protons. outside the nucleus. It took another twenty years before James Chadwick Table 2.2. The main ideas in Ernest Rutherford’s 1911 atomic model. (Figure 2.12), another Englishman, discovered the neutron. Before World War I, Chadwick studied under Rutherford (at Cambridge, of course). Then the war began. Not only did the war interrupt the progress of the research in general, but Chadwick was a prisoner of war in Germany. Working back in England after the war, he discovered the neutron in 1932 and received the Nobel Prize in Physics for his discovery in 1935. Chadwick’s discovery of the neutron enabled physicists to fill in a lot of blanks in their understanding of the basic structure of atoms. But years before Chadwick made his discovery, Rutherford’s atomic model was already being taken to another level through the work of Niels Bohr. We explore Bohr’s atomic model, and the quantum model to which it led, in the next chapter.

2.2

Types of Substances

A substance is anything that contains matter. There are Figure 2.12. English physicist James several major classifications of substances, but as shown in Chadwick (1891–1974). Figure 2.13, they all fall into two major categories, pure substances and mixtures.

2.2.1 Pure Substances: Elements and Compounds There are two kinds of pure substances, elements and compounds. We will discuss elements first. In previous science classes you may have seen or studied the Periodic Table of the Elements, which lists all the known elements. This famous table plays a major role in the study of chemistry and is shown in Figure 2.14 on the next page. We dive into the periodic table in detail in Chapter 4, but I bring up the table here to assist in our discussion of elements. Elements

The characteristic that defines each element in the periodic table is the number of protons the element has in each of its atoms, Substances a number called the atomic number. The elements are ordered in the Mixtures Pure Substances periodic table by atomic number. For example, carbon is element numHeterogeneous Homogeneous Compounds ber six in the periodic Elements Mixtures Mixtures table. This means that an atom of carbon has six Figure 2.13. Classifications of substances. 49

87.62 56

Barium

137.327 88

Radium

226.0254

85.468 55

Cesium

132.905 87

Francium

223.0197

40.078 38

39.098 37

Calcium

Strontium

Potassium

24.3050 20

22.9898 19

Rubidium

Magnesium

Sodium

9.0122 12

6.941 11

Beryllium

Lithium

232.0381

227.0278

231.0359

Protactinium

Thorium

Actinium

140.9077 91

Cerium

140.115 90

Lanthanum

138.9055 89

Praseodymium

Dubnium

180.9479 105

Tantalum

92.9064 73

Niobium

50.9415 41

Vanadium

262.114

Rutherfordium

178.49 104

Hafnium

91.224 72

Zirconium

47.88 40

Titanium

261.11

262.11

Lawrencium

174.967 103

Lutetium

88.9059 71

Yttrium

44.9559 39

Scandium

238.0289

Uranium

144.24 92

Neodymium

263.118

Seaborgium

183.85 106

Tungsten

95.94 74

Molybdenum

51.9961 42

Chromium

Promethium

237.0482

Neptunium

144.9127 93

262.12

Bohrium

186.207 107

Rhenium

98.9072 75

Technetium

54.9380 43

Manganese

244.0642

Plutonium

150.36 94

Samarium

(265)

Hassium

190.2 108

Osmium

101.07 76

Ruthenium

55.847 44

Iron

Europium

243.0614

247.0703

Curium

157.25 96

Gadolinium

(281)

Darmstadtium

195.08 110

Platinum

106.42 78

Palladium

58.6934 46

Nickel

Am Cm Americium

151.965 95

(266)

Meitnerium

192.22 109

Iridium

102.9055 77

Rhodium

58.9332 45

Cobalt

Silver

63.546 47

Copper

Terbium

247.0703

Berkelium

158.9253 97

(281)

Roentgenium

196.9665 111

Gold

107.8682 79

radioactive

251.0796

Californium

162.50 98

Dysprosium

(285)

Copernicium

200.59 112

Mercury

112.411 80

Cadmium

65.39 48

Zinc

Holmium

252.083

Einsteinium

164.9303 99

(284)

Nihonium

204.3833 113

Thallium

114.82 81

Indium

69.723 49

Gallium

26.9815 31

Aluminum

10.811 13

Boron

1.0079 3

Hydrogen

Erbium

257.0951

Fermium

167.26 100

(289)

Flerovium

207.2 114

Lead

118.710 82

Tin

72.61 50

Germanium

28.0855 32

Silicon

12.011 14

Carbon

Thulium

258.10

Mendelevium

168.9342 101

(288)

Moscovium

208.9804 115

Bismuth

121.76 83

Antimony

74.9216 51

Arsenic

30.9738 33

Phosphorus

14.0067 15

Nitrogen

Ytterbium

259.1009

Nobelium

173.04 102

(293)

Livermorium

208.9824 116

Polonium

127.60 84

Tellurium

78.96 52

Selenium

32.066 34

Sulfur

15.9994 16

Oxygen

(294)

Tennessine

209.9871 117

Astatine

126.9045 85

Iodine

79.904 53

Bromine

35.4527 35

Chlorine

18.9984 17

Fluorine

(294)

Oganesson

222.0176 118

Radon

131.29 86

Xenon

83.80 54

Krypton

39.948 36

Argon

20.1797 18

Neon

4.0026 10

Helium

liquid at room temperature

Note: The upper set of group numbers has been recommended by the International Union of Pure and Applied Chemistry (IUPAC) and is now in wide use. The lower set of numbers is still in common use in America.

18 8A

Figure 2.14. The Periodic Table of the Elements.

Chapter 2

Atoms and Substances protons—all carbon atoms have six protons. If an atom does not have six protons, it is not a carbon atom, and if an atom does have six protons, it is a carbon atom. An element is therefore a type of atom, classified according to the number of protons the atom has. A lump of elemental carbon—which could be graphite, diamond, coal, or several other varieties of pure carbon—is any lump of atoms that contain only six protons apiece. Oxygen (element 8) is another example of an element. Pure oxygen is a gas (ordinarily) that contains only atoms with eight protons each, because oxygen is element number eight. Other examples of elements you have heard of are iron, gold, silver, neon, copper, nitrogen, lead, and many others. For every element, there is a chemical symbol that is used in the periodic table and in the chemical formulas for compounds. For a few elements, a single upper-case letter is used, such as N (7) for nitrogen and C for carbon. But for most elements, an upper-case letter is followed by one lower-case letter, such as Na (11) for sodium and Mg (12) for magnesium. The three-letter symbols for elements 113, 115, 117, and 118 are placeholders until official names and two-letter symbols are selected by the appropriate governing officials. Some of the chemical symbols are based on the Latin names of elements, such as Ag for silver, from its Latin name argentum. Other examples are Au for gold, from the Latin aurum, and Pb for lead, from the Latin plumbum (everyone’s favorite Latin name). The most common representations of the periodic atomic number table show four pieces of information for each element, chemical symbol indicated in Figure 2.15. At the top of the cell is the atomic number, symbolically represented as Z. Again, element name Chlorine this number indicates the number of protons in each atomic mass 35.4527 atom of the element and is the number used to order the elements in the periodic table. Below the atomic Figure 2.15. The basic information in each number are the element’s chemical symbol and name. cell of the periodic table. At the bottom of the cell is the atomic mass. We look at the atomic mass more closely in Section 2.3.

Compounds

As the name implies, a compound is formed when two or more different elements are chemically bonded together. This bonding is always the result of a chemical reaction. A chemical reaction is any process in which connecting bonds between atoms are formed or broken. Once bonded together chemically, the elements in a compound can only be separated by chemical means. In other words, it takes a different chemical reaction to break atoms apart. The elements or compounds that go into a chemical reaction are called the reactants. The compounds formed by the reaction are called the products. The physical and chemical properties of a compound are completely different from the properties of any of the elements in the compound. For example, consider oxygen, hydrogen, and water. Hydrogen and oxygen react with a boom (Figure I.8) to form water, according to this chemical equation: 2H2 + O2 → 2H2O Oxygen is an invisible gas that we breathe in the air and that supports combustion. Hydrogen is an invisible, flammable gas. Water is composed of oxygen atoms bonded to hydrogen atoms, but one cannot breathe water, nor does water combust or support combustion. Or consider the sodium and chlorine in sodium chloride. These elements combine according to this chemical equation: Na + Cl → NaCl 51

Chapter 2 We all require salt in our diets, and we find it tasty. But both sodium and chlorine, the two elements of which sodium chloride is composed, are deadly dangerous in their pure, elemental forms. Sodium is a shiny, peach-colored metal, pictured in Figure 2.16. Sodium slices just like cheddar cheese and the photo shows me slicing a chunk of sodium open to show its beautiful, pinkish shiny color. Chlorine is a greenish gas, pictured in a glass bottle in Figure 2.17. As you can see, the properties of these substances are nothing at all like the properties of the salt they form when they react together. We discuss physical and chemiFigure 2.16. Sodium metal. cal properties in more detail a bit later. When atoms bond together to form a compound, the atoms in the compound can be arranged in either of two basic types of structures. In many cases, the atoms join together to form molecules. In every molecule of a given substance, the atoms bond together in the same whole-number ratio—a perfect example of the important general principle that we first encountered in the Introduction. In the Introduction are shown computer models of water (H2O), carbon dioxide, CO2 ammonia (NH3), and methane (CH4) molecules. A few other well-known Figure 2.17. Chlorine gas. molecular substances are carbon dioxide (CO2), propane (C3H8), and ozone (O3), all represented by space-filling models in Figure 2.18. The standard color coding used in computer models is white for hydrogen, black or charcoal gray for carbon, and red for oxygen. It is important to note that in any chemical formula for a molecular substance, the formula indicates the number of each type of atom in the molecule. Propane, C3H8, has three carbon atoms and eight hydrogen atoms in each molecule. The subscripts are only shown when the quantity of atpropane, C3H8 oms of an element is greater than one. The formula for sucrose—table sugar—is C12H22O11. Each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. The other common way atoms combine is by forming a continuous geometric arrangement. These compounds are called crystals, and the structure the atoms in the compound make when they join together is called a crystal lattice. The number of different arrangements ozone, O3 atoms can make in a lattice is endless, and these arrangements are responsible for many of the unusual Figure 2.18. More common molecular properties crystals possess. But what all lattices have in substances. Hydrogen atoms are white, common is the regular arrangement of the atoms into carbon atoms are black, and oxygen atoms are red.

Atoms and Substances repeating, geometrical patterns. A space-filling model of the very simple crystal structure for sodium chloride, NaCl (table salt), is shown in Figure 2.19. The formula tells us that the ratio of sodium atoms to chlorine atoms in the crystal is 1 : 1. The model shows that the atoms are bonded together in a simple alternating arrangement. We have seen several space-filling models so far. Another common type of computer model is the ball-and-stick model. Figure 2.20 shows a ball-and-stick model of the somewhat more complex crystal structure of copper(II) chloride, CuCl2. As the formula indicates, in the lattice structure there are two chlorine atoms for Figure 2.19. A space-filling model each copper atom. of the sodium chloride lattice structure. Sodium atoms are purple and chlorine atoms are So far we have been discussing one major category of sub- green.

2.2.2 Mixtures

stances—pure substances. Elements and compounds are pure substances. The other major category is mixtures. Anytime substances are mixed together without a chemical reaction occurring, a mixture is formed. Remember—if a chemical reaction occurs, compounds are formed, not mixtures. If you toss vegetables in a salad, you’ve made a mixture. If you put sugar in your tea or milk in your coffee, you’ve made a mixture. If you mix up a batch of chocolate chip cookie dough, a bowl of party mix, or the batter for a vanilla cake, you’ve made a mixture. In contrast to compounds, when a mixture is formed the individual substances in the mixture retain their physical and chemical properties. If you mix salt in water, you’ve made a mixture. The salt is still there and tastes salty. The water is still there too, Figure 2.20. A ball-and-stick model of the copper(II) and tastes watery. Also in contrast to com- chloride crystal structure. Copper atoms are copper colored and chlorine atoms are green. pounds, the substances in a mixture can be separated by physical means such as filtering, boiling, freezing, or settling. Again, in compounds, the original properties of the elements in the compound are chemically changed into the properties of the new chemical substance that is formed by means of a chemical reaction. Further, the elements in a compound cannot be separated by physical means. They can only be separated by the same means that brought them together in the first place—a chemical reaction. The distinguishing features of mixtures and compounds are summarized in Table 2.3. There are two classes of mixtures—homogeneous mixtures and heterogeneous mixtures. We examine these next. Homogeneous Mixtures

Homogeneous mixtures have uniform composition down to, but not including, the groups of atoms at the atomic level. The individual particles of the different substances in a homogeneous mixture cannot be seen with the eye, not even with the most powerful microscope. Particles at the atomic level—atoms, molecules, ions—are too small to reflect visible light, and so cannot be seen in the ordinary way, regardless of the magnification.

Chapter 2 Mixtures

Compounds

• Formed when substances combine without • Elements combine chemically to form a a chemical reaction occurring. new substance—a compound. • The individual substances in the mixture retain their physical and chemical properties.

• The physical and chemical properties of the compound are completely different from those of the reactants that formed the compound.

• The substances in mixtures can be separated by physical means such as filtering, boiling, freezing, or settling.

• The elements in a compound can only be separated by chemical means.

Table 2.3. Summary of the distinctions between mixtures and compounds.

The implication of this definition is that homogeneous mixtures are identical with solutions, mixtures in which one pure substance (or more than one) is dissolved in another pure substance. Solutions are so important in chemistry that we devote an entire chapter to the topic later on. For now, we move on to the other types of mixtures. Heterogeneous Mixtures In contrast to homogeneous mixtures, in a heterogeneous mixture there are lumps of different substances mixed together. You might be able to see the lumps with the naked eye, as with the mixture of spices in meat seasoning. Or the different substances may be visible only under a microscope, such as microscopic particulates in well water. Either way, if the different substances can be seen, the mixture is heterogeneous. In addition to the examples of salads and so on I mentioned above, there are two classes of heterogeneous mixtures that we encounter every day: suspensions and colloidal dispersions. Suspensions A suspension is formed when particles of size approximately 1 micrometer (1 μm) or larger are dispersed in a fluid (liquid or gas) medium. Particles this large do not remain in suspension indefinitely; they eventually settle out due to gravity. Figure 2.21 shows an example of a suspension. At room temperature, corn starch is not soluble in water; that is, it does not dissolve in water. But if corn starch is stirred into water, a suspension is formed, as shown in the upper photo in the figure. After a few minutes, it is evident that the starch does not dissolve in the water, and after a few hours the starch particles are all at the bottom of the container, as shown in the lower photo. Muddy water is another example of a suspension. If left to stand, the particles of sand, soil, and other organic matter eventually settle out, leaving the water quite transparent. Colloidal Dispersions

Figure 2.21. Corn starch mixed in water forms a suspension (top). The starch particles eventually settle out of the liquid and fall to the bottom of the beaker (bottom).

Another class of heterogeneous mixtures is the colloidal dispersions, usually referred to simply as colloids. A colloid is formed when microscopic particles, ranging in 54

Atoms and Substances Dispersing Medium State

Dispersed Particle State

Solid

Liquid sol blood, paint

solid emulsion butter, cheese

Gas

Solid

solid sol cobalt glass, cranberry glass

solid aerosol smoke, fine airborne dust

Liquid

gel gelatin, hair gel

liquid emulsion mayonnaise, milk

liquid aerosol fog, hair spray, insect repellent

Gas

solid foam Styrofoam, foam rubber, marshmallow

foam shaving cream, whipped cream

(none)

Table 2.4. Types of colloids with examples.

size from 1 nm to 1,000 nm, are dispersed throughout a dispersing medium. Unlike the particles in suspensions, these particles do not settle out. Forces from Brownian motion (see the box on the next page) keep the particles dispersed in the dispersing medium. Table 2.4 lists the different types of colloids, distinguished by the states of the dispersed particles and the dispersing medium.2 Since most of these substances are probably familiar to you, although perhaps not by name, I will just point out a few things about the information in the table. You may or may not have seen items made of cobalt glass or cranberry glass, both examples of solid sols and shown in Figure 2.22. Glass is silicon dioxide, SiO2, just like sand. Cobalt glass is silicon dioxide with cobalt salts such as cobalt(II) aluminate, CoAl2O4, dispersed within the molten glass. Cranberry glass contains particles of gold(III) oxide in the molten glass. Once the glass hardens, the colloid that remains consists of solid particles dispersed within a solid medium. Two different terms are used for colloids consisting of solid particles dispersed within a liquid Figure 2.22. Cobalt glass (left) and cranberry glass medium, sol and solid emulsion. As the examples (right), also known as “gold ruby.” in the table show, these two classes of colloids are rather different. In a gel, the solid particles link together in long molecules known as polymers, and these linked solids form a network or matrix throughout the substance. Finally, there are no colloids consisting of gas particles dispersed in other gas particles. This is because when no reaction occurs, all gases are miscible in each other. That is, the particles of the gases mix together in all proportions. Thus, mixed gases form solutions, not colloids. We conclude this section with two more interesting facts about colloids. First, although the dispersed particles in a colloid are smaller than the particles in a suspension, they are still not 2 Sources differ to a frustrating degree on the definitions of some of the terms and on the identifications of some of the substances. If you go digging around in different texts or online sources, prepare to get confused. 55

Chapter 2

Hmm... Interesting.

Brownian motion

In 1827, Scottish botanist Robert Brown (1773–1858) was using a microscope to examine tiny particles from pollen grains. The particles were suspended in water, and Brown noticed that they jittered back and forth and moved through the water in an erratic and apparently random path. The motion of those particles is now known as Brownian motion, although no one knew what caused it until 1905. That was the year Albert Einstein published a paper demonstrating that the motion was caused by a particle being bombarded on all sides by moving water molecules. On this topic, still pictures just don’t cut it. There are lots of animations online, but nothing beats the real thing. Go to commons.wikimedia.org, search under Brownian motion, and watch the clip with the file name Brownianmotion beads in water spim video.gif (sic). small enough to pass through a filtration medium. In a true solution, the individual particles are ions or molecules, and their diameters are down in the sub-one nanometer range. Particles this small can pass right through a filtration medium such as filter paper. (So don’t bother trying to separate salt from water by passing saltwater through a filter.) Colloidal particles are much larger than this and cannot pass through a filter. This is why people can wear dust masks to filter out pollens and dust (but not water vapor) from breathing air. For colloids in fluid media, there is a test that distinguishes between solutions and colloids. Figure 2.23 illustrates the so-called Tyndall effect in a colloid. In the upper photograph, a flashlight beam is passed from right to left through a solution of saltwater. The light from the beam cannot be seen passing through the solution because the particles in the solution are so small they do not scatter light. In the lower photo, the beam is passed through a mixture of gelatin and water. The particles of gelatin form a colloidal dispersion, and are large enough that they do scatter the light. This is demonstrated by the fact that the beam is visible, by the illumination of the liquid, and by the brightness of the white markings on the beaker. The two photos were taken under the same lighting conditions. The light in the room was very dim except for the flashlight. The Tyndall effect is what makes the beams from a car’s headlights visible in foggy conditions. Fog is a colloid of liquid particles of water in air (a liquid aerosol), and the liquid particles scatter the light. And by the way, true water vapor—which is what steam is—is an invisible gas, just as the Figure 2.23. A salt water solution (top) humidity in the air is invisible. If you can see the moisture, allows the flashlight beam to pass through without scattering the light. A you are seeing light scattered by liquid water particles in air, colloidal dispersion (bottom) scatters and you are looking at a colloid, a mist of tiny condensed the light, making the beam visible, water droplets, not at steam. illuminating the surrounding liquid and the markings on the beaker from behind.

Atoms and Substances

2.2.3 Physical and Chemical Properties All substances have certain properties. We divide up the different properties substances can possess into two broad classes. Some properties have to do with the physical characteristics of the substance, such as color, shape, size, phase, boiling point, texture, thermal conductivity, electrical conductivity, opacity, and density. These properties are called physical properties. Below is a list of example statements about the physical properties of substances. Consider how each one relates to the definition of physical properties just given. • • • • • • • • • • • • • • • • • •

Iron is gray in color. Copper has a high electrical conductivity. Mica is shiny. Glass is smooth, but has sharp edges. Aluminum has a high thermal conductivity. Ethyl alcohol is transparent and colorless. Chlorine is greenish-yellow gas at atmospheric pressure and room temperature. Helium is a gas at atmospheric pressure and room temperature. At standard pressure, water freezes at 0°C. Milk is opaque. Oil is slippery. Clay brick is ochre in color. At 4.0°C, the density of water is 1.0 g/cm3. Gold is malleable and ductile. Cast iron is not malleable. Glass is not malleable. Play-Doh is malleable, but not ductile. Jello is not ductile.

You may have noticed a couple of unfamiliar terms in the above list. The terms malleable and ductile are used to describe two important properties possessed by many metals. A substance is malleable if it can be hammered into different shapes, or hammered flat into sheets. A substance is ductile if it can be drawn into a wire. Wire drawing is a process of making wire by pulling the metal through a small hole in a metal block called a die. Usually the metal is already formed into a wire of larger diameter. The end of this larger diameter wire is hammered down or filed to get it through the hole in the die, and then a machine pulls the wire through the die to make the new, smaller diameter wire. Substances that can be drawn through a die like this without simply snapping are said to be ductile. Notice from the above examples that a good student of science should be careful when describing physical properties. We need to make sure our statements are accurate in cases where temperature or pressure affect the property in question. For example, it is inaccurate to say that H2O is a liquid. A more accurate statement is to say that one of the physical properties of water is that it is a liquid at temperatures between 0°C and 100°C. The statement is made even more accurate by specifying that the preceding sentence is correct at atmospheric pressure because at other pressures the boiling and freezing points of water are different. The second broad class of properties has to do with the kinds of chemical bonds a substance forms, that is, the chemical reactions a substance does or does not participate in. These properties are called chemical properties. We have not yet studied chemical reactions, so you may not know that much about them. However, there are two common chemical reactions that you are quite familiar with—burning and rusting. Both of these are chemical reactions in which a substance combines with oxygen, and both are examples of a type of reaction called oxidation. Fiery explosions are simply combustions that happen very rapidly. But whether the combustion hap57

Chapter 2 pens slowly, as with a log on a fire, or rapidly, as with a firecracker, combustion is a chemical reaction with oxygen. Substances that react with oxygen in this way are said to be flammable or combustible. (Oddly, inflammable means the same thing!) Iron oxidizes to form compounds known as oxides. There are several different forms of iron oxide, colored red, yellow, brown, and black. Other metals oxidize as well. When copper oxidizes, it can form two different oxides, one red and one black. This is why copper objects exposed to the air turns dark brown or black. (Over a longer period of time the copper oxide forms other compounds, such as copper carbonate, which give the copper its pretty blue-green color. The Statue of Liberty is made of copper, and has been there for a long time. It is essentially covered with a layer of copper carbonate and other copper compounds.) Aluminum also oxidizes. Aluminum oxide is dark gray, and anyone who has done a lot of hand work with aluminum parts has noticed his or her hands blackened by the particles of aluminum oxide building up. Figure 2.24 shows a few different oxides. Here are some examples describing chemical properties of substances: • • • • • • • • • • •

Hydrogen is combustible. Figure 2.24. Oxides: red iron oxide Aluminum oxidizes to form aluminum oxide. (top), yellow iron oxide (center), Water is not flammable. and black copper oxide (bottom). Platinum does not oxidize. (This is why it is so valuable. It stays shiny as a pure element.) Baking soda (sodium bicarbonate, NaHCO3) reacts with vinegar (acetic acid, CH3COOH). Iron oxidizes to form iron oxide, or rust. Sodium reacts violently with water. Hydrogen reacts with a number of different polyatomic ions to form acids. Dynamite is explosive. Sodium hydroxide reacts with aluminum. Sulfuric acid reacts with many metals.

We discuss the physical and chemical properties of substances further in the coming chapters. The two broad classes of properties we have been discussing, physical properties and chemical properties, are related to two broad classes of changes that substances undergo. If a substance experiences a change with respect to one of its physical properties, we call this a physical change. When a physical change occurs, the substance is still the same substance, it just looks or feels different. If a chemical reaction occurs to a substance, this is called a chemical change. Chemical properties basically describe the kinds of chemical changes a substance can undergo. When a chemical change occurs, the original substances that go into the reaction—the reactants— are converted into new substances—the products—with totally different physical and chemical properties. When asked to describe a given change as physical or chemical, ask yourself if the substance is still the same substance, or if it has actually gone through a chemical reaction to become a different substance. Table 2.5 lists some examples of physical and chemical changes, with comments explaining why the type of change is physical or chemical.

Atoms and Substances Process

Change

Comments

glass breaking

physical

The broken glass is still glass, it just changes shape.

firecracker exploding

chemical

This explosion is a combustion. All combustions are chemical reactions. The substances in the firecracker react to form new substances such as ash and various gases.

mercury boiling

physical

The mercury is still mercury, it just changes from the liquid state to the vapor state.

copper turning dark brown or black

chemical

This occurs because the copper is oxidizing and forming copper oxide, a new substance. This is a chemical reaction.

iron pipes corroding

chemical

Corrosion is a chemical reaction. In this case, the iron reacts with the substances surrounding the pipes to form a new substance.

water evaporating

physical

The substance is still H2O, it simply changes state from liquid to vapor.

mixing cake batter

physical

The eggs and flour and so on form a mixture, but no chemical change (reaction) occurs.

baking cookies

chemical

The heat causes a chemical reaction to occur in the dough. The substance is no longer cookie dough. It is cookie.

spilled pancake batter drying out

physical

No chemical reaction occurs. Dried batter is still batter. (If you want a pancake you have to cook it, which causes a chemical change.)

molten lead hardening

physical

The lead is still lead, it just changes state from liquid to solid.

balloon popping

physical

The balloon material is still the same material, it is just in shreds now. The air inside the balloon is at a lower pressure and is not contained in the balloon any longer, but it is still air.

Table 2.5. Examples of physical and chemical changes.

2.3

Isotopes and Atomic Masses

2.3.1 Isotopes As you know, the atomic number (Z) of an element designates the number of protons in the nucleus of an atom of that element. For a given element, the atomic number is fixed: if an atom has a different number of protons, it is an atom of a different element. But the number of neutrons that may be present in the nucleus is not always the same for atoms of a given element. For most elements, there are variations in the number of neutrons that can be present in the nucleus. These varieties are called isotopes. For most elements, there is one isotope that is the most abundant in nature and several other isotopes that are also present but in smaller quantities. The general term for any isotope of any element is nuclide. Isotopes are designated by writing the name of the element followed by the number of nucleons (protons and neutrons) in the isotope. The number of nucleons in a nucleus is called 59

Chapter 2 the mass number. For example, the most common isotope of carbon is carbon-12, accounting for about 98.9% of all the naturally occurring carbon. In the nucleus of an atom of carbon-12 there are six protons and six neutrons. There are two other naturally occurring carbon isotopes. Carbon-13 with seven neutrons accounts for about 1.1% of natural carbon. Atoms of carbon-14, of which only a trace exists in nature, have eight neutrons in the nucleus.

2.3.2 The Unified Atomic Mass Unit The mass of a single atom is an extremely small number. But so much of our work in chemistry depends on atomic masses that scientists having been using units of relative atomic mass for a long time—all the way back to John Dalton, before actual masses of atoms were even known. Prior to the discovery of isotopes in 1912, the so-called atomic mass unit (amu) was defined as 1/16 the mass of an oxygen atom. After the discovery of isotopes, physicists defined the amu as 1/16 the mass of an atom of oxygen-16, but the definition used by chemists was 1/16 the average mass of naturally occurring oxygen, which is composed of several isotopes. To eliminate the confusion resulting from these conflicting definitions, the new unified atomic mass unit (u) was adopted in 1961 to replace them. Many texts continue to use the amu as a unit, but they define it as the u is defined. Strictly speaking, the amu is an obsolete unit that has been replaced by the u, now also called the dalton (Da). The u and the Da are alternative names (and symbols) for the same unit. The use of the dalton has increased in recent years, particularly in molecular biology. The unified atomic mass unit, u, is defined as exactly 1/12 the mass of an atom of carbon-12. Table 2.6 lists a few nuclides and their atomic masses using the u as a unit of mass. All the elements listed exist as other isotopes in addition to those shown, but as you see from the percentage abundances, the ones shown are the major ones for the elements represented Z isotope mass (u) abundance (%) in the table. 1 hydrogen-1 1.0078 99.9885

2.3.3 Atomic Masses

hydrogen-2

2.0141

0.0115

In addition to the atomic number, the Periodic Table of the Elements lists the atomic mass in unified atomic mass units (u) for each element. But since there are multiple isotopes for just about every element, the atomic mass values in the periodic table represent the weighted average of the masses of naturally occurring isotopes. An example of a weighted average is the average age of the students in the junior class at your school. Let’s say there are 47 juniors, 40 of whom are 16 years old and 7 of whom are 17 years old at the beginning of the school year. To determine the average age of these students, let’s first determine the proportion of the students at each age.

carbon-12

12.0000

98.93

carbon-13

13.0034

1.078

silicon-28

27.9769

92.223

silicon-29

28.9765

4.685

silicon-30

29.9738

3.092

chlorine-35

34.9689

75.76

chlorine-37

36.9659

24.24

calcium-40

39.9626

96.941

calcium-42

41.9586

0.647

iron-54

53.9396

5.845

iron-56

55.9349

91.754

iron-57

56.9354

2.119

iron-58

57.9333

0.282

copper-63

62.9296

69.15

copper-65

64.9278

30.85

uranium-235

235.0439

0.7204

uranium-238

238.0508

99.2742

40 = 0.851 (85.1%) 47 7 = 0.149 (14.9%) 47 60

Table 2.6. Major isotopes for a few elements.

Atoms and Substances To calculate the average age, we first multiply each student age by the proportion of students of that age to find the contribution to the average from each age group. Then we add the contributions together to find the weighted average age for the junior class. 16 years⋅0.851 = 13.6 years +17 years⋅0.149 = 2.53 years = 16.1 years We perform a similar calculation when computing the average atomic mass of an element from the masses of its isotopes, as shown in the following example. Example 2.1 Given the isotope masses and abundances for copper-63 and copper-65 in Table 2.6, determine the atomic mass for naturally occurring copper. We need to multiply each isotope’s mass by its abundance to get the isotope’s contribution to the average mass of the element, which is the atomic mass. Then we add together the contributions from each isotope. The data from the table are: copper-63: mass = 62.9296 u, abundance = 69.15% copper-65: mass = 64.9278 u, abundance = 30.85% 62.9296 u ⋅0.6915 = 43.51 u + 64.9278 u ⋅0.3085 = 20.03 u = 63.55 u Compare this value to the value shown in the periodic table in Figure 2.14 or inside the back cover of the text.

The unified atomic mass unit, u, is defined as 1/12 the mass of an atom of carbon-12. Although the value of this mass is quite close to the masses of the proton and neutron, it is not exact because of the mass of the electrons in atoms of carbon-12, and also because of the massenergy involved in binding the nucleus of the atom together. (The mass of nucleons bound together in a nucleus does not equal the sum of their individual masses.) Table 2.7 shows the masses of the three basic subatomic particles in unified atomic mass units. Still, the proton and neutron masses are very close to unity (one) and the electron mass is extremely small. This means that for elements with a very large abundance of one isotope we can use the atomic mass and atomic particle mass number in the periodic table to determine the numbers of proproton 1.007276 u tons and neutrons in the nucleus of the most common isotope. neutron 1.008665 u For example, from Table 2.6, the mass of uranium-238 is very close to 238 u. Since an atom of uranium-238 has 92 protons, electron 0.0005486 u the balance of the mass is essentially all neutrons. Thus, there are Table 2.7. Masses in u of the three 238 – 92 = 146 neutrons in uranium-238. basic subatomic particles.

Chapter 2

2.4

Density and Quantity of Substances

2.4.1 Density Density is a physical property of substances. Density is a measure of how much matter is packed into a given volume for different substances. No doubt you are already familiar with the concept of density. You know that if you hold equally sized balloons in each hand, one filled with water and one filled with air, the water balloon weighs more because water is denser than air. You know that for equal weights of sand and Styrofoam packing peanuts the volume of the packing material is much larger because the packing material is much less dense. And you probably also know that objects less dense than water float, while objects denser than water sink. The equation for density is

ρ=

m V

(2.1)

where the Greek letter ρ (spelled rho and pronounced “row,” which rhymes with snow) is the density in kg/m3, m is the mass in kg, and V is the volume in m3. These are the variables and units in the MKS unit system. However, since laboratory work typically involves only small quantities of substances, it is more common in chemistry for densities to be expressed in g/cm3 (for solids) or g/mL (for liquids). In the examples that follow, I illustrate the use of g/cm3 and kg/m3. Since 1 mL = 1 cm3 (see Table A.3 in Appendix A), calculations of densities in g/mL are essentially the same as those solving for densities in g/cm3. If all you are doing is using the density equation, then any of these units of measure is fine. One final item for you to note is that the density of water at room temperature is

ρw = 0.998

g cm3

(22.0°C)

This value is useful to know because water comes up in many different applications. The densities at other temperatures, along with other properties of water, are listed in Table A.5 of Appendix A. Example 2.2 The density of germanium is 5.323 g/cm3. A small sample of germanium has a mass of 17.615 g. Determine the volume of this sample. Begin by writing the given information.

ρ = 5.323

g cm3

m = 17.615 g V =? Now write Equation (2.1) and solve for the volume.

Atoms and Substances

ρ=

m V

ρ ⋅V = m V=

m ρ

Next, insert the values and compute the result. V=

m 17.615 g = = 3.309 cm3 ρ 5.323 g cm3

This value has four significant digits, as it should based on the given information. Example 2.3 Determine the density of a block of plastic that has a mass of 1,860 g and dimensions 4.0 in × 2.5 in × 9.50 in. State your result in kg/m3. To solve this problem, we use the given dimensions to calculate the volume of the block. Then we use Equation (2.1) to calculate the density. Always begin your problem solutions by writing down the given information and performing any necessary unit conversions. Since the units of measure required for the result are kg/m3, we convert the mass to kilograms and the lengths to meters. When solving problems requiring unit conversions like this, write down the given information on separate lines down the left side of your page. Then perform the unit conversions by multiplying the conversion factors out to the right. From the given information, our result must have two significant digits. This means we must perform the unit conversions and volume calculation with three significant digits (one more than we need), and round to two digits when we get our final result. m = 1860 g ⋅ l = 4.0 in⋅

1 kg = 1.86 kg 1000 g

2.54 cm 1 m ⋅ = 0.102 m 1 in 100 cm

w = 2.5 in⋅ h = 9.5 in⋅

2.54 cm 1 m ⋅ = 0.0635 m 1 in 100 cm

2.54 cm 1 m ⋅ = 0.241 m 1 in 100 cm

Now that the units are squared away, let’s determine the volume of the block. V = l ⋅w ⋅h = 0.102 m⋅0.0635 m⋅0.241 m = 0.00156 m 3

Chapter 2 Finally, using Equation (2.1) the density is

ρ=

m 1.86 kg kg = = 1192 3 3 V 0.00156 m m

Rounding to two significant digits we have 1200

kg m3

2.4.2 The Mole and the Avogadro Constant When solving problems in chemistry, we are generally working with chemical reactions in which huge numbers of atoms are involved, including all the naturally occurring isotopes, so performing reaction calculations with the masses of individual atoms is not practical. However, the average mass of a given multiple of some kind of atom is simply that multiple times the atomic mass. The mass of one million atoms of aluminum is 1,000,000 times the atomic mass of aluminum. In chemistry, the standard bulk quantity of substance used in calculations is the mole (mol), the SI base unit for quantity of substance (Table 1.1). The mole is a particular number of particles of a substance, just as the terms dozen, score, and gross refer to specific numbers of things (12, 20, and 144, respectively). A mole is exactly 6.02214076 × 1023 particles of a substance. This value is known today as the Avogadro constant, NA. More formally, the Avogadro constant is defined as exactly: N A = 6.02214076 × 1023 mol −1

(2.2)

Usually, we just round this value to 6.022 × 1023 mol–1. In the next section, I’ll describe why this value is what it is, instead of being a more convenient round number. For the moment, let’s focus on what it means. Now, don’t freak out over the unit of measure. Allow me to explain. Raising a unit of measure to the power –1 is mathematically equivalent to placing the unit in 1 a denominator because x −1 = . In other words, Equation (2.2) is the same thing as saying x “6.02214076 × 1023 per mole.” To make things even clearer, it’s okay to say it this way: “NA is about 6.022 × 1023 particles per mole.” This is the way I like to think of it when performing unit conversions, as we do quite a lot in coming chapters. Without the units of measure, the value 6.02214076 × 1023 is called Avogadro’s number. With the units, it is called the Avogadro constant. Using this terminology, the mole can be defined this way: A mole is the amount of a pure substance (element or compound) that contains Avogadro’s number of particles of the substance. Let’s now consider what we mean when we refer to particles of a substance. For substances that exist as molecules, the particles are the molecules. Examples of these are the molecular substances we encountered back in Figure 2.18. For substances that exist as individual atoms, the particles are the individual atoms. Metals are like this, since a pure metal is composed of individual atoms of the same element joined together in a crystal lattice. The noble gases are also like this. The noble gases are located in the far right-hand column of the Periodic Table of the Elements. As I discuss more in coming chapters, atoms of noble gases are almost completely

Atoms and Substances unreactive—they don’t bond with other atoms at all. At ordinary temperature and pressure, the noble gases are gases composed of individual atoms. For crystalline compounds, the “particles” in a mole of the substance are the formula units in the crystal lattice. A formula unit is one set of the atoms represented by the chemical formula of the compound. For example, the chemical formula for calcium carbonate is CaCO3. One formula unit of calcium carbonate includes one calcium atom, one carbon atom, and three oxygen atoms. The value of the Avogadro constant was determined approximately by French Physicist Jean Perrin (Figure 2.25) in the early 20th century. Perrin determined the value of the constant through several different experimental methods. In the 19th century, many scientists did not yet accept the existence of atoms as a scientific fact and Perrin’s research put the atomic nature of matter beyond dispute. For this work, he received the Nobel Prize in Physics in 1926. Perrin proposed naming the constant after Amedeo Avogadro, a 19th-century Italian scientist who was the first to propose that the volume of a gas at a given temperature and pressure is proportional to the number of particles of the gas (atoms or molecules), regardless of the identity of the gas. In fact, at 0°C and atmospheric pressure, one mole of any gas occupies a volume of 22.4 L.

2.4.3 Molar Mass and Formula Mass

Figure 2.25. French physicist Jean Perrin (1870–1942).

Since 2019, the value of the Avogadro constant in Equation (2.2) is exact by definition. But the number has the value it does because it was originally chosen so that the average atomic mass in u of a molecule of a compound, as computed from the mass values in the periodic table, would be numerically equivalent to the mass of one mole of the compound in grams per mole. (We address these calculations below.) Now, recall that the definition of the unified atomic mass unit (or dalton) is such that an atom of carbon-12 has a mass of exactly 12 u. According to the original definition of Avogadro’s number, there were also exactly 12 grams of carbon-12 in one mole of carbon-12. So according to these definitions, an atom of carbon-12 has a mass of exactly 12 u, and a mole of carbon-12 had a mass of exactly 12 grams. This quantity, the mass of one mole of a substance, is called the molar mass. Because of the way the molar and atomic masses were defined, the molar mass for an atom was numerically equivalent to the atomic mass. As a result of the 2019 redefinition of Avogadro’s number, the atomic mass in u and the molecular mass in g/mol are no longer exactly equivalent. However, they are extremely close and may still be treated as equal for practical purposes. (The difference is a factor of only about 4 × 10–10.) Even though the exact equivalence ended in 2019, these are still very handy definitions! For example, from the periodic table we find that the average mass of one atom of silicon (Z = 14) is 28.0855 u. This also tells us that the mass of one mole of silicon is 28.0855 g, so the molar mass of silicon is 28.0855 g/mol. Likewise, from the periodic table we find that the average mass of one atom of copper (Z = 29) is 63.546 u. This also tells us that the mass of one mole of copper is 63.546 g, so the molar mass of copper is 63.546 g/mol. For the elements that exist as single atoms, the molar mass in g/mol and the atomic mass in u are numerically equivalent (almost). From the periodic table, we can also determine the molar mass of compounds—the mass of one mole of the compound. We simply add up the molar masses for the elements in the chemical formula, taking into account any subscripts present in the formula, and we have the molar mass for the compound in g/mol. If we add up the element atomic masses in unified atomic mass

Chapter 2 Quantity

Units

Definition

molar mass

g/mol

The mass of one mole of a substance, approximately equal to the sum of the atomic masses of the elements in a chemical formula, taking into account the subscripts indicating atomic ratios in the compound.

formula mass

The mass of one formula unit of a substance. Numerically nearly equivalent to the molar mass.

molecular mass

The average mass of a single molecule of a molecular substance. Numerically equivalent to the formula mass. (May also be converted to grams and expressed in grams, see Section 2.4.4.)

Table 2.8. Definitions and units for molar mass, formula mass, and molecular mass.

units, we obtain what is called the formula mass of the compound in u (or Da). If the compound is molecular, then the formula mass may also be referred to as the molecular mass, the average mass of a single molecule of the substance. The details of these three different mass terms are summarized in Table 2.8. Example 2.4 Determine the molar mass and formula mass for water, H2O. Note that since water is composed of molecules, the formula mass may also be called the molecular mass. From the periodic table, the atomic masses of hydrogen (H) and oxygen (O) are: H: 1.0079 u O: 15.9994 u Written as molar masses, these are: g mol g O: 15.9994 mol H: 1.0079

The formula for water, H2O, tells us there are two hydrogen atoms and one oxygen atom in each molecule of water, so we multiply these numbers by the molar masses and add them up to get the molar mass of H2O. g ⎞ g g ⎞ ⎛ ⎛ ⎟⎠ + ⎜⎝ 1×15.9994 ⎟⎠ = 18.0152 ⎜⎝ 2 ×1.0079 mol mol mol The calculation of the formula mass is identical, except we use units of u instead of g/mol. From the periodic table, the atomic masses of hydrogen (H) and oxygen (O) are: H: 1.0079 u O: 15.9994 u There are two hydrogen atoms and one oxygen atom in each molecule, so we multiply these numbers by the element atomic masses and add them up to get the formula mass of H2O. 66

Atoms and Substances

( 2 ×1.0079 u ) + (1×15.9994 u ) = 18.0152 u Thus, the formula mass for water is 18.0152 u. This value is also the molecular mass of water.

Example 2.5 Nitrogen is one of several elements that exist in nature as diatomic gases. (The others include hydrogen, oxygen, fluorine, and chlorine.) Determine the molar mass of nitrogen gas, N2. From the periodic table, the atomic mass of nitrogen (N), written as a molar mass for individual nitrogen atoms, is: N: 14.0067

g mol

There are two nitrogen atoms in each molecule of N2, so we multiply the molar mass of N by two to get the molar mass of N2. g g ⎞ ⎛ ⎟ = 28.0134 ⎜⎝ 2 ×14.0067 mol ⎠ mol

Example 2.6 Determine the mass in grams of 2.5 mol sodium bicarbonate, NaHCO3 (baking soda). In any problem like this, we first find the molar mass of the given compound. Then we simply use that molar mass to compute the mass of the given quantity. From the periodic table, the atomic masses of the elements in the compound are: Na: 22.9898

g mol

g mol g C: 12.011 mol g O: 15.9994 mol H: 1.0079

The oxygen appears three times in the formula, so its mass must be multiplied by three and added to the others. 22.9898

g g g ⎛ g ⎞ g +1.0079 +12.011 + ⎜ 3×15.9994 ⎟ = 84.007 mol mol mol ⎝ mol ⎠ mol

This value is the molar mass for NaHCO3. To find the mass of 2.5 mol we multiply:

Chapter 2 2.5 mol⋅84.007

g = 210 g mol

Notice that the significant digits in the molar mass are obtained by the addition rule. The atomic mass of carbon, 12.011 u, only goes to three decimal places, so the molar mass of the compound goes to three decimal places. The final result is obtained by multiplying the compound’s molar mass by the quantity 2.5 mol, which only has two significant digits. Thus, the result must have two significant digits as well.

Example 2.7 A scientist measures out 125 g of potassium chloride (KCl). How many moles of KCl does this quantity represent? First, determine the molar mass of KCl. From the periodic table: g mol g Cl: 35.4527 mol K: 39.098

The formula includes one atom of each, so we add them to obtain the molar mass: 39.098

g g g + 35.4527 = 74.551 mol mol mol

Beginning now, always think of the molar mass of any substance as a conversion factor that can be written right side up or upside down to convert grams to moles or vice versa. For KCl, 74.551 g is equivalent to 1 mol, so these quantities can be written as conversion factors, like this: 74.551 g 1 mol = 1 mol 74.551 g This makes the last step of this problem easy. Just select the way of writing the molar mass conversion factor that cancels out the given units (g) and gives the units required (mol). This is nothing but a unit conversion. 125 g ⋅

1 mol = 1.68 mol 74.551 g

The photograph in Figure 2.26 shows one mole of each of four substances. The first is one mole of copper, equal to 63.5 g. The second is a 250-mL beaker containing one mole of water. As you can see, this is not much water—only 18 mL. At the upper right is a weigh tray Figure 2.26. Clockwise from left are shown

1 mole of copper, 1 mole of water, 1 mole of table salt, and 1 mole of baking soda.

Atoms and Substances containing one mole of sodium chloride, 40.0 g. (This is just under 1/4 cup.) Finally, one mole of baking soda, 84.1 g. (This is right at 1/3 cup.) Example 2.8 Calculate the number of water molecules in a 1.00-liter bottle of water. The logic of this problem, in reverse, is as follows: To calculate a number of molecules, we must use the Avogadro constant. To use the Avogadro constant, we need to know the number of moles of water we have. To determine the number of moles, we need to know both the molar mass and the mass of the water. To determine the mass from a volume, we use the density equation. So we begin with the given information and the density equation to determine the mass of water we have. The given information and unit conversions are as follows: V = 1.00 L⋅

ρ = 0.998

1000 cm3 = 1.00 ×103 cm3 1L

g cm3

m=? Now we write down Equation (2.1) and solve for the mass:

ρ=

m V

m = ρ ⋅V = 0.998

g ⋅1.00 ×103 cm3 = 998 g cm3

Next we need the molar mass of water. We calculated this in Example 2.4 and obtained 18.0152 g/mol. We use this molar mass as a conversion factor to convert the mass of water into a number of moles of water: 998 g ⋅

1 mol = 55.40 mol 18.0152 g

This intermediate result has four significant digits—one more than we need in the final result. Finally, with the number of moles in hand we use the Avogadro constant to determine how many particles of water this is, which is identical to the number of water molecules. 55.40 mol⋅

6.022 ×1023 particles = 3.34 ×1025 particles mol

2.4.4 Gram Masses of Atoms and Molecules The atomic mass from the periodic table and the Avogadro constant can be used to calculate the mass in grams of an individual atom. Recall that the atomic mass value in the periodic table gives both the average atomic mass in u, and the molar mass in g/mol. Knowing the molar mass in g/mol we can simply divide by the number of atoms there are in one mole to find the mass of 69

Chapter 2 one atom in grams. Although this kind of calculation is quite simple, I have found that it is very easy for students to get confused and not be able to determine whether one should multiply or divide or what. So here’s a problem solving tip: let the units of measure help you figure out what to do. If you include the units of measure in your work and pay attention to how the units cancel out or don’t cancel out, these calculations are pretty straightforward. Keep this principle firmly in mind throughout your study of chemistry! Units of measure are not an annoying burden; they are the student’s friend. Example 2.9 Determine the average mass in grams of an atom of boron. From the periodic table, we find that the molar mass of boron is 10.811 g/mol. One mole consists of Avogadro’s number of atoms of boron, so if we divide the molar mass by the Avogadro constant, we have the mass of a single atom of boron. Let’s begin by setting up the division I just described, and then use the old invert-and-multiply trick for fraction division to help with the unit cancellations. g g 1 mol mol = 10.811 ⋅ 23 particles mol 6.0221×10 particles 6.0221×1023 mol 10.811

10.811 g g = 1.7952 ×10−23 23 6.0221×10 particle particle

So the average mass of one boron atom is 1.7952 × 10–23 g. Note that I use five digits in the value of the Avogadro constant to preserve the precision we have in the molar mass.

For molecular substances, the molar mass is used to compute the molecular mass in grams— the average mass of one molecule. This is done by first computing the molar mass of the compound, just as we did before. Then we simply divide by the Avogadro constant to obtain the mass of a single molecule. Like the atomic mass, the molecular mass is an average mass, since the atomic masses used in calculating the molar mass are all based on the average mass of different isotopes with their abundances taken into account. The molecular mass for a specific molecule must be calculated using the specific masses of the nuclides in the molecule. Example 2.10 Determine the mass in grams of one molecule of carbon tetrachloride, CCl4. From the periodic table we find that the molar masses of carbon and chlorine are 12.011 g/mol and 35.4527 g/mol, respectively. From this we calculate the molar mass of CCl4: g ⎞ ⎛ g ⎞ g ⎛ ⎟ + ⎜ 4 × 35.4527 ⎟ = 153.822 ⎜⎝ 1×12.011 mol ⎠ ⎝ mol ⎠ mol With this molar mass we can use the Avogadro constant to get the molecular mass in grams. This time, instead of writing the Avogadro constant in the denominator of a big fraction, I simply treat it as a conversion factor and write it in the equation such that the mole units cancel 70

Atoms and Substances out. (This is the way I always perform such calculations.) I also use six digits in the Avogadro constant to preserve the precision we have in the molar mass. 153.822

g 1 mol g ⋅ = 2.55427 ×10−22 mol 6.02214 ×1023 particles particle

Chapter 2 Exercises SECTION 2.1

1. Write paragraphs describing the experiments performed by J.J. Thomson, Robert Millikan, and Ernest Rutherford. 2. Describe the main points or features in the atomic models proposed by John Dalton, J.J. Thomson, and Ernest Rutherford. 3. Explain why Ernest Rutherford found the reflection of alpha particles off gold foil so astonishing. SECTION 2.2

4. Write paragraphs distinguishing between these pairs of terms: a. compounds and elements b. mixtures and compounds c. heterogeneous mixtures and homogeneous mixtures d. suspensions and colloids 5. Classify each of the following as element, compound, homogeneous mixture, or heterogeneous mixture. a. water

b. cesium chloride

c. pond water

d. methane

e. a soft drink

f. nitric acid

g. black coffee

h. argon

i. air

j. hydrogen nitrate

k. exhaust fumes

l. quartz

m. brass

n. hydrogen gas

o. hydrogen cyanide

p. mouthwash

q. platinum

r. dirt

s. radon

t. a smoothie

6. Explain why salt water and sugar water are homogeneous mixtures while automotive paint, which contains invisible particulates, is not. 7. Write a paragraph describing the two basic types of structures atoms can take when bonding together. 8. Select three pure substances not mentioned in the chapter. For each substance, list at least eight physical properties and three chemical properties. 9. Explain why colloids reflect light. What is this effect called?

Chapter 2 10. Identify each of the following as a physical change or a chemical change. For each, explain your choice. a. an avalanche

b. a cigar burning

c. spilling a glass of milk

d. digesting your food

e. swatting a fly

f. stirring cream into coffee

g. firing a pop gun

h. firing a real gun

i. boiling mercury

j. welding steel

k. filling a helium balloon

l. allowing molten iron to harden

m. frying chicken

n. snow melting

o. a car exhaust pipe rusting

p. paint “drying”

q. wood rotting

r. a ball rolling down a hill

SECTION 2.3

11. How is the unified atomic mass unit, u, defined? 12. Referring to Table 2.6, calculate the atomic mass for silicon, calcium, and uranium. Compare your results to the values shown in the periodic table. 13. Which two nuclides in Table 2.6 have 20 neutrons? 14. In Table 2.6, how many protons, neutrons, and electrons are there in the heaviest nuclide listed? How many protons, neutrons, and electrons are there in the lightest nuclide listed? 15. As mentioned in the chapter, the sum of the masses of the particles in an atom does not equal the mass of the atom. Some of the mass of the individual particles is converted to energy, and the atom weighs less than the sum of the weights of its parts. How much mass is converted into energy when the individual protons, neutrons, and electrons are assembled to form an atom of uranium-238? SECTION 2.4

16. What is the density of carbon dioxide gas if 0.196 g of the gas occupies a volume of 100.1 mL? 17. Oil floats because its density is less than that of water. Determine the volume of 550 g of a particular oil with a density of 955 kg/m3. State your answer in mL. 18. A factory orders 15.7 kg of germanium. The density of germanium is 5.32 g/cm3. Calculate the volume of this material and state your answer both in m3 and cm3. 19. A graduated cylinder contains 23.35 mL of water. An irregularly shaped stone is placed into the cylinder, raising the volume to 27.79 mL. If the mass of the stone is 32.1 g, what is the density of the stone? 20. A standard 55-gallon drum is 34.5 inches tall and 24 inches in diameter. Consider a 55-gallon drum filled with kerosene. Using the dimensions in inches to calculate the volume, determine the mass of kerosene that fills this drum, given that the density of kerosene is 810 kg/m3. 21. Iron has a density of 7,830 kg/m3. An iron block is 2.1 cm by 3.5 cm at the base and has a mass of 94.5 g. How tall is the block? 22. A student measures out 22.5 mL of mercury and finds the mass to be 306 g. Determine the density of mercury, and state your answer in kg/m3. 23. A large contemporary water tower holds over 3 million gallons of water. Determine the mass in kilograms of 3.0 × 106 gallons of water. 24. The famous Kon-Tiki was a raft sailed by Norwegian explorers in 1947 from South America to the Polynesian islands in the South Pacific. (Thor Heyerdahl’s book about it is a great read.) The trip took three and a half months and covered 4,300 miles in the Pacific Ocean. The crew of six men built the main section of the raft out of 9 massive balsa trees, each 2.0 ft 72

Atoms and Substances in diameter. Assume for simplicity that the balsa trees had a typical balsa wood density of 160 kg/m3, and the logs were each 45 ft in length. Determine the total mass in kilograms of these logs used to build the raft. 25. How is the mole defined? 26. Determine the number of atoms in each of the following. a. 73.2 g Cu

b. 1.35 mol Na

c. 1.5000 kg W

27. Determine the mass in grams for each of the following. a. 6.022 × 1023 atoms K

b. 100 atoms Au

c. 0.00100 mol Xe

d. 2.0 mol Li

e. 4.2120 mol Br

f. 7.422 × 1022 atoms Pt

28. Determine the number of moles present in each of the following. a. 25 g Ca(OH)2

b. 286.25 g Al2(CrO4)3

c. 2.111 kg KCl

d. 47.50 g LiClO3

e. 10.0 g O2

f. 1.00 mg C14H18N2O5

29. Calculate the molar mass for each of the following compounds or molecules. a. ammonia, NH3

b. carbon dioxide, CO2

c. chlorine gas, Cl2

d. copper(II) sulfate, CuSO4

e. calcium nitrite, Ca(NO2)2

f. sucrose, C12H22O11

g. ethanol, C2H5OH

h. propane, C3H8

i. glass, SiO2

30. Determine the formula masses for these compounds: a. MgCl2

b. Ca(NO3)2

c. (SO4)2– (The 2– indicates this is an ion with an electrical

d. CuSO4

e. BF3

f. CCl4

charge of –2. The charge does not affect your calculation.)

31. Determine the mass in grams of 2.25 mol silver nitrate, AgNO3. 32. Given 2.25 kg CCl4, answer these questions: a. How many moles CCl4 are present? b. How many carbon atoms are present? c. Approximately how many carbon-13 atoms are present? 33. Given 1.00 gal H2O at 4°C, answer the questions below. (Hint: You must use the appropriate volume conversion and the density of water to determine the mass of 1.00 gal H2O. See the information in Tables A.3 and A.5 in Appendix A.) a. How many moles H2O are present? b. How many hydrogen atoms are present? c. Approximately how many deuterium (hydrogen-2) atoms are present?

Chapter 7 Chemical Reactions and Stoichiometry

These test tubes contain various solutions and reaction products involving copper(II) nitrate, Cu(NO3)2. In some cases, precipitates formed; in others, the products remained in solution. We use some of these reactions as examples in this chapter.

168

Chemical Reactions and Stoichiometry

Objectives for Chapter 7 After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary. SECTION 7.1

1. State the law of conservation of mass for chemical reactions. 2. Recognize, interpret, and use the standard symbols involved in chemical equations. 3. Given a verbal description of the reactants and products in a chemical reaction, write the formula equation and balance it. 4. Define the terms precipitate, oxidation, reduction, oxidizer, oxidation state, and oxidation number. 5. Use the standard oxidation state guidelines from memory to determine the oxidation states of atoms in pure elements, in polyatomic ions, and in compounds. SECTION 7.2

6. Classify chemical reactions as synthesis, decomposition, single replacement, double replacement, combustion, acid-base neutralization, or oxidation-reduction. Also, identify when a reaction may be classified as two or more of these types simultaneously. 7. Use the activity series of metals to predict whether a given single replacement reaction with metals will occur. 8. Recognize and identify common hydrogen compounds as acids and common hydroxide compounds as bases. 9. Explain what happens in an acid-base neutralization reaction and use this information to form a definition for the term salt. 10. Explain the term redox reaction and describe what happens in such reactions. SECTION 7.3

11. Describe the origin of the term stoichiometry. 12. Given a mole or mass quantity of one of the reactants or products in a specified chemical reaction, perform stoichiometric calculations to determine the quantities required or produced, in moles or mass, of any of the other reactants or products in the reaction. 13. Define the terms limiting reactant and limiting reagent. 14. Use a balanced chemical equation and available quantities of reactants to identify the limiting reactant in the reaction. 15. Use the available quantity of a limiting reactant to determine the theoretical yield for the reaction. 16. Use the theoretical yield and actual yield in a reaction to determine the percent yield for a given reaction product.

169

Chapter 7

7.1

Introduction to Chemical Equations

7.1.1 Fascinating Chemistry Chemical reactions take place around us—and in us— all the time, but unless a fire or explosion occurs, we often don’t notice. You probably already know that fires and explosions are chemical reactions, but so are the drying of paint, the filling of the air bags in a car during a collision, and the digestion of your food. Chemical reactions are of supreme importance in industry, and without them you and I might have electricity and running water, but we would be without nearly all the products we enjoy. Without the chemical research of the 19th and 20th centuries, we would not have computers, plastic products, batteries, synthetic fibers (such as nylon), or just about anything else composed of nonmetallic man-made materials. There are several tell-tale signs that a chemical reaction is taking place. The first is the release of heat and light. When you are burning logs at a camp fire, you can be assured a chemical reaction is taking place. The second is the production of a gas, such as the carbon dioxide that fizzes out of a soft drink after the can has been opened. A third is a color change. Figure 7.1 shows the dramatic color change that occurs when beautiful blue-green copper carbonate is heated Figure 7.1. Copper carbonate decomposes to copper oxide when to above 290°C. The copheated. per carbonate decomposes to dark brown copper oxide, releasing carbon dioxide gas. Finally, a precipitate may form. A precipitate is any solid substance that forms when two solutions are combined. Figure 7.2 shows a solid copper(II) hydroxide precipitate coming out of solution because of the reaction that occurs when aqueous solutions of CuSO4 and NaOH are combined. Now, you may be of the opinion that chemistry is really only enjoyed by science nerds, but I suggest it should be otherwise. Because chemistry is so important in contemporary society, and because so many of the processes and products around us involve chemical reactions that are easily underFigure 7.2. Solid copper(II) hydroxide stood, I would expect anyone with a normal, healthy level precipitates out of solution when of curiosity about the world to be interested—if not fasciaqueous solutions of CuSO4 and NaOH nated—to learn about it. In this chapter, we begin diving in are combined. to some of these ubiquitous chemical reactions.

7.1.2 The Law of Conservation of Mass in Chemical Reactions French chemist Antoine Lavoisier (Figure 7.3) is regarded as the father of modern chemistry. He was the first to understand that combustion was a chemical reaction involving oxygen. He was also responsible for helping to turn chemistry from a descriptive science into a quantitative one. This was because of the famous tin experiments Lavoisier conducted in 1774. In his tin experiments, Lavoisier investigated the formation of tin oxide from tin, and in the process he 170

Chemical Reactions and Stoichiometry discovered that when tin turns to tin oxide, it does so by reacting with the oxygen in the air. Before the experiments, no one knew what air was composed of. Through the experiments, Lavoisier demonstrated that air is composed of more than one gas, and that oxygen accounts for about 20% of it. Also in 1774, English chemist Joseph Priestley (Figure 7.4) was the first to isolate oxygen, which he accomplished by heating mercury oxide. Priestley assumed he had produced a particularly pure form of air. Lavoisier identified the gas that Priestley had produced as being the same as the gas that reacted with the tin and named the gas oxygen. In the tin experiments, Lavoisier heated tin in sealed vessels so that he could carefully weigh the contents before and after the reaction. He found the weights before Figure 7.3. French chemist and after to be the same. As a re- Antoine Lavoisier (1743–1794). sult of his measurements, Lavoisier demonstrated conclusively the principle now known as the law of conservation of mass in chemical reactions: in any chemical reaction, the total mass of the products equals the total mass of the reactants. This law is the basis for the methods we use to balance chemical equations, a critical step that undergirds all the calculations involving chemical reactions. We consider the procedure for balancing chemical Figure 7.4. English chemist Joseph Priestley (1733–1804). equations momentarily. First, we consider the notation used in writing chemical equations.

7.1.3 Reaction Notation The chemical reaction Joseph Priestley used to isolate oxygen is the following: Δ

2HgO( s ) → 2Hg ( l ) + O2 (g ) To perform this reaction, solid mercury(II) oxide, shown in Figure 7.5, is heated to produce liquid mercury and gaseous diatomic oxygen. The formulas for each of these compounds and elements are written in the equation with their subscripts. Subscripts in chemical equations are part of the chemical formulas that identify the specific compounds involved in a reaction. The coefficients in front of the compound formulas are determined during the process of balancing the equation. The symbols used in a chemical equation convey a lot of information about the reaction. The physical states of each of the compounds are shown in parentheses after the chemical symbol or formula. In addition to s (solid), l (liquid), and g (gas) in the equation above, (aq) is used to indicate a substance in aqueous solution. The Δ symbol (delta, the Greek upper-case D) over the “yields” arrow indicates heating, and if heating to a particular temperature is required, this temperature is written below Figure 7.5. Mercury(II) the arrow. Other symbols may also be placed over the arrow to indicate oxide, HgO. 171

Chapter 7 Symbol

Meaning

→

yields

reaction is reversible; proceeds in both directions

solid

liquid

gas

in aqueous solution

heat

→ or → Δ

→

290°C

2.5 atm

→ Fe

→

heating heat to 290°C reaction occurs at a pressure of 2.5 atm chemical symbol or formula for a catalyst that must be present

reaction conditions. The symbols commonly used are listed in Table 7.1.

7.1.4 Balancing Chemical Equations As you will see throughout this chapter, the chemical equation is a very important tool for analysis of chemical reactions. To write a chemical equation, first the formulas of the reactants and products are written. This is called the formula equation. Using Priestley’s mercury(II) oxide reaction again as an example, the formula equation is Δ

HgO( s ) → Hg ( l ) + O2 ( g ) (formula equation) Once the correct formulas for the reactants and products are present, the equation must be balanced so that it conforms correctly to the law of conservation of mass in chemical reactions. As it stands, the formula above has HgO on the left side, a formula indicating one atom of oxygen. But the O2 on the right indicates a molecule containing two oxygen atoms. The balanced equation for the mercury(II) oxide reaction is Δ

2HgO( s ) → 2Hg ( l ) + O2 (g ) (balanced equation)

The coefficients in front of the HgO and Hg apply to the entire compound they precede. Thus, 2HgO means “two units of HgO, and thus two mercury atoms and two oxygen atoms.” Note that this is similar in meaning to the mathematical expression 2(a + b), but it is completely different in meaning from the expression 2ab. While balancing equations, it is most convenient to read a chemical equation as representing numbers of atoms. Read this way, we can simply count atoms of each element on both sides of the equation and manipulate the coefficients until the equation is balanced. But very importantly, the equation can also be read in terms of the bulk quantity moles. And since most reactions involve bulk quantities of atoms, this is the usual way to read an equation once the balancing is complete. Read this way, the balanced mercury(II) oxide equation reads, “two moles of solid mercury(II) oxide yield two moles of liquid mercury and one mole of diatomic oxygen gas.” Just in case you’ve never seen liquid mercury before, it is shown in Figure 7.6. It may look hot, as if it were molten metal, but it is not. It is one of only two elements that are liquids at room temperature. (The other is bromine.) When I was in high school, we were allowed to play with Figure 7.6. Pouring liquid mercury. Table 7.1. Symbols used in chemical equations.

172

Chemical Reactions and Stoichiometry mercury in our hands. It is almost as dense as lead (look at the periodic table), so a blob of mercury feels unbelievably heavy in your hand. Alas, however, playing with mercury is no longer allowed. Once scientists discovered that it is absorbed through the skin, is quite toxic, can enter the body by breathing the vapor, and that its effects are cumulative, the party was over. With prolonged exposure, mercury destroys brain cells and as a result, people lose brain function. Thus the expression, “mad as a hatter.” Hat makers in the old days used mercury to shape the felt hats they made. Not good. Oh, and please make sure you and your family don’t toss used fluorescent lamps or compact fluorescent lamps in the trash. They all contain mercury. Take them to a recycle center such as one of those giant hardware stores. Not kidding. Now, back to balancing equations. There are some key points to attend to when writing down the formula equation. Once those are satisfied, then the equation may be balanced. 1. All formulas for compounds must be correct, with the Figure 7.7. A sample of sulfur (above) and correct subscripts. a model of cyclooctosulfur, S8. 2. For pure elements in an equation, there are several cases when a pure element appears in molecular form, and has a subscript. Seven of these are the seven diatomic substances: H2, N2 O2, F2, Cl2, Br2, and I2. Sulfur and phosphorus form molecules with themselves in the natural state and appear as the molecules S8 and P4. The most common form of sulfur, for example, takes the form of the eight-atom ring called cyclooctosulfur. A sample of sulfur and the cyclooctosulfur ring are shown in Figure 7.7. And since we are talking about sulfur, I can’t resist showing the beautiful images of burning sulfur in Figure 7.8. When it burns, sulfur melts into a blood-red liquid. The flame is bright blue, but cannot be seen in the daytime photo. The second photo, made in the dark, shows it well. 3. The equation must contain all the reactants and products that are part of the reaction that actually occurs. The process of balancing a formula equation is the process of adding coefficients in front of either reactants, products, or both until the equation displays equal numbers of atoms of each element involved on both sides of the equation. For the relatively simple reactions we consider in this chapter, balancing is performed by inspection and trial and error, although there are some guidelines that help expedite the process. Let’s consider the guidelines, and then move to examples. Here they are: 1. If an element appears in an odd number on one side of the equation and an even number on the other, you need to make them both even. Start by placing a coefficient Figure 7.8. Burning sulfur in the daylight (usually 2) on the formula with the odd number. (above) and in the dark (below).

173

Chapter 7 2. Usually, treat polyatomic ions in formulas as single units. It is easier to count sulfates than it is to count individual sulfur and oxygen atoms. However, if the polyatomic ion does not appear on both sides of the equation, you must count individual atoms. 3. Tackle the more complicated formulas first. Save placing coefficients on elements that appear alone in the equation (such as the Hg and O2 terms in the mercury(II) oxide reaction) until last. Placing a coefficient on one element doesn’t affect anything else. Placing coefficients on compounds affects two or more elements simultaneously. 4. When you finish balancing, make sure there is no common multiple (other than one) in all the coefficients in an equation. So for example, if all your coefficients are even, divide them all by two. We now illustrate the process of balancing equations with a number of examples. Example 7.1 Write the formula equation and balanced equation for the following reaction: zinc metal is combined with hydrochloric acid in aqueous solution. The reaction produces a solution of zinc chloride and hydrogen gas. We begin by writing the formula equation. Hydrochloric acid is HCl. From Figure 4.9, the oxidation state of zinc is Zn2+, so the formula for zinc chloride is ZnCl2. Zn(s) + HCl(aq) → ZnCl 2 (aq) + H 2 (g )

(formula equation)

Notice that in the reactants there is a single atom of chlorine and a single atom of hydrogen, but in the products there are two of each. We resolve this odd-even problem by placing a coefficient of 2 on the HCl. Zn(s) + 2HCl(aq) → ZnCl 2 (aq) + H 2 (g )

(balanced equation)

This appears to have balanced the equation, but count to make sure. Zn: one on the left, one on the right. H: two on the left, two on the right. Cl: two on the left, two on the right. The equation is balanced.

Example 7.2 One of the forms rust can take is iron(III) oxide (Figure 7.9). Write the formula equation and balanced equation for the combination of iron with oxygen to produce rust. Oxygen is a diatomic gas, so it appears as O2 in the reactants. Fe(s) + O2 (g ) → Fe 2O3 (s)

(formula equation)

We have the odd-even problem with both Fe and O. Oxygen, with two on the left and three on the right, is more compli- Figure 7.9. Rust on iron. cated, so we start there. To make the oxygen match, we need the least common multiple of two and three, which is six. To get six on each side, we need a coefficient of 3 on the left and 2 on the right. Note that the coefficient on the right goes in front of the entire compound formula. 174

Chemical Reactions and Stoichiometry Fe(s) + 3O2 (g ) → 2Fe 2O3 (s) The oxygens are balanced now, but the iron is not. There are four iron atoms represented in the product. Since iron is by itself in the reactants, a simple coefficient on the iron finishes the balancing to give us 4Fe(s) + 3O2 (g ) → 2Fe 2O3 (s)

(balanced equation)

We now double check the count of each atom to verify that the equation is balanced.

Example 7.3 Aluminum sulfate (Figure 7.10) is used in water treatment plants to remove impurities in the water. When combined in aqueous solution with calcium hydroxide, the reaction produces aluminum hydroxide and calcium sulfate, neither of which are soluble in water. The two solids that are formed attract particulate impurities in the water, and as they settle, the impurities are taken with them. Write the formula equation for this reaction and balance it. The formula equation is straightforward, given the oxidation states of aluminum and calcium (Figure 4.9) and the Figure 7.10. Aluminum sulfate. polyatomic ion list (Table 5.6). Al 2 ( SO4 )3 ( aq ) + Ca ( OH )2 ( aq ) → Al( OH )3 ( s ) + CaSO4 ( s ) (formula equation) There are two polyatomic ions that appear on both sides of this equation. We treat these as single objects during the balancing. The aluminum sulfate is as good a place to start as any. There are three sulfates on the left and one on the right, so we place a coefficient of 3 on the calcium sulfate. Al 2 ( SO4 )3 ( aq ) + Ca ( OH )2 ( aq ) → Al( OH )3 ( s ) + 3CaSO4 ( s ) Calcium is now out of balance, so we address that with a coefficient of 3 on the calcium hydroxide. Al 2 ( SO4 )3 ( aq ) + 3Ca ( OH )2 ( aq ) → Al( OH )3 ( s ) + 3CaSO4 ( s ) Now we look at the hydroxides. We have six on the left, so a coefficient of 2 on the aluminum hydroxide gives us six on the right. Al 2 ( SO4 )3 ( aq ) + 3Ca ( OH )2 ( aq ) → 2Al( OH )3 ( s ) + 3CaSO4 ( s ) (balanced equation) By adding the coefficient of 2 on the aluminum hydroxide we see that the aluminum becomes balanced at the same time as the hydroxide. We now double check the count of each atom and polyatomic ion to verify that the equation is balanced.

175

Chapter 7 Example 7.4 On the opening page of this chapter, the first test tube contains a solution of white copper(I) iodide precipitate in an aqueous solution of iodine (shown again in Figure 7.11). These products are actually formed in a second reaction that is part of a two-step reaction sequence. The sequence begins with combining aqueous solutions of copper(II) nitrate and potassium iodide to produce an aqueous solution of potassium nitrate and solid copper(II) iodide. The copper(II) iodide immediately decomposes to form white copper(I) iodide precipitate and iodine in aqueous solution. Write the formula equations for each of these reactions and balance them. The formula equation for the first reaction is as follows: Cu(NO3 )2 (aq) + KI(aq)→ KNO3 (aq) + CuI2 (s)

(formula equation)

We start with the nitrate. There are two on the left and one on the right, so we add a coefficient of 2 on the potassium nitrate. Cu(NO3 )2 (aq) + KI(aq)→ 2KNO3 (aq) + CuI2 (s) Figure 7.11. White

Now we see that the potassium and iodine are both single on the left and copper(I) iodide in a double on the right. Adding a coefficient of 2 on the potassium iodide takes yellow iodine solution. care of both of them and gives us the balanced equation. Cu(NO3 )2 (aq) + 2KI(aq)→ 2KNO3 (aq) + CuI2 (s)

(balanced equation)

Be sure to double check the count of each atom and polyatomic ion to verify that the equation is balanced. The formula equation for the second reaction is CuI2 (s) → CuI(s) + I2 (aq)

(formula equation)

With three iodines on the right, the thing to do is double the single one so we have an even number on the right. This gives CuI2 (s) → 2CuI(s) + I2 (aq) We now have four iodines on the right and two on the left. Placing a coefficient of 2 on the CuI2 balances the iodine and the copper at the same time. 2CuI2 (s) → 2CuI(s) + I2 (aq)

(balanced equation)

Finally, double check the count of each atom and polyatomic ion to verify that the equation is balanced.

176

Chemical Reactions and Stoichiometry

7.1.5 Oxidation States We introduced the concept of oxidation states back in Chapter 4 in the context of ionization. It is now time to develop this further so that you are able to determine the oxidation state of any element in any compound. When an atom, such as a metal, loses one or more electrons, we say it has been oxidized. When an atom gains one or more electrons we say it has been reduced. The terms oxidation and reduction go back to the days of Antoine Lavoisier, just a few years before the French Revolution. At the time, it was thought that oxygen was the only element that caused oxidation, hence the name of the process. Lavoisier coined the term reduction, referring to the weight loss of metals during smelting. In this process, ore is converted to pure metal by the removal of oxides and other substances (giving the metal atoms their electrons back). The resulting pure metal weighs less than the ore, so it has been “reduced.” We will use sodium and chlorine to illustrate oxidation and reduction. If an atom of sodium is ionized and loses its one valence electron to a chlorine atom, it is oxidized to become Na+ and its oxidation state is +1. Chlorine is the oxidizer or oxidizing agent in the process. When the chlorine atom receives that electron, the chlorine atom is reduced to become Cl– and its oxidation state is –1. Sodium is the reducer or reducing agent. These two ions join together by electrostatic attraction to become part of the crystal lattice of NaCl. If this chemistry is driven in the opposite direction (which can be done by running an electric current through molten sodium chloride), the Cl– ion loses an electron to become Cl; the chlorine ion is now the one that is oxidized. The sodium ion receives that electron to become Na, so the sodium ion is reduced. This is what happens during smelting: the metal atoms in the ionic compounds in the rock get their electrons back, cease to be ions, and become elemental metal. A tried and true mnemonic to help you remember oxidation and reduction is shown in Figure 7.12. The oxidation state of an element is indicated by an oxidation number such as +2, +1, 0, –2, –1, and so on. The oxidation state of any neutral unbonded atom is 0. As illustrated just above, in ionic compounds the oxidation state of an element is directly related to the number of electrons the element has gained or lost, and the oxidation number is simply the charge on the ion. This charge is usually obvious because of the element’s position in the periodic table, or because it is indicated in the name of the compound. In magnesium chloride, MgCl2, the oxidation number of magnesium is +2 because as an alkaline-earth element, magnesium ionizes by losing two electrons. The oxidation number of chlorine is –1 because chlorine, like all the halogens, ionizes by gaining one electron. In copper(I) iodide, CuI, the Roman numeral in the name indicates the oxidation state of the metal (+1). However, in covalent bonds where electrons are being shared, determining the oxidation state of an element is not as simple as keeping track of gaining or losing electrons. Instead, we use electronegativities and a list of rules of thumb to determine oxidation states. In such compounds, we think of the oxidation state as indicating partial gain or loss of electrons. Because of this, it is probably more helpful to you if you think of oxidation and LEO the lion says GER. reduction this way: oxidation occurs any time the oxidation state of an element increases; reduction occurs any time the oxLEO: Lose Electrons Oxidize idation state of an element decreases. This definition also aids GER: Gain Electrons Reduce your memory—you can associate the term reduction with a decrease in the oxidation state (and the oxidation number). Here are the rules we follow to determine the oxidation Figure 7.12. A classic memory aid for state of elements in covalent compounds: oxidation and reduction. Say it over and over to yourself. Say it now.

177

Chapter 7 1. The operating rule is this: in a molecular compound, the oxidation states of the elements involved must add up to the charge on the molecule. In a neutral molecule, the oxidation numbers must add up to zero. In a polyatomic ion, they must add up to the charge on the ion. 2. In pure elements in their natural state, the oxidation number is 0. This also applies to diatomic molecules such as H2 and the other elements that naturally make molecules with themselves, S8 and P4. 3. The most electronegative element in the molecule has the oxidation state it would have if it were an anion. Since fluorine has the highest electronegativity, its oxidation state is always –1. Oxygen has the second-greatest electronegativity, so it is usually assigned an oxidation state of –2. One exception is when oxygen forms peroxides as the molecule O22–, in which case its oxidation state is –1. A common peroxide is hydrogen peroxide, H2O2. Another exception is when oxygen bonds with fluorine to form OF2, in which case its oxidation state is +2 (since F is always –1). 4. The oxidation number for halogens is –1. The exception is when they bond with oxygen to form polyatomic ions with negative charge (oxyanions). In that case, they have positive oxidation states (except for fluorine). 5. When hydrogen forms a compound with an element that is more electronegative than itself (a nonmetal), its oxidation state is +1. In compounds with metals, the oxidation state of H is –1. Example 7.5 Determine the oxidation states for each atom in the following compounds or ions: MgCl2, SF6, BaO2, HClO3, PO43–. MgCl2 is an ionic compound. The oxidation numbers for Mg and Cl are thus +2 and –1, respectively. These add up to zero: 2 + 2(–1) = 0. In SF6, fluorine has an oxidation number of –1. Since there are six fluorine atoms in the molecule, the contribution to the overall charge of the molecule is 6(–1) = –6. Since the overall charge on the molecule is zero, the oxidation numbers must add up to zero. Thus, the oxidation number of sulfur is +6. Barium peroxide, BaO2, is a nontypical ionic compound—it is a peroxide. We would normally expect Ba and O to form BaO since their typical ionizations are Ba2+ and O2–. But in the compound BaO2, the oxidation state of the metal is the one that runs the show. The oxidation number for Ba is +2 so the oxidation number for O must be –1. Oxygen is not very happy in peroxides, so the compounds are unstable and like to go boom. See the accompanying box for more about that exciting subject. Chlorine and oxygen are both more electronegative than hydrogen, so in HClO3 the oxidation state of H is +1. The chlorate ion, ClO3–, is an oxyanion, so we expect the chlorine to have a positive oxidation state. The oxidation state of oxygen is –2 and there are three of them, so the charge contribution from oxygen is 3(–2) = –6. Adding the hydrogen gives –6 + 1 = –5. For the oxidation numbers in the molecule to add to zero, the oxidation state of chlorine must be +5.

178

Chemical Reactions and Stoichiometry

Hmm... Interesting.

Why nitrates and nitros blow up

To a chemist, any substance that causes an element to oxidize is an oxidizing agent, or oxidizer. The oxygen in air causes iron to oxidize and form rust, Fe2O3, and is an oxidizing agent in this general chemical sense. But a more narrow definition of an oxidizing agent is used in classifying substances as dangerous materials. In such classifications, an oxidizing agent is a substance that causes other substances to combust rapidly. In barium peroxide, BaO2, the oxidation state of oxygen is –1, whereas oxygen’s preferred oxidation state, the state in which its valence shell is full, is –2. This means the compound is unstable and the oxygen in BaO2 aggressively reacts with other substances. BaO2 is used in fireworks to produce an intense green color. Hydrogen peroxide is also an aggressive oxidizer and is sometimes used as a rocket fuel. It was also the oxidizer used in the 2005 London bombings that killed 52 people. But don’t worry about the hydrogen peroxide solution you may have in the medicine chest at home. That’s only a 3% aqueous solution of H2O2. (Researchers now say that while hydrogen peroxide solution is useful for disinfecting inanimate objects and removing blood stains, it is not the best for disinfecting wounds because it causes tissue damage and thus slows healing.) Potassium nitrate, KNO3, is the oxidizer used in gunpowder. Nitrate compounds in general are aggressive oxidizers. Ammonium nitrate, NH4NO3, a common fertilizer, was used as an explosive in the 1995 Oklahoma City bombing (which killed 168 people) and the 1993 World Trade Center bombing (which killed 6 people but injured several thousand). (The WTC bombing occurred before the WTC was destroyed by terrorists on September 11, 2001.) Ammonium nitrate is also the substance that exploded and CH3 destroyed a fertilizer plant in West, Texas in 2013, killing 15 people. In NO2 the 1947 Texas City disaster, a ship containing 2,300 tons of ammonium O2N nitrate detonated, killing 581 people—the deadliest industrial accident in U.S. history. When it comes to explosive compounds, so-called nitro compounds NO2 are the most explosive of all. Nitro compounds include TNT, nitroglycerine (which can explode just by being jostled), and the stabilized form of nitroglycerine called dynamite. Nitro compounds contain carbon rings with nitronium ions (NO2+) attached, as illustrated by the Lewis structure for TNT, trinitrotoluene, shown above. Nitro compounds are not oxidizers, but are explosive for the same reason the nitrates are: because of the nitrogen and oxygen atoms they contain. When the nitrogen atoms are freed from the molecule, they form diatomic molecules of nitrogen gas, N2. The oxygen atoms also form gases during reactions, typically CO2, but also CO. The N2 (and CO) molecules contain triple bonds, which means they are very strong and release a lot of energy while forming (see Table 5.10). The presence of a lot of heat while gases are forming means the gases expand very rapidly. Put all this together and you have huge explosions. The reason the nitro compounds explode more vigorously than the nitrate oxidizers is that the nitro compounds have multiple carbons (which get oxidized in the explosion, releasing additional heat), nitrogens, and oxygens in the same molecule. You can’t get any closer together—thus faster—than that. One more interesting tidbit: dynamite was invented by Alfred Nobel, and the fortune he made from it was used to create the Nobel Foundation, which awards the annual Nobel Prizes. 179

Chapter 7 Oxygen is more electronegative than phosphorus, so its oxidation state in PO43– is –2. The four oxygens make a charge contribution of 4(–2) = –8. The oxidation numbers must add up to the charge on the ion, which is –3. Thus, the oxidation state of phosphorus is +5.

7.2

General Types of Chemical Reactions

7.2.1 Synthesis Reactions In synthesis reactions, separate elements or compounds are combined to form a single, new compound. Synthesis reactions are described by the general equation form A + B → AB The classic example of a synthesis reaction is the formation of water by the combustion of hydrogen: 2H 2 (g ) + O 2 (g ) → 2H 2O(g ) Another example is the production of calcium hydroxide from calcium oxide: CaO(s) + H 2O(l) → Ca(OH)2 (s) Calcium hydroxide, shown in Figure 7.13, is known as “slaked lime” for the same reason people use the phrase Figure 7.13. Calcium hydroxide. “slake your thirst”: both are accomplished by adding water. The production of Ca(OH)2 from CaO is an important step in the curing of concrete. Because of its environmental importance, an important synthesis reaction to know about involves the production of acids in the atmosphere, which leads to acid rain. Sulfur dioxide, SO2, is a common product of the combustion of coal in electric power stations. In the atmosphere, SO2 combines with moisture to produce sulfur trioxide, SO3. This compound reacts easily with moisture in the air to produce sulfuric acid: SO3 (g ) + H 2O(l) → H 2SO4 (aq)

Figure 7.14. A forest destroyed by acid rain.

180

Acidic rainfall was first detected back in the 17th century, but after the Industrial Revolution, atmospheric levels began to skyrocket. By the late 1960s, forests like the one in Figure 7.14 were being destroyed, as well as outdoor stone sculptures and architectural works made of marble or stone. New legislation to control emissions of SO2 and other pollutants from power stations went into effect in the 1980s and 1990s, and as a result acid rain levels have dropped by 65% since 1976 (for a total cost much lower than originally predicted).

Chemical Reactions and Stoichiometry

7.2.2 Decomposition Reactions In synthesis reactions, compounds or elements are brought together; in decomposition reactions, compounds are taken apart. The general equation form is as follows: AB → A + B The decomposition reaction you see happening most often is probably the decomposition of carbonic acid, H2CO3: H 2CO3 (aq) → H 2O(l) + CO 2 (g ) Carbonic acid, an oxyacid, is added to soft drinks to make them fizzy when the can is opened. At atmospheric pressure, carbonic acid spontaneously decomposes to water and carbon dioxide. The CO2 that evolves (i.e., is given off) during the reaction is the gas that makes the drink fizzy. The heating of mercury oxide by Joseph Priestley, mentioned at the beginning of the chapter, is a decomposition reaction in which oxygen evolves: Δ

2HgO( s ) → 2Hg ( l ) + O2 (g ) A third example goes back to the calcium oxide synthesis reaction mentioned above. To obtain the CaO in the first place, limestone, which is calcium carbonate, CaCO3, is roasted (that’s what they call it) at high temperature, producing the following decomposition reaction: Δ

CaCO3 (s) → CaO(s) + CO2 (g )

This is the starting point for the manufacture of “Ordinary Portland Cement,” the most common cement in the world.

7.2.3 The Activity Series of Metals To understand single replacement reactions (coming up next), you need to know about the activity series of metals, a list of metals in order of their chemical activity, that is, how aggressively they react relative to one another. The activity series is shown in Table 7.2. The metals in this list are listed from top to bottom in order of their chemical activity, with the most chemically aggressive metals at the top and the metals that react the least at the bottom. In aqueous solutions of compounds of these metals, any metal on the list replaces any metal below it in a solution. Figure 7.15 illustrates this with a solution of silver nitrate, AgNO3, which is colorless. When a coil of copper wire is placed in the solution, the copper begins going into solution, which turns the solution blue-green. The silver comes out of solution, precipitating in fluffy tufts onto the copper wire. The chemical activity of the metals was determined empirically, that is, through experiments. Using the information on the list, you can determine if certain reactions occur. Table 7.2 also indicates the reactivity of the metals with water, oxygen, and acids. Active metals like Figure 7.15. In a solution of silver nitrate, the lithium and sodium react very vigorously in water. All copper replaces the silver in solution. 181

Chapter 7 Activity of Metals

Activity with Water

Activity with Oxygen

Activity with Acids

Example 7.6

Rb K Ba

reaction

reaction (to form oxides)

reaction (replacing hydrogen)

Ca Na Mg Al Mn Zn Cr

reaction with steam, but not cold water

reaction

Fe Cd Co Ni Sn

no reaction

reaction

no reaction

reaction

no reaction

Pb Sb Bi Cu Hg no reaction

no reaction

Au Table 7.2. The activity series of metals and other activity information.

empty test tube with nail

182

What do you expect to happen if a steel nail is placed in a container of copper(II) chloride solution? Steel is mostly iron and iron is above copper in the activity series of metals. When placed in the solution, ions of iron begin going into solution and copper ions begin precipitating out of solution by attaching to the nail.

The phenomenon described in the previous example is shown in the sequence of images in Figure 7.16. After about 30 minutes, the solution becomes colorless, indicating that all the copper ions have been reduced to copper atoms and have precipitated out of solution. The reaction is a single replacement reaction, our next topic.

7.2.4 Single Replacement Reactions

Ag Pt

metals except the most unreactive react with oxygen to form oxides. Most metals also react with acids.

CuCl2 solution added

In single replacement reactions, a lone element replaces another in a Figure 7.16. Iron displaces the copper from solution, and the nail is buried in copper.

Cu precipitation underway Cu precipitation complete

Chemical Reactions and Stoichiometry compound. Typically, this happens in aqueous solutions, and may be generically represented by the equation A + BC → B + AC Three common single replacement reactions are: one metal replacing another in solution, as in the copper(II) chloride example above; a metal replacing some of the hydrogen atoms in water to form a hydroxide solution; and a metal replacing the hydrogen atoms in an acid to form a salt. Iron replacing copper in solution in Figure 7.16 is a typical example of the first type. The equation for the reaction is as follows: Fe( s ) + CuCl 2 ( aq ) → Cu ( s ) + FeCl 2 ( aq ) In this single replacement reaction, the copper(II) chloride is in aqueous solution. As we explore further in Chapter 10, this means the copper and chlorine ions have dissociated in the water— the crystal comes apart and the individual ions are separated from each other. Iron is a more active metal than copper, so when the steel nail is placed in the solution, the Fe atoms are oxidized to become Fe2+ ions in solution. The Cu2+ ions in the solution are reduced to become Cu atoms, copper metal. The steel nail provides a landing place where atoms of solid Cu can attach to a crystal lattice as they precipitate. The reaction of sodium in water is a well-known example of the second type of single replacement reaction. The equation is: 2Na ( s ) + 2H 2O( l ) → 2NaOH ( aq ) + H 2 ( g )

Figure 7.17. Magnesium replaces the hydrogen in the hydrochloric acid molecules, forming a Sodium is high in the activity series, and reacts very salt and evolving hydrogen gas. The photo on vigorously in water. The hydrogen gas evolves quick- the left is at the beginning of the reaction, when ly and a lot of heat is released—so much, in fact, that H2 is rapidly evolving. The photo on the right the hydrogen easily combusts, causing an explosion. was taken after about 5 minutes.

The third common single replacement reaction is when a metal replaces the hydrogen in an acid. A typical example is the formation of the metal salt magnesium chloride when magnesium is placed in hydrochloric acid: Mg ( s ) + 2HCl ( aq ) → MgCl 2 ( aq ) + H 2 ( g ) This reaction is shown in Figure 7.17. The bubbles you see are the evolving hydrogen gas. As you see from Table 7.2, every metal in the activity series from lithium down through lead reacts this way with acids. The last seven metals listed generally do not. However, both platinum and gold dissolve in a mixture of acids known from the medieval period as aqua regia, Latin for “royal water.” This extremely corrosive substance, shown in Figure 7.18, is composed of one part nitric acid and two or three parts hydrochloric acid.

Figure 7.18. Aqua regia.

183

Chapter 7

Hmm... Interesting.

A story about aqua regia

After 1935, Adolph Hitler prohibited Germans from accepting Nobel Prizes. German physicists Max von Laue and James Franck sent their Nobel Prize gold medals to Niels Bohr at the Bohr Institute in Denmark for safe keeping. In 1940, Germany invaded Denmark. Fearing that the Nobel Prize medals might be confiscated by the Nazis, Bohr directed Hungarian chemist George de Hevesy to dissolve the medals in aqua regia, and place the flask of red liquid with the gold in it on a shelf in the lab. The Nazis never knew what was in that flask and after the war it was still sitting there. The Nobel Foundation took back the gold and recast it into two Nobel Prize medals. Here is one final type of single replacement reaction: the replacement of one halogen by another. The chemical activity of the halogens is highest with fluorine and decreases down Group 17 in the periodic table. Thus, any halogen replaces any of the ones below it. This is illustrated by fluorine gas replacing the chlorine atoms in sodium chloride: F2 ( g ) + 2NaCl ( aq ) → 2NaF ( aq ) + Cl 2 ( g ) When the fluorine replaces the chlorine in solution, chlorine gas evolves.

7.2.5 Double Replacement Reactions Double replacement reactions may be represented by the following generic equation: AB + CD → AD + CB In this type of reaction, the four ions in two compounds all “switch partners.” Notice that in the product ionic compounds in this generic equation, the cations A and C are still written first. Double replacement reactions often result in the formation of a precipitate or the evolution of a gas when product compounds are insoluble in water. One example is the reaction discussed in Example 7.4. Another, also involving copper(II) nitrate, is the following reaction: Cu ( NO3 )2 ( aq ) + H 2S ( aq ) → CuS( s ) + 2HNO3 ( aq ) The copper monosulfide1 product is a black powder that is insoluble in water, so it forms a precipitate that turns the solution black, as shown in the third test tube from the left in the image on the opening page of this chapter, and in Figure 7.19. The evolution of a gas in a double replacement reaction is illustrated by the reaction of iron(II) sulfide and hydrochloric acid: FeS( s ) + 2HCl ( aq ) → H 2S ( g ) + FeCl 2( aq ) Notice that here, just as in the Mg and HCl single replacement reaction, a metal replaces hydrogen in an acid. In the single replacement reaction, a sample of pure metal is placed in the acid solution. Here, an ionic compound of the metal is placed in the acid solution. 1 There are quite a few different copper sulfide compounds. To distinguish CuS from the others, the prefix mono- is added to the sulfide in the naming of the compound. 184

Chemical Reactions and Stoichiometry

7.2.6 Combustion Reactions Combustion reactions involve the reaction of a fuel with oxygen to produce carbon dioxide and water. Combustion reactions are not actually a separate class of reactions; they are oxidation-reduction reactions (discussed below), and some are also synthesis reactions. But they are so common and important that they warrant their own section here. The natural gas commonly burned in homes for heating and cooking is primarily composed of methane, CH4. The methane combustion reaction is as follows: CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H 2O( g ) The combustion of most fuels, such as methane, propane, and all other hydrocarbon fuels, produces carbon dioxide and water. One important exception is the combustion of hydrogen, a synthesis reaction: 2H 2 ( g ) + O2 ( g ) → 2H 2O( g ) Burning hydrogen produces nothing at all but pure water vapor, and this makes hydrogen the ultimate clean fuel. If we could develop efficient and safe ways to prepare, transport, and burn hydrogen, pollution and CO2 emissions from cars could be eliminated (although the air in cities would be somewhat more humid).

7.2.7 Acid-Base Neutralization Reactions

Figure 7.19. The black precipitate copper monosulfide forms when copper nitrate combines with hydrogen sulfide.

In Chapter 11, we explore what acids and bases are and look with more detail into acid-base chemistry. However, acid-base neutralization reactions are so important that they warrant a brief mention here. Careful definitions for the terms acid and base must wait until Chapter 11. However, we can say roughly that acids are recognizable as compounds with hydrogen and a halogen, or hydrogen and one of the polyatomic ions. Examples are HCl and H2SO4. Equally roughly, bases are recognizable as compounds composed of a metal and hydroxide such as NaOH and Ca(OH)2. Though it may be tricky to define acids and bases, it is not tricky to define what you get when an acid reacts with a base: you get a salt, and often nothing else but water. In fact, that is the definition for a salt—the product of an acid-base reaction. Acid-base chemistry typically takes place in aqueous solution, and is so indicated in the chemical equations. The reaction between an acid and a base is referred to as neutralization. Both acids and bases are corrosive. But when an acid and a base are combined in chemically equivalent ratios, their corrosive properties are neutralized as they are converted into a salt and water. A good example of an acid-base reaction involves the acid in the human stomach, gastric acid, HCl. Gastric acid plays a key role in food digestion by causing the large protein molecules in the foods we eat to unravel so that the digestive enzymes in the stomach can break down the molecules in the food. However, an unbalanced diet sometimes leads to excess gastric acid, a condition known as acid indigestion or heartburn. For minor symptoms, over-the-counter antacids can be taken to neutralize excess gastric acid. Rolaids, a popular antacid, contains magnesium hydroxide. The gastric acid neutralization reaction from taking Rolaids is as follows: 2HCl ( aq ) + Mg ( OH )2 ( aq ) → MgCl 2 ( aq ) + 2H 2O( l ) 185

Chapter 7 The magnesium chloride salt formed from this reaction is composed of the cation from the base and the anion from the acid. This is always the case in acid-base neutralization reactions. Tums, another popular antacid, contains calcium carbonate, which, you may recall, is the same compound that composes chalk and limestone. Even though CaCO3 does not contain the hydroxide ion, it still acts as a base to neutralize the HCl in the stomach. The gastric acid neutralization reaction from taking Tums is as follows: 2HCl ( aq ) + CaCO3 ( aq ) → CaCl 2 ( aq ) + H 2O( l ) + CO2 ( g ) The carbonate base in this reaction results in the production of CO2.

7.2.8 Oxidation-Reduction Reactions As with acid-base reactions, oxidation-reduction reactions, or redox reactions as they are fondly known, are so important that we treat them separately in a later chapter. But since we are talking about different types of chemical reactions here, we need to include a brief look at redox reactions. Any time an element’s oxidation state increases, oxidation occurs. And any time oxidation occurs, reduction occurs as well. These two sentences apply in reverse as well. If an element’s oxidation state decreases, reduction occurs, and oxidation also occurs at the same time. Combustion reactions are always redox reactions because oxidation always occurs. (The oxidation is caused by the oxidizer, or oxidizing agent, which we discussed a few pages back.)For example, here is the reaction for the combustion of propane, C3H8: C 3H8 ( g ) + 5O2 ( g ) → 3CO2 ( g ) + 4H 2O( g ) Let’s look at the oxidation states of the carbon in propane and carbon dioxide. The electronegativity of carbon is higher than that of hydrogen, so hydrogen’s oxidation state is +1 in propane. There are eight hydrogen atoms in propane, giving a total charge contribution from hydrogen of 8(+1) = +8. The oxidation numbers must all add up to zero, which means carbon’s oxidation state in propane, x, must solve the equation: 3x + 8 = 0, giving an oxidation state of –8/3. In carbon dioxide, the oxidation state of oxygen is –2, so the oxidation state for carbon must be +4. Thus, the oxidation state of carbon increases from –8/3 to +4. The carbon is oxidized. The oxidizing agent in this reaction is O2. In redox reactions, the oxidizing agent is the compound that causes the oxidation of another compound in the reaction. The result for the oxidizer itself is that it is reduced. We can easily see this in the chemical equation because as O2, the oxidation state of oxygen is 0, but its oxidation state in both CO2 and H2O is –2. The oxygen is reduced. Note that although the carbon is the element oxidized in this reaction, it is common parlance to say that the propane is oxidized. I know this sounds confusing, but it is just the way everyone talks, so students must simply get used to it. At first, this kind of language will drive you crazy because you will be wondering about specifically which element is the one that is oxidized in the reaction. To preserve your sanity, just sit down and figure it out like we did above with the carbon in propane. It takes a lot of practice for one’s intuitions about oxidation and reduction to become automatic. Just stay with it. Tossing a metal into an acid solution also produces a redox reaction. Let’s look again at the Mg and HCl reaction from a few pages back: Mg ( s ) + 2HCl ( aq ) → MgCl 2 ( aq ) + H 2 ( g )

186

Chemical Reactions and Stoichiometry Here the pure metal Mg is oxidized to Mg2+ to form the ionic compound MgCl2. The oxidizing agent is the hydrochloric acid, but here we should note that more specifically it is the hydrogen in the HCl that is reduced. In HCl, hydrogen’s oxidation state is +1; in H2 it is 0. Thus, the hydrogen is reduced. The number of different redox reactions is beyond telling. We look at a few more of them in the redox chapter.

7.3

Stoichiometry

Everyone who has ever taken a chemistry class remembers performing stoichiometric calculations. The term stoichiometry comes from the Greek words stoicheion, meaning “element,” and metron, meaning “measure.” When we do stoichiometry, we are measuring the elements. There are many different basic types of calculations we perform in chemistry, but stoichiometric calculations might just be the most basic of the basic. And here’s the good news: even though this section has a long, strange foreign name, the calculations themselves are easy. You really have to know only four things to do stoichiometric calculations. First, you need to be able to set up and solve a proportion, a basic math skill you no doubt learned how to do years ago in Prealgebra. Second, you need to be able to compute the molar mass of a compound. This topic is covered in Chapter 2 and should not be difficult at this point. Third, you need to be able to convert from moles to grams and vice versa. This, too, is covered in Chapter 2 and is not difficult. Finally, you need to be able to set up and balance a chemical equation, which we have just covered and which you have (no doubt) just mastered. So now, with your confidence meter reading “high,” let us proceed.

7.3.1 Stoichiometric Calculations Stoichiometry is all about calculating the quantities of compounds involved in chemical reactions. For example, just look back at previous page at the reaction of C3H8 and O2. If we have, say, 10.0 kg of propane, how much oxygen is required to burn it all? How much CO2 and water are formed in the process? These calculations are standard stoichiometry. Let’s begin with RULE NUMBER 1. RULE NUMBER 1 Perform stoichiometric calculations in moles. If the problem asks for masses in grams, or supplies you with masses in grams, that’s fine. If your given quantities are masses in grams, you simply convert given masses into number of moles, as you already know how to do. If you are required to state your answer as a mass in grams, you just convert to grams at the end. But the stoichiometric calculation is performed in moles. Now for RULE NUMBER 2. RULE NUMBER 2 The mole ratios for performing the stoichiometric calculations come from the coefficients in the balanced chemical equation. That’s right. That’s why this is easy. Remember that while we are balancing a chemical equation, we think of the coefficients as helping us to figure numbers of atoms. But remember also that earlier in this chapter, in Section 7.1.4 on balancing chemical equations, I write that chemical reactions happen with bulk quantities of atoms, and in chemistry our favorite bulk quantity is moles. So think of the coefficients in the balanced equation as telling us the numbers of moles of compounds or elements that participate in the reaction. Now, thinking in moles, look again at that propane equation on the previous page. The coefficients in the equation say this: one mole of propane reacts with five moles of oxygen to produce 187

Chapter 7 three moles of carbon dioxide and four moles of water. That’s how you do it. The rest is just the details, which I demonstrate below in a few examples. Before we start in on the examples, let’s briefly consider how to use mole ratios to perform calculations when the quantities you are given do not match the coefficients. (They never do.) Looking again at the propane equation, the equation says that for every five moles of oxygen involved, four moles of water form. If you are given a quantity of oxygen and asked to determine how much water forms, you simply use the 5 : 4 ratio as a conversion factor to figure it out. We can write this conversion factor two ways: 5 mol O2 4 mol H 2O or 4 mol H 2O 5 mol O2 Now here’s how you use one of these ratios to solve a problem. Suppose the problem is to determine how many moles of water are produced by the reaction of 13.55 mol O2 with propane, assuming an unlimited supply of propane. Just take your given quantity and multiply it by the mole ratio that appropriately cancels with the given quantity to give you the quantity you need: 13.55 mol O2 ⋅

4 mol H 2O 13.55⋅4 = mol H 2O = 10.84 mol H 2O 5 mol O2 5

Like I said, easy. Now two more comments before we proceed. First, it should be pretty obvious to you that I write the mole ratio in that computation as I do because I am given moles of oxygen and I want that to cancel to give me moles of water. That’s why the oxygen is on the bottom in the conversion factor. If I am given moles of water and asked to find moles of oxygen, I flip the conversion factor the other way. Second, the coefficients in the chemical equation are exact; they are not approximations. When the equation says 5 mol O2 react to produce 4 mol H2O, these figures are exact. Accordingly, they play no part in determining the significant digits to use in your results. In the computation above, the given quantity of oxygen has four significant digits. That’s why the result is written with four significant digits. Now we are ready to dive into some examples. The following examples should illustrate everything you need to know. Keep Rule Number 1 in mind: perform the computations in moles. If you are given a quantity as a mass, use the compound’s molar mass to convert it to number of moles for the calculation. If you are asked for a result to be stated as a mass, use the molar mass to convert it at the end after you determine the result in moles. Example 7.7 In submarines and spacecraft, the breathing air has to be “scrubbed” of excess carbon dioxide to remove the carbon dioxide continuously being exhaled by those on board. One way to purify the breathing air of excess CO2 is with the compound lithium hydroxide, shown in Figure 7.20. Solid lithium hydroxide reacts with carbon dioxide to produce solid lithium carbonate and water. If the average amount of carbon dioxide exhaled each day is 880 g/person, determine the number of moles of lithium hydroxide consumed per person per day in the reaction. First, the given quantity is in grams, so we convert this to number of moles. Referring to the periodic table, we determine the molar mass of CO2, which is 44.010 g/mol. Converting the given quantity to moles, we have 880 g CO2 ⋅

188

1 mol = 20.0 mol CO2 44.010 g

Chemical Reactions and Stoichiometry We know our final answer must be stated with two significant digits because of the given quantity. I have kept an extra significant digit in the gram/mole conversion because it is an intermediate result. Final rounding is performed at the end. Next, write the formula equation: LiOH( s ) + CO2 ( g ) → Li 2CO3 ( s ) + H 2O( l )

Figure 7.20. Lithium hydroxide.

Then, balance the equation. 2LiOH( s ) + CO2 ( g ) → Li 2CO3 ( s ) + H 2O( l ) Now we are ready for the stoichiometry. In the problem, we are given a quantity of CO2 and asked to find the corresponding quantity of LiOH in the reaction. From the coefficients in the balanced equation, we see that the mole ratio of LiOH to CO2 in the reaction is 2 : 1. So we use this mole ratio as a conversion factor to convert the mole quantity of CO2 into a corresponding mole quantity of LiOH. 20.0 mol CO2 ⋅

2 mol LiOH = 40.0 mol LiOH 1 mol CO2

This is our final result, but it is stated with three significant digits. However, we require two significant digits in our result, and the only way to round 40.0 down to two significant digits is to write it in scientific notation. Thus, our answer is 4.0 × 101 mol LiOH

Example 7.8 One of the body’s metabolic processes—from which we get our fuel—is the oxidation of solid glucose by reaction with the oxygen we get from breathing.2 Glucose, C6H12O6, is one of the fundamental carbohydrates we take in from eating starch foods such as potatoes and rice. The products of the oxidation reaction are carbon dioxide, which we exhale, and water. (We eliminate more water than we take in; the extra water comes from the oxidation of glucose.) Determine the mass of water produced by the oxidation of 2.00 g glucose. Again, we are given a quantity in grams, so we convert this to moles right at the beginning. Using data from the periodic table, the molar mass of glucose is found to be 180.157 g/mol. Converting the given mass of glucose into moles, we have 2.00 g C 6H12O6 ⋅

1 mol C 6H12O6 = 0.01110 mol C 6H12O6 180.157 g C 6H12O6

2 Here is another instance of using the term oxidation in reference to an entire molecule. If you look at the oxidation states, you see that it is the carbon in the glucose that is being oxidized. 189

Chapter 7 Again, I have an extra significant digit in this intermediate calculation to avoid the build up of rounding error. I round to the required three significant digits at the end. Next, we write the formula equation for the oxidation of glucose: C 6H12O6 ( s ) + O2 ( g ) → CO2 ( g ) + H 2O( l ) Now we balance the equation: C 6H12O6 ( s ) + 6O2 ( g ) → 6CO2 ( g ) + 6H 2O( l ) This problem gives us a quantity of glucose and asks about a quantity of water, so these are the two compounds to look at in the equation. From the coefficients in the balanced equation, we see that the mole ratio of glucose to water in this reaction is 1 : 6. Now we write down the given quantity of glucose (in moles) and use the mole ratio to convert this into a mole quantity of water. 0.01110 mol C 6H12O6 ⋅

6 mol H 2O = 0.06660 mol H 2O 1 mol C 6H12O6

Now that we have the amount of water produced, we simply need to use the molar mass of water to convert this into grams as the problem requires. From the periodic table, we determine that the molar mass of water is 18.02 g/mol. Using this value as a conversion factor, we determine the mass of water produced by the reaction: 0.06660 mol H 2O⋅

18.02 g H 2O = 0.06660⋅18.02 g H 2O = 1.20 g H 2O 1 mol H 2O

Notice, finally, that this result is rounded to three significant digits as the problem requires.

It should be apparent from these two examples that a stoichiometric calculation can be performed using any or all of the products or reactants in a reaction. It doesn’t matter which ones are asked about in the problem; you simply set up mole ratios using two compounds (or elements) at a time. We use this principle in our final example. Example 7.9 The photosynthesis reaction in plants is identical to the oxidation of glucose running in reverse. In the plant, CO2 from the air and water from the soil are converted (using sunlight as an energy source, of course) into glucose, which becomes fuel and building material for the plant, and oxygen, which is released into the atmosphere. For each 5.00 g H2O consumed by a plant, determine the masses of the CO2 consumed, the O2 released, and the glucose produced. Since this reaction is identical to the one in the previous problem, only running in reverse, we can write down the balanced equation immediately: 6CO2 ( g ) + 6H 2O( l ) → C 6H12O6 ( s ) + 6O2 ( g )

190

Chemical Reactions and Stoichiometry In this example, we are required to calculate everything, so we may as well calculate all four molar masses right now. Since the mass given in the problem has three significant digits, we write down each of the molar masses with four significant digits. CO2:

44.01 g/mol

H2O:

18.02 g/mol

C6H12O6:

180.2 g/mol

O2:

32.00 g/mol

Next, let’s convert the given quantity from grams to moles. 5.00 g H 2O⋅

1 mol H 2O = 0.2775 mol H 2O 18.02 g H 2O

(This value has the extra significant digit required for an intermediate calculation.) We are given an amount of H2O in the problem, and from this quantity we must calculate quantities of three other substances. So we need mole ratios for water and the three other substances. From the coefficients in the balanced equation, these ratios are H2O : CO2

6 : 6, which is equal to 1 : 1

H2O : C6H12O6

6:1

H2O : O2

6 : 6, which is equal to 1 : 1

Now we just use the given amount of water, in moles, with one of these ratios to compute the quantities of the other three substances. 0.2775 mol H 2O⋅

1 mol CO2 = 0.2775 mol CO2 1 mol H 2O

0.2775 mol H 2O⋅

1 mol C 6H12O6 = 0.04625 mol C 6H12O6 6 mol H 2O

0.2775 mol H 2O⋅

1 mol O2 = 0.2775 mol O2 1 mol H 2O

The last step is to use the molar masses to convert each of these mole quantities into grams.

191

Chapter 7 0.2775 mol CO2 ⋅

44.01 g CO2 = 12.2 g CO2 1 mol CO2

0.04625 mol C 6H12O6 ⋅

0.2775 mol O2 ⋅

180.2 g C 6H12O6 = 8.33 g C 6H12O6 1 mol C 6H12O6

32.00 g O2 = 8.88 g O2 1 mol O2

These quantities are all rounded to three significant digits as required. We have completed the problem, and there are a lot of calculations involved. A great way to perform an overall error check on the calculations is to verify that the law of conservation of mass in chemical reactions is satisfied. From the equation, the masses of the water and carbon dioxide must add up to equal the masses of the glucose and oxygen. mass CO2 + mass H2O = 12.2 g + 5.00 g = 17.2 g mass C6H12O6 + mass O2 = 8.33 g + 8.88 g = 17.21 g Rounding that second value to three significant digits, we see that the reaction consumes 17.2 g of reactants and produces 17.2 g of products. Mass conservation is confirmed, so it is highly likely that our calculations are correct.

7.3.2 Limiting Reactant Let’s say you have a portable propane burner and a small canister of propane to fuel it. We know that the combustion of propane is a chemical reaction with oxygen that produces carbon dioxide and water. In the combustion of the propane in this canister, what is the limiting factor that determines how much water and carbon dioxide are produced? By limiting factor, what I really mean with this question is limiting reactant—the reactant that runs out first, once the reaction starts. Clearly, the oxygen available for this reaction, which is in the atmosphere, is available without limit. Thus, the limiting reactant is the propane, and when the propane is completely consumed, the reaction ceases. Every chemical reaction has a limiting reactant. One of the reactants is in shorter supply than the others and runs out first. When it does, the reaction ceases. The quantities of products produced by the reaction are determined by the quantity of the limiting reactant available for the reaction. The limiting reactant is also often called the limiting reagent (pronounced ree-A-gent), a reagent simply being one of the compounds consumed in the reaction. If you have known quantities available for a chemical reaction, you can determine which reactant is the limiting reactant simply by looking at the mole quantities available and the mole ratios from the balanced chemical equation. Once you know which compound is the limiting reactant, you can calculate the quantities of the products produced by the reaction based on the quantity of the limiting reactant supplied. As with the stoichiometric calculations we just reviewed, all the calculations associated with determining the limiting reactant are performed in moles. When you are given mass quantities, begin your solution by converting masses into numbers of moles. 192

Chemical Reactions and Stoichiometry Example 7.10 One of the most important industrial chemical processes in the world is the Haber-Bosch process for producing ammonia from nitrogen in the air. Grains need nitrogen as a nutrient, but cannot get it from the air because the nitrogen molecules are held together by the strong nitrogen triple bond. Bacteria in the soil produce the “fixed” nitrogen that plants use, but bacteria cannot supply the large quantities of nitrogen needed for industrial scale agriculture. The nitrogen-based fertilizers used are made from ammonia. Some have estimated that one third of the world’s population is sustained by food grown using fertilizer produced by ammonia made with the Haber-Bosch process. The process occurs at extremely high pressures—around 200 atmospheres, or 3,000 psi—in reactors like the historical 1921 reactor shown in Figure 7.21. For years chemists thought the reaction was impossible, but German chemist Fritz Haber solved the problem in 1909, and with engineering assistance from Carl Bosch, had the commercial production of ammonia up and running by the end of 1910. The reaction is as follows: N 2 ( g ) + 3H 2 ( g ) → 2NH3 ( g ) Consider a test run of this process, with 652 kg N2 and 175 kg H2 available for the reaction. Determine whether the N2 or the H2 is the limiting reactant. Then assuming that the limiting reactant is completely consumed in the reaction, calculate the mass of ammonia produced by the time the reaction ceases. We begin by converting our mass quantities from kilograms to grams, and from grams to moles. We use the periodic table to determine the molar masses of H2 and N2, and then we use these as conversion factors to get Figure 7.21. This highpressure reactor from mole quantities: 175,000 g H 2 ⋅

1 mol H 2 = 86,810 mol H 2 2.016 g H 2

652,000 g N 2 ⋅

1 mol N 2 = 23,280 mol N 2 28.01 g N 2

1921 is on display at the Karlsruhe Institute of Technology in Germany.

These mole quantities are each written with four significant digits, one more than required in the final result. Looking now at the balanced equation for the reaction, we see that the ratio of hydrogen to nitrogen required is 3 : 1. In order to consume the 23,280 moles of nitrogen we have available, we need 3(23,280) = 69,840 moles of hydrogen. We have a lot more hydrogen than this available, so nitrogen is the limiting reagent. Now we perform a standard stoichiometric calculation to determine the amount of ammonia produced from 23,280 mol N2. The mole ratio of N2 to NH3 is 1 : 2, so we use this ratio as a conversion factor to determine the amount of NH3 produced: 23,280 mol N 2 ⋅

2 mol NH3 = 46,560 mol NH3 1 mol N 2

193

Chapter 7 Finally, we convert this mole quantity to mass as the problem requires. This requires calculating the molar mass of NH3 to use as a conversion factor. 46,560 mol NH3 ⋅

17.03 g NH3 1 kg = 792,900 g NH3 ⋅ = 793 kg NH3 1 mol NH3 1000 g

This final result is rounded to three significant digits as the given information requires.

Example 7.11 A 5.00-g strip of magnesium metal is placed in an aqueous solution containing 15.0 g ZnCl2, causing the following reaction: Mg ( s ) + ZnCl 2 ( aq ) → Zn( s ) + MgCl 2 ( aq ) Determine (a) the limiting reactant, (b) the mass of MgCl2 produced, and (c) the mass of the excess reactant that remains. Converting the given quantities to moles, we have: 5.00 g Mg ⋅

1 mol Mg = 0.2057 mol Mg 24.305 g Mg

15.0 g ZnCl 2 ⋅

1 mol ZnCl 2 = 0.1101 mol ZnCl 2 136.3 g ZnCl 2

The mole ratio for Mg and ZnCl2 in the balanced equation is 1 : 1, so the ZnCl2 is the limiting reactant. The mole ratio of ZnCl2 and MgCl2 is also 1 : 1, so 0.1101 mol ZnCl2 produces 0.1101 mol MgCl2, giving a mass of 0.1101 mol MgCl 2 ⋅

95.21 g MgCl 2 = 10.5 g MgCl 2 1 mol MgCl 2

Since the mole ratio for Mg and ZnCl2 is 1 : 1, 0.1101 mol ZnCl2 consume 0.1101 mol Mg. The reaction began with 0.2057 mol Mg, so 0.2057 – 0.1101 mol Mg = 0.0956 mol Mg are left over. This converts to a mass of 0.0956 mol Mg ⋅

24.305 g Mg = 2.32 g Mg 1 mol Mg

7.3.3 Theoretical Yield and Percent Yield In the example of the Haber-Bosch process to produce ammonia, the 793 kg of ammonia we calculated that are produced by the reaction is called the theoretical yield. Let’s now assume we perform the reaction using the quantities of reactants listed in that example and find that 194

Chemical Reactions and Stoichiometry we actually end up with 742 kg NH3 when all the N2 is consumed and the reaction ceases. This quantity is called the actual yield. We use this quantity to calculate the percent yield as follows: percent yield =

actual yield × 100% theoretical yield

Accordingly, the percent yield associated with an actual yield of 742 kg NH3 is: percent yield =

742 kg NH3 × 100% = 93.6% 793 kg NH3

When chemists develop procedures to synthesize compounds, the percent yield is a measure of the effectiveness of the procedure. Yields above 80% are considered very good.

Chapter 7 Exercises SECTION 7.1

1. Butane, C4H10, is the fuel used in disposable lighters. Butane burns in oxygen to produce carbon dioxide and water. Write the formula equation for this reaction and balance it. 2. Iron(II) sulfide reacts with hydrochloric acid to produce hydrogen sulfide and iron(II) chloride. Write the formula equation for this reaction and balance it. 3. Acetylene, C2H2, is a gas that produces an extremely hot flame when combusting with oxygen, so hot that cutting torches use this gas to cut through steel. The products of the combustion are carbon dioxide and water. Write the formula equation for this reaction and balance it. 4. With the aid of a catalyst, liquid methanol, CH3OH, is produced from a reaction of carbon monoxide gas and hydrogen gas. Write the formula equation for this reaction and balance it. 5. Potassium carbonate is used to neutralize the sulfuric acid in well water caused by acid rain. The products of the neutralization reaction are potassium sulfate, carbon dioxide, and water. Write the formula equation for this reaction and balance it. 6. The tungsten metal used as filaments in light bulbs is produced by reacting tungsten oxide with hydrogen gas to produce tungsten metal and water. Write the formula equation for this reaction and balance it. 7. Sodium azide, NaN3, is the original compound formerly used in automobile air bags. Other compounds were used to initiate the reaction, but the bags were filled by the conversion of solid sodium azide into sodium metal and nitrogen gas. Write the formula equation for this reaction and balance it. 8. Identify the oxidation state for each element in the following compounds, elements, and ions: a. PBr3

b. As2O5

c. ClO4–

d. F2

e. UF6

f. CO3

g. H3PO4

h. Zn(NO3)2

i. C4H10

j. Cr2O7

k. KH

l. Fe

2–

9. Why must chemical equations be balanced?

195

Chapter 7 SECTION 7.2

10. For the reactions in exercises 1–7 above, identify the reactions that can be classified as synthesis, decomposition, single replacement, or double replacement. 11. In each of the following reactions, determine which element is oxidized and which is reduced. Also, where possible, identify the reaction as synthesis, decomposition, single replacement, or double replacement. a. 3Fe ( NO3 )2 ( aq ) + 2Al( s ) → 3Fe( s ) + 2Al ( NO3 )3 ( aq ) b. PbS( s ) + 4H 2O2 ( aq ) → PbSO4 ( s ) + 4H 2O( l ) c. N 2( g ) + 3H 2 ( g ) → 2NH3 ( g ) d. Cl 2( aq ) + 2NaI( aq ) → I2 ( aq ) + 2NaCl ( aq ) 12. Explain what a redox reaction is. 13. For each of the following reactions, write the formula equation, balance it, and identify the reaction by type. a. magnesium hydroxide yields magnesium oxide and water b. sodium chloride and sulfuric acid yield sodium sulfate and hydrogen chloride gas c. calcium oxide and water yield calcium hydroxide d. ammonium bicarbonate and sodium chloride yield sodium bicarbonate and ammonium chloride 14. Use the activity series of metals to determine which of the following reactions occur. For those that do, complete the equation and balance it. a. Ca ( s ) + O2 ( g ) → b. Fe( s ) + ZnCl 2 ( aq ) → c. Pt ( s ) + CuSO4 ( aq ) → d. Al( s ) + H 2SO4 ( aq ) → e. Ni ( s ) + CuCl 2 ( aq ) → f. Ba ( s ) + H 2O( l ) → g. Cu ( s ) + FeSO4 ( aq ) → h. Al( s ) + Pb( NO3 )2 ( aq ) → i.

Li ( s ) + KI( aq ) →

Au ( s ) + O2 ( g ) →

15. Assuming the following compounds are the result of decomposition reactions, identify the reactant, write the formula equation, and balance it. a. lithium chloride and oxygen (Hint: The chlorine and oxygen are in an oxyanion.) 196

Chemical Reactions and Stoichiometry b. copper(II) oxide and water (Hint: The hydrogen and oxygen are in an oxyanion.) c. carbon dioxide and water (Hint: Think about soft drinks.) SECTION 7.3

16. Silver bromide is used on film for producing holograms. If 3.5 mol silver nitrate react in a double replacement reaction with excess sodium bromide to produce silver bromide, what mass of silver bromide is formed? 17. Hydrogen, the perfect clean fuel, combusts with oxygen to produce water vapor. If 1,575 mol H2 react, determine the mole quantities of oxygen required and water produced. 18. Aluminum sulfide reacts with water to produce dihydrogen sulfide (usually just called hydrogen sulfide) and aluminum hydroxide. If 2.290 kg aluminum sulfide react completely with excess water, what mass of aluminum hydroxide is produced? 19. Aluminum hydroxide is a popular compound for use in antacid tablets. This compound reacts with stomach acid, HCl, to produce the salt aluminum chloride and water. If 750 mg aluminum hydroxide are consumed, determine: a. the number of moles of hydrochloric acid neutralized. b. the mass of water produced. 20. Iron ore contains iron(III) oxide along with several other substances. The iron is reduced by heating it to about 1,250°C in the presence of carbon monoxide (a so-called reducing atmosphere) in a blast furnace. This process, which has been in use since around 1491, is shown to the right. The reaction produces iron metal and carbon dioxide. What is the theoretical yield in kilograms of iron from this reaction if 2.50 × 104 kg iron(III) oxide react with excess carbon monoxide? 21. A common acid-base reaction is between sulfuric acid and sodium hydroxide to produce sodium sulfate and water. If 29.55 g sodium hydroxide react with 44.11 g sulfuric acid, determine: a. the limiting reactant. b. the mass of sodium sulfate produced. c. the mass of water produced. 22. Octane, C8H18, is the major component of gasoline. The combustion reaction between octane and oxygen is: 2C 8H18 ( l ) + 25O2 ( g ) → 16CO2 ( g ) + 18H 2O( g ) a. How many moles of oxygen gas are required to burn 350.0 moles of octane? b. How many kilograms of water are produced if the combustion consumes 3.5 kg O2?

197

Chapter 7 c. The density of octane is 0.692 g/mL. Determine the mass of oxygen required to burn 11.5 gal octane. (Recall, conversion factors are in Appendix A, Table A.3.) 23. Nitric acid is used in the process of nitration, in which organic molecules are converted into explosive compounds. The first step in the industrial production of nitric acid involves the following reaction between ammonia and oxygen to produce nitric oxide and water: cat

4NH3 ( g ) + 5O2 ( g ) → 4NO( g ) + 6H 2O( g ) Assume that 855 g NH3 are combined with 1,750 g O2 in this reaction. Determine: a. the limiting reagent. b. the theoretical yield of nitric oxide, assuming all the limiting reagent is consumed. c. the percentage yield of NO, assuming 1,272 g NO are actually produced. 24. Aluminum hydroxide undergoes a double replacement reaction with sulfuric acid. Assume that 31.8 g sulfuric acid are combined with 25.4 g aluminum hydroxide with 100% yield. Determine the following: a. the limiting reactant. b. the mass of excess reactant remaining. c. the mass of each product formed. 25. Dinitrogen tetroxide, N2O4, is an important rocket fuel. Its structure is pictured to the right. On one of the Apollo lunar missions, N2O4 powered the lunar landing module as it ascended from the surface of the moon. During the ascent, 1.200 × 103 kg N2H4, 1.000 × 103 kg (CH3)2N2H2, and 4.500 × 103 kg N2O4 were available for the following reaction: 2N 2H 4 + ( CH3 )2 N 2H 2 + 3N 2O4 → 6N 2 + 2CO2 + 8H 2O a. Which of the compounds was consumed first in this reaction? b. How many kilograms of waste water were spewed into space by the reaction? c. How many nitrogen atoms were left in space by the reaction? 26. For reasons we explore in a later chapter, hydrofluoric acid is classified as a weak acid. Nevertheless, HF is so corrosive it eats right through glass, silicon dioxide, SiO2, and is used for glass etching. (Needless to say, HF is not stored in glass bottles.) HF has also been used for etching silicon wafers in the process of making semiconductors, although the use of HF has waned in recent years. The reaction between HF and SiO2 is SiO2 ( s ) + 6HF ( aq ) → H 2SiF6 ( aq ) + 2H 2O( l )

a. How many moles of HF are needed to react with 542 g SiO2? b. How many grams of H2SiF6 form when 4.25 mol HF react completely? c. How many moles of HF are required to produce 2.0 gallons of water? (Remember, the density of water is 0.998 g/mL at room temperature.) d. If 2.50 kg HF are available to react with 1.205 kg SiO2, what is the limiting reactant? e. What is the theoretical yield of H2SiF6 if the limiting reagent is completely consumed, assuming the reactant quantities listed in part (d)? 198

Chemical Reactions and Stoichiometry f. Using the theoretical yield from the previous question, what is the percentage yield of H2SiF6 if 2.657 kg H2SiF6 are produced? 27. When benzene, C6H6, reacts with bromine (a type of reaction called halogenation), the reaction produces bromobenzene, C6H5Br, and hydrogen bromide. a. If 45.0 g of benzene reacts with 97.5 g of bromine, what is the theoretical yield of bromobenzene? b. If the actual yield of bromobenzene is 63.25 g, what is the percent yield of bromobenzene? 28. To monitor levels of ozone, O3, in the air, an air sample is processed by an instrument that causes ozone to react with sodium iodide in the following reaction: O3 ( g ) + 2NaI( aq ) + H 2O( l ) → O2 ( g ) + I2 ( s ) + 2NaOH ( aq ) a. Determine the number of moles of sodium iodide required to process 2.85 × 10–6 moles of ozone. b. Determine how many grams of sodium iodide are required to process 3.00 grams of ozone. c. If 1,455 g NaI are available to process a sample of 250.0 g O3 in a reaction with excess water, determine the limiting reactant and the theoretical yield of I2. d. From part (c), determine the number of iodine atoms present in the theoretical yield of I2. GENERAL REVIEW EXERCISES

29. Melatonin is a compound secreted naturally in the brain when it is time to go to sleep. The molar mass of melatonin is 232.281 g/mol. A sample of 25.0 mg of melatonin is found to contain 16.81 mg C, 1.736 mg H, 3.015 mg N, and 3.445 mg O. a. Determine the percent composition of melatonin. b. Determine the molecular formula for melatonin. c. Melatonin is sold as an over-the-counter sleeping aid. Each tablet contains 3.0 mg of melatonin. Determine the number of carbon atoms in each tablet of melatonin. 30. Mercury-vapor lamps are widely used for general lighting in gymnasiums, exhibit halls, and other large spaces. One of the wavelengths in the emission spectrum of mercury is a bright green line with an energy of 2.271 eV. Determine the wavelength of this line. 31. Write the condensed electron configurations for yttrium, tin, titanium, and iodine. 32. Write explanations in one or two sentences describing the Aufbau principle, the Madelung rule, and Hund’s Rule. 33. Use the information from Table 2.6 to calculate the atomic mass of silicon. 34. Based on your knowledge of general trends in the periodic table, place the following in order of increasing electronegativity: Fe, F, Se, Fr, Ba, and Cl. 35. Draw the Lewis structures for the following molecules and polyatomic ions: a. CCl2S

b. PO43–

c. NO2+

e. O3

f. SF6

g. CO

d. N2H2

h. H3O+

36. Identify the electron domain geometry and molecular geometry for each molecule listed in the previous exercise. 199

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ACCELERATED CHEMISTRY Accelerated Chemistry is designed for accelerated students in high school. This course is typically taught in 10th grade, while students are concurrently enrolled in Algebra 2 and after students have completed Accelerated Studies in Physics and Chemistry (9th). Here we highlight a few distinctive features of this text. •

Accelerated Chemistry supports our signature philosophy of science education based on wonder, integration, and mastery. We always seek to build on and stimulate the student’s innate sense of wonder at the marvels found in nature. Integration refers to the epistemology, mathematics, and history embedded in the text, and a curriculum designed to help develop the student’s facility with using language to communicate scientific concepts. Finally, Accelerated Chemistry is designed to be used in a mastery-learning environment, using the mastery teaching model John developed and describes in From Wonder to Mastery: A Transformative Model for Science Education. When teachers use this mastery-learning model, the result is high student achievement and exceptional long-term retention for all students.

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The text treats advanced topics in detail. Since this course is designed to follow the rigorous introduction to chemistry included in Accelerated Studies in Physics and Chemistry (ASPC), Accelerated Chemistry skips the usual introductory material on units of measure, unit conversions, and the history of atomic models. Instead, the text dives right into more advanced topics and includes chapters on Thermochemistry and Kinetics as well as Chemical Equilibrium. For students who did not take ASPC but are otherwise qualified for this course, an introduction to unit conversions (dimensional analysis) is included in the appendices.

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The text includes a bonus chapter. Just for fun, we included a bonus chapter introducing interested students to organic chemistry. This chapter would not normally be taught as part of a one-year course, but some students are likely to be so interested in this subject that they want more!

ACCELERATED CHEMISTRY Support Resources A number of support resources are available to accompany Accelerated Chemistry. They include: The Student Lab Report Handbook Students should begin writing their lab reports from scratch in 9th grade. This popular manual gives them everything they need. We recommend supplying this handbook to every freshman so they can refer to it throughout high school. Read more about this resource on pg. 6. Solutions Manual to Accompany Accelerated Chemistry This book contains complete written solutions for all the computations in the chapter exercises. Chemistry Experiments for High School This student manual includes 20 excellent chemistry experiments that illustrate concepts and foster real-world lab skills. Using this book requires a standard high school laboratory facility. Digital Resources The following materials are accessible exclusively through Centripetal Press: • a full year of chapter exams • two short quizzes for each chapter • two semester exams • a document containing answer keys for all the quizzes and tests and sample answers for all the verbal questions in the text • a document with recommendations for teaching the course • a lesson list and example calendar Tips and Tools A variety of tips and tools are available at www.centripetalpress.com. Here, teachers will find scores of images and short videos organized chapter by chapter. Teachers are free to use any of this content as they wish to assemble vivid and fascinating lessons.

ACCELERATED CHEMISTRY Contact Us Our team is here to assist as you choose curricula that will be the best fit for your classroom. We encourage you to take advantage of the following resources: Questions & Guidance We look forward to answering any questions you might have regarding our curriculum, placement, or pricing. Please contact us! Email: info@classicalsubjects.com Phone: (717) 730-0711 Quotes and Vendor Applications We are able to provide formal quotes for all of our materials, and can also submit applications required to become an approved vendor for your school district. Inquire at info@classicalsubjects.com. Get the Latest Information from Centripetal Press and Classical Academic Press For more information on Centripetal Press products, visit www.centripetalpress.com.

Sample Chapters The following pages contain samples from the text. The Table of Contents is shown, as well as Chapters 1, 2, and 10. Note that the sample chapters shown are from the new edition planned for 2022. The new edition has been revised to accommodate the extensive changes to the metric system that went into effect in 2019. The edition currently in print looks exactly the same but uses the former definitions for the Avogadro constant, the mole, and other constants.

Accelerated Chemistry A Mastery-Oriented Curriculum

Second Edition

John D. Mays

Camp Hill, Pennsylvania 2020

Accelerated Chemistry © Classical Academic Press®, 2020 Edition 2.0 ISBN: 978-0-9972845-6-0 All rights reserved. Except as noted below, this publication may not be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior written permission of Classical Academic Press. All images attributed to others under any of the Wikimedia Commons licenses, such as CCBY-SA-3.0 and others, may be freely reproduced and distributed under the terms of those licenses. Classical Academic Press 515 S. 32nd Street Camp Hill, PA 17011 www.ClassicalAcademicPress.com/Novare/ Cover design by Nada Orlic, http://nadaorlic.info/ ISP.04.20

Contents Preface For Teachers

Preface For Students Introduction What is Chemistry All About? I.1 A Few Major Themes

I.1.1 Chemistry Is All About Electrons I.1.2 Chemistry Is All About Electrical Forces

Hmm... Interesting. Why water forms beads

I.1.3 Chemistry Is All About Minimizing Energy I.1.4 Chemistry Is All About Whole Number Ratios of Atoms I.1.5 Chemistry Is All About Modeling

I.2 Conclusion Introduction Study Questions Chapter 1 Atomic Structure 1.1 Atomic Spectra

1.1.1 The Electromagnetic Spectrum 1.1.2 Energy in Atoms

Hmm... Interesting. Neon signs and phonons

1.1.3 The Hydrogen Atom

1.2 The Bohr Model of the Atom 1.3 The Quantum Model of the Atom

1.3.1 Schrödinger and Pauli 1.3.2 Shells, Subshells, and Orbitals 1.3.3 The Aufbau Principle, the Madelung Rule, and Hund’s Rule

1.4 Electron Configurations

1.4.1 Electron Configurations and Orbital Diagrams 1.4.2 Condensed Electron Configurations 1.4.3 Anomalous Electron Configurations

1.5 Isotopes and Atomic Masses

1.5.1 Isotopes 1.5.2 The Unified Atomic Mass Unit 1.5.3 Atomic Masses 1.5.4 The Mole and the Avogadro Constant

xiv xiv xv xvi xviii xix xx 2 3 3 4 6 6 9 10 12 13 14 16 16 18 19 21 22 23 23 23 27 29 29 32 33 34 34 34 34 36 v

Contents 1.5.5 Molar Mass and Formula Mass 1.5.6 Gram Masses of Atoms and Molecules 1.5.7 Percent Composition and Empirical Formulas 1.5.8 Determining a Molecular Formula from an Empirical Formula 1.5.9 Significant Digit Rules for Addition

Chapter 1 Exercises

Chapter 2 The Periodic Law 2.1 The Periodic Table of the Elements 2.2 Periodic Table Nomenclature 2.3 Periodic Physical Properties

2.3.1 Atomic Radius and Bonding Atomic Radius 2.3.2 Ionic Radius

2.4 Periodic Chemical Properties

2.4.1 Core and Valence Electrons 2.4.2 Effective Nuclear Charge 2.4.3 Ionization Energy 2.4.4 Electron Affinity 2.4.5 Electronegativity

Hmm... Interesting. Hydrogen in space 2.5 A Few Notes About Hydrogen Chapter 2 Exercises Chapter 3 Chemical Bonding 3.1 Preliminaries

3.1.1 Types of Substances 3.1.2 Chemical Possibilities 3.1.3 The Octet Rule

3.2 Ionic Bonding

3.2.1 Ionic Bonds and Crystals 3.2.2 Naming Ionic Compounds 3.2.3 Energy in Ionic Bonds 3.2.4 Hydrates 3.2.5 Intensive and Extensive Properties 3.2.6 Physical Properties of Ionically Bonded Substances

3.3 Covalent Bonding

3.3.1 Covalent Bonds and Molecules 3.3.2 Polyatomic Ions 3.3.3 Ionic Compounds with Polyatomic Ions 3.3.4 Polyatomic Ion Names 3.3.5 Naming Acids 3.3.6 Lewis Structures 3.3.7 Exceptions to the Octet Rule 3.3.8 Resonance Structures 3.3.9 Naming Binary Covalent Compounds 3.3.10 Energy in Covalent Bonds

37 41 42 44 45 46 50 52 54 55 55 57 59 59 60 61 64 66 68 68 69 72 74 74 75 75 76 76 78 80 82 83 83 84 84 86 87 87 88 89 93 94 95 96

Contents 3.3.11 Bond Number 3.3.12 Physical Properties of Covalently Bonded Substances

97 98 3.4 Electronegativity, Polarity, and Bond Character 99 3.4.1 Polarity and Dipoles 99 3.4.2 The Nature of the Bond 99 Chapter 3 Exercises 100 Hmm... Interesting. The molecular structure of glass and quartz 101

Chapter 4 Molecular Theory and Metallic Bonding 4.1 Molecular Structure

4.1.1 Covalent Bond Theory 4.1.2 Valence Shell Electron Pair Repulsion (VSEPR) Theory 4.1.3 The Effect of Nonbonding Domains on Bond Angle 4.1.4 Orbital Hybridization Theory 4.1.5 Valence-Bond Theory

4.2 Metallic Bonding

4.2.1 Metallic Lattices

Hmm... Interesting. Tin pest

4.2.2 Physical Properties of Metals

4.3 Intermolecular Forces

4.3.1 Bonding Forces 4.3.2 Intermolecular Forces 4.3.3 Hydrogen Bonding 4.3.4 Van der Waals Forces

Chapter 4 Exercises

Chapter 5 Chemical Reactions and Stoichiometry 5.1 Introduction to Chemical Equations

5.1.1 Fascinating Chemistry 5.1.2 The Law of Conservation of Mass in Chemical Reactions 5.1.3 Reaction Notation 5.1.4 Balancing Chemical Equations 5.1.5 Oxidation States

Hmm... Interesting. Why nitrates and nitros blow up 5.2 General Types of Chemical Reactions 5.2.1 Synthesis Reactions 5.2.2 Decomposition Reactions 5.2.3 The Activity Series of Metals 5.2.4 Single Replacement Reactions 5.2.5 Double Replacement Reactions

Hmm... Interesting. A story about aqua regia

5.2.6 Combustion Reactions 5.2.7 Acid-Base Neutralization Reactions 5.2.8 Oxidation-Reduction Reactions

5.3 Stoichiometry

5.3.1 Stoichiometric Calculations

104 106 106 106 110 111 114 114 114 115 116 117 117 117 118 119 121 124 126 126 126 127 128 132 135 136 136 137 137 138 140 140 141 141 142 143 143 vii

Contents 5.3.2 Limiting Reactant 5.3.3 Theoretical Yield and Percent Yield

Chapter 5 Exercises

Chapter 6 Kinetic Theory and States of Matter 6.1 Temperature, Kinetic-Molecular Theory, and Pressure 6.1.1 Temperature and Molecular Energy 6.1.2 Velocity Distribution of Gases 6.1.3 The Kinetic-Molecular Theory of Gases 6.1.4 Gas Pressure

6.2 States of Matter

6.2.1 The Four Basic States of Matter

Hmm... Interesting. How barometers work

6.2.2 Solids 6.2.3 Liquids 6.2.4 Gases

Hmm... Interesting. Gas diffusion

6.2.5 Plasmas 6.2.6 Phase Transitions and Phase Diagrams 6.2.7 Heat Capacity, Heat of Fusion, and Heat of Vaporization 6.2.8 Evaporation 6.2.9 Vapor Pressure

Chapter 6 Exercises

Chapter 7 The Gas Laws 7.1 Early Formulations of the Gas Laws 7.1.1 Boyle’s Law 7.1.2 Charles’ Law 7.1.3 Avogadro’s Law

7.2 The Ideal Gas Law

7.2.1 Standard Temperature and Pressure 7.2.2 The Ideal Gas Law

Hmm... Interesting. The gas laws as models

7.2.3 Using the Ideal Gas Law to Find Molar Mass and Density

7.3 The Law of Partial Pressures

7.3.1 Dalton’s Law of Partial Pressures 7.3.2 Collecting a Gas Over Water

7.4 Stoichiometry of Gases and Effusion 7.4.1 Stoichiometry of Gases

viii

148 150 151 156 158 158 158 160 160 161 161 162 164 165 167 168 169 170 172 176 178 179 182 184 184 185 188 188 188 189 190 196 198 198 201 203 203

Contents 7.4.2 Gas Diffusion and Effusion

Hmm... Interesting. Uranium enrichment Chapter 7 Exercises Chapter 8 Solution Chemistry 8.1 What Is and Is Not a Solution 8.1.1 Suspensions 8.1.2 Colloidal Dispersions

Hmm... Interesting. Brownian motion 8.2 Dissolution 8.2.1 The Process of Dissolving 8.2.2 Enthalpy of Solution 8.2.3 Entropy and Free Energy 8.2.4 Electrolytes

8.3 Solubility

8.3.1 Ionic Solids in Water 8.3.2 Ionic Solids in Nonpolar Solvents 8.3.3 Polar Liquids 8.3.4 Nonpolar Liquids 8.3.5 Solutions of Solids

Hmm... Interesting. How soap works

8.3.6 Gases in Liquid Solutions: Le Châtelier’s Principle and Henry’s Law 8.3.7 The Effect of Temperature on Solubility

8.4 Quantifying Solution Concentration 8.4.1 Molarity 8.4.2 Molality

8.5 Compounds in Aqueous Solution

8.5.1 Ionic Equations and Precipitates 8.5.2 Net Ionic Equations and Spectator Ions

8.6 Colligative Properties of Solutions

8.6.1 Vapor Pressure Lowering 8.6.2 Freezing Point Depression and Boiling Point Elevation 8.6.3 Osmotic Pressure

Chapter 8 Exercises

Chapter 9 Acids and Bases 9.1 Properties and Nomenclature of Acids and Bases 9.1.1 Introduction 9.1.2 Properties of Acids and Bases 9.1.3 Acid Names and Formulas

9.2 Acid-Base Theories

9.2.1 Arrhenius Acids and Bases 9.2.2 Brønsted-Lowry Acids and Bases

Hmm... Interesting. What is an alkali?

9.2.3 Lewis Acids and Bases 9.2.4 Strength of Acids and Bases

205 207 209 214 216 216 216 217 218 218 221 222 223 225 225 225 226 227 227 228 230 231 233 233 235 236 236 237 239 239 241 246 247 252 254 254 254 256 257 258 260 261 264 265 ix

Contents

9.3 Aqueous Solutions and pH

9.3.1 The Self-ionization of Water 9.3.2 Calculating [H3O+] and [OH–] 9.3.3 pH as a Measure of Ion Concentration and Acidity 9.3.4 pH Measurement, pH Indicators, and Titration 9.3.5 Titration Procedure 9.3.6 Determining [H3O+] or [OH–] from Titration Data

Chapter 9 Exercises

Chapter 10 Thermochemistry and Kinetics 10.1 Energy in Chemical Reactions

10.1.1 Introduction 10.1.2 Enthalpy 10.1.3 Understanding Enthalpy and Energy 10.1.4 Enthalpy of Combustion 10.1.5 Enthalpy of Formation

10.2 Calculating Enthalpy of Reaction and Enthalpy of Formation 10.2.1 Hess’s Law 10.2.2 Hess’s Law and the General Enthalpy Change Equation

10.3 Free Energy

10.3.1 Brief Review of Enthalpy and Entropy 10.3.2 More on Entropy 10.3.3 Gibbs Free Energy

10.4 Reaction Kinetics

10.4.1 Collision Theory 10.4.2 Factors Influencing Reaction Rate 10.4.3 Reaction Mechanisms 10.4.4 Activation Energy and the Activated Complex 10.4.5 Reaction Rate Laws 10.4.6 Rate Laws and Reaction Mechanisms

Chapter 10 Exercises

Chapter 11 Chemical Equilibrium 11.1 Physical and Chemical Equilibrium 11.1.1 Equilibria We Have Seen So Far

Hmm... Interesting. Is dissolution a chemical change?

11.1.2 Dynamic Chemical Equilibrium 11.1.3 The Law of Chemical Equilibrium and the Equilibrium Constant 11.1.4 Le Châtelier’s Principle and Equilibrium Shifts 11.1.5 Reactions That Go To Completion 11.1.6 The Common-Ion Effect

11.2 Acid-Base Equilibrium

11.2.1 The Acid Dissociation Constant 11.2.2 The Base Dissociation Constant 11.2.3 Buffered Solutions

268 268 268 269 274 278 280 281 286 288 288 288 292 294 295 297 297 302 303 303 304 305 308 308 309 311 312 314 316 318 322 324 324 325 326 327 330 333 334 335 335 337 338

Contents 11.2.4 Hydrolysis of Salts

Hmm... Interesting. Buffering in blood 11.3 Solubility Equilibria

11.3.1 The Solubility Product 11.3.2 Calculating Ksp from Concentration or Solubility Data 11.3.3 Calculating Solubility from Ksp 11.3.4 Using Ksp to Predict Precipitation

Chapter 11 Exercises

Chapter 12 Redox Chemistry 12.1 Oxidation and Reduction

12.1.1 Introduction to Redox Reactions 12.1.2 Oxidation States 12.1.3 Strengths of Oxidizing and Reducing Agents

12.2 Redox Reaction Equations

12.2.1 Redox Half-Reactions 12.2.2 Balancing Redox Equations

12.3 Electrochemistry

12.3.1 Copper and Zinc Redox 12.3.2 Electricity Instead of Heat 12.3.3 Electrochemical Cells

Hmm... Interesting. How are salt bridges made?

12.3.4 Electrode Potentials 12.3.5 Electrochemical Applications

Chapter 12 Exercises

Chapter 13 Organic Chemistry—An Introduction 13.1 The Chemistry of Carbon 13.1.1 Introduction 13.1.2 Carbon Molecular Geometry 13.1.3 Carbon Allotropes

13.2 Hydrocarbons 13.2.1 Alkanes

Hmm... Interesting. The end of CFCs (almost)

13.2.2 Alkyl groups 13.2.3 IUPAC Naming Standards 13.2.4 Functional Groups 13.2.5 Alkenes and Alkynes

Hmm... Interesting. Fractional distillation

13.2.6 Aliphatic Cyclo– Hydrocarbons 13.2.7 Aromatic Hydrocarbons

13.3 Other Organic Compounds

13.3.1 Alcohols and Ethers 13.3.2 Aldehydes and Keytones 13.3.3 Carboxylic Acids and Esters

341 341 343 343 344 346 347 349 354 356 356 356 360 362 362 366 372 372 373 374 378 380 383 388 392 394 394 394 395 397 397 400 401 402 406 406 409 411 412 415 415 417 418

Contents 13.3.4 Nitrogen Compounds: Amines and Amides

13.5 A Wee Bit of Biochemistry Hmm... Interesting. Recycling codes Chapter 13 Exercises

419 420 422 422 423 423 426 426 427

Glossary

432

Answers to Selected Exercises

452

Hmm... Interesting. Trans fats 13.4 Polymers 13.4.1 Types of Polymers 13.4.2 Polymerization 13.4.3 A Short Polymer Showcase

Appendix A Units, Unit Conversions, Significant Digits, and Scientific Notation A.1 Units of Measure

Appendix A Exercises

466 466 466 466 468 470 470 472 475 476 476 477 483 483 485 486

Appendix B Reference Data

488

Appendix C Scientists to Know About

496

Appendix D Memory Requirements

498

References and Citations

499

Image Credits

501

Index

503

A.1.1 The US Customary System A.1.2 The SI Unit System A.1.3 Metric Prefixes

A.2 Converting Units of Measure

A.2.1 Basic Principles of Unit Conversion Factors A.2.2 Tips for Converting Units of Measure A.2.3 Converting Temperature Units

A.3 Accuracy and Precision

A.3.1 Distinguishing Between Accuracy and Precision A.3.2 Significant Digits

A.4 Other Important Math Skills

A.4.1 Scientific Notation A.4.2 Calculating Percent Difference

xii

Chapter 1 Atomic Structure

Ne When excited by heat or electricity, the atoms of each element emit a specific, unique set of wavelengths of light—the atomic spectrum for that element. The visible spectra emitted by mercury and neon are shown above. Atomic spectra were known and studied in the 19th century, but there was no theory at that time that could explain the source of the colors different elements emit. Then in 1913, Niels Bohr published his new model of the atom, locating the electrons in atoms in specific energy levels. Bohr theorized that when excited, electrons jump to higher energy levels, and that to drop back down to a lower energy level an electron emits a packet of electromagnetic energy—what we now call a photon. In 1901, Max Planck had published the equation relating specific amounts of energy to specific wavelengths (colors) of light. Bohr’s successful explanation for atomic spectra opened the door for detailed study of the internal structure of atoms. The two spectra shown above were imaged in the Laser Optics Lab at Regents School of Austin in Austin, Texas.

Atomic Structure

1. Describe the electromagnetic spectrum and state approximate wavelengths for the ends of the visible spectrum. 2. Define quantum and explain what it means for energy to be quantized. 3. Given the Planck relation and Planck’s constant, determine the energy of a photon of a given wavelength and vice versa. SECTION 1.2

4. Describe the Bohr model of the atom and relate how the model explained the phenomenon of atomic spectra. SECTION 1.3

5. Distinguish qualitatively between the orbital energies in the hydrogen atom and those of other atoms. 6. Describe the two main ways that atoms can possess energy. 7. State and describe the four quantum numbers required to describe the quantum state of an electron. 8. For the first three principle quantum numbers, describe the orbitals available for electrons. 9. State the Aufbau principle, the Madelung rule, and Hund’s rule, and relate them to the way electrons are located in atoms. 10. State the Pauli exclusion principle and explain its relationship to the placement of electrons in atoms. SECTION 1.4

11. Given the periodic table, write electron configurations (full and condensed) and draw orbital diagrams for all elements in the first five periods (including the nine d-block elements with anomalous configurations). SECTION 1.5

12. Given the periodic table, determine the number of protons, electrons, and neutrons in the atoms of a given nuclide. 13. Given isotope mass and abundance data, calculate the atomic mass of an element. 14. Define the unified atomic mass unit, u, and state two definitions for the mole. 15. Calculate the molar mass of a compound or molecule, and calculate the mass in grams of a given mole quantity of a compound or molecule, or vice versa. 16. Calculate the number of atoms or molecules in a given quantity of substance. 17. Given mass data for an unknown compound, determine the percent composition and empirical formula of the compound. 18. Use the percent composition along with the molar mass or molecular mass of an unknown compound to determine the molecular formula for the compound. 19. Correctly use the rules for significant digits in computations, including the addition rule.

Chapter 1

1.1

Atomic Spectra

1.1.1 The Electromagnetic Spectrum

longer wavelength

380 nm

450 nm

495 nm

570 nm

590 nm

620 nm

750 nm

Understanding our present theory of atomic structure and the story of how it unfolded requires a basic understanding of the electromagnetic spectrum. We thus begin this chapter with a brief review of this topic. The spectrum of visible light is shown in Figure 1.1. The visible spectrum runs through the colors of the rainbow—red, orange, yellow, green, blue, violet—and includes wavelengths from about 750 nm (red) down to about 400 nm (violet).

shorter wavelength

Figure 1.1. Colors and approximate wavelength ranges in the visible portion of the electromagnetic spectrum.

Figure 1.2. The electromagnetic spectrum.

1 pm

0.01 nm

1 nm

0.01 μm

0.1 μm

0.1 nm

X-R ays

ays

rar ed Lig Vis ht ibl Ult e L rav igh iol t et Lig ht

0.01 mm

0.1 mm

1 mm

1 μm

I nf

cro wa ves 1 cm

dio

10 m

100 m

AM 1 km

0.1 m

dio

Visible light is just a small portion of a vast spectrum of electromagnetic radiation that occurs in nature. Figure 1.2 shows the most important regions of the electromagnetic spectrum, from radiation with wavelengths in the range of 1 km, the region of AM Radio waves, down to the high-energy Gamma Rays, with wavelengths in the range of 1 picometer (pm). (The metric prefix pico–, which may be new to you, means 10–12. One picometer is one thousandth of a nanometer.) As you see from the figure, the solar emission spectrum runs from wavelengths of about 1 mm down to wavelengths of about 0.1 μm. The solar spectrum includes the infrared, visible, and ultraviolet regions, and is strongest in the middle of the visible spectrum. The contemporary theory of light (a shorter term for electromagnetic radiation in general) holds that light exhibits both wave-like properties and particle-like properties. Since light behaves like waves, we can refer to the wavelengths of particular colors. But light also behaves like particles. We call these particles photons, and each phosolar emission spectrum ton represents a single packet of energy. The packet of energy in a photon is also called a quantum of energy. radiation wavelength (The plural is quanta.)

Atomic Structure When thinking of light as waves, we characterize those waves by the wavelength. When considering light as discrete packets of energy, we tend to think of the amount of energy in each packet (each photon). It turns out there is a simple equation relating these together. It was in 1901 that German physicist Max Planck (Figure 1.3) conceived of treating energy as if it were quantized. He was working on a different problem at the time (the so-called blackbody radiation problem), and he did not imagine that energy really is quantized. However, he introduced what he thought was a mathematical trick—the quantization of energy—and in the process, quantum theory was born. Planck won the Nobel Prize in Physics for this work in 1918. (It’s ironic, isn’t it, to win the Nobel Prize for a major discovery that the scientist thinks is just a mathematical trick?) Four years later in 1905, German physicist Albert Einstein (Figure 1.4) proposed that energy really is quantized and used this idea to solve another problem (explaining Figure 1.3. German the photoelectric effect). For this, Einstein won the physicist Max Planck Nobel Prize in Physics in 1921. (1858–1947). The equation Planck introduced is called the Planck relation. This important equation is: E = hf

(1.1)

In this equation, E is the energy in the photon in joules (J). The next term in the equation, h, is a constant known as the Planck constant. The value of h is defined as exactly h = 6.62607015 ×10−34 J⋅s

Figure 1.4. German physicist Albert Einstein (1879–1955).

(1.2)

The last term in the Planck relation, f, is the frequency of the wave. The frequency and wavelength of a wave are related by the equation v=λf

(1.3)

In this equation, v is the velocity of the wave, which is the speed of light in this case (2.9979 × 108 m/s). The wavelength is represented by the Greek letter λ (lambda, the Greek lower-case letter l). If we solve Equation (1.3) for the frequency and insert it into the Planck relation, we have E=

hv λ

(1.4)

With this equation, we can compute the energy in a single photon of light at any wavelength, or vice versa. To illustrate such a calculation, we have our first example problem. As you see below, example problems in this text are set off by two red triangles. In presenting this example, I am assuming you are: • • •

familiar with the SI System of units and unit prefixes proficient at performing unit conversions proficient at applying the basic rules for the use of significant digits in measurements and computations (except for the addition rule, covered later in the chapter). 17

Chapter 1 If you are lacking skills in any of these areas, please refer to Appendix A for a tutorial. Example 1.1 The bright blue line in the mercury vapor spectrum (see the upper image on the opening page of this chapter) has a wavelength of 435.8 nm. Determine the energy contained in a single photon of this blue light. We begin by writing down the given information and the unknown we seek to find.

λ = 435.8 nm E=? Next, we convert the given wavelength into the MKS1 length unit, meters.

λ = 435.8 nm⋅

1m = 4.358 ×10−7 m 109 nm

From this value, along with the Planck constant and the speed of light, we calculate the energy of a photon with this wavelength. hv ( E= = λ

m⎞ 6.626 ×10−34 J⋅s )⎛⎜ 2.9979 ×108 ⎟ ⎝ s⎠ = 4.558 ×10−19 J −7 4.358 ×10 m

This is an extremely small amount of energy, less than a billionth of a billionth of a joule. The given wavelength and the value for the Planck constant each have four significant digits. Thus, the result is stated with a precision of four significant digits.

1.1.2 Energy in Atoms As mentioned in the caption on the opening page of this chapter, the atoms of every element emit a specific set of colors when excited. In the context of atomic theory, the term excitation refers to the absorption of energy by atoms, either from electromagnetic radiation (light) or from collisions with other particles. Let’s spend a moment considering the ways an individual atom can possess energy. There are two basic mechanisms by which atoms can possess energy. First, all atoms possess kinetic energy, the energy associated with motion. Kinetic energy in atoms is illustrated in Figure 1.5. In solids, the atoms are fixed in place and are not free to move around, so the kinetic energy is manifest in the atoms’ vibrations. In liquids and gases (fluids), atoms are free to move around, so the energy possessed by atoms in fluids is in their translational kinetic energy. Also, when atoms in fluids are bound together in molecules, the molecules can tumble and rotate, so some of their kinetic energy is in the energy of rotation. Atoms in molecules also vibrate, just as balls attached to one another by springs can wiggle back and forth. In all these cases, the kinetic energy in atoms and

1 MKS stands for meter-kilogram-second. The MKS system is a subset of the metric or SI System of units. Using MKS units for computations is always wise practice because the units of measure in the computation will all be consistent with each other, and the result of the computation always comes out in MKS units. For more on this, see Appendix A. 18

Atomic Structure molecules correlates directly to their temperature. The hotter they are, the more vigorously they vibrate and the faster they move. The second basic way an atom can possess energy is in the energies of the atom’s electrons. As mentioned in the Introduction and Chapter 1 (and discussed in detail later in this chapter), the electrons in atoms are located in various orbitals, and different orbitals are associated with different amounts of electron energy. Atoms can absorb quanta of energy from the photons of electromagnetic radiation and from collisions with other particles, such as ions and free electrons. When an atom absorbs energy in this way, the quantum of energy absorbed by the atom is manifest in one or more of the atom’s electrons moving into higherenergy orbitals. This is atomic excitation. When an atom’s electrons are all in their lowest-energy orbitals, the atom is said to be in the ground state. Excitation occurs when an atom absorbs a quantum of energy causing an electron to move to a higherenergy orbital. When this happens, the atom is said to be in an excited state. Atoms tend not to remain in excited states. Instead, after becoming excited an atom typically heads straight back to the ground state, generally by emitting the energy it absorbed in the form of one or more new photons.2 A newly emitted photon may not have the same amount of energy Figure 1.5. The atoms in as the original quantum of energy the atom absorbed. To explain this, the solid crystal at the top we need to introduce a commonly used graphical representation of the are blurred to illustrate different energies electrons can possess. For now, let’s call these energy their vibrations. The gas molecules at the bottom are levels. We will relate these more carefully to the energies of electrons in atoms in the next few sections. translating and tumbling. Figure 1.6 is a diagram representing four different energy levels actually available to an electron in a hydrogen atom, labeled in the figure n = 1, n = 2, and so on. I explain this diagram carefully below, but first we need to pause here to revisit one of the points made in the Introduction. In Section I.3, I note that when an electron is 2 In some substances, electrons in atoms can remain in excited states for an extended period of time. As the atoms in such a substance return to the ground state over time the substance gradually radiates the energy away. This is the way phosphorescence (glowing in the dark) works.

Hmm... Interesting.

Neon signs and phonons

When excited atoms in gases return to the ground state they do so by emitting photons. Neon signs are tubes of gas excited by high-voltage electricity. Their glowing colors are caused by the atoms returning to the ground state. When excited atoms in solids (and some liquids) return to the ground state, they can do so by emitting photons as gases do, but they can also emit phonons, packets of vibrational enλ ergy. Phonons can travel as waves through the crystal lattice in a solid, displacing the atoms from their equilibrium positions. In the image on the right, the wavelength of the emitted energy is shown in red (and the displacement of the atoms is greatly exaggerated). 19

Chapter 1 held at a certain distance away from the positive nucleus, it has a high pon=4 –0.14 × 10–18 J tential energy. As the electron is aln=3 –0.24 × 10–18 J lowed to move closer to the nucleus, emission of 4.09 × 10–19 J = its potential energy decreases just as 486 nm (blue) the gravitational potential energy of n=2 –18 –0.55 × 10 J an object above the ground decreases with decreasing height. Since the elecemission of –18 1.63 × 10 J = tron’s potential energy decreases as the 122 nm electron moves closer to the nucleus, (ultraviolet) the electron releases energy as it gets closer and closer to the nucleus. absorption of 2.04 × 10–18 J = In discussions of the energies of 97.3 nm (ultraviolet) electrons in atoms, it is customary to assign a reference value of 0 joules to the energy an electron has when it is completely free from the nucleus—in other words, when it is very far away. Then, since the electron’s potential energy decreases as it gets closer to the nucleus, the energy an electron has is expressed as a negative value. This Lowest Possible n=1 happens because instead of setting the –2.18 × 10–18 J Energy potential energy to be zero at the nuFigure 1.6. An energy-level diagram illustrating how quanta cleus, we set the zero energy reference of energy are absorbed and released by electrons in to be when the electron is far away hydrogen atoms. from the nucleus. So don’t let the negative energy values bother you. It makes sense that the electron’s energy is set to zero when it is far away from the nucleus because at that point the electron really has nothing to do with that nucleus. But to be consistent with what we know about potential energy, the electron’s energy must decrease as it enters the area near the nucleus, so its energy takes on increasingly negative values relative to the zero-energy reference. Going back now to Figure 1.6, let’s assume that an electron in a hydrogen atom is in the ground state, which means it has the lowest possible energy. Since hydrogen atoms only have one electron, this places the electron at the bottom of the figure at the first energy level, n = 1. An energy of –2.18 × 10–18 J is the energy an electron has when it is in this first energy level. Assume now that this electron absorbs a quantum of energy equal to 2.04 × 10–18 J from an incoming photon, indicated by the arrow pointing upward on the left side of the figure. From the Planck relation, you can verify that this corresponds to a wavelength of 97.3 nm, placing this photon in the ultraviolet region of the electromagnetic spectrum. The electron is now in the fourth energy level. If you subtract the energy of n = 1 from that of n = 4, the difference is the amount of energy the electron absorbed, 2.04 × 10–18 J. Remember, energy in atoms is quantized. Electrons can only have certain specific values of energy, and the permissible values of energy an electron in a hydrogen atom can have (for the first four energy levels) are the energies listed down the left side of Figure 1.6. Very quickly the atom emits this energy in the form of new photons and the electron drops back down to the ground state. But as you can see from the right side of the figure, the electron in the hydrogen atom has four different ways of doing this. 0J

Highest Possible Energy

Atomic Structure First, the electron can release the smallest permissible amount of energy each time it emits a photon. This causes it to drop down one energy level at a time, emitting three separate photons on its way back to the ground state, as shown by the sequence of three downward pointing arrows leading from n = 4 to n = 1. Second, the electron can first drop from n = 4 to n = 3 and then drop to n = 1. Third, the electron can first drop from n = 4 to n = 2 and then drop to n = 1. For this possibility, the emitted amounts of energy are shown in the figure. The energy emitted when dropping from n = 4 to n = 2 is 4.09 × 10–19 J. Using the Planck relation, you can calculate the wavelength of a photon with this energy. Doing so gives a wavelength of 486 nm, which is in the visible portion of the electromagnetic spectrum. A check of Figure 1.1 indicates that this wavelength corresponds to blue light. The drop from n = 2 to n = 4 is a much larger energy drop, 1.63 × 10–18 J. This energy corresponds to a wavelength of 122 nm, which is in the ultraviolet region and is not visible. Finally, the electron can drop back to n = 1 by emitting 2.04 × 10–18 J, the same amount of energy it absorbed in the first place. This energy produces a new ultraviolet photon with the same wavelength as the photon originally absorbed by the atom. Note from Equation (1.4) that energy and wavelength are inversely proportional. Longer wavelengths represent lower energies; infrared ultraviolet visible shorter wavelengths represent higher energies. This relationship is illustrated in Figure 1.7. The spectrum of wavelengths emitted longer wavelength shorter wavelength by each element is unique, which means lower energy higher energy that light spectra can be used to identify the element’s presence in a gas or solu- Figure 1.7. Wavelengths and energies in and near the tion. The science of such identifications is visible spectrum. called spectroscopy.

1.1.3 The Hydrogen Atom The hydrogen atom is the simplest atom, with only one electron, and thus it has been studied extensively. The wavelengths for the possible electron energy transitions in the first six energy levels of the hydrogen atom are shown in Figure 1.8. All the arrows in this diagram are shown pointing in both directions because the wavelengths shown can represent either the absorption or emission of energy. These are the energies hydrogen atoms can absorb and emit. In 1885, Swiss mathematician and physicist Johann Balmer discovered the formula that predicts the lines in the visible hydrogen spectrum. This series of lines is now called the Balmer series. In 1888, Swedish physicist Johannes Rydberg worked out the more general formula for all the hydrogen wavelengths. The ultraviolet and infrared lines in the hydrogen spectrum were not known initially (because they are invisible). But in 1906, American physicist Theodore Lyman observed the ultraviolet series that bears his name, and in 1908, German physicist Friedrich Paschen observed the infrared series of lines in the hydrogen spectrum. The Rydberg formula that predicts all these wavelengths has an interesting mathematical structure, and is worth showing here. Here it is: ⎛ 1 1⎞ 1 = R⎜ 2 − 2 ⎟ λ ⎝ n1 n2 ⎠

(1.5)

The R in this equation is the so-called Rydberg constant (1.097 × 107 m–1), and n1 and n2 represent the numbers for the two energy levels in question.

Chapter 1 Lyman Series (ultraviolet)

Balmer Series (visible)

Paschen Series (infrared) n=6

95 nm 97 nm

434 nm 486 nm

1282 nm 1875 nm

n=4 1094 nm n=3

410 nm 103 nm

656 nm n=2

n=5

I love pointing out to students that mathematical relationships are all over the place in nature—from the orbits of the planets, to the structure of gravity, to the ways atoms emit wavelengths of light. This is one of the reasons the natural world is so full of wonder. It is also wonderful that humans are able to identify and understand these mathematical relationships.

1.2 The Bohr Model of the Atom As mentioned on the opening page of this chapter, Danish physicist Niels Bohr introduced his new model of the atom in 1913. This new atomic model was of tremendous importance in the development of atomic theory. Rydberg’s formula predicting the lines in the hydrogen spectrum had been known since 94 nm 1888, but until Bohr’s model there was no theoretical basis for the observed spectrum. 122 nm In Bohr’s atomic model, the electrons orbit the nucleus like planets orbiting the sun. In the model, the electrons have fixed energies, the same energies as those shown in Figures 1.6 and 1.8. These different energy levels correspond to different orbits around the nucleus. Bohr correctly described the cause of the specific lines in the emission spectra of atoms—electrons absorbing energy and moving to higher energy levels and then releasing photons at specific energies as they move back to lower energy levels. Another significant feature of the Bohr n=1 model is Figure 1.8. Wavelengths for electron transitions in the the number n=4 (18) first six energy levels of the hydrogen atom. n=3 of electrons (8) that he per(8) n=2 mitted at each energy level. These numbers are shown in Figure 1.9. If you compare these numbers to the Periodic Table of the Elements shown on the inside back cover of this text, you see that the number of electrons in each energy level corresponds to the number of elements in each period (row) of the table: two in the first, eight in the second, eight in the third, and 18 in the fourth, etc. As powerful as it is, the Bohr model was known to have weaknesses from the start. For one thing, there is no explana- (2) n=1 tion for why electrons are able to stay in their orbits. Electrons Figure 1.9. Number of permissible moving in circles radiate energy, so one would think electrons electrons in the first four energy in orbits would gradually lose energy, slow down, and spiral in levels of the Bohr model of the atom. 22

Atomic Structure to the nucleus. Another issue is that for atoms other than hydrogen, the energies between the energy levels do not match up precisely with the observed wavelengths in the emission spectra of elements.

1.3

The Quantum Model of the Atom

1.3.1 Schrödinger and Pauli In 1926, Austrian physicist Erwin Schrödinger (Figure 1.10) published what is now called the Schrödinger equation. This was a landmark achievement and one of the hallmarks of 20th-century physics. For this work, Schrödinger received the Nobel Prize in Physics in 1933. Solutions to the Schrödinger equation are now understood to provide us with the details of the internal structure of energy levels in atoms. With the arrival of the Schrödinger equation, the quantum model of the atom began to unfold. The history of quantum physics is still being written. There are many mysteries associated with the behavior of electrons as described by the quantum model. But quantum theory has a colossally impressive string of achievements, and its success Figure 1.10. Austrian in predicting atomic behavior is undeniable. We know the quantum physicist Erwin Schrödinger model will continue to evolve, and may some day even be replaced. (1887–1961). But the details we consider in this section are now generally accepted as correct. Remember, chemistry is all about modeling—developing theories. Theories are explanations, and the quantum model is widely accepted as our best explanation of how atoms are structured. In addition to Schrödinger’s equation, there is one other theoretical milestone that we need to have in hand to understand the details to follow. In 1925, while Schrödinger was working on his equation, another Austrian physicist, Wolfgang Pauli (Figure 1.11) formulated what is now known as the Pauli exclusion principle. In short, the Pauli exclusion principle holds that no two electrons in the same atom can occupy the same quantum state. We will unpack this further as we go along. For this important contribution to quantum theory, Pauli won the Nobel Prize in Physics in 1945.

1.3.2 Shells, Subshells, and Orbitals The quantum state of an electron in an atom—its unique address, we might say, within the atomic quantum realm—is specified Figure 1.11. Austrian physicist by four different quantum numbers. According to the Pauli exclusion Wolfgang Pauli (1900–1958). principle, every electron in an atom has a unique quantum state. This is one of the laws of nature governing the way atoms are structured. This situation in atoms is analogous to postal addresses. Every postal customer in the U.S. has a unique address. For a house, this unique address requires four pieces of information—the street number, street name, city, and state. For an apartment complex, an apartment number is also required. (The zip code doesn’t have any additional location information in it; it just helps speed things up.) The physics behind these quantum numbers is quite complicated, and as an introductory chemistry student you would not normally be required to get much into that. However, introductory chemistry classes do generally now require students to learn the arrangement of shells, subshells, and orbitals in atoms for “energy levels” n = 1 through n = 4 because knowing this structure allows us to specify where the electrons are in an atom. And as you recall from the In23

Chapter 1 troduction, chemistry is all about electrons! So here we go. There is a lot of detail in this section, and it is all important. The phrase energy levels is in quotes just above for an important reason. We are transitioning now from the energy levels in Bohr’s atomic model to those of the far more accurate quantum model. In Bohr’s model, and in the hydrogen atom as we still understand it, there is only one energy level for each value of n. The quantum model is quite different, as we will see. Recall from Figure 1.9 that the numbers of electrons permitted in the first four levels of Bohr’s model are 2, 8, 8, and 18. These numbers correspond to the number of elements in the first four periods of the periodic table. Bohr was on the right track, but did not initially perceive the correct pattern. We now refer to n as the principle quantum number, and in every atom except hydrogen there are multiple energy levels associated with each value of n. As explained in detail below, the number of electrons allowed for each value of n is actually 2n2. This gives us 2, 8, 18, and 32 electrons in the various energy levels associated with n = 1 through n = 4. The clusters of energy levels associated with each value of n are commonly called shells. As I state just above, the quantum state of an electron in an atom, including its energy, is specified by four quantum numbers; the principle quantum number—the shell number—is the first of them. So beginning with the principle quantum number you are already familiar with, here is a list of the names and other details for the four quantum numbers: 1. Principle Quantum Number, n Values for n are the integers 1, 2, 3, 4, 5, ... These are the main clusters of energy levels in the atom, also called shells. So far as we know, there is no highest value for n. 2. Azimuthal Quantum Number, l Within each shell except the first one (n = 1), there are subshells. The number of subshells in a shell is equal to the principle quantum number. For example, for n = 3 there are three subshells. Values for l are integers ranging from 0 to (n – 1). Typically, these subshells are referred to by the letters s, p, d, f, and g rather than by the values of l. These common letter designations are shown Table 1.1. The azimuthal quantum numbers describe specific types of subshell configurations. So for example, within any shell the s subshell is always structured the same way. Likewise, the p subshell has the same general structure in every shell except n = 1 (since n = 1 doesn’t have a p subshell). Again, the number of subshells in a given shell is equal to the principle quantum number. So, in the first shell there is one subshell, denoted as 1s. In the n = 2 shell, there are two subshells, denoted as 2s and 2p, and so on. (Note: The azimuthal quantum number is also sometimes called the angular momentum quantum number.)

l value

Common Letter Designation

Table 1.1. Letters used to designate values of the azimuthal quantum number, l.

3. Magnetic Quantum Number, ml Within each subshell (numbered l), the possible values for ml are the integers ranging from –l to l. So, in a subshell with l = 2, the values for ml are –2, –1, 0, 1, and 2. The magnetic quantum number is associated with specific shapes and orientations of orbitals within a subshell. A important point to note is that any orbital in an atom can hold at most two electrons. 4. Spin Projection Quantum Number, ms As you recall, the Pauli exclusion principle requires every electron in an atom to be in a unique quantum state. That is, each electron has a unique set of quantum numbers. And since each orbital can hold two electrons, we need one more piece of information to distinguish from one another the quantum states of the two electrons. This char24

Atomic Structure acteristic is called spin. Unfortunately, it’s a very misleading term because electrons aren’t really spinning. In fact, it’s pretty hard to say exactly what they are doing. But anyway, accepting spin as a real property analogous in some way to spinning, any two electrons in the same orbital have opposite spins. The two possible values for electron spin are ms = +1/2 and ms = –1/2, and we call these “spin up” and “spin down.” At this point in your career you really don’t need to worry about what these strange names and numbers mean. The fact is, if there are two electrons in the same orbital (and there can be at most two) one has spin up and one has spin down. This final quantum specification allows each electron in every atom to inhabit a unique quantum state. All this information pertaining to the first three quantum numbers is summarized in Table 1.2. Hopefully your understanding of all these shells, subshells and orbitals will be enhanced by looking at images of computer models of the orbitals. Let’s be a bit clearer about what these orbitals are: they represent the solutions to the Schrödinger equation for electrons in atoms with different energies. Table 1.3 depicts the orbitals in the various subshells associated with the first three shells, n =1 through n = 3. Note first that in each shell there is an s orbital. These are spherical in shape. The models shown depict the sphere cut in half so you can see the relative sizes. If you look carefully at the 2s orbital, you can see the tiny 1s orbital inside it. And inside the 3s orbital you can see both the 1s and 2s orbitals inside it. We are coming back to electron energy soon, but for now note that within any shell, the s orbital is the lowest energy orbital in that shell. Note also that for all these orbital arrangements, the atomic nucleus is at the center. All orbitals are symmetric about the nucleus. Beginning with n = 2, there is a p subshell in each shell, and beginning with n = 3 there is also a d subshell in each shell. The orbitals in the p subshell are usually described as resembling “dumbbells” because of their twin lobes. There are three of these twin-lobed orbitals in each p subshell, each oriented at right angles to the other two. For this reason, they are designated the px, py, and pz orbitals—they can be thought of as lined up along the x, y, and z axes in a threedimensional Cartesian coordinate system as depicted in Figure 1.12. (I explain the elongated appearance of the orbitals shown in Figure 1.12 shortly.) In Table 1.3, in order to make the s orbitals large enough to see and still have room to fit the p and d orbitals on the page, the p and d orbitals are shown much smaller than their actual size relative to the s orbitals. Looking now at the n = 3 orbitals in Table 1.3, note that the 3p orbitals are shown surrounding the 2p orbitals. The three 3p orbitals are superimposed on each other just as the 2p ones are n

Possible Values of l

Subshell Name

Possible Values of ml (Each value corresponds to one orbital.)

Number of Orbitals in the Subshell

Total Number of Orbitals in the Shell (= n2)

–1, 0, 1

–2, –1, 0, 1, 2

–1, 0, 1

–2, –1, 0, 1, 2

–3, –2, –1, 0, 1, 2, 3

4 9

Table 1.2. Subshells and orbitals for the n = 1 through n = 4 shells.

Chapter 1 (Figure 1.12). Finally, you can see that shapes of the five 3d orbitals are pretty bizarre. These orbitals are also superimposed on each other, and the whole bunch of them is superimposed on top of all the other orbitals in the table. Then, of course, there are all the orbitals for higher principle quantum numbers superimposed on top of them. Beginning with the n = 4 shell (not shown in the table) there is an f subshell in each shell. There are seven orbitals in each f subshell, Shell n=1 s orbitals

n=2

n=3

3px

2px

2p y

3p y

2pz

3pz

Subshell

p orbitals

3dz 2 d orbitals

3dxz

3dxy Table 1.3. Shapes of s, p, and d orbitals for the n = 1, n = 2, and n = 3 shells.

3d yz

3dx 2 − y 2

Atomic Structure and they sort of resemble the d orbitals, only with six or eight lobes instead of four. Recall that each orbital can house a maximum of two electrons (with opposite spins). For example, just to be clear, the 2px orbital with its two lobes is a single orbital (even though Table 1.3 shows the two lobes in different colors). Likewise, the 3dz 2 orbital with its two lobes and doughnut around the middle is also a single orbital. With a maximum of two electrons in each orbital, you can see that the first shell, n = 1, can hold at most two electrons, both in the 1s orbital. The n = 2 shell can hold a maximum of eight electrons: two in the 2s orbital, and two in each of the three 2p orbitals. The n = 3 shell can hold up to 18 electrons: two in the 3s orbital, six in the 3p orbitals, and a total of 10 in the 3d orbitals. Figure 1.12. The elongated “probability Before we move on and get back to talking about en- distributions” of the 2s and 2p orbitals ergy, one more important point should be made about or- shown together. (2s = green; 2px = bitals. As noted above, the orbitals shown in Table 1.3 are yellow, 2py = blue, 2pz = red) the solutions to the Schrödinger equation. However, it is not correct to think of these shapes as locating where the electrons are. (Remember, the world of quantum mechanics is weird.) But it turns out that if we square the solutions to the Schrödinger equation we get shapes indicating probabilities of where the electrons are. This is what is depicted in Figure 1.12. Squaring the 2p solutions elongates the shapes of the orbitals. These orbitals that come from squaring solutions to the Schrödinger equation are called probability distributions. They should be envisioned as fuzzy at the edges and denser in the middle, indicating a lower probability that an electron is at the edge of the orbital and a higher probability that an electron is in the center part of the orbital. Note just one more feature of the orbital arrangements: just because an orbital has more than one part—like the two lobes of a p orbital—does not mean that one electron is in one lobe and the other electron is in the other lobe. Instead, both electrons inhabit both lobes. Even stranger, to pass from one lobe to the other the electron somehow passes right through the atomic nucleus. (Don’t hurt your brain by trying too hard to understand this. No one else understands it either! Electrons are very strange.)

1.3.3 The Aufbau Principle, the Madelung Rule, and Hund’s Rule Now that you know how the orbitals are arranged, we return to the topic of electron energies. Let’s begin by recalling how one knows how many electrons an atom has. Unless it has ionized, an atom has the same number of electrons as protons, and the number of protons is given by the atomic number (Z). For example, if you check the periodic table inside the back cover of the book, you see that iron is element 26. This means an atom of iron has 26 protons and 26 electrons. The protons are all in the nucleus with the neutrons. The electrons are distributed around in various orbitals. Figure 1.13 is another type of energy level diagram and illustrates the energies associated with the different orbitals. In this diagram, each little square represents an orbital, and each string of connected squares represents a subshell. On the left are the orbital energies for the hydrogen atom. As you see, all orbitals associated with a given principle quantum number have the same energy. These are the energies shown in Figures 1.6 and 1.8. On the right side of Figure 1.13 is a general arrangement depicting the energies for atoms other than hydrogen. Here the energies go up with each subshell. For example, subshell 4f has 27

Hydrogen Atoms

increasing energy

Chapter 1

5f 5d

4d 3d

3s 2p 2s

Other Atoms

Figure 1.13. In hydrogen atoms, all orbitals within a given shell are at the same energy level. In atoms of other elements, orbital energy increases with increasing azimuthal quantum number, and the sequence of energies follows the Madelung rule.

a higher energy than 4d, which has a higher energy than 4p, which has a higher energy than 4s. Also, note especially that the energies associated with different principle quantum numbers (shells) overlap. Thus, subshell 4s has a lower energy than subshell 3d. An important point to note about the right side of Figure 1.13 is that the exact energies associated with various subshells are different for every atom. With only one electron, the orbitals in an energy level of a hydrogen atom are basically all the same. But with multiple electrons repelling each other in an atom, the subshells begin spreading upward and each subshell is at a different energy. The amount of spread—and thus the exact energy associated with each subshell—is different for every atom. The important consequence of this for what we have covered so far in this chapter pertains to atomic spectra. The energy released by an electron transition from, say, a 5d orbital to a 4p orbital depends on the atom—that is, the element—involved. As you know, the energy in an emitted photon determines its wavelength and color (the Planck relation). The fact that the energies for the different orbitals depend on the element is the reason why spectroscopy can be used to identify the presence of elements in a sample. Each atom emits its own spectrum of wavelengths corresponding to the unique energy differences between the orbitals in that particular kind of atom. The colors emitted by excited atoms in two metals are illustrated in Figure 1.14. The images show lithium and copper wires heated in a stove-top gas flame, causing electrons in the metal atoms to absorb photons of heat energy (electromagnetic radiation in the infrared region). As the electrons return to the ground state, they emit photons of visible light, and the colors produced depend on the energies of the subshells in the atoms of the respective metals. If the flames are observed through a prism, the colors in the flames are separated into a line spectrum and the inFigure 1.14. Lithium (top) and copper flame tests.

Atomic Structure dividual color wavelengths can be identified. A test like this that uses flame as the energy source for exciting the metal is called a flame test. With the energy picture under our belts, we are finally ready to describe how electrons are arranged in atoms. We are describing here the electron positions when atoms are in the ground state. You already know that when atoms are excited, electrons jump from ground state energies up to higher energies. There are three principles involved in determining electron arrangement in ground state atoms. The first is the Aufbau principle, named after a German word meaning “building up.” The Aufbau principle states that electrons fill places in orbitals in order of increasing energy, starting from the lowest energy orbital and going up from there. Remember: chemistry is all about minimizing energy. Electrons in a ground-state atom go into the lowest energy orbitals available. The second principle is the Madelung rule. This principle specifies the order of the shells and orbitals for increasing energy. On the right side of Figure 1.13, the sequence the orbitals are in as energy increases follows the Madelung rule. Another common way of depicting the sequence of energies according to the Madelung rule is shown in Figure 1.15. If you start at the top and follow the arrows in descending order, you get the 1s same sequence of orbitals as shown on the right side of Figure 1.13. The 2s 2p mathematical principle involved is that each arrow in Figure 1.15 repre3s 3p 3d sents a particular value of the sum of the principle quantum number and 4s 4p 4d 4f the azimuthal quantum number, n + l. For example, look at the arrow starting at the 3d orbital. For 3d, n = 3 5s 5p 5d 5f and l = 2, so n + l = 5. (The values of l are shown in Table 1.1.) For 4p, n = 4 6s 6p 6d (etc.) and l = 1, and n + l = 5. For 5s, n = 5 and l = 0, so n + l = 5. So the subshells 7s 7p fill up in the sequence shown in Figure 1.15. The third principle involved in electron arrangements is Hund’s rule, Figure 1.15. Since each which applies to the case of subshells that are only partially filled. Hund’s red arrow represents a rule states that if orbitals of equal energy are available within a subshell, particular value of n + electrons fill them all up singly before they begin doubling up in orbitals. For l, this diagram shows the order in which example, as you can see from the right side of Figure 1.13, the 3d subshell the subshells fill with contains five orbitals of equal energy. According to Hund’s rule, if there are electrons, according to electrons in this subshell, but not enough electrons to fill the subshell, the the Madelung rule. electrons go into the orbitals as one electron per orbital until each of the five orbitals has one electron in it. After that, any remaining electrons go in as the second electron in each orbital until each electron has a place. And again, remember that all orbitals can hold at most two electrons. The principle at work behind Hund’s rule is again energy minimization. Spreading single electrons in the orbitals of unfilled subshells is a lower energy configuration than putting pairs of electrons together when other orbitals remain empty. Minimizing the energy this way also makes the atom more stable, just as the cone on its side in Figure I.6 is more stable than the cone on its point.

1.4

Electron Configurations

1.4.1 Electron Configurations and Orbital Diagrams You may be pleased to know that the ocean of information described in the previous section will be a lot easier to remember after you have had a bit of practice writing electron configurations to indicate where all the electrons are in an atom of a given element. The electron configuration for a given element is a list, written in a particular format, of all the subshells in use in an atom and how many electrons are in each one. As an example, consider iron, atomic number 26 29

Chapter 1 (Z = 26). There are 26 electrons in an atom of iron. The subshells required to hold them all, in order of increasing energy according to the Madelung rule, are as follows: 1s

holds 2 electrons

holds 6 electrons

holds 2 electrons

holds 6 electrons

holds 2 electrons

holds 6 electrons in 10 places

The electron configuration is formed simply by chaining these together, placing the numbers of electrons as superscripts on the subshells they go with, without any punctuation. In front of the electron configuration, it is customary to place the element’s chemical symbol followed by a colon. So, the electron configuration for iron is written as follows: Fe: 1s22s22p63s23p64s23d 6 Electron configurations only indicate subshells; they do not indicate which orbitals electrons are in inside the subshells. But we can use an orbital diagram similar to Fig3p ure 1.13 to show more precisely where the electrons are. Remember, Hund’s 3s rule comes into play, requiring that orbitals of equal energy each receive one 2p spin-up electron before any of them take a second spin-down electron. A great metaphor for this was first used by Wolfgang Pauli, who formulated 2s the Pauli exclusion principle. Pauli said that when filling up the orbitals in a subshell, electrons are like passengers filling a bus. Each takes a seat by himself until every seat has one person in it. After that, people start doubling up. 1s Figure 1.16 shows the electron arrangement for phosphorus, Z = 15. Each of the little arrows represents one electron, with upward arrows representing spin up and downward arrows representing spin down. Notice that Figure 1.16. The the three electrons in the 3p subshell are placed so that each orbital contains electron arrangement for phosphorus, with 15 one spin-up electron, as Hund’s rule requires. electrons. To make an orbital diagram, you simply show the orbitals in order, side by side, and put in the arrows representing the electrons. Thus, the orbital diagram and electron configuration for phosphorus are as follows: Orbital Diagram 1s 2s

Electron Configuration

P: 1s22s22p63s23p3

Here are three more examples: sodium (Na) with 11 electrons, chlorine (Cl) with 17 electrons, and nickel (Ni) with 28 electrons: Z

Orbital Diagram

1s 2s

Electron Configuration

Na:

1s22s22p63s1

Cl:

1s22s22p63s23p5

Ni:

1s22s22p63s23p64s23d 8

Note in each case that the superscripts add up to the number of electrons being represented in the notation, 11, 17, and 28 in the three examples above. 30

Atomic Structure 1 1 1

s-block elements (includes He) 2

p-block elements 13

d-block elements 3

2 17

f-block elements

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

Figure 1.17. Common terms applied to groups of elements based on the type of suborbital that is being filled as we move through the block in a given period (row). Group numbers shown are the numbers typically shown on the standard table with the f-block elements taken out.

To write an electron configuration, you need to know the atomic number (Z) to get the number of electrons, the number of electrons that can reside in each of the types of subshells (s, p, d, and f; we won’t deal with g subshells), and the energy sequence according to the Madelung rule. The periodic table itself is the best aid to writing an electron configuration. Figure 1.17 is a depiction of the periodic table in its full, long form. The rare-earth elements, normally shown separated out beneath the main table, are shown in their rightful place.3 Notice the captions on the different blocks of elements in the table. There are two groups (columns) in the s block, because s subshells can only hold two electrons. The p subshells can hold six electrons because there are three orbitals in each p subshell, and the p block is six groups wide. In the same way, the d block is 10 groups wide, and the f block is 14 groups wide. (See Table 1.2 to confirm the number of orbitals in a subshell, keeping in mind that each orbital can hold up to two electrons.) Now look what happens as we move in order through elements in the table: • •

•

Hydrogen (H) has one electron, helium (He) has two. In the element at the end of the 1st period (row)—helium—the 1s subshell is full. Lithium (Li) has three electrons, beryllium (Be) has four. With Be, the 1s and 2s subshells are full. The 5th electron possessed by boron (B) goes into the 2p subshell. So do all the additional electrons added in elements 6 through 10. Neon (Ne) has 10 electrons, so in the element at the end of the 2nd period (neon), subshells 1s, 2s, and 2p are all full. Sodium (Na) has 11 electrons, and the 11th one goes in the 3s subshell. Magnesium (Mg) has 12, and the 12th one fills the 3s subshell. Aluminum (Al) has 13, and the 13th one goes in the 3p subshell. In the element at the end of the third period—argon (Ar)— the 1s, 2s, 2p, 3s, and 3p subshells are full. The 4th period begins with potassium (K) and calcium (Ca), in which new electrons are placed in the 4s subshell. Then look what happens next: the next element, scandium (Sc), has 21 electrons and the 21st one goes into a d subshell, the 3d subshell, in fact. Each new electron for elements 21 through 30 goes into the 3d subshell. Then the 31st electron in gallium (Ga) is placed in the 4p subshell. In the element at the end of the 4th period—krypton (Kr)—the 1s, 2s, 2p, 3s, 3p, 4s, 3d, and 4p subshells are full.

3 The f-block elements shown are also called the inner transition metals or rare-earth elements. They are usually removed from the table and shown beneath it for the simple reason that with them in place, the table is inconveniently wide. Since most of our work in chemistry is with s-, p-, and d-block elements, the removal of the f-block elements in the standard representation of the periodic table doesn’t cause much trouble. 31

Chapter 1 This pattern continues, and the order of new shells coming into use continues to follow the Madelung rule. Now you can see why the blocks of elements in Figure 1.16 are identified the way they are. Moving from left to right across any period in the table, the additional electron for atoms of the next element goes into a subshell of the type indicated by the name of the block the element is in. Be careful when writing electron configurations for elements at the beginning of the d and f blocks. The first d subshell that occurs in the sequence of shell filling is the 3d subshell, even though the elements that fill it are in the 4th period. Table 1.2 and Figure 1.15 will help remind you that the number of subshells for a particular principle quantum number n is equal to n.

1.4.2 Condensed Electron Configurations You may already know that the elements in Group 18, the noble gases, are very unreactive. In fact, these elements eluded discovery by researchers for a long time. Since they don’t form compounds, scientists didn’t even know they existed! 5f The reason the noble gases are so nonreactive is that their 7s 6p 5d electron arrangements are very stable, low-energy configura4f 6s 5p tions. Obviously, each of the noble gases is at the end of a 4d 5s period in the periodic table (Group 18). Figure 1.18 illustrates 4p 3d the pattern that occurs in the orbital filling of these elements. 4s 3p In each case, all orbitals are filled up to but not including the 3s s orbital of the next principle quantum number. 2p Since the noble gases are so stable, occurring as they do Xe at the end of each period, the chemical symbols of the noble 2s Kr gases are used to form the so-called condensed electron conAr Ne figurations. The condensed electron configuration is a shorter, more convenient form. He Here’s an example to show how this works. A glance at the 1s periodic table shows that the only difference in the electron Figure 1.18. The orbital filling pattern configurations of, say, titanium (Ti, Z = 22) and argon (Ar, of the noble gases. Z = 18), is that titanium has four extra electrons. The electron configurations for argon and titanium are: Ar:

1s22s22p63s23p6

Ti:

1s22s22p63s23p64s23d 2

The condensed electron configuration for any element is written by using the chemical symbol of the noble gas in the previous period to represent all the filled orbitals up to that point, and then just adding on the orbitals in the period where the element is. The noble gas chemical symbol is always written in square brackets. The condensed electron configuration for titanium is written as Ti:

[Ar]4s23d2

As one more example, the condensed electron configuration for phosphorus (Z = 15) is written as P:

[Ne]3s23p3

In the condensed electron configuration, the inner-shell electrons lumped together under the noble gas symbol are called the core electrons. 32

Atomic Structure

1.4.3 Anomalous Electron Configurations There are a few elements with electron configurations that are notable exceptions to the ordinary rules. All these exceptions are d-block or f-block elements, as shown in Figure 1.19. We would expect the condensed electron configuration for chromium (Z = 24) to be Cr:

[Ar]4s23d 4

However, it is not! One of the 4s electrons goes into the 3d subshell instead, giving Cr:

[Ar]4s23d 9

Instead, it is Cu:

Nb Mo

[Ar]4s13d 5

Copper (Z = 29), does the same thing. We would expect the condensed electron configuration to be Cu:

Cr 41

Ru Rh Pd Ag 78

La Ce 89

Gd 91

Ac Th Pa

Figure 1.19. Elements with anomalous electron configurations.

[Ar]4s13d10

The electron configurations Atomic Condensed Element Symbol for all the anomalous elements are Number, Z Configuration shown in Table 1.4. Notice that pal24 Cr Cr: [Ar]4s13d 5 ladium (Z = 46) is so anxious to get chromium electrons into the 4d subshell that copper 29 Cu Cu: [Ar]4s13d10 it steals both the electrons that or- niobium 41 Nb Nb: [Kr]5s14d 4 dinarily would be in the 5s subshell molybdenum 42 Mo Mo: [Kr]5s14d 5 and places them in the 4d subshell ruthenium 44 Ru Ru: [Kr]5s14d 7 to fill it up. The electron configurations for rhodium 45 Rh Rh: [Kr]5s14d 8 these elements are not difficult to palladium 46 Pd Pd: [Kr]4d10 remember. For the d-block excep47 Ag Ag: [Kr]5s14d10 tions, except for palladium, they silver 57 La La: [Xe]6s25d1 move one s-subshell electron into a lanthanum d subshell at the highest energy in cerium 58 Ce Ce: [Xe]6s24f 15d1 the atom. Palladium is the exception gadolinium 64 Gd Gd: [Xe]6s24f 75d1 among exceptions: it moves two. 78 Pt Pt: [Xe]6s14f 145d 9 The f-block exceptions move one platinum f subshell electron into a d subshell gold 79 Au Au: [Xe]6s14f 145d10 at the highest energy in the atom. actinium 89 Ac Ac: [Rn]7s26d1 As with the d-block exceptions, thorium 90 Th Th: [Rn]7s26d2 there is one exception among the 91 Pa Pa: [Rn]7s25f 26d1 f-block anomalies: thorium moves protactinium two f-subshell electrons into the d uranium 92 U U: [Rn]7s25f 36d1 subshell. neptunium 93 Np Np: [Rn]7s25f 46d1 Although the details are com96 Cm Np: [Rn]7s25f 76d1 plex, the bottom line for these ex- curium ceptions to the normal rules is that Table 1.4. Elements with anomalous electron configurations. 33

Chapter 1 there are other factors coming into play in their atomic structure. At high energy levels, the orbital energies are very close together. The anomalous configurations represent the arrangement of electrons that minimizes the energy in the atom.

1.5

Isotopes and Atomic Masses

1.5.1 Isotopes As you know, the atomic number (Z) of an element designates the number of protons in the nucleus of an atom of that element. For a given element, the atomic number is fixed: if an atom has a different number of protons, it is an atom of a different element. But the number of neutrons that may be present in the nucleus is not fixed. For most elements, there are variations in the number of neutrons that can be present in the nucleus. The varieties are called isotopes. For most elements, there is one isotope that is the most abundant in nature and several other isotopes that are also present but in smaller quantities. The general term for any isotope of any element is nuclide. Isotopes are designated by writing the name of the element followed by the number of nucleons (protons and neutrons) in the isotope. The number of nucleons in a nucleus is called the mass number. For example, the most common isotope of carbon is carbon-12, accounting for about 98.9% of all the naturally occurring carbon. In the nucleus of an atom of carbon-12 there are six protons and six neutrons. There are two other naturally occurring carbon isotopes. Carbon-13 with seven neutrons accounts for about 1.1% of natural carbon. Atoms of carbon-14, of which only a trace exists in nature, have eight neutrons in the nucleus.

1.5.2 The Unified Atomic Mass Unit The mass of a single atom is an extremely small number. But so much of our work in chemistry depends on atomic masses that scientists having been using units of relative atomic mass for a long time—all the way back to John Dalton, before actual masses of atoms were even known. Prior to the discovery of isotopes in 1912, the so-called atomic mass unit (amu) was defined as 1/16 the mass of an oxygen atom. After the discovery of isotopes, physicists defined the amu as 1/16 the mass of an atom of oxygen-16, but the definition used by chemists was 1/16 the average mass of naturally occurring oxygen, which is composed of several isotopes. To eliminate the confusion resulting from these conflicting definitions, the new unified atomic mass unit (u) was adopted in 1961 to replace them. Many texts continue to use the amu as a unit, but they define it as the u is defined. Strictly speaking, the amu is an obsolete unit that has been replaced by the u, now also called the dalton (Da). The u and the Da are alternative names (and symbols) for the same unit. The use of the dalton has increased in recent years, particularly in molecular biology. The unified atomic mass unit, u, is defined as exactly 1/12 the mass of an atom of carbon-12. Table 1.5 lists a few nuclides and their atomic masses using the u as a unit of mass. All the elements listed exist as other isotopes in addition to those shown, but as you see from the percentage abundances, the ones shown are the major ones for the elements represented in the table.

1.5.3 Atomic Masses In addition to the atomic number, the Periodic Table of the Elements lists the atomic mass in unified atomic mass units (u) for each element. But since there are multiple isotopes for just about every element, the atomic mass values in the periodic table represent the weighted average of the masses of naturally occurring isotopes. An example of a weighted average is the average age of the students in the sophomore class at your school. Let’s say there are 47 sophomores, 40 of whom are 15 years old and 7 of whom are 16 years old at the beginning of the school year. To determine the average age of these students, 34

Atomic Structure let’s first determine the proportion of the students at each age.

Nuclide

Mass (u)

Abundance (%)

hydrogen-1

1.0078

99.9885

40 = 0.851 (85.1%) 47

hydrogen-2

2.0141

0.0115

carbon-12

12.0000

98.93

7 = 0.149 (14.9%) 47

carbon-13

13.0034

1.078

silicon-28

27.9769

92.223

silicon-29

28.9765

4.685

To calculate the average age, we first multiply each student age by the proportion of students of that age to find the contribution to the average from each age group. Then we add the contributions together to find the weighted average age for the sophomore class. 15 years⋅0.851 = 12.8 years +16 years⋅0.149 = 2.38 years =15.2 years

silicon-30

29.9738

3.092

chlorine-35

34.9689

75.76

chlorine-37

36.9659

24.24

calcium-40

39.9626

96.941

calcium-42

41.9586

0.647

iron-54

53.9396

5.845

iron-56

55.9349

91.754

iron-57

56.9354

2.119

iron-58

57.9333

0.282

29 copper-63 62.9296 We perform a similar calculation 29 copper-65 64.9278 when computing the average atomic 92 uranium-235 235.0439 mass of an element from the masses 92 uranium-238 238.0508 of its isotopes, as shown in the folTable 1.5. Major isotopes for a few elements. lowing example.

69.15 30.85 0.7204 99.2742

Example 1.2 Given the isotope masses and abundances for copper-63 and copper-65 in Table 1.5, determine the atomic mass for naturally occurring copper. Multiply each isotope’s mass by its abundance to get the isotope’s contribution to the average atomic mass of the element. Then add together the contributions from each isotope. The data from the table are: copper-63: mass = 62.9296 u, abundance = 69.15% copper-65: mass = 64.9278 u, abundance = 30.85% 62.9296 u ⋅0.6915 = 43.51 u + 64.9278 u ⋅0.3085 = 20.03 u = 63.55 u Compare this value to the value shown in the periodic table inside the back cover of the text.

Chapter 1 ergy involved in binding the nucleus of the atom together. (The mass of nucleons bound together in a nucleus does not equal the proton 1.007277 u sum of their individual masses.) Table 1.6 shows the masses of neutron 1.008665 u the three basic subatomic particles in unified atomic mass units. Still, the proton and neutron masses are very close to unity electron 0.0005486 u (one) and the electron mass is extremely small. This means that Table 1.6. Masses in u of the three for elements with a very large abundance of one isotope we can basic subatomic particles. use the atomic mass and atomic number in the periodic table to determine the numbers of protons and neutrons in the nucleus of the most common isotope. For example, from Table 1.5, the mass of uranium-238 is very close to 238 u. Since an atom of uranium-238 has 92 protons, the balance of the mass is essentially all neutrons. Thus, there are 238 – 92 = 146 neutrons in uranium-238. Particle

Mass

1.5.4 The Mole and the Avogadro Constant When solving problems in chemistry, we are generally working with chemical reactions in which huge numbers of atoms are involved, including all the naturally occurring isotopes, so performing reaction calculations with the masses of individual atoms is not practical. However, the average mass of a given multiple of some kind of atom is simply that multiple times the atomic mass. The mass of one million atoms of aluminum is 1,000,000 times the atomic mass of aluminum. In chemistry, the standard bulk quantity of substance used in calculations is the mole (mol). The mole is a particular number of particles of a substance, just as the terms dozen, score, and gross refer to specific numbers of things (12, 20, and 144, respectively). A mole is exactly 6.02214076 × 1023 particles of a substance. This value is known today as the Avogadro constant, NA. More formally, the Avogadro constant is defined as exactly: N A = 6.02214076 × 1023 mol −1

(1.6)

Usually, we just round this value to 6.022 × 1023 mol–1. In the next section, I’ll describe why this value is what it is, instead of being a more convenient round number. For the moment, let’s focus on what it means. Now, don’t freak out over the unit of measure. Allow me to explain. Raising a unit of measure to the power –1 is mathematically equivalent to placing the unit in 1 a denominator because x −1 = . In other words, Equation (1.6) is the same thing as saying x “6.02214076 × 1023 per mole.” To make things even clearer, it’s okay to say it this way: “NA is about 6.022 × 1023 particles per mole.” This is the way I like to think of it when performing unit conversions, as we do quite a lot in coming chapters. Without the units of measure, the value 6.02214076 × 1023 is called Avogadro’s number. With the units, it is called the Avogadro constant. Using this terminology, the mole can be defined this way: A mole is the amount of a pure substance (element or compound) that contains Avogadro’s number of particles of the substance. Let’s now consider what we mean when we refer to particles of a substance. For substances that exist as molecules, the particles are the molecules. For substances that exist as individual atoms, the particles are the individual atoms. Metals are like this, since a pure metal is composed of individual atoms of the same element joined together in a crystal lattice. The noble gases are also like this. The noble gases are located in the far right-hand column of the Periodic Table of the Elements. As I discuss more in coming chapters, atoms of noble gases are almost completely

Atomic Structure unreactive—they don’t bond with other atoms at all. At ordinary temperature and pressure, the noble gases are gases composed of individual atoms. For crystalline compounds, the “particles” in a mole of the substance are the formula units in the crystal lattice. A formula unit is one set of the atoms represented by the chemical formula of the compound. For example, the chemical formula for calcium carbonate is CaCO3. One formula unit of calcium carbonate includes one calcium atom, one carbon atom, and three oxygen atoms. The value of the Avogadro constant was determined approximately by French Physicist Jean Perrin (Figure 1.20) in the early 20th century. Perrin determined the value of the constant through several different experimental methods. In the 19th century, many scientists did not yet accept the existence of atoms as a scientific fact and Perrin’s research put the atomic nature of matter beyond dispute. For this work, he received the Nobel Prize in Physics in 1926. Perrin proposed naming the constant after Amedeo Avogadro, a 19th-century Italian scientist who was the first to propose that the volume of a gas at a given temperature and pressure is proportional to the number of particles of the gas (atoms or molecules), regardless of the identity of the gas. In fact, at 0°C and atmospheric pressure, one mole of any gas occupies a volume of 22.4 L. Figure 1.20. French physicist Jean

1.5.5 Molar Mass and Formula Mass

Perrin (1870–1942).

Since 2019, the value of the Avogadro constant in Equation (2.2) is exact by definition. But the number has the value it does because it was originally chosen so that the average atomic mass in u of a molecule of a compound, as computed from the mass values in the periodic table, would be numerically equivalent to the mass of one mole of the compound in grams per mole. (We address these calculations below.) Now, recall that the definition of the unified atomic mass unit (or dalton) is such that an atom of carbon-12 has a mass of exactly 12 u. According to the original definition of Avogadro’s number, there were also exactly 12 grams of carbon-12 in one mole of carbon-12. So according to these definitions, an atom of carbon-12 has a mass of exactly 12 u, and a mole of carbon-12 had a mass of exactly 12 grams. This quantity, the mass of one mole of a substance, is called the molar mass. Because of the way the molar and atomic masses were defined, the molar mass for an atom was numerically equivalent to the atomic mass. As a result of the 2019 redefinition of Avogadro’s number, the atomic mass in u and the molecular mass in g/mol are no longer exactly equivalent. However, they are extremely close and may still be treated as equal for practical purposes. (The difference is a factor of only about 4 × 10–10.) Even though the exact equivalence ended in 2019, these are still very handy definitions! For example, from the periodic table we find that the average mass of one atom of silicon (Z = 14) is 28.0855 u. This also tells us that the mass of one mole of silicon is 28.0855 g, so the molar mass of silicon is 28.0855 g/mol. Likewise, from the periodic table we find that the average mass of one atom of copper (Z = 29) is 63.546 u. This also tells us that the mass of one mole of copper is 63.546 g, so the molar mass of copper is 63.546 g/mol. For the elements that exist as single atoms, the molar mass in g/mol and the atomic mass in u are numerically equivalent (almost). From the periodic table, we can also determine the molar mass of compounds—the mass of a mole of the compound. We simply add up the molar masses for the elements in the chemical formula, taking into account any subscripts present in the formula, and we have the molar mass for the compound in g/mol. If we add up the element atomic masses in unified atomic mass 37

Chapter 1 Quantity

Units

Definition

molar mass

g/mol

formula mass

The mass of one formula unit of a substance. Numerically nearly equivalent to the molar mass.

molecular mass

The average mass of a single molecule of a molecular substance. Numerically equivalent to the formula mass. (May also be converted to grams and expressed in grams, see Section 2.4.4.)

Table 1.7. Definitions and units for molar mass, formula mass, and molecular mass.

units we obtain what is called the formula mass of the compound in u. If the compound is molecular, then the formula mass may also be referred to as the molecular mass, the average mass of a single molecule of the substance. The details of these three different mass terms are summarized in Table 1.7. Example 1.3 Determine the formula mass and molar mass for water, H2O. We note that since water is composed of molecules, the formula mass may also be called the molecular mass. From the periodic table, the atomic masses of hydrogen (H) and oxygen (O) are: H: 1.0079 u O: 15.9994 u There are two hydrogen atoms and one oxygen atom in each water molecule, so we multiply these numbers by the element atomic masses and add them up to get the formula mass of H2O.

( 2 ×1.0079 u ) + (1×15.9994 u ) = 18.0152 u Thus, the formula mass for water is 18.0152 u. This is also the molecular mass. The calculation of the molar mass is identical, except we use units of g/mol instead of u. From the periodic table, the molar masses of hydrogen (H) and oxygen (O) are: g mol g O: 15.9994 mol H: 1.0079

There are two hydrogen atoms and one oxygen atom in each water molecule, so we multiply these numbers by the element masses and add them up to get the molar mass of H2O. g ⎞ g g ⎞ ⎛ ⎛ ⎟⎠ + ⎜⎝ 1×15.9994 ⎟⎠ = 18.0152 ⎜⎝ 2 ×1.0079 mol mol mol

Atomic Structure Example 1.4 Determine the molar mass for nitrogen gas, N2. From the periodic table, the atomic mass of nitrogen (N) is: N: 14.0067

g mol

There are two nitrogen atoms in each molecule, so we multiply the atomic mass by two to get the molar mass of N2. g g ⎞ ⎛ ⎟⎠ = 28.0134 ⎜⎝ 2 ×14.0067 mol mol

Example 1.5 Determine the mass in grams of 2.5 mol sodium bicarbonate, NaHCO3 (baking soda). In any problem like this, we first find the molar mass of the given compound. Then we simply use that molar mass to compute the mass of the given quantity. From the periodic table, the atomic masses of the elements in the compound are: Na: 22.9898

g mol

g mol g C: 12.011 mol g O: 15.9994 mol H: 1.0079

The oxygen appears three times in the formula, so its mass must be multiplied by three and added to the others. 22.9898

g g g g ⎛ g ⎞ = 84.007 +1.0079 +12.011 + ⎜ 3×15.9994 ⎟ mol mol mol ⎝ mol ⎠ mol

This value is the molar mass for NaHCO3. To find the mass of 2.5 mol we multiply: 2.5 mol⋅84.007

g = 210 g mol

Example 1.6 A scientist measures out 125 g of potassium chloride (KCl). How many moles of KCl does this quantity represent?

Chapter 1 First, determine the molar mass of KCl. From the periodic table: g mol g Cl: 35.4527 mol K: 39.098

The formula includes one atom of each, so we add them to obtain the molar mass: 39.098

g g g + 35.4527 = 74.551 mol mol mol

1 mol = 1.68 mol 74.551 g

The photograph in Figure 1.21 shows one mole of each of four substances. The first is one mole of copper, equal to 63.5 g. The second is a 250-mL beaker containing one mole of water. As you can see, this is not much water—only 18 mL. In the upper right is a weigh tray Figure 1.21. Clockwise from left are shown containing one mole of sodium chloride, 40.0 g. (This 1 mole of copper, 1 mole of water, 1 mole of table salt, and 1 mole of baking soda. is just under 1/4 cup.) Finally, one mole of baking soda, 84.1 g. (This is right at 1/3 cup.) Example 1.7 Calculate the number of water molecules in a 1.00-liter bottle of water. The logic of this problem, in reverse, is as follows: To calculate a number of molecules, we must use the Avogadro constant. To use the Avogadro constant, we need to know the number of moles of water we have. To determine the number of moles, we need to know both the molar mass and the mass of the water. To determine the mass from a volume, we use the density equation. So we begin with the given information and the density equation to determine the mass of water we have. The given information and unit conversions are as follows:

Atomic Structure V = 1.00 L⋅

ρ = 0.998

1000 cm3 = 1.00 ×103 cm3 1L

g cm3

m=? Now we write down the density equation and solve for the mass:

ρ=

m V

m = ρ ⋅V = 0.998

g ⋅1.00 ×103 cm3 = 998 g cm3

Next we need the molar mass of water. We calculated this in Example 1.3 and obtained 18.0152 g/mol. We use this molar mass as a conversion factor to convert the mass of water into a number of moles of water: 998 g ⋅

1 mol = 55.40 mol 18.0152 g

6.022 ×1023 particles = 3.34 ×1025 particles mol

1.5.6 Gram Masses of Atoms and Molecules The molar mass from the periodic table and the Avogadro constant can be used to calculate the mass in grams of an individual atom. Recall that the atomic mass value in the periodic table gives both the average atomic mass in u, and the molar mass in g/mol. Knowing the molar mass in g/mol we can simply divide by the number of atoms there are in one mole to find the mass of one atom in grams. Although this kind of calculation is quite simple, I have found that it is very easy for students to get confused and not be able to determine whether one should multiply or divide or what. So here’s a problem solving tip: let the units of measure help you figure out what to do. If you include the units of measure in your work and pay attention to how the units cancel out or don’t cancel out, these calculations are pretty straightforward. Keep this principle firmly in mind throughout your study of chemistry! Units of measure are not an annoying burden; they are the student’s friend. Example 1.8 Determine the average mass in grams of an atom of boron. From the periodic table we find that the molar mass of boron is 10.811 g/mol. One mole consists of Avogadro’s number of atoms of boron, so if we divide the molar mass by the Avogadro 41

Chapter 1 constant, we will have the mass of a single atom of boron. Let’s begin by setting up the division I just described, and then use the old invert-and-multiply trick for fraction division to help with the unit cancellations. g g 1 mol mol = 10.811 ⋅ 23 23 particles mol 6.0221×10 particles 6.0221×10 mol 10.811

10.811 g g = 1.7952 ×10−23 6.0221×1023 particle particle

So the average mass of one boron atom is 1.7952 × 10–23 g. Note that I use five digits in the value of the Avogadro constant to preserve the precision we have in the molar mass.

For molecular substances, the molar mass can be used to compute the molecular mass in grams, the average mass of one molecule. This is done by first computing the molar mass of the compound, just as we did before. Then we simply divide by the Avogadro constant to obtain the mass of one molecule. Like the atomic mass, the molecular mass is an average mass, since the atomic masses used in calculating the molar mass are all based on the average mass of different isotopes with their abundances taken into account. The molecular mass for a specific molecule would have to be calculated based on the specific masses of the nuclides in the molecule. Example 1.9 Determine the mass in grams of one molecule of carbon tetrachloride, CCl4. From the periodic table we find that the molar masses of carbon and chlorine are 12.011 g/mol and 35.4527 g/mol, respectively. From this we calculate the molar mass of CCl4: g ⎞ g g ⎞ ⎛ ⎛ ⎟⎠ + ⎜⎝ 4 × 35.4527 ⎟⎠ = 153.822 ⎜⎝ 1×12.011 mol mol mol With this molar mass we use the Avogadro constant to get the molecular mass in grams. This time, instead of writing the Avogadro constant in the denominator of a big fraction, I simply treat it as a conversion factor and write it in the equation such that the mole units cancel out. (This is the way I always perform such calculations.) I also use six digits in the Avogadro constant to preserve the precision we have in the molar mass. 153.822

g 1 mol g ⋅ = 2.55427 ×10−22 mol 6.02214 ×1023 particles particle

1.5.7 Percent Composition and Empirical Formulas Laboratory chemical analysis of a substance enables a chemist to determine the percent composition of the substance. When a new compound is discovered, chemists place a high priority on determining the percentages, by mass, of each element in the substance. This is the percent 42

Atomic Structure composition. From the percent composition a so-called empirical formula for the substance can be worked out—a formula that represents the ratios of the elements in the substance. For example, suppose laboratory analysis of a 221.6-g sample of ascorbic acid (vitamin C) results in the following mass data: H: 10.15 g C: 90.68 g O: 120.8 g If we divide each of the mass values by the total mass of the sample we have the percent composition of the sample: H:

10.15 g = 0.04580 221.6 g

90.68 g = 0.4092 221.6 g

120.8 g = 0.5451 221.6 g

Thus, the percent composition is 4.58% hydrogen, 40.92% carbon, and 54.51% oxygen. Note that we expect these percentages to add up to 100%, but due to limits on the precision of the data they add to 100.01%. Given either the percent composition or the actual masses from a sample we can determine the empirical formula for a substance. The empirical formula may differ from the actual molecular formula of the substance. An empirical formula represents the smallest whole number ratios of the elements in the substance, while the molecular formula represents the actual numbers of each element in the molecule. For example, hydrogen peroxide, H2O2, is a common household disinfectant. Molecules of hydrogen peroxide contain two atoms of hydrogen and two atoms of oxygen, so the molecular formula for this substance is H2O2. But the empirical formula is HO, because the empirical formula contains the smallest whole number values that can represent the ratios in the compound. Since each molecule of H2O2 contains two atoms of H and two atoms of O, the ratio of H to O in the molecule is 1 : 1, giving an empirical formula of HO. In many cases, the empirical and molecular formulas are identical. The molecular formula for methane, for example, is CH4. This formula indicates a ratio of carbon to hydrogen atoms in the molecule of 1 : 4. This same formula is the empirical formula, because 1 and 4 are the smallest whole numbers that can represent this ratio. To determine the empirical formula from percent composition, assume you have a sample of the substance with a mass of exactly 100 g. Use the percent composition to determine the masses of each element in the 100-g sample, then use the mass data from the periodic table to convert each of these masses to numbers of moles. Finally, divide each of the mole values by the smallest number of moles to determine the whole number ratios in the formula. An example illustrates the calculation. Example 1.10 Given percent composition data for ascorbic acid (see above), determine the empirical formula for this substance.

Chapter 1 We assume a sample with a mass of exactly 100 g. We begin by using the percent composition to obtain masses in grams for each element in the substance. Assuming a 100-g sample just makes this easy. Since hydrogen is 4.58% of the 100-g sample, the mass of the hydrogen in the sample is 4.58 g. Similarly, the masses of the carbon and oxygen are 40.92 g and 54.51 g, respectively. Next, we use the molar masses for each element to convert each of these masses to number of moles. We use the molar mass as a conversion factor, just as we have before. 4.58 g H⋅

1 mol = 4.54 mol H 1.0079 g

40.92 g C⋅

1 mol = 3.407 mol C 12.011 g

54.51 g O⋅

1 mol = 3.408 mol O 15.9994 g

Next, to determine the ratios of elements in the substance, divide each of these mole amounts by the smallest of them. 4.54 mol = 1.33 3.407 mol 3.407 mol = 1.00 3.407 mol 3.408 mol = 1.00 3.407 mol These values tell us that the ratio of hydrogen to carbon to oxygen in ascorbic acid is 1.33 : 1.00 : 1.00. Now, we need the smallest whole numbers that preserve this same ratio. Noting that the value 1.33 is very close to 4/3, we multiply all the values by 3 to get whole number ratios of 4 : 3 : 3 for hydrogen : carbon : oxygen. Finally, we use these ratios to write the empirical formula. In formulas containing these three elements it is traditional to write the elements in the formula in the order C—H—O. Doing so gives us C3H4O3.

1.5.8 Determining a Molecular Formula from an Empirical Formula The empirical formula determined in the previous example relates to the molecular formula by some simple multiple. Recall that the subscripts in the molecular formula of hydrogen peroxide, H2O2, are simply double the subscripts in the empirical formula, HO. We can determine the molecular formula for a compound from the empirical formula if we have access to the molecular mass of the compound. We do this by computing the formula mass for the empirical formula and comparing this to the molecular mass to see what the multiple is between the empirical

Atomic Structure formula mass and the molecular mass. Then we can multiply the subscripts in the empirical formula by the same multiple to get the molecular formula. In other words, whole number multiple =

molecular mass empirical formula mass

This calculation is illustrated in the following example. Note that although this example uses atomic masses and molecular mass in u, the same computation can be performed using molar masses in g/mol. Example 1.11 The experimentally determined molecular mass for ascorbic acid is 176.1 u. Use this value and the empirical formula from Example 1.9 to determine the molecular formula for ascorbic acid. We begin by determining the formula mass for the empirical formula, C3H4O3. C: 12.011 u H: 1.0079 u O: 15.9994 u

( 3×12.011 u ) + ( 4 ×1.0079 u ) + ( 3×15.9994 u ) = 88.063 u Next we calculate the whole number ratio by dividing the molecular mass by the empirical formula mass: whole number multiple =

176.1 u = 2.000 88.063 u

Finally, we multiply all the subscripts in the empirical formula by this multiple to obtain the molecular formula: C6H8O6.

1.5.9 Significant Digit Rules for Addition As noted just before Example 1.1, I have been assuming in this chapter that students using this text are already familiar with the use of significant digits in scientific measurements and computations. If you are not, then now is the time to study the tutorial on the subject in Appendix A. If you have used the significant digits rules prior to this course, your experience with the use of significant digits may be limited to computations involving multiplication and division. The rule for these kinds of computations is based on the number of significant digits in the values used in the computation: the result must have the same number of significant digits as the least precise value in the computation. With this rule, the limitation on the result is the number of significant digits in the least precise value used in the computation. For addition, a completely different rule applies. If you pay attention to the significant digits in Example 1.3, you notice that our result contains six significant digits, even though one of the values used in the computation has only five significant digits. This is a result of the addition rule. 45

Chapter 1 When performing addition, it is not the number of significant digits that governs the precision of the result. Instead, it is the place value of the last digit that is farthest to the left in the numbers being added that governs the precision of the result. To illustrate, consider the following addition example: 13.65 1.9017 + 1,387.069 1,402.62 Of the three values being added, 13.65 has digits out to the hundredths place, the second number goes out to the ten thousandths place, and the last number goes out to the thousandths place. Looking at the final digits of these three, you can see that the final digit farthest to the left is the 5 in 13.65, which is in the hundredths place. This is the digit that governs the final digit of the result. There can be no digits to the right of the hundredths place in the result. The justification for this rule is that one of our measurements is precise only to the nearest hundredth, even though the other two are precise to the nearest thousandth or ten thousandth. Since one of our values is precise only to the nearest hundredth, it makes no sense to have a result that is precise to a place more precise than that, so hundredths are the limit. Correctly performing addition problems in science (where nearly everything is a measurement) requires that you determine the place value governing the precision of your result, perform the addition, then round the result. In the above example, the sum is 1,4602.6207. Rounding this value to the hundredths place gives 1,4602.62. Going back to Example 1.3, performing the multiplications gives the following addition problem: 2.0158 + 15.9994 18.0152 Both values are precise to the nearest ten thousandth, and so is the result. In this case, we gain precision because now we have a value with six significant digits. The same thing occurs in the illustration above. One of the values in the addition has only four significant digits, but the result has six.

Chapter 1 Exercises SECTION 1.1

1. Determine the energy in a photon of light from a green laser with a wavelength of 543 nm. 2. An atom absorbs a photon, causing one of its electrons to move to an orbital associated with 2.2718 × 10–19 J higher energy. Determine the wavelength of the absorbed photon and state what region of the electromagnetic spectrum it is in. 3. For a single photon to ionize a ground-state hydrogen atom, its energy has to raise the energy of the atom’s electron to 0 J. What wavelength of light does this and what part of the electromagnetic spectrum is it in? 4. Calculate the energies for the four lines in the visible spectrum of the hydrogen atom. SECTION 1.2

5. What are two of the limitations of the Bohr model of the atom? 6. In the Bohr model, how many electrons would you expect the 5th energy level to be able to hold? Explain your response. 46

Atomic Structure SECTION 1.3

7. A certain atom is in the ground state. The 3p subshell of this atom is 2/3 full. a. Identify the element this atom represents. b. How many unpaired electrons are there in the atom? (A paired electron is one in an orbital with another one possessing opposite spin.) 8. In a certain ground-state atom, the 4d subshell has two electrons in it. a. Identify the element this atom represents. b. How many unpaired electrons are there in the atom? 9. How do the values of the azimuthal quantum number and magnetic quantum number relate to the principle quantum number? 10. Demonstrate mathematically that the 4f subshell can accommodate 14 electrons. 11. Generally speaking, what is the explanation for an atom’s electron configuration not following the sequence described by the Madelung rule? SECTION 1.4

12. For each of the following elements, draw the orbital diagram and write the full-length electron configuration. a. chlorine

b. oxygen

c. ruthenium

d. potassium

e. vanadium

f. bromine

13. For each of the following elements, write the condensed electron configuration. a. chlorine

b. nitrogen

c. aluminum

d. yttrium

e. strontium

f. tungsten

g. cesium

h. iodine

i. neodymium

14. Compare the electron configurations for beryllium, magnesium, and calcium. Formulate a general rule for the condensed electron configuration of a Group 2 element. 15. For which group of elements does the electron configuration always end with np2? Explain how you know. 16. Write the condensed electron configurations for ytterbium, einsteinium, and nobelium. SECTION 1.5

17. Which two nuclides in Table 1.5 have 20 neutrons? 18. In Table 1.5, how many neutrons are there in the heaviest nuclide listed? How many neutrons are there in the lightest nuclide listed? 19. Determine the number of atoms in each of the following. a. 73.2 g Cu

b. 1.35 mol Na

c. 1.5000 kg W

20. Determine the mass in grams for each of the following. a. 6.022 × 1023 atoms K

b. 100 atoms Au

c. 0.00100 mol Xe

d. 2.0 mol Li

e. 4.2120 mol Br

f. 7.422 × 1022 atoms Pt

21. Determine the number of moles present in each of the following.

Chapter 1 a. 25 g Ca(OH)2

b. 286.25 g Al2(CrO4)3

c. 2.111 kg KCl

d. 47.50 g LiClO3

e. 10.0 g O2

f. 1.00 mg C14H18N2O5

22. As mentioned in the text, the sum of the masses of the particles in an atom does not equal the mass of the atom. Some of the mass of the individual particles is converted to energy, and the atom weighs less than the sum of the weights of its parts. How much mass is converted into energy when the individual protons, neutrons, and electrons are assembled to form an atom of uranium-238? 23. Referring to Table 1.5, calculate the atomic mass for silicon, calcium, iron, and uranium. Compare your results to the values shown in the periodic table. 24. Calculate the molar mass for each of the following compounds or molecules. a. ammonia, NH3

b. carbon dioxide, CO2

c. chlorine gas, Cl2

d. copper(II) sulfate, CuSO4

e. calcium nitrite, Ca(NO2)2

f. sucrose, C12H22O11

g. ethanol, C2H5OH

h. propane, C3H8

i. glass, SiO2

25. Determine the formula masses for these compounds: a. MgCl2

b. Ca(NO3)2

c. (SO4)2– (The 2– indicates this is an ion with an electrical

d. CuSO4

e. BF3

f. CCl4

charge of –2. The charge does not affect your calculation.)

26. Determine the mass in grams of 2.25 mol silver nitrate, AgNO3. 27. Given 2.25 kg CCl4, answer these questions: a. How many moles CCl4 are present? b. How many carbon atoms are present? c. Approximately how many carbon-13 atoms are present? 28. Given 1.00 gal H2O at 4°C, answer the questions below. (Hint: You must use the appropriate volume conversion and the density of water to determine the mass of 1.00 gal H2O. See the information in Tables B.3 and B.5 in Appendix B.) a. How many moles H2O are present? b. How many hydrogen atoms are present? c. Approximately how many deuterium (hydrogen-2) atoms are present? 29. Automobile antifreeze is composed of ethylene glycol. This green liquid is 38.7% C, 9.7% H, and 51.6% O by mass. The molecular mass is 62.1 u. Determine the empirical formula and the molecular formula for ethylene glycol. 30. A scientist isolates 47.593 g of a new, unidentified substance. The scientist also determines the following masses for the elements in the substance: carbon: 43.910 g; hydrogen: 3.683 g. Finally, the scientist is also able to determine the molecular mass of the substance to be 78.11 u. From these data, determine: a. the percent composition b. the empirical formula c. the molecular formula

Atomic Structure 31. Hydrogen chlorate, HClO3, is a molecular substance that becomes chloric acid when dissolved in water. Determine the number molecules present in 125.0 g HClO3. 32. Determine the percentage composition of these compounds: a. sodium bicarbonate, NaHCO3

b. sodium oxide, Na2O

c. iron(III) oxide, Fe2O3

d. silver nitrate, AgNO3

e. calcium acetate, Ca(CH3COO)2

f. aspirin, C9H8O4

33. A hydrate is a compound with water molecules trapped in the crystal lattice. Determine the mass percentage of water in zinc sulfate septahydrate, ZnSO4∙7H2O. (The coefficient on the H2O indicates the number of water molecules present for each unit of ZnSO4.) 34. The results of quantitative analysis show that a compound contains 22.65% sulfur, 32.38% sodium, and 44.99% oxygen. Determine the empirical formula for this compound. 35. A compound has an empirical formula of CH2O and a molar mass of 120.12 g/mol. Determine the molecular formula for this compound. 36. Determine the empirical and molecular formulas for each of the following: a. caffeine, which contains 49.5% C, 5.15% H, 28.9% N, and 16.5% O by mass, and has a molecular mass of 195 u. b. ibuprofen, which contains 75.69% C, 8.80% H, and 15.51% O by mass, and has a molar mass of 206 g/mol. c. propane, which contains 81.71% C and 18.29% H by mass, and has a molar mass of 44.096 g/mol. d. aspartame, a sugar-free sweetener, which contains 57.14% C, 6.16% H, 9.52% N, and 27.18% O, and has a molecular mass of 294.302 u. e. acetylene, a gas used in cutting torches, which contains 92.26% C and 7.74% H, and has a molar mass of 26.038 g/mol. 37. Toluene is a solvent commonly found in chemistry labs. An analysis of a 10.5-g sample shows that the sample contains 9.581 g carbon and 0.919 g hydrogen. If the molar mass of toluene is 92.140 g/mol, determine the percent composition, empirical formula, and molecular formula.

Chapter 2 The Periodic Law

The patterns among the elements have tempted many scientists to try their hands at developing different forms of the Periodic Table of the Elements. At the top is the original periodic table, published by Russian scientist Dmitri Mendeleev in 1869.

The Periodic Law

1. Describe the general structure and arrangement of the Periodic Table of the Elements. SECTION 2.2

2. Identify the names and locations of the major regions in the periodic table. 3. Identify the names and locations of the elements in Groups 1, 2, 3–12, 16, 17, and 18. Also identify the groups known as the rare-earth elements and the names of the two rows of these elements. 4. State the two collective names for Groups 1–2 and 13–18. 5. State the chief chemical property that distinguishes the metals from the nonmetals. SECTION 2.3

6. Define the length unit known as the angstrom (Å). 7. Use diatomic bond length and atomic radius data to estimate bond lengths in molecules. 8. Describe the trends in atomic radii in the periodic table across periods and down groups. 9. Explain three factors influencing the trends in atomic radii in the periodic table. 10. Compare ionic radii to atomic radii for metals and nonmetals. 11. Use your knowledge of atomic and ionic size trends in the periodic table to arrange lists of elements and ions in order by size. SECTION 2.4

12. Distinguish between core electrons and valence electrons. 13. Estimate effective nuclear charge, Zeff. 14. Define ionization energy, describe the trends for ionization energy across periods and down groups in the periodic table, and use Zeff and other factors to account for these trends. 15. Write regular and condensed electron configurations for an ion in a given oxidation state. 16. Use Zeff to explain the large difference in ionization energy between an atom’s core electrons and its valence electrons. 17. Predict oxidation states for metals in Groups 1–4 and nonmetals in Groups 15–17. 18. Define electron affinity, use Zeff to account for the high electron affinity of elements in Groups 16 and 17, and explain why electron affinity values are not available for the noble gases. 19. Define electronegativity. 20. Describe the trends for electronegativity across periods and down groups in the periodic table. Name the elements with the lowest and highest electronegativity values. 21. Use Zeff to account for the trend in electronegativity across periods. SECTION 2.5

22. Explain why hydrogen is located in Group 1 and why it acts like the Group 17 elements.

Chapter 2

2.1

The Periodic Table of the Elements

The contemporary Periodic Table of the Elements is shown again in Figure 2.1 and in the inside the rear cover of the text. Discovering new elements and figuring how they relate to one another was one of the hottest issues in science in the 19th century. By the 1860s, several scientists had noticed periodicities in the properties of the known elements. A periodicity is a regular, cyclic variation of some kind. These scientists noticed, for example, that when elements were listed in order according to atomic weight, the physical property of density increased and decreased in a cyclic fashion. In 1864, German chemist Lothar Meyer published a paper describing cyclic variation in the chemical property known as valence, an important property related to the number of bonds an atom makes with other atoms to form compounds. Credit for the discovery of the periodic law and the development of the first Periodic Table of the Elements is generally given to Russian scientist Dmitri Mendeleev, who published his table of the elements in 1869. Mendeleev (Figure 2.2) had not only noticed the periodicities, but he also arranged some 67 elements into a table and predicted the existence of several unknown elements based on gaps in the table as he had organized it. The elements Mendeleev predicted included those now known as gallium, germanium, technetium, and others. The properties Mendeleev predicted for these elements included valence, density, atomic weight, and color. His predictions of not only the existence of these elements but also their properties is the reason for the general credit Mendeleev gets for discovering the periodic law. As Mendeleev organized the elements, he ordered them by atomic weight and aligned them into rows and columns based on their chemical and physical properties. Today, just as back then, one of the most important things to know Figure 2.2. Russian scientist Dmitri about the periodic table is that elements in the same group Mendeleev (1834–1907). (column) exhibit very similar chemical properties. Mendeleev’s original table is shown on the opening page of this chapter. In that image, just above the center and a bit to the left you can see gaps at the atomic mass values 68 and 72. These are the positions now occupied by gallium and germanium, discovered in 1875 and 1886, respectively. Mendeleev was a brilliant scientist. His fields of expertise included physics, chemistry, and a host of areas of technology. Mendeleev taught in St. Petersburg at several different institutions, and because of his work there St. Petersburg became internationally known for prominence in chemical research. There were some debates in the 19th century regarding proper placement for four of the elements. If you look at tellurium (Z = 52) and iodine (Z = 53) in the periodic table to the right, you see that the atomic mass of tellurium is the larger of the two. Because of this, many scientists felt that iodine should come before tellurium in the table. But iodine exhibits all the properties of the Group 17 elements and Mendeleev argued that it should come after tellurium. This problem was resolved when scientists realized that the atomic number was the correct parameter to use for ordering the elements in the table. Mendeleev had ordered them by atomic mass (known then as atomic weight); today the atomic number governs the order. A similar debate over placement surrounded the elements cobalt (Z = 27) and nickel (Z = 28). The noble gases were not known in Mendeleev’s time and he provided no place for them in his table. Interestingly, when they were eventually discovered Mendeleev was resistant to accept the discovery because the new elements didn’t fit into his table. The solution to this little problem was simply to add another column for them, now known as Group 18. 52

87.62 56

Barium

137.327 88

Radium

226.0254

85.468 55

Cesium

132.905 87

Francium

223.0197

40.078 38

39.098 37

Calcium

Strontium

Potassium

24.3050 20

22.9898 19

Rubidium

Magnesium

Sodium

9.0122 12

6.941 11

Beryllium

Lithium

232.0381

227.0278

231.0359

Protactinium

Thorium

Actinium

140.9077 91

Cerium

140.115 90

Lanthanum

138.9055 89

Praseodymium

Dubnium

180.9479 105

Tantalum

92.9064 73

Niobium

50.9415 41

Vanadium

262.114

Rutherfordium

178.49 104

Hafnium

91.224 72

Zirconium

47.88 40

Titanium

261.11

262.11

Lawrencium

174.967 103

Lutetium

88.9059 71

Yttrium

44.9559 39

Scandium

238.0289

Uranium

144.24 92

Neodymium

263.118

Seaborgium

183.85 106

Tungsten

95.94 74

Molybdenum

51.9961 42

Chromium

Promethium

237.0482

Neptunium

144.9127 93

262.12

Bohrium

186.207 107

Rhenium

98.9072 75

Technetium

54.9380 43

Manganese

244.0642

Plutonium

150.36 94

Samarium

(265)

Hassium

190.2 108

Osmium

101.07 76

Ruthenium

55.847 44

Iron

Europium

243.0614

247.0703

Curium

157.25 96

Gadolinium

(281)

Darmstadtium

195.08 110

Platinum

106.42 78

Palladium

58.6934 46

Nickel

Am Cm Americium

151.965 95

(266)

Meitnerium

192.22 109

Iridium

102.9055 77

Rhodium

58.9332 45

Cobalt

Silver

63.546 47

Copper

Terbium

247.0703

Berkelium

158.9253 97

(281)

Roentgenium

196.9665 111

Gold

107.8682 79

radioactive

251.0796

Californium

162.50 98

Dysprosium

(285)

Copernicium

200.59 112

Mercury

112.411 80

Cadmium

65.39 48

Zinc

Holmium

252.083

Einsteinium

164.9303 99

(284)

Nihonium

204.3833 113

Thallium

114.82 81

Indium

69.723 49

Gallium

26.9815 31

Aluminum

10.811 13

Boron

1.0079 3

Hydrogen

Erbium

257.0951

Fermium

167.26 100

(289)

Flerovium

207.2 114

Lead

118.710 82

Tin

72.61 50

Germanium

28.0855 32

Silicon

12.011 14

Carbon

Thulium

258.10

Mendelevium

168.9342 101

(288)

Moscovium

208.9804 115

Bismuth

121.76 83

Antimony

74.9216 51

Arsenic

30.9738 33

Phosphorus

14.0067 15

Nitrogen

Ytterbium

259.1009

Nobelium

173.04 102

(293)

Livermorium

208.9824 116

Polonium

127.60 84

Tellurium

78.96 52

Selenium

32.066 34

Sulfur

15.9994 16

Oxygen

(294)

Tennessine

209.9871 117

Astatine

126.9045 85

Iodine

79.904 53

Bromine

35.4527 35

Chlorine

18.9984 17

Fluorine

(294)

Oganesson

222.0176 118

Radon

131.29 86

Xenon

83.80 54

Krypton

39.948 36

Argon

20.1797 18

Neon

4.0026 10

Helium

liquid at room temperature

18 8A

Figure 2.1. The Periodic Table of the Elements.

The Periodic Law

Chapter 2 The first 92 elements in the periodic table are found in nature; elements 93–118 have been synthesized in laboratories. The “discovery” (by synthesis) of elements 114 and 116 was confirmed in 2011, and in January 2016 official confirmation of elements 113, 115, 117, and 118 was announced. The reason confirmations for the last few elements took so long is that once the nucleus of one of these heavy elements is assembled it doesn’t stay around very long—far less than one second.

2.2

Periodic Table Nomenclature

The columns in the periodic table are called groups and the rows are called periods. The images in Figures 2.3 and 2.4 identify several different specific regions of elements in the periodic table. In the long form of the table shown in Figure 2.3, the elements are classified as metals, nonmetals, and metalloids. Note that hydrogen (H) is classified as a nonmetal, even though it is positioned with the metals in Group 1. Hydrogen’s location in Group 1 is due to the fact that hydrogen has one valence electron, which I address in more detail later in the chapter. As you probably know, metals possess a number of properties in common. Common physical properties include high electrical and thermal conductivity, malleability, ductility, and shininess or luster. Chemically, the metals are known for ionizing by losing electrons to form positive ions, known as cations (pronounced cat-ion). As positive ions, they bond with negative ions to form ionic compounds. (Again, we address compounds in detail later.) People commonly think of metals as shiny conductors of electricity. Chemists think of them as elements that form positive ions. The metalloids possess properties that are neither clearly metallic nor clearly nonmetallic. For example, under some conditions they conduct electricity and under other conditions they don’t. This property is the reason why some of the metalloids are the elements used to manufacture computer “semiconductors.” The nonmetals have their own distinguishing properties, such as ionization by gaining electrons to form negative ions, called anions. They also bond with each other—something metals almost never do. In the standard form periodic table of Figure 2.4 are shown the common names for several specific groups (columns) of elements. All these names are used frequently in scientific discourse and you need to commit them to memory. Groups 1–2 and 13–18 are also collectively referred to as the main group elements or representative elements. You probably noticed that in both the figures elements 113–118 are not included with the rest of the elements in the different classes or groups. This is due to the fact that they have only existed for extremely short periods of time in laboratories and little is known about their properties. By the way, you don’t need to feel sorry for Groups 13–15 not having nicknames. They do, 1 1

nonmetals

metalloids

1 2

18 2

metals

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

Figure 2.3. The long form of the periodic table indicating the three major classes of elements.

The Periodic Law noble gases 1

chalcogens, aka oxygen group

alkaline-earth metals

1 1

halogens, aka fluorine group

alkali metals

transition metals

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

100

101

102

lanthanides, aka lanthanide series actinides, aka actinide series

inner transition metals, aka rare-earth elements

Figure 2.4. The standard form table indicating many of the common names for particular groups of elements.

but the names are seldom used. (Okay, Group 13 is called the icosagens, Group 14 the crystallogens, and Group 15 the pnictogens. See? Well, now you know.)

2.3

Periodic Physical Properties

Physical properties of substances are properties such as color, density, boiling point, electrical conductivity, malleability, and many others. Many of these properties exhibit strong periodicity—cyclic rises and falls with cycles that correspond to the periods in the periodic table. In this section, we look at one of the most important of these—atomic size. As we look at periodic trends in this section and the next, you should closely compare the charts presented with the periodic table in Figure 2.1.

2.3.1 Atomic Radius and Bonding Atomic Radius Defining the size of atoms is a bit tricky. The nucleus of the atom is extremely small, and virtually all the space an atom takes up is defined by the electrons in their orbitals surrounding the nucleus. The size of the orbitals themselves is defined by the probability distribution of where electrons may be found, and this probability does not drop cleanly to zero at the edge of the orbital. Instead, it fades toward zero, making the atomic radius fuzzy. Still, the electrons in the orbitals create a negatively charged shell around the atom that strongly resists penetration by the shells around other atoms. Thus scientists can measure atomic radius by firing two atoms at one another and examining how closely the two atoms come together before bouncing apart. 55

Chapter 2 The electron clouds repel each other so stiffly that atoms bounce apart as if they were steel spheres. Figure 2.5 shows two atoms bonded together in a molecule, such as the diatomic oxygen molecule, O2. As you see, when atoms are bonded together this way the electron shells do interpenetrate one another. The bond atomic length, defined as the distance between nuclei, is less than radius twice the atomic radius. Accordingly, we define the bonding atomic radius as half the bond length and use this bond length value to estimate the bond length in various molecules. When discussing atomic size, a convenient length unit Figure 2.5. Comparison of atomic radius is the angstrom (Å), which is equal to 10–10 m, a tenth of and bonding atomic radius. a nanometer (0.1 nm or 100 pm). The following example illustrates the approximation of bond length in a molecule. electron shell

bonding atomic radius

atomic nucleus

Example 2.1 Figure 2.6 depicts a molecule of sulfur dioxide, SO2, one of the compounds emitted into the atmosphere by burning fossil fuels. The atomic radii of oxygen and sulfur atoms are 0.64 Å and 1.04 Å, respectively. The bond lengths in O2 and S2 molecules are 1.21 Å and 1.89 Å, respectively. Estimate the S—O bond length in the SO2 molecule.

1.43 Å

1.04 Å sulfur

oxygen atom

atom We expect the bonding atomic radius of oxygen to be half the O2 bond length, or about 0.61 Å. Similarly, we expect the bonding atomic radius of sulfur to be half the S2 bond length, 0.64 Å or about 0.95 Å. Adding these two bonding atomic radii gives 1.56 Å. This figure compares pretty well with the actual bond Figure 2.6. Comparison of atomic length in the SO2 molecule of 1.43 Å (about 9% difference).

radius and bonding atomic radius in an SO2 molecule.

Figure 2.7 shows the presently accepted values for the atomic radii of the first 100 elements. As you see, typical radii are in the range of 1–2 Å, or 0.1–0.2 nm. This means that typical atomic diameters are in the range of 0.2–0.4 nm. The most striking things about the graph are the strong peak that occurs at the beginning of each new period and the gradual decrease in atomic size from one element to the next within the period. There are at least three different effects present governing the size of atoms and leading to these patterns. The first is the number of subshells in use holding the atoms’ electrons. At the beginning of each period, an electron appears with a new, higher principle quantum number, n, and a new s subshell associated with that principle quantum number. This new shell allows for electrons to be much farther from the nucleus and is the major cause of the peak in atomic radius at the beginning of each period, as well as the overall upward trend in atomic size. The second effect is the increasing attraction between the nucleus and the electrons as we move from left to right across the periodic table in any given period, or from top to bottom in a given group. Moving left to right across the periodic table, with each new element comes a new proton in the nucleus. The increasing positive electrical charge at the center of the atom tends to pull the atom’s electrons in tighter and tighter. Thus, with a few exceptions, atomic size decreases

The Periodic Law 2.5 2.3

atomic radius (Å)

1.8

1.3

2.0

1.5

Na Li

Xe Kr

1.0 0.8

0.5

0.3 0

100

atomic number atomic number Figure 2.7. Atomic radius values. The six peaks are the alkali metals in Periods 2–7.

from left to right in a period. In the chart of Figure 2.7, this decrease in size after the start of a new period is quite pronounced. The third effect is called the shield effect or atomic shielding. The electrons in shells with lower values of n effectively form an electrical screen around the nucleus and to some extent shield off the attraction of the positive nucleus for electrons in higher shells. To quantify this screening effect, chemists use a parameter called the effective nuclear charge (Zeff ). The details of the effective nuclear charge are easier to explain a bit later, so we return to it in Section 2.4.2. A second feature to note from Figure 2.7 is that the radius of each of the alkali metals is larger than the one just above it in the periodic table. The same holds for each of the noble gases. In fact, this same trend is present in every group in the table. Down every group, the atomic radius of each element is greater than that of the element above it, and this results in the overall upward trend in atomic size shown in the chart. In summary, atomic size generally decreases from left to right in a period and always increases from top to bottom in a group.

2.3.2 Ionic Radius Figure 2.8 depicts the sizes of atoms and their ions in four of the groups of the representative elements. The blue disks are the atomic sizes and the yellow disks are the ionic sizes, both in angstroms. Metals ionize by losing electrons to form cations (positive ions). The loss of an electron leaves an atom with more protons than electrons, and thus the atom possesses a net positive charge. As we will discuss more in Section 2.4, Group 1 metals always ionize by losing one electron to become ions with a charge of +1. With sodium, for example, we write this ion as Na+. Group 2 metals always ionize by losing two electrons to become ions with a charge of +2. Thus, for the calcium ion we write Ca2+.1 Nonmetals ionize by gaining electrons to become anions (negative ions). Group 17 elements ionize by gaining one electron to become ions with a net charge of –1. Group 16 elements ionize by gaining two electrons to become ions with a net charge of –2. The blue circles in Figure 2.8 represent the atomic sizes, and yellow circles represent ionic sizes. Since the metals always ionize by losing the electrons in their highest s orbital, their di1 The most common convention is to write the sign of the charge after the value of the charge on the chemical symbol for an ion. 57

Chapter 2 ameters decrease considerably when they ionize. The opposite happens to the nonmetals. Gaining electrons adds to the 0.59 Be 0.64 O Li 0.60 F mutual electron repulsion in 0.90 0.99 1.26 1.19 the highest-energy orbitals, in1.30 creasing atomic size. Just as with neutral atoms, Na Mg S Cl ionic sizes decrease in a period 0.86 1.04 1.00 1.16 from left to right in the periodic 1.60 1.40 1.70 1.67 table and increase going down a group. Further, we can summarize the paragraphs above by saying that cations are smaller K Ca Se Br than their neutral atoms and 1.14 1.18 1.17 1.52 anions are larger than their neu1.82 1.74 1.84 2.00 tral atoms. One final ionic trend needs to be mentioned: among neutral atoms or ions with the same Rb Sr Te I number of electrons, atomic 1.32 1.37 1.36 1.66 size always decreases as the 1.90 2.06 2.07 2.15 number of protons increases. Based on what we have seen so Figure 2.8. Atomic radii (blue) and ionic radii (yellow) for the Group 1 far, this should make perfect and 2 positive ions and Group 16 and 17 negative ions. All values are in angstroms. sense. Among atoms with the same number of electrons—a so-called isoelectronic series—there are no new shells to consider and there are no variations in the amount of shielding produced by changing electron configurations. Thus, the only variation is in the positive charge in the nucleus, and as this goes up, the attraction of the nucleus for the electrons goes up, pulling the electrons in tighter and making the atom smaller. Group 1 (+1)

Group 2 (+2)

Group 16 (–2)

Group 17 (–1)

Example 2.2 Based on your knowledge of trends in the periodic table, place the following atoms and ions in order of decreasing size: S, Se2–, O, and S2–. Anions are larger than their neutral atoms, so S2– > S. Elements and ions farther down in a group are larger than those above, so S > O. Also, Se2– is larger than S2–, so Se2– > S2– > S > O.

Example 2.3 Arrange the following ions in order of size from largest to smallest: O2–, Na+, F–, and Mg2+. We can address this question without any published data on atomic size. Referring to the periodic table, you see that each of the ions listed has 10 electrons. For example, magnesium (Mg) is element 12 and has 12 electrons before ionizing. But the charge on the ion is +2, which means the magnesium atom has lost two electrons and now has 10. Therefore, the order from largest 58

The Periodic Law to smallest follows the atomic number order from smallest to greatest, giving the sequence O2–, F–, Na+, Mg2+.

2.4

Periodic Chemical Properties

2.4.1 Core and Valence Electrons Chemistry is all about electrons, and when it comes to the chemical bonding that occurs in chemical reactions, the valence electrons of the elements involved determine what kinds of compounds form. To illustrate, consider the elements sulfur (Z = 16) and cobalt (Z = 27) located in the partial periodic tables of Figure 2.9. The electron configurations for these elements are as follows: Z

Electron Configuration

Condensed Electron Configuration

1s22s22p63s23p4

[Ne]3s23p4

Co: 1s22s22p63s23p64s23d7

Co: [Ar]4s23d7

As we saw in the previous chapter, when moving from one element to the next in the periodic table, we add one electron and one proton for each new element. As we look at an element’s position in the periodic table, we can think of each of the elements before it as representing the position of one of the element’s electrons, arranged according to the Madelung rule and Hund’s rule. Now consider Figure 2.9 and the electron configuration for sulfur. The first 10 electrons in a sulfur atom go to completely filling the n = 1 and n = 2 shells. The next six electrons go into the 1

16 3

13 3

Sulfur

2 2

9 27

Cobalt

Figure 2.9. Core electrons (green) and valence electrons (yellow) for sulfur, cobalt, and bromine.

2 2

12 35

Bromine

Chapter 2 n = 3 shell. This shell can hold eight electrons, so it is only partially full. The electrons in the full shells are called core electrons. The electrons in the partially filled shell are called valence electrons, and the shell they are in is called the valence shell. The core electrons are the ones involved in screening the nucleus from the outer electrons in the valence shell. These core electrons have more negative energies than the valence electrons, which means they are more tightly bound to the nucleus. As a result, the core electrons are not involved in the electron swapping and sharing that takes place in chemical reactions. That involvement is limited to the valence electrons. When you look at sulfur’s position in the periodic table, just a quick glance indicates that sulfur has six valence electrons, all of them in the third shell, n = 3. It’s as simple as observing that sulfur is in the third period and counting columns from the left over to where sulfur is. Notice that in the condensed electron configuration notation, all of sulfur’s core electrons are represented by [Ne]. The valence electrons are still shown explicitly as 3s23p4. Looking now at cobalt, the element positions colored in green represent cobalt’s core electrons. The first 18 electrons are the core electrons, represented by [Ar] in the condensed electron configuration. The next nine electrons are the valence electrons, easily seen from the fact that cobalt is in the fourth period and the ninth column from the left side of the periodic table. Finally, note that for p-block elements in Periods 4–7, the electrons in filled, lower d subshells do not act as valence electrons. The only valence electrons in these atoms are those in the s and p subshells of the unfilled shell. This is illustrated by the diagram for bromine at the bottom of Figure 2.9.

2.4.2 Effective Nuclear Charge Now we return to the idea of effective nuclear charge, Zeff, mentioned a few pages back in Section 2.3.1. Effective nuclear charge is a useful concept when it comes to explaining some of the atomic size variations we see in the periodic table. Zeff is calculated as Z eff = Z − S where Z is the atomic number and S is a value called the screening constant. The idea here is that to a large extent the core electrons screen off the positive charge of the nucleus, reducing the actual electrical pull on the valence electrons. For a rough approximation of the effect of this screening, the screening constant can be taken to be the number of core electrons in the atom. As examples, let’s consider again sulfur and cobalt. For sulfur (Figure 2.10), Zeff = Z – S ≈ 16 – 10 ≈ +6, and for Cobalt, Zeff = Z – S ≈ 27 – 18 ≈ +9. Compare these two values to the values for the Group 1 metals in the same periods. For sodium, Zeff = Z – S ≈ 11 – 10 ≈ +1 and for potassium, Zeff = Z – S ≈ 19 – 18 ≈ +1. The alkali metals only have one valence electron, so Zeff comes out to +1. Moving to the right across the period, additional protons are nuclear charge = Z = 16 accrued in the nucleus but the number of core electrons remains the same as it is for the Group 1 metal at the beginning of S ≈ shielding core the period. The increase in Zeff from left electrons = 10 to right in a period explains the fact that atomic radii decrease from left to right. Zeff = Z – S ≈ +6 The value of Zeff can also be used to explain the trends in ionization energy we look at next. A word of caution is in order here. The calculation outlined above for Zeff is Figure 2.10. Effective nuclear charge for sulfur. quite simplified, and the results it gives 60

The Periodic Law are only approximate. Simply using the number of core electrons for S gives Zeff = +1 for each of the alkali metals. In fact, the values of Zeff obtained from more sophisticated calculation methods for the first three alkali metals, lithium (Z = 3), sodium (Z = 11), and potassium (Z = 19), are +1.3, +2.5, and +3.5, respectively.

2.4.3 Ionization Energy Ionization energy is defined as the amount of energy required to remove a ground-state electron from an isolated, gaseous atom. Recall from our discussion in the previous chapters that adding an electron to an atom releases energy because the electron is going into a lower energy state. Conversely, removing an electron from an atom requires an input of energy, which is the work required to pull the electron from its negative energy state up to zero energy, where it is free from the nuclear attraction of the atom. Using hydrogen as an example, the removal of the electron from an atom of hydrogen is modeled by the following ionization equation: H(g )→ H + + e −

ΔE = 2.18 ×10−18 J

(2.1)

This expression shows the neutral hydrogen atom on the left, with (g) indicating that the atom is in the gaseous state. On the right, H+ indicates a hydrogen ion with a charge of +1, and e– indicates a free electron. In an equation like this, it is customary to write the energy change (ΔE) that occurred in the atomic system (the atom and its electron) during the process. Notice that ΔE is positive, meaning that a certain amount of energy has to be added in order to accomplish the ionization. This amount of energy is the ionization energy. Because the amounts of energy involved are so small, it is more convenient in discussions of this sort to use an energy unit called the electron volt (eV). The electron volt is defined as 1 eV = 1.60218 × 10–19 J

(2.2)

For the ionization of hydrogen, converting the energy above into eV gives ΔE = 2.18 ×10−18 J⋅

1 eV = 13.6 eV 1.602 ×10−19 J

The energy required to remove one electron from a neutral atom is called the first ionization energy. The additional energy required to remove a second electron is called the second ionization energy, and so on. Figure 2.11 charts the first ionization energies of the elements. The strong upward trend in each period is easily accounted for by the increasing value of Zeff moving from left to right in the period. The higher Zeff is, the stronger the nucleus attracts the outermost electrons and the greater the energy required to remove one of them. Notice from the Group 1 and Group 18 elements labeled in the figure that the trend down a group is for the ionization energy to decrease, even though Zeff actually increases down the group, as we saw previously. This is accounted for by the fact that the elements in each new period have their valence electrons in a shell with a higher principle quantum number. These electrons are thus farther from the nucleus, have a less negative energy, and are easier to remove. Writing electron configurations for ions requires you to keep a somewhat surprising rule in mind: the electrons an element loses during ionization come from the orbital with the highest principle quantum number, n. Further, among the orbitals associated with the highest principle quantum number, the electrons come from the orbitals with the highest value of the azimuthal quantum number, l. This means, for example, that sodium ionizes (as we would expect) by losing 61

Chapter 2 27.0

24.0

ionization energy (eV)

21.0 18.0

Ar 15.0 12.0

Kr Xe

9.0 6.0

3.0 0

100

atomic number

Figure 2.11. First ionization energies.

its 3s electron. However, scandium ionizes by losing a 4s electron, not a 3d electron. The following example illustrates further. Example 2.4 Write the electron configurations for Sn4+ and Cr3+. Sn is a p-block metal in Period 5 and Group 14 with configuration Sn: [Kr]5s24d105p2 Its valence electrons are therefore in the 5s and 5p subshells. It first loses its two 5p electrons, because these have n = 5 and l = 1. The next two to go are the 5s electrons which have n = 5 and l = 0. The electron configuration of the ion is: Sn4+: [Kr]4d10 Cr is a transition metal with the anomalous electron configuration: Element

1st

2nd

3rd

5.38

75.64

122.45

9.32

18.21

153.90

217.72

8.30

25.15

37.93

259.38

340.23

5th

6th

7th

11.26

24.38

47.89

64.49

392.09

489.99

14.53

29.60

47.45

77.47

97.89

552.07

667.05

13.62

35.12

54.94

77.41

113.90

138.12

739.29

17.42

34.97

62.71

87.14

114.24

157.17

185.19

21.56

40.96

63.45

97.12

126.21

157.93

207.28

Table 2.1. Ionization energies (eV) for Period 2 elements.

4th

The Periodic Law Cr: [Ar]4s13d5 To ionize to Cr3+, the atom first loses the 4s electron, which has n = 4 and l = 0. Then it loses two of its 3d electrons, which have n = 3 and l = 2. Thus, the configuration of the ion is: Cr3+: [Ar]3d3

Table 2.1 lists the ionization energies up through the 7th ionization energy for the Period 2 elements. Values without shading represent the energies required to remove valence electrons. The yellow shading indicates where core electrons are being removed. Notice the whopping increase in the ionization energy once we start getting into an atom’s core electrons. This is explained by the shield effect we discussed previously. Core electrons shield the nucleus from exerting its full attraction on the valence electrons, an effect quantified by the value of Zeff. But inside the core, the full attraction of the nucleus is felt and the energy required to remove an electron is dramatically greater. This discussion about ionization is an appropriate time to show representative values for ionizations exhibited by the main group elements and transition metals. These values, displayed in Figure 2.12, are called oxidation states. I address the origin and use of the term “oxidation” in a later chapter. In this context, it just means that when atoms of a given element ionize, these are the values of charge they typically acquire. The first and most important thing to notice is that on both ends of the periodic table, elements ionize so as to end up with either an empty valence shell or a full one, depending on whether the valence shell was closer to being full or empty to start with. All Group 1 metals and hydrogen ionize by losing their only valence electron and retaining their core electrons. After this ionization, the Group 1 metals are left with one more proton than electron, so they are cations with a charge of +1. Similarly, Group 2 metals lose two electrons and Group 3 metals lose three electrons to end up as cations with charges of +2 or +3, respectively. On the other end of the table, the noble gases don’t readily ionize, so no oxidation states are shown. The nonmetals ionize by taking on electrons to become anions. The halogens all have p subshells containing five electrons, so they ionize by gaining one electron to fill up the p sub1

H+

Li+

Na+ Mg2+

Be2+

B3+

C4+ C4–

N3–

O2–

F–

Al3+

Si4+

P3–

S2–

Cl–

Cr6+ Mn4+ Fe3+ Co3+ Cu2+ Ge4+ As3+ Ni2+ Zn2+ Ga3+ Se2– Cr3+ Mn2+ Fe2+ Co2+ Cu+ Ge2+ As5+

Br–

V5+ V4+

K+

Ca2+

Sc3+

Ti4+

Rb+

Sr2+

Y3+

Zr4+ Nb5+

Mo6+ Tc7+ Ru4+ Pd4+ Rh3+ Ag+ Cd2+ Mo4+ Tc4+ Ru3+ Pd2+

Cs+ Ba2+

Hf4+

W6+

Ta5+

Re7+ Os4+ Re6+

Ir4+ Ir3+

Pt4+ Au3+ Hg4+ Pt2+ Au+ Hg2+

In3+

Sn4+ Sb5+ Te2– Sn2+ Sb3+

I–

Tl+

Pb4+ Pb2+

Bi5+ Bi3+

Po4+

Figure 2.12. Representative oxidation states.

Chapter 2 shell, becoming anions with a charge of –1 in the process. Similarly, Group 16 nonmetals gain two electrons and Group 15 nonmetals gain three electrons to become anions with charges of –2 or –3, respectively. All the transition metals have multiple oxidation states, and the oxidation state they assume depends on what other elements are around to swap electrons with. The values shown are the most common states. This chart makes it easy to visualize the metalloids as the boundary between the cations and the anions. In the cases of lutetium (Lu) and astatine (At), these elements have multiple oxidation states with none preferred, and for this reason none are shown. It should be clear now that although ionization energy is defined as the energy required to remove an electron from an atom, which is what happens to metals when they ionize, the nonmetals ionize by gaining electrons to form anions. The energy involved in a neutral atom gaining an electron is called the electron affinity, and is our next topic.

2.4.4 Electron Affinity Electron affinity is defined as the amount of energy released when adding an electron to a ground-state, isolated, gaseous atom. The term “affinity” indicates that electron affinity is a measure of how eager the atoms of a given element are to take on another electron, an important measure when trying to understand or predict chemical reactions. Here is yet another instance in which energy relationships are crucial for understanding chemical behavior. Most atoms release energy when an electron is added, and accept the additional electron into an orbital to become an anion with a charge of –1. The major exceptions to know about are the noble gases. These elements do not release energy when an electron is added. Instead, energy is required to attach the electron and as a result the atom is not stable. Such an ion immediately rejects the electron and returns to its neutral state. This is because the valence shells in the noble gas atoms are full—the most stable electron configuration there is. Using sulfur as an example, the gaining of an electron is modeled with an ionization equation as follows: S(g )+ e − → S − (g )

ΔE = 2.077 eV

3.7

3.3

(2.3)

Br I

electron affinity (eV) electron affi nity (eV)

2.9

2.6

2.2

1.4 1.0

0.7

0.3 -0.1

Na B

Al Ca

Figure 2.13. Electron affinities.

1.8

Cs Sr

Ba 50

atomic number atomic number

Yb 70

100

The Periodic Law As with ionization Equation (2.1), the (g) indicates that the sulfur is in the gaseous state, and the energy change is stated to the right of the equation. It is important for you to distinguish clearly in your mind between ionization energy and electron affinity. Ionization energy is the energy required to remove an electron from a neutral atom (creating a cation). This energy is positive for every element. Electron affinity is the energy released when an electron is added to a neutral atom (creating an anion). Defined this way, this energy is also positive for every element except for the noble gases and a few other elements. Figure 2.13 shows the electron affinities for the stable atoms. As with other properties we have addressed, there is an obvious periodicity in the affinity values corresponding to the periods in the periodic table. We can again explain the large values associated with the Group 16 and Group 17 elements in terms of the effective nuclear charge, Zeff. In these two groups, the value of Zeff is as high as it gets, so the nuclei of these elements exert the highest possible attraction on a free electron, causing the atom/electron system to release as much energy as possible when the electron drops down into the lower energy state in one of the atom’s orbitals. The breaks between connected dots in Figure 2.13 indicate that several other elements are missing besides the noble gases. These are Be, N, Mg, Mn, and Zn. All these elements exhibit the same unstable behavior the noble gases do when an electron is added to them. Most of these can be at least partially explained by referring to the periodic table. Beryllium has two electrons in the 2s subshell. An added electron would go into the 2p subshell, where its energy is too high for the small nucleus to hold it. The same is the case for magnesium. Nitrogen has three electrons in the 2p subshell—a stable, low-energy state. Adding a fourth electron results in repulsion between two electrons in the px orbital, and once again the small nucleus does not have a strong enough attraction to wrangle those electrons and keep them both there. Similar considerations apply for manganese and zinc. Manganese has five electrons in the 3d subshell, so the 3d subshell is exactly half full with one electron in each orbital. This puts manganese in a situation similar to that of nitrogen. Zinc has 10 electrons in the 3d subshell and none in the 4p subshell, so zinc is in a situation similar to that of beryllium and magnesium. There is one element with a very slightly negative value for electron affinity that will still accept an electron. This element is ytterbium (Yb, Z = 70). Ytterbium is at the far right of the

Term

Definition Used In This Text

If the energy quantity ΔE is positive, it means:

ionization energy

energy required to remove an electron

energy flows into the atomic system from outside

electron affinity

energy released when adding an electron

energy flows out of the atomic system

Term

Alternative Definition

If the energy quantity ΔE is positive, it means:

ionization energy

energy involved when removing an electron

energy flows into the atomic system from outside

electron affinity

energy involved when adding an electron

energy flows into the atomic system from outside (thus, ΔE and the electron affinity values are nearly always negative)

Table 2.2. Definitions used here and elsewhere for ionization energy and electron affinity.

Chapter 2 rare-earth elements, in Period 6. Its 4f subshell is full and an additional electron goes into the 5d subshell. There is one final point to make before we move on. I have used care in defining ionization energy and electron affinity so that it is clear which way the energy is going when energy values are positive or negative. But you should be aware that some texts and other sources use definitions that may not be as clear, possibly leading to confusion. Instead of using the terms required or released to define the direction of positive energy flow, some sources use the term involved, and define positive energy flow in terms of energy going into or out of the atomic system. Both definitions are valid, as long as one specifies the direction of energy flow that constitutes a positive value of energy. The definitions we have seen in the last two sections are fairly standard. But to help you avoid confusion, Table 2.2 spells out my definitions and the alternative definitions side by side.

2.4.5 Electronegativity American chemist Linus Pauling (Figure 2.14) was one of the most important chemists of the 20th century. His work in the 1930s, 1940s, and 1950s on the nature of chemical bonds remains foundational for our understanding of chemistry to this day. For his work, Pauling won the Nobel Prize in Chemistry in 1954. Beginning in the mid-1940s, Pauling showed a deep concern for the negative health effects due to nuclear fallout from nuclear weapons testing, and he presented a petition signed by 11,000 scientists to the United Nations in 1958 to urge a ban on nuclear weapons testing. A scientific study that came out in 1961 showed that radioactivity contamination was indeed widespread in the population, leading to nuclear test ban treaties with the Soviet Union. All this led to Pauling winning the Nobel Peace Prize in 1962. To this day, Pauling remains the only person ever to win two unshared Nobel Prizes. In 1932, as part of his efforts to understand chemical bonding, Figure 2.14. American Pauling introduced what is now called the Pauling electronegativity chemist Linus Pauling scale. The electronegativity scale uses a dimensionless quantity called (1901–1994). electronegativity, with values running from 0.7 (francium) to 3.98 (fluorine). Electronegativity is a measure of how strongly atoms attract the electrons shared between atoms inside molecules. The higher an element’s electronegativity relative to the other elements in a molecule, the more an atom attracts the shared electrons in the molecule toward itself. Once again, the importance of electrical attraction in the atomic world is crucial. Recall from the Introduction that many of the important properties of water are – due to the difference between the electronegativities of oxygen (3.44) and hydrogen (2.20) in the water molecule. The oxygen atom attracts shared electrons more strongly than the hydrogen atoms do, and as a result the four bonding electrons in the molecule crowd over toward + + the oxygen atom. The result is that the oxygen region of the water molecule is more electrically negative and the hydrogen regions are more electrically positive, so water molecules are polar—negative at the elbow and positive at the ends, as shown in Figure 2.15. In this diagram, Figure 2.15. The higher electronegativity of oxygen the arrows point from the positive region of the molecule toward the atoms compared to hydrogen negative region of the molecule. We consider electronegativity further atoms results in the polar water in the next chapter in the context of covalent bonding. molecule.

The Periodic Law

4.0

3.5

electronegativity

3.0

2.5

2.0 1.5

1.0

0.5 0

100

atomic number

Figure 2.16. Electronegativity values, according to the Pauling electronegativity scale.

Figure 2.16 shows the periodicity of electronegativity and the trend of increasing electronegativity from left to right in the periodic table. There are gaps between the halogens and the alkali metals because the noble gases are not included. This is because electronegativity measures atomic attraction within molecules, and the noble gases don’t form bonds with other elements, molecular or otherwise, except under the extreme and unusual conditions that can be produced in a specialized laboratory. As with electron affinity and ionization energy, the trends in electronegativity can be understood in terms of the effective nuclear charge, Zeff , which increases from left to right as the number of protons in the nucleus increases while the number of core electrons screening the nucleus remains constant. A larger value of Zeff means not only that an atom attracts its own valence electrons more tightly. It also means the atom attracts the valence electrons shared with neighboring atoms more strongly relative to the attraction of neighboring atoms. The diagram in Figure 2.17 summarizes the general trend of electronegativity values over the periodic table from the lowest value held by francium, to the highest value held by fluorine. You 9 F may need various electronegativity values for exercises in this and later chapters, so they are all shown in Figure 2.18 and inside the back cover. If you look at the values toward the right end of the transition metals, you see that around Groups 87 10–13 there is a small decline in the Fr electronegativities, going against the general upward trend in values Figure 2.17. The arrows show the trends of increasing electronegativity from the lowest value at francium to the highest from lower left to upper right. value at fluorine.

Chapter 2 1 1

18 2

2 4

13 5

14 6

15 7

16 8

17 9

3.44 16

3.98 17

Mg 1.31 20

3 21

4 22

5 23

6 24

7 25

8 26

9 27

10 28

11 29

12 30

Sc 1.36 39

1.54 40

1.63 41

Sr 0.95 56

1.22 71

Os 2.2 108

2.2 109

0.98 11 3

0.93 19

0.82 37 5

0.82 55

0.79 87

2.20 3

0.7

1.57 12

1.00 38

0.89 88 0.9

2.04 13

1.0 103

1.33 72 1.3 104

1.6 73

1.5 105

1.66 42 2.16 74 1.7 106

1.55 43 2.10 75 1.9 107

1.83 44 2.2 76

1.88 45 2.28 77

2.19 33

2.58 34

Cd 1.69 80

1.78 81

Te 2.1 84

2.66 85

2.20 78 2.2 110

1.93 79 2.4 111

1.65 48

1.9 112

1.81 49

1.8 113

2.01 50 1.96 82 1.8 114

1.90 32

1.90 47

3.04 15

1.61 31

1.91 46

2.55 14

2.18 51 2.05 83 1.9 115

2.55 52

2.0 116

3.16 35

2.96 53

2.2 117

118

Figure 2.18. Electronegativity values, according to the Pauling electronegativity scale.

2.5

A Few Notes About Hydrogen

Hydrogen is located in Group 1 for a couple of reasons. First, it is an s-block element. Its lone electron is in a partially filled s subshell, just like the alkali metals. Second, hydrogen usually ionizes by losing an electron to form a cation, like all metals. In fact hydrogen forms the most basic of all cations—a lone proton. Third, in an aqueous solution (a solution with water as the solvent), acids, which are formed with hydrogen, dissociate (come apart) just as ionic compounds do. And just as the metal in a soluble ionic compound is a cation, so is the hydrogen

Hmm... Interesting.

Hydrogen in space Hydrogen is the most abundant element in the universe, making up about 75% of the mass of all matter. The image below, taken by the Hubble Space Telescope, is of the NGC 604 nebula, an enormous region of ionized hydrogen gas (a plasma) in the constellation Triangulum. The gas cloud is about 1,500 light years across, and has been called a “nursery of new stars” because of all the new stars formed within it. At the center are over 200 hot stars, each 10–15 times the size of our sun. These hot stars excite the hydrogen atoms, causing them to fluoresce, and heat the nebula to 10,000 kelvins—about twice the temperature at the surface of our sun.

The Periodic Law from a dissolved acid. Hydrogen’s atomic structure and its normal ionization as a cation with a charge of +1 indicate that hydrogen belongs at the top of Group 1. But though hydrogen’s structure assures that its place in Group 1 is probably not going to change, hydrogen’s chemical behavior is more like a halogen than an alkali metal. For one thing, hydrogen shares electrons in covalent bonds with other elements to form molecules, just as nonmetals do. In contrast, when metals bond they form crystals. A second factor is that hydrogen atoms bond to themselves to form molecules of H2, something no metal does. (An exception is the ion Hg22+.) Third, hydrogen can ionize by gaining an electron to fill the 1s subshell and become H–, an anion known as hydride. With its only subshell half full with one electron, easily emptied by losing one electron and easily filled by gaining one electron or sharing one pair of electrons (details next chapter), hydrogen is unique among the elements in the periodic table.

Chapter 2 Exercises SECTION 2.1

1. What entire group of elements did not appear in Mendeleev’s original periodic table? Why were they left out and how were they put in? 2. Write a paragraph explaining the general structure and arrangement of the periodic table. SECTION 2.2

3. State the chief chemical property that distinguishes the metals from the nonmetals. 4. Distinguish between cations and anions. 5. What is the “long form” of the periodic table, and why are there two forms? SECTION 2.3

6. Use the data in the table below to estimate the following bond lengths. In each case, determine the percent difference between the accepted value given and your estimate (see Section A.4.2 in Appendix A). a. the N—H bond length in a molecule of ammonia, NH3 (accepted value: 1.012 Å). b. the C—O bond length in carbon monoxide, CO (accepted value: 1.128 Å). c. the C—O bond length in carbon dioxide, CO2 (accepted value: 1.160 Å). d. the P—F bond length in phosphorus trifluoride, PF3 (accepted value: 1.570 Å). 7. Describe the trend in atomic radius going down a group and across a period. 8. Using the concept of effective nuclear charge, Zeff , write a description accounting for the trends in atomic size in the periodic table. 9. From your knowledge of the periodic table, put the elements rubidium (Rb), silver (Ag), xenon (Xe),

Element

Atomic Radius (Å)

Bonding Atomic Radius of Diatomic Molecule (Å)

hydrogen

0.32

0.741

carbon

0.75

1.242

nitrogen

0.71

1.098

oxygen

0.64

1.208

fluorine

0.60

1.413

phosphorus

1.09

1.893

sulfur

1.04

1.889

chlorine

1.00

1.987

selenium

1.18

2.166

bromine

1.17

2.281

iodine

1.36

2.666 69

Chapter 2 and yttrium (Y) in order of increasing atomic radius. Explain your order by referring to trends in the periodic table. 10. From your knowledge of the periodic table, put the elements sodium (Na), barium (Ba), cesium (Cs), and magnesium (Mg) in order of increasing atomic radius. Explain your order by referring to trends in the periodic table. 11. Based on your knowledge of trends in the periodic table, place the following atoms and ions in order of decreasing size: Be2+, Mg, Ca, and Mg2+. 12. Based on your knowledge of trends in the periodic table, arrange the following atoms and ions in order of size from largest to smallest: S2–, Ar, K+, Cl–, and Ca2+. SECTION 2.4

13. Aluminum and scandium both ionize to +3, even though scandium is in Group 3 and aluminum is in Group 13. Explain why this is. 14. Estimate the effective nuclear charge, Zeff , for vanadium (V), magnesium (Mg), chlorine (Cl), and arsenic (As). 15. Define ionization energy and describe the trends for ionization energy in the periodic table across periods and down groups. 16. Referring again to Table 2.1, explain the large increase in ionization energy that occurs in the yellow shaded region of that table. 17. Write a description accounting for the trends in ionization energy in terms of effective nuclear charge, Zeff , and other factors. 18. Write the condensed electron configurations for Cu2+, As5+, Ag+, and Au3+. 19. Why are ionization energies so much higher when core electrons are involved than they are when only valence electrons are involved? 20. Based on your knowledge of trends in the periodic table, place the following atoms in order of increasing ionization energy: Ar, Sr, P, Mg, and Ba. 21. Distinguish between ionization energy and electron affinity. 22. Based on your knowledge of trends in the periodic table, place the following atoms in order of increasing electron affinity: Br, Rb, and S. 23. Explain what the electronegativity scale is used for and how it arose. 24. Of the following cations, which is least likely to form: Ca3+, Mg2+, K+? Explain your response. 25. Why do the chalcogens form ions with a charge of –2? 26. Distinguish between electron affinity and electronegativity. 27. Referring to the periodic table, describe the chemical properties of potassium (K), sulfur (S), xenon (Xe), iodine, (I), and manganese (Mn). 28. Which of the following is likely to have the greatest difference between the third and fourth ionization energies: Cl, Sc, Na, C? 29. Using only the periodic table as a reference (without electronegativity data), predict the relative electronegativities of these elements and put them in order from least to greatest: Ni, Ta, Se, F, Cs, Cl. 30. Develop an explanation for why the electron affinity values for the chalcogens are each significantly lower than those of their halogen neighbors. 70

The Periodic Law 31. Why don’t the noble gases have electronegativity values listed in Figure 2.16? SECTION 2.5

32. Describe the chemical properties that place hydrogen in Group 1, and the chemical properties hydrogen shares with Group 17 elements. GENERAL REVIEW EXERCISES

33. Determine the energy released by the largest electron transition in the Lyman series (see Figure 1.8). State your answer in eV. 34. If a beam of laser light consists of photons with energies of 5.09 × 10–19 J, is the light visible? Explain your response. 35. Identify the block, period, and group for the elements represented by each of the following condensed electron configurations: a. [Ne]3s23p3 b. [Xe]6s24f145d106p1 c. [Kr]5s14d5 d. [Ar]4s23d3 36. How many orbitals are there in the shell associated with n = 4? How many electrons can this shell hold? 37. Determine the number of carbon atoms present in 112 g CO2. 38. Given 35.0 g H2SO4 (hydrogen sulfate, which is called sulfuric acid in aqueous solution), determine the percent composition. Then determine numbers of hydrogen, sulfur, and oxygen atoms present. 39. Identify some specific differences between the chemical properties of the alkali metals and those of the transition metals. 40. Naturally occurring bromine consists of the two isotopes bromine-79 and bromine-81. Given bromine’s atomic mass of 79.904 u, is the percentage of bromine-79 found in nature likely to be closer to 25%, 52%, 67%, or 80%? Explain your response. 41. How many grams of calcium are there in 3.00 mol CaBr2? 42. Analysis of a certain sample finds that the sample consists of 53.64% chlorine and 46.36% tungsten. Determine the empirical formula for this compound. 43. What is meant by the phrase, “chemistry is all about modeling”? 44. Why is it that scientists, when they are being accurate in their speech, avoid using the term truth? What are they likely to say instead? 45. Is there a difference between scientific facts and historical facts? If so, what is it? 46. What is the difference between molar mass and molecular mass? 47. Why must a calculation of molecular mass based on periodic table data necessarily be an average mass and not the mass of a specific molecule? 48. Distinguish between the two definitions for the mole.

Chapter 10 Thermochemistry and Kinetics

In the 19th century, physicists learned how to quantify the amount of heat released or absorbed in chemical reactions, such as in the beautiful exothermic reaction pictured above. For reactions that occur at constant pressure like the one pictured above, this quantity of heat is called the enthalpy of reaction. In the 1870s, American scientist Josiah Willard Gibbs combined enthalpy, entropy, and temperature together in an equation that defines a quantity called the Gibbs free energy. The Gibbs free energy is a powerful theoretical tool used to predict if reactions will occur at stated conditions. After studying this chapter, you will be able to look at the reaction pictured above and state whether the Gibbs free energy for the reaction is positive or negative.

286

Thermochemistry and Kinetics

Objectives for Chapter 10 After studying this chapter and completing the exercises, you should be able to do each of the following tasks, using supporting terms and principles as necessary. SECTION 10.1

1. Define enthalpy of reaction and relate this term to the energy pathways for exothermic and endothermic reactions. 2. Interpret enthalpy diagrams and thermochemical equations. 3. Use the First Law of Thermodynamics to distinguish between enthalpy and energy. 4. Define enthalpy of combustion and enthalpy of formation. SECTION 10.2

5. State Hess’s law. 6. Use enthalpy of combustion and enthalpy of formation data with the principle of Hess’s law to calculate the enthalpy of reaction and enthalpy of formation for given reactions. SECTION 10.3

7. Describe the relationships between entropy and temperature, entropy and phase, and entropy and the number of moles or particles of a substance. 8. Define the Gibbs free energy. 9. Outline the four possibilities for ΔH and ΔS in a chemical reaction, describe the resulting possibilities for ΔG, and describe the conditions under which a reaction occurs spontaneously. 10. Calculate the Gibbs free energy two different ways and use the result to determine if a reaction occurs spontaneously. SECTION 10.4

11. Describe collision theory and identify the two requirements for a collision between molecules of reactants to be effective in leading to the formation of product species. 12. Describe four factors scientists can control that influence the rate of reactions. 13. Explain the related concepts of reaction mechanism and intermediates. 14. Explain the related concepts of activation energy and the activated complex. 15. Using the Haber-Bosch process as an example, explain how metal catalysts work. 16. Distinguish between heterogeneous and homogeneous reactions. 17. Given a reaction mechanism with reaction speeds labeled, identify the rate-determining step and determine the rate law for the reaction. 18. Given a rate law, determine the effect on the reaction rate of altering the concentrations of reactants by specific proportions. 19. Explain how an experimentally determined rate law can provide clues to help chemists determine a reaction’s mechanism. 20. Identify the order of a rate law and the order of a reaction with respect to the concentrations in the rate law.

287

Chapter 10

10.1

Energy in Chemical Reactions

10.1.1 Introduction Energy plays a huge role in chemical reactions. The purpose for many chemical reactions is specifically to deliver energy. Examples of reactions like this are metabolism in animals, electrochemical reactions in batteries to power electronic devices, and burning of fuels to power machines. Figure 10.1 shows the test firing of one of the Space Shuttle main engines. The fuel for this engine is liquid hydrogen and liquid oxygen, which means there is nothing coming out of this engine except water vapor and heat. And yet there is enough energy released by this reaction to power a vehicle weighing over 2,000 tons. Throughout this text, we have seen that one of the major principles of nature is that processes tend to go in directions that minimize energy. But in Section 8.2.1 we saw that some processes, such as the dissolution of an ionic crystal, entail an energy increase. The melting of an ice cube is another example of a process that spontaneously goes in a direction leading to higher particle energy. In Section 8.2.3, we saw that the reason such processes can occur is that there is another factor involved in natural processes besides energy—entropy. The entropy of a system is a measure of the disorder present in the system. Just as spontaneFigure 10.1. The chemical reaction between liquid ous processes tend to go in a direction that minihydrogen and liquid oxygen is the energy source mizes energy, they also tend to go in a direction used by the Space Shuttle main engine. The white that maximizes entropy. Several of these examples formation under the engine at the bottom is are discussed in Section 8.2.3. Together, the enerwater vapor that has condensed to liquid droplets, gy and the entropy pathways open to a substance allowing the water vapor to become visible. or combination of substances determine what the substances do. Applying these topics to the study of chemical reactions is the subject of a branch of chemistry known as thermochemistry. A tour through the basic principles of thermochemistry will occupy us for the first three sections of this chapter. We begin with a study of energy in reactions. In Section 10.3.3, we come back to the topic of entropy, and we combine energy and entropy together into a single reaction-governing principle known as the Gibbs free energy. Before proceeding further in this chapter, you may wish to reread the paragraphs in Section 8.2.3 on entropy.

10.1.2 Enthalpy Recall from Section 8.2.2 that the enthalpy in a chemical system is related to the heat flow that occurs during chemical processes. The net amount of heat that flows into or out of a chemical system during a reaction at constant pressure is called the change in enthalpy, ΔH. By change in enthalpy we mean the difference between the enthalpy of the reaction products and the enthalpy of the reactants, as shown in the following equation: ΔH = H products − H reactants 288

(10.1)

Thermochemistry and Kinetics reactants

products

ΔH (negative)

energy

products

ΔH (positive)

reactants

time

exothermic reaction pathway

endothermic reaction pathway

Figure 10.2. Energy pathways for exothermic and endothermic reactions.

The change in enthalpy represents the heat that flows into (is absorbed by) or out of (is released from) a system during a process occurring at a constant pressure. During any process, the change in enthalpy is either positive or negative, as illustrated in Figure 10.2. In this figure, the red curves illustrate the energy pathway during the course of a reaction. As the reaction proceeds, the substances involved either release heat or absorb heat. As shown on the left side of the figure, if ΔH is negative, heat is released during the process, and the process is exothermic. In this case, the energy in the reaction products, Ef , is less than the energy in the reactants, Ei. For processes at constant pressure, this change in energy is equal to the change in enthalpy, ΔH. If ΔH is positive, as shown on the right, heat is absorbed during the process and the process is endothermic. In this case, the chemical potential energy in the reaction products is higher than the energy in the reactants. The enthalpy, H, of the reactants or products in a reaction cannot be measured directly. Only changes in enthalpy can be measured. As mentioned in Section 8.2.2, we are not going to concern ourselves too much with what enthalpy itself is, although in Section 10.1.3 we briefly explore how enthalpy relates to the internal energy in a system and the ways energy can flow into or out of a system. Our main concern here is with what the change in enthalpy represents. The general term for the change in enthalpy during a chemical reaction is enthalpy of reaction, also called the heat of reaction. The enthalpy of reaction is often listed with a chemical equation to form a thermochemical equation. For example, we have regularly encountered the reaction between hydrogen gas and oxygen gas to form water vapor: 2H 2 ( g ) + O2 ( g ) → 2H 2O( g ) In this reaction, two moles of hydrogen react with one mole of oxygen to produce two moles of water vapor, and in the process 483.6 kJ of heat are released. We write the thermochemical equation this way: 2H 2 ( g ) + O2 ( g ) → 2H 2O( g )

ΔH = −483.6 kJ

(10.2)

Again, ΔH is negative because the energy in the reaction products is less than the energy in the reactants and heat is released during the reaction. The information in Equation (10.2) can also be represented on an enthalpy diagram, as shown in Figure 10.3. The horizontal lines in this dia289

Chapter 10

enthalpy

gram indicate the enthalpies of the reactants and the products. The 2H2(g) + O2(g) reaction goes in the direction of the red arrow, releasing 483.6 kJ of heat at constant pressure. Several comments are in order about thermochemical equations such as Equation (10.2). First, enthalpy is an extensive property. This ΔH = –483.6 kJ means that ΔH depends on the amount of the substances present in the reaction. For this reason, when interpreting thermochemical equations we always take the coefficients in the equation to represent numbers of moles rather than numbers of atoms. In fact, sometimes fractional coefficients are used in thermochemical equations. (We 2H2O(g) see why in Section 10.2.1.) Thus, for the reaction of hydrogen and Figure 10.3. Enthalpy diagram oxygen involving half as many moles as shown in Equation (10.2), showing the enthalpy of we write the thermochemical equation this way: reaction for the combustion of 1 H 2 ( g ) + O 2 ( g ) → H 2O ( g ) 2

hydrogen.

ΔH = −241.8 kJ

enthalpy

When half as many moles are involved in the reaction, the enthalpy of reaction is half as much. Second, it is conventional to state enthalpy values in thermochemical equations at 25°C (and, again, constant pressure). This means that the reactants begin at 25°C and after the reaction are returned to 25°C. Equation (10.2) is written according to this convention, even though the H2O is shown as a vapor in the 2H2(g) + O2(g) equation. As written, the equation states that when the reaction occurs and the H2O vapor is returned to 25°C, the total energy released is 483.6 kJ. ΔHT = ΔH1 = Third, as you can see from the previous comment, it is im–571.6 kJ –483.6 kJ portant always to indicate the states for each reactant and product in thermochemical equations. In the hydrogen reaction shown in Equation (10.2), the enthalpy of reaction is different 2H2O(g) if H2O is in the liquid state instead of the vapor state shown ΔH2 = in the equation. This is because energy must be removed from –88.0 kJ H2O vapor in order to condense it to liquid. At 25°C, the molar 2H2O(l) heat of vaporization for water is 44.0 kJ/mol, so to condense two Figure 10.4. Enthalpy diagram moles of H2O, 88.0 kJ of heat must be removed. The thermo- showing the enthalpy of reaction chemistry of this condensation can be represented as for the combustion of hydrogen 2H 2O( g ) → 2H 2O( l )

ΔH = −88.0 kJ

to produce water vapor, and the additional enthalpy change required to condense the water.

So then, to write Equation (10.2) showing water in the liquid state, we must include the enthalpy change associated with condensing two moles of H2O, an additional decrease of 88.0 kJ. Including this quantity of heat in ΔH gives: 2H 2 ( g ) + O2 ( g ) → 2H 2O( l )

ΔH = −571.6 kJ

(10.3)

The combined process of combustion and condensation is represented in the enthalpy diagram of Figure 10.4. In this diagram, the enthalpy change in the combustion step is labeled ΔH1 and the enthalpy change for the condensation step is labeled ΔH2. The total enthalpy change is denoted by ΔHT.

290

Thermochemistry and Kinetics Finally, thermochemical equations are often written in reverse, as we see in Section 10.2.1. When we do this, the enthalpy change for the reverse reaction has the opposite sign. For example, the thermochemical equation for the combustion of methane is: CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H 2O( l )

ΔH = −890.8 kJ

(10.4)

This is an exothermic reaction that releases 890.8 kJ of heat for each mole of methane burned. If we imagine the reverse reaction of combining water and carbon dioxide to form methane, we have CO2 ( g ) + 2H 2O( l ) → CH 4 ( g ) + 2O2 ( g )

ΔH = 890.8 kJ

This reverse reaction is endothermic, and the same amount of energy has to be supplied as is released in the forward reaction. Accordingly, the sign on ΔH is reversed to show that energy is absorbed by the reactants. ΔH is now a positive quantity, as it always is for an endothermic reaction. Example 10.1 Hydrogen peroxide, H2O2, is a compound that decomposes to water and oxygen according to the following reaction: 2H 2O2 ( l ) → 2H 2O( l ) + O2 ( g )

ΔH = −196 kJ

Answer the following questions about this process: 1. Is this process exothermic or endothermic? 2. What is the change in enthalpy for each mole of hydrogen peroxide that decomposes? 3. If 25.0 g of H2O2 decomposes at 25°C, determine the net amount of heat released or absorbed. First, since ΔH is a negative quantity, the products of the reaction have less energy than the reactants, so heat is released during the reaction. The reaction is exothermic. Second, the heat of reaction shown is for two moles of H2O2, as the coefficient in the thermochemical equation indicates. The change in enthalpy for one mole—half this amount—is half as much, or –98.0 kJ. Third, we begin by determining how many moles of H2O2 we have. The molecular mass is 34.01 g/mol. Calculating the number of moles from this, we have 25.0 g ⋅

1 mol = 0.7351 mol 34.01 g

Finally, the given equation tells us that 196 kJ are released for every two moles of H2O2 involved. We use this ratio as a conversion factor and apply it to the number of moles we have. 0.7351 mol⋅

196 kJ = 72.0 kJ 2 mol

We find that 72.0 kJ of heat is released by the decomposition of 25.0 g H2O2.

291

Chapter 10

10.1.3 Understanding Enthalpy and Energy Before we move on, you may be wondering why we need to use the term enthalpy change to describe the heat being absorbed or released by a chemical reaction. The term enthalpy is probably unfamiliar to you and may sound strange or conceptually difficult. So why can’t we just use a term such as energy change or internal energy change, since we are talking about heat energy entering or leaving the thermochemical system? The answer has to do with the fact that there is more than one way for energy to enter or leave a thermochemical system. Thus, in general, the change in enthalpy may or may not be equal to the change in the internal energy of the substances in the reaction or the heat absorbed or released. Let’s consider these energy possibilities. First, energy can enter or leave the system through the absorption or release of heat. This can occur by means of an exothermic chemical reaction releasing heat into the system or an endothermic reaction absorbing heat from the system. Heat exchange can also occur by means of heating or cooling the system. Next, energy can enter or leave by means of work being done by or on the system. For example, imagine a chemical reaction—such as an explosive combustion of fuel—that occurs inside a cylinder, as depicted in Figlarger gas volume, hot, expanding ure 10.5. This is, in fact, exactly lower pressure reaction products what happens when a mixture of gasoline and air is ignited in one of the cylinders of a car engine. The explosion products are gas molecules at high pressure. This high pressure pushes the piston, which is mechanically connected to the machinery that makes Figure 10.5. Energy leaves a system when a gas does mechanical the wheels of the car turn. Pushwork by expanding inside a cylinder and pushing a piston. ing the piston is an example of mechanical work: force is applied to an object and moves it a certain distance. Work is energy, and when the expanding gases push the piston, the energy involved—the work—leaves the thermochemical system (the gases in the cylinder) and is converted into the kinetic energy of the moving car. This is called expansion work, and is energy that leaves the thermochemical system by a mechanism other than heat. Third, there are other forms of work by which energy can enter or leave the system, such as electromagnetic interactions, or mechanical parts driven from the outside that stir (accelerate) the gases in the system and thus increase their internal energy. For the rest of this discussion, we assume these play no role in our system and that only expansion work is involved. In general, the change in enthalpy in a system is a function of pressure and entropy. But if the pressure is held constant, then the mathematics allows us to describe the change in enthalpy more simply as follows: ΔH = ΔU + PΔV

(10.5)

In this equation, ΔU is the change in internal energy in the system (see Section 6.1.1), P is the system pressure, and ΔV is the change in the system volume. Again, Equation (10.5) can only be written because we are holding the pressure constant.1 Now to see how this change in enthalpy can be equal to the heat absorbed or released, we start with the so-called first law of thermodynamics, which is: 1 An example of a reaction at constant pressure is a chemical reaction that occurs in the open atmosphere. The pressure is atmospheric pressure throughout the reaction. 292

Thermochemistry and Kinetics Q = ΔU +W

(10.6)

In this equation, Q is the heat entering or leaving the system, ΔU is the change in the internal energy of the molecules as before, and W is the work done by (or on) the system. Now, since we are considering only the expansion work, you may recall from an earlier physics class that the work associated with pushing an object is defined as W = F ⋅d

(10.7)

In this equation, F represents a force (in newtons) and d represents a distance (in meters). We can easily relate this expression for work to the pressure and volume of a gas. Recall from Section 6.1.4 that pressure is defined as force per unit area, or P=

F A

(10.8)

Solving this equation for the force, we have F = P⋅A

(10.9)

Substituting this expression into Equation (10.7) gives us W = P ⋅ A⋅d

(10.10)

Figure 10.6 relates these ΔV quantities to a piston in a cylpiston in inder. The force applied to the piston in position 2 position 1 end of the piston is equal to the force = F pressure in the cylinder times the surface area of the end of the piston. The force causes area = A d the piston to move a distance pressure = P = F/A d. (Again, we assume this happens in a way that allows the Figure 10.6. Given W = F ∙d, F = P ∙A, and A∙d = ΔV, we get W = P ΔV. pressure to remain constant.) When this happens, the volume change, ΔV, in the cylinder, shown in blue, is equal to the surface area of the end of the piston, A, times the distance it moves, d, or ΔV = A⋅d

(10.11)

Substituting this expression into Equation (10.10) gives us W = PΔV

(10.12)

Finally, substituting this expression for the expansion work into Equation (10.6) gives: Q = ΔU + PΔV

(10.13)

Comparing this equation to Equation (10.5), you see that ΔH = Q, and thus the change in enthalpy equals the heat entering or leaving the system. This result depends on the pressure being constant and on the expansion work being the only form of work affecting the system. As a 293

Chapter 10 quick application, the chemical reaction of an outdoor explosion releases heat and produces expansion work as the hot gases formed by the explosion push back the atmosphere. In this case, the change in enthalpy of the chemical system is equal to the heat released by the reaction.

10.1.4 Enthalpy of Combustion The heat released by the complete combustion of one mole of a substance at a constant pressure is called the molar enthalpy of combustion (or heat of combustion). The enthalpy of combustion is a special case of the enthalpy of reaction we have been discussing. Conventionally, enthalpy of combustion values are listed for controlled conditions in which reactants and products are in their so-called standard states at standard conditions (T = 25°C and P = 1 bar). The standard state is the state compounds are normally in at standard conditions. Table 10.1 lists enthalpy of combustion values for a number of representative substances. Notice in the table that the value for hydrogen is listed as –285.8 kJ/mol. For two moles, this is –517.6 kJ, which agrees with Equation (10.3). We use a special notation to designate enthalpy of combustion. Using hydrogen as an example, the enthalpy of combustion is ΔH c! = −285.8

kJ mol

The subscript c designates combustion, and the superscript ° indicates that the value pertains to the reactants and products in their standard states. As another example, the thermochemical equation for the combustion of butane is C 4H10 ( g ) +

13 O2 ( g ) → 4CO2 ( g ) + 5H 2O( l ) 2

ΔH c! = −2877.6

kJ mol

Enthalpy of combustion values have been measured for many common substances. As we see in Section 10.2, these values may be used along with other data to perform thermochemical Substance

Formula

ΔH c! (kJ/mol)

Substance

Formula

ΔH c! (kJ/mol)

acetic acid (l)

CH3COOH –874.2

methane (g)

CH4

–890.8

acetone (l)

C3H5OH

–1789.9

ethyne (acetylene) (g)

C2H2

–1301.1

ammonia (g)

NH3

–382.8

ethene (ethylene) (g)

C2H4

–1411.2

carbon (graphite) (s)

–393.5

ethane (g)

C2H6

–1560.7

carbon monoxide (g)

–283.0

propane (g)

C3H8

–2219.2

ethanol (l)

C2H5OH

–1366.8

butane (g)

C4H10

–2877.6

ethylene glycol (l)

C2H6O2

–1189.2

pentane (l)

C5H12

–3509.0

formic acid (l)

CH2O2

–254.6

benzene (l)

C6H6

–3267.6

glucose (s)

C6H12O6

–2805

hexane (l)

C6H14

–4163.2

hydrogen (g)

–285.8

toluene (l)

C7H8

–3910.3

methanol (l)

CH3OH

–726.1

heptane (l)

C7H16

–4817.0

sucrose (s)

C12H22O11

–5640.9

naphthalene (s)

C10H8

–5156.3

Table 10.1. Standard molar enthalpy of combustion values for representative substances at T = 25°C and P = 1 bar. Common hydrocarbons are shown in the right column.

294

Thermochemistry and Kinetics calculations that can be used to predict how chemical reactions would proceed without the need for additional experimentation. Being able to predict the results of a reaction without actual experimental data is a powerful analytical tool. Scientists measure the enthalpy of combustion using a device called—believe it or not—a bomb calorimeter. In general, a calorimeter is an insulated container with accommodations for measuring the temperature of the contents. Substances at different temperatures are placed inside the calorimeter, and allowed to reach thermal equilibrium without any heat flowing into or out of the calorimeter from the outside. Combined with the known masses of the substances inside the calorimeter, the temperature at thermal equilibrium may be used to determine thermal properties (such as the molar heat capacity) Figure 10.7. A bomb calorimeter. of one of the substances inside. Calculations like this are based on the heat calculations we encountered in Chapter 6 and are not difficult. For simple experiments, a calorimeter can be as rudimentary as an insulated coffee cup with a lid and a thermometer. The bomb calorimeter used for measuring enthalpy of combustion values is significantly more sophisticated, as the photo in Figure 10.7 shows. The sample to be burned is placed in the steel canister, called a bomb. Inside the canister is an electronic ignition source. After being sealed, the air in the canister is replaced with oxygen and the canister is lowered into the larger chamber shown in the figure. The larger chamber contains a precisely measured amount of water at a precisely measured temperature and is insulated to prevent heat flowing in from or out to the outside. The ignition source is used to ignite the sample under test, which burns completely in the presence of pure oxygen. The energy released by the reaction is calculated from the temperature rise of the water in the outer chamber.

10.1.5 Enthalpy of Formation The reaction shown in Equation (10.2) describes the formation of water from its elements, hydrogen and oxygen. In a manner similar to the enthalpy of combustion, the molar enthalpy of formation is defined as the amount of heat released or absorbed when one mole of a substance is formed from its elements. As with enthalpy of combustion, enthalpy of formation values are stated assuming that the reactants and products are at standard conditions (25°C, 1 bar) and in their standard states. The notation used for designating enthalpy of formation is ΔH !f In this notation, the subscript f denotes formation. For most substances, the formation of the substance from its elements is an exothermic reaction, resulting in a negative heat of formation. The enthalpy of formation for water is the same as the enthalpy of combustion of hydrogen (since hydrogen and oxygen are the elements composing water), and so for water ΔH !f = −285.8

kJ mol 295

Chapter 10 Substance

Formula

ΔH !f (kJ/mol)

Substance

Formula

ΔH !f (kJ/mol)

acetylene (g)

C2H2

227.4

hydrogen chloride (g)

HCl

–92.3

ammonia (g)

NH3

–45.9

hydrogen fluoride (g)

–273.3

benzene (l)

C6H6

49.1

hydrogen iodide (g)

26.5

calcium chloride (s)

CaCl2

–795.4

methane (g)

CH4

–74.6

carbon monoxide (g)

–110.5

methanol (l)

CH3OH

–239.2

carbon dioxide (g)

CO2

–393.5

propane (g)

C3H8

–103.8

ethane (g)

C2H6

–84.0

sodium chloride (s)

NaCl

–411.2

ethanol (l)

C2H5OH

–277.6

sucrose (s)

C12H22O11 –2226.1

ethylene (g)

C2H4

52.4

sulfur dioxide (g)

SO2

–296.8

glucose (s)

C6H12O6

–1273.3

water (l)

H2O

–285.8

–36.3

water (g)

H2O

–241.8

hydrogen bromide (g) HBr

Table 10.2. Standard molar enthalpy of formation values for representative substances at T = 25°C and P = 1 bar. See Appendix B, Table B.7 for a more extensive list.

Considering all we have discussed so far in this text about energy, it should make sense to you that substances with a large negative value for the enthalpy of formation are very stable because a lot of energy is required to undo the formation of the substance. Substances with a small negative value for the enthalpy of formation are relatively unstable, since only a small amount of energy can disrupt the structure of the atoms in such a substance. Table 10.2 contains enthalpy of formation values for a number of representative substances. As you see, some substances have a positive value for ΔH !f . The formation of such compounds is endothermic, and the compounds are quite unstable. These compounds can spontaneously decompose or react violently. As examples, hydrogen iodide spontaneously decomposes when stored at room temperature. Acetylene is the gas used with oxygen in the cutting torches welders use to cut through steel. A lot of energy has to be supplied to form acetylene, so the molecules are unstable and combust very violently, releasing a lot of energy as heat in the process. Although it is not listed in the table, mercury(II) fulminate, Hg(CNO)2, has a high positive enthalpy of formation, ΔH !f = 270 kJ/mol. This substance is very sensitive to shock, which makes it a great detonator to set off larger charges of more environmentally safe explosives (safer, that is, because they don’t contain mercury). Interestingly, although mercury(II) fulminate was first prepared in 1800, its crystal structure was not known until 2007. A space-filling model of a portion of the crystal lattice for this compound is shown in Figure 10.8. This compound has made an appearance in a number of popular movies and TV series. The enthalpy of formation for pure elements in their standard states is defined as ΔH !f = 0. This value applies to molecules for elements that are naturally molecular at standard conditions. Thus, for H2, O2, and the other diatomic elements, ΔH !f = 0. However, the enthalpy of formation for atomic hydrogen, H, is ΔH !f = 218.0 kJ/mol. To separate an H2 molecule, enFigure 10.8. Crystal structure of mercury(II) ergy must be supplied to break the covalent bond in fulminate. Mercury atoms are shown in gray.

296

Thermochemistry and Kinetics H2. If you refer back to Table 3.9, you see that the bond energy for one mole of H2 is –436.0 kJ. Since two atoms of H are in each H2 molecule, the heat of formation for a mole of individual H atoms is half this amount (and positive). The implication of defining the enthalpy of formation for pure elements to be zero is that substances with a negative value for ΔH !f are more stable than the elements from which they are formed. Compounds with a positive value for the molar enthalpy of formation are less stable than the elements from which they are formed.

10.2

Calculating Enthalpy of Reaction and Enthalpy of Formation

10.2.1 Hess’s Law Hess’s law2 states that the enthalpy change in a chemical reaction is independent of the pathway the reaction takes. Another way of stating this principle is that the enthalpy change in a reaction is the same whether the reaction takes place in one step or in many steps. So if we can imagine a sequence of reaction steps that leads, in principle, from the same reactants to the same products as for a single-step reaction, the sum of the enthalpy changes for the individual steps must equal the enthalpy change for the single-step reaction. This important principle can be used to determine the enthalpy of reaction in cases where measuring ΔH is difficult or even impossible. We can use Hess’s law to solve for any unknown enthalpy of reaction so long as we know either the enthalpy of combustion values for all the reactants and products, or the enthalpy of formation values for all the reactants and products. To illustrate the use of Hess’s law for such calculations, we calculate the enthalpy of formation for methane gas, CH4. The elements composing methane are solid carbon (graphite) and hydrogen gas. This means the equation showing the formation of methane from its elements is C( s ) + 2H 2 ( g ) → CH 4 ( g )

ΔH !f = ?

(10.12)

For this illustration, we derive the enthalpy of formation for methane using only enthalpy of combustion values for the three reactants and products (C, H2, and CH4) in Equation (10.12). These three thermochemical equations, with enthalpy values from Table 10.1, are as follows: C( s ) + O2 ( g ) → CO2 ( g ) H2 ( g ) +

1 O 2 ( g ) → H 2O( l ) 2

CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H 2O( l )

ΔH c! = −393.5 kJ/mol ΔH c! = −285.8 kJ/mol ΔH c! = −890.8 kJ/mol

Notice that the equation for the combustion of hydrogen is written to represent the combustion of one mole of H2 because the enthalpy of combustion value from Table 10.1 is for a single mole. Writing the equation this way requires the use of a fractional coefficient on O2. Our plan now is essentially to add these equations together to generate the formation equation for methane, Equation (10.12). To do this, we must arrange the equations so that the desired compounds are on the correct side of the equation in each case. Unwanted terms in the addition cancel out if we have the same term (with the same coefficient) appearing on both sides. 2 American orthography would normally have us write Hess’ law. However, the British orthography with the final s added is used universally when referring to Hess’s law. 297

Chapter 10 To assure proper cancellation, it may be necessary to multiply one or more of the equations by a constant. So we are about to make these two preparatory maneuvers before adding the equations: (1) reverse equations and signs on the enthalpy as necessary to get compounds on the correct side according to Equation (10.12), and (2) multiply equations by constants as necessary so that when the equations are added together Equation (10.12) results without any extra terms. In Equation (10.12), carbon and hydrogen are on the left side. In the first two combustion equations, carbon and hydrogen are on the left, so these equations are fine in this respect. However, methane is on the right in Equation (10.12), so we need to reverse the methane combustion equation to show methane on the right. Recall from earlier in the chapter that when we reverse a thermochemical equation, the sign on the enthalpy of reaction changes. CO2 ( g ) + 2H 2O( l ) → CH 4 ( g ) + 2O2 ( g )

ΔH ! = 890.8 kJ/mol

(Note that the enthalpy is now written ΔH !, without the subscript, because written in reverse this is no longer a combustion reaction.) Now let’s list our equations in preparation for adding them together. C( s ) + O2 ( g ) → CO2 ( g ) H2 ( g ) +

1 O 2 ( g ) → H 2O( l ) 2

CO2 ( g ) + 2H 2O( l ) → CH 4 ( g ) + 2O2 ( g )

ΔH c! = −393.5 kJ/mol ΔH c! = −285.8 kJ/mol ΔH ! = 890.8 kJ/mol

When we add these together, we need all the CO2, H2O, and O2 terms to cancel out because none of these compounds appears in Equation (10.12). Let’s look at what it takes for this to happen in the case of O2. We see that if we multiply the entire second equation by 2, the 12 O2 term becomes just O2, and the O2 terms on the left add to be equal the 2O2 shown on the right, enabling them to cancel out. It turns out that doubling the coefficients on the second equation enables everything to cancel out that needs to cancel out. Accordingly, we multiply this entire equation by 2, including the enthalpy. Doing so and canceling out terms that appear on both sides gives us the following set of equations. These we now add together to give the equation at the bottom, which is identical to Equation (10.12). C( s ) + O2 ( g ) → CO2 ( g )

ΔH c! = −393.5 kJ/mol

2H 2 ( g ) + O2 ( g ) → 2H 2O( l )

2ΔH c! = 2( −285.8 kJ/mol )

CO2 ( g ) + 2H 2O( l ) → CH 4 ( g ) + 2O2 ( g )

ΔH ! = 890.8 kJ/mol

C( s ) + 2H 2 ( g ) → CH 4 ( g )

ΔH !f = −74.3 kJ/mol

Adding together the enthalpy values gives us the desired result, the enthalpy of formation of methane, –74.3 kJ/mol. Comparing this value to the one given in Table 10.2, we see a slight difference. This is probably due to rounding in the values stated in Tables 10.1 and 10.2. 298

Thermochemistry and Kinetics

enthalpy (kJ/mol)

Figure 10.9 is an enthalpy diagram that compares the artificially constructed multistep reaction pathway we used in our calculation (Path 1) to the single-step reaction pathway we sought to compute (Path 2). The two pathways are indicated in the figure by yellow circles. Hess’s law says that the net enthalpy change is the same for all such paths. This is why we can use the ΔH values for the sequence of three reactions in Path 1 to determine the ΔH for the desired singlestep reaction in Path 2. The diagram looks complicated—but ignore the dashed arrows for the moment as we walk through it. Both pathways begin with solid carbon, C(s), at a reference enthalpy of 0 kJ/mol. If you follow the red arrows and yellow circles for each path, you see they both end up at CH4(g) at H = –74.3 kJ/mol. The simplest path is Path 2. The enthalpy change for the path is what we sought to compute (the unknown). We computed it using Hess’s law applied to the sequence of known reactions with known enthalpy of combustion values in Path 1. To follow any path, keep this in mind: the reactants in the reaction we sought to compute are C(s) and 2H2(g). We can make use of these reactants in the path whenever we need to. But whenever a path requires other reactants that have not been produced yet, we “borrow” them from the environment for the purpose of our calculation, knowing that we can pay them back later in the reaction sequence. (Since this computation is a mathematical model, we 1 2 should not be bothered if we need to “borrow” some oxygen for a moment.) C(s) + O2(g) C(s) + 2H2(g) 0 Path 1 begins with a reΔH = –74.3 kJ/mol action between C(s) and CH4(g) + 2O2(g) CH4(g) –74.3 O2(g) because this is the first reaction listed in our computation. (We could take 1.1Combustion of C(s) these reactions in any seΔH = –393.5 kJ/mol quence, but I am just taking them in the same order as they are listed in the computation.) So, we borrow one CO2(g) –393.5 mole of O2 to begin. This 2H2(g) + O2(g) reaction produces CO2(g), and the change in enthalpy 1.3Combustion of CH4(g) is –393.5 kJ/mol. The sec(in reverse) ΔH = 890.8 kJ/mol ond reaction is between 2H2(g) and O2(g). Thus, we now borrow another mole of 1.2Combustion of 2H2(g) O2 in order to proceed. We ΔH = –571.6 kJ/mol don’t yet need our CO2 from 1 the previous reaction, so we will simply bring it with us for later use. This is indicated by the dashed blue arrow. The product of the second 1 CO2(g) + 2H2O(l) 2H2O(l) CO2(g) reaction is 2H2O(l), with an –965.1 additional enthalpy change Figure 10.9. Enthalpy diagram comparing the net enthalpy of reaction for of –571.6 kJ/mol, bringing two separate pathways leading from C(s) + 2H (g) to CH4(g). According to the total enthalpy change so Hess’s law, the net enthalpy change is the same2 for all such paths. 299

Chapter 10 far to –965.1 kJ/mol. For the third reaction in Path 1, we take the CO2(g) and 2H2O(l) produced by the first two reactions, and together they undergo an endothermic reaction to produce CH4(g) and 2O2(g). This reaction entails an increase in enthalpy of 890.8 kJ/mol, bringing us up to an enthalpy level of –74.3 kJ/mol. Here we are at the end of Path 1. We arrived at CH4(g), and created two moles of O2(g) along the way. As shown by the orange dashed arrows, we use these to pay back what we borrowed, leaving us with CH4(g) and a net enthalpy change of ΔH !f = −74.3 kJ/mol Now notice what we just did. We calculated the enthalpy of formation for CH4 using only enthalpy of combustion data. In the next example, we calculate an enthalpy of combustion using only enthalpy of formation data. Example 10.2 Use Hess’s law to determine the enthalpy of combustion for propane, given the enthalpy of formation for each of the reactants and products in the thermochemical equation for the combustion of propane: C 3H8 ( g ) + 5O2 ( g ) → 3CO2 ( g ) + 4H 2O( l )

ΔH c! = ?

(10.13)

The thermochemical equations for the formation of the other three reactants and products are as follows. There is no formation equation for O2 because the enthalpy of formation for pure elements is zero. Enthalpy of formation values are from Table B.7 in Appendix B. 3C( s ) + 4H 2 ( g ) → C 3H8 ( g )

ΔH !f = −103.8 kJ/mol

C( s ) + O2 ( g ) → CO2 ( g )

ΔH !f = −393.5 kJ/mol

H2 ( g ) +

1 O 2 ( g ) → H 2O( l ) 2

ΔH !f = −285.8 kJ/mol

To prepare for adding these together, we must get each compound on the correct side and adjust coefficients so the addition cancels out all unwanted terms. In Equation (10.13), propane is on the left, so we reverse the first equation and change the sign on the enthalpy. Carbon dioxide and water are both on the right in Equation (10.13) and they are also on the right in the formation equations. Reversing the first equation gives us C 3H8 ( g ) → 3C( s ) + 4H 2 ( g )

ΔH ! = 103.8 kJ/mol

C( s ) + O2 ( g ) → CO2 ( g )

ΔH !f = −393.5 kJ/mol

H2 ( g ) +

1 O 2 ( g ) → H 2O( l ) 2

ΔH !f = −285.8 kJ/mol

Looking now at the coefficients, Equation (10.13) has no C(s) or H2(g) in it. Multiplying the second equation by three enables the carbons to cancel, and multiplying the third equation by 300

Thermochemistry and Kinetics four enables the hydrogens to cancel. Remember, when multiplying an equation by a constant, the enthalpy must be multiplied as well. The following equations show the new coefficients and enthalpies after the multiplications, the cancellations of terms, and the summed equation at the bottom. C 3H8 ( g ) → 3C( s ) + 4H 2 ( g )

ΔH ! = 103.8 kJ/mol

3C( s ) + 3O2 ( g ) → 3CO2 ( g )

3ΔH !f = 3( −393.5 ) kJ/mol

4H 2 ( g ) + 2O2 ( g ) → 4H 2O( l )

4ΔH !f = 4( −285.8 ) kJ/mol

C 3H8 ( g ) + 5O2 ( g ) → 3CO2 ( g ) + 4H 2O( l )

ΔH c! = −2219.9 kJ/mol

The bottom equation is identical to Equation (10.13), as required. We calculate the enthalpy of combustion for propane to be –2219.9 kJ/mol. Comparing this to the value listed in Table 10.1, we again see a small difference in the last digit due to rounding.

enthalpy (kJ/mol)

Notice this important tip from this example: CO2 and H2O are compounds that appear in the final equation and each appears only once in the formation equations. When this happens, simply multiply their equations by the coefficients you need in the final equation. Every1 3C(s) + 4H2(g) 4H2(g) thing else works out. (This 103.8 tip will come in handy in the 3C(s) + 3O2(g) 1.1Decomposition of C3H8(g) exercises.) ΔH = 103.8 kJ/mol 1 C3H8(g) An enthalpy diagram for 0 the reaction sequence in ExC3H8(g)+ 5O2(g) 2 ample 10.2 is shown in Fig1.2Formation of 3CO2(g) ΔH = –1180.5 kJ/mol ure 10.10. Path 2 represents the combustion of propane, the enthalpy of combustion we sought to compute. Path 1 is the sequence of three reactions we put together us3CO2(g) 4H2(g) ing Hess’s law. Reactants we –1076.7 can use whenever we need 4H2(s) + 2O2(g) them are the C3H8(g) and 5O2(g). As before, the blue 1 dotted arrows indicate reac1.3Formation of 4H2O(l) ΔH = –1143.2 kJ/mol

Figure 10.10. Enthalpy diagram comparing the net enthalpy of reaction for two separate pathways leading from C3H8(g) + 5O2(g) to 3CO2(g) + 4H2O(l).

–2219.9

3CO2(g) + 4H2O(l)

301

Chapter 10 tion products that we carry along from one step to another until they are needed. No “borrowing” is needed this time.

10.2.2 Hess’s Law and the General Enthalpy Change Equation Recall that Equation (10.1) defines the change of enthalpy for a reaction as the difference between the enthalpy of the products and the enthalpy of the reactants. This relationship may be expressed using the enthalpy of formation values, ΔH !f , for products and reactants as follows: ΔH ! = ΣΔH !f , products − ΣΔH !f , reactants

(10.14)

In this expression, the upper case Greek letter sigma, Σ, means “sum.” So this expression tells us that we can calculate any enthalpy of reaction by summing the enthalpy of formation values for the products, summing the enthalpy of formation values for the reactants, and then taking the difference between these two sums. This is a powerful computational tool. The entire process of manipulating reactions we just described boils down to the difference between two sums of enthalpy of formation values. To illustrate, we will revisit Example 10.2, noting that Table B.7 in Appendix B contains an extended list of enthalpy of formation values. The goal is to calculate the enthalpy of reaction for the following reaction: C 3H8 ( g ) + 5O2 ( g ) → 3CO2 ( g ) + 4H 2O( l ) We tabulate the ΔH !f values for the reactants and for the products, recalling that the enthalpy of formation for a pure element in its standard state (such as O2) is zero. We need to take into account the number of moles for each reactant and product in the equation. Pulling the ΔH !f values from Appendix B, Table B.7, we have: Reactants

Products

ΔH ! (kJ/mol)

ΔH !f (kJ/mol)

C3H8(g)

–103.8

3CO2(g)

3(–393.5) = –1180.5

5O2(g)

4H2O (l)

4(–285.8) = –1143.2

Sum (Σ)

–103.8

–2323.7

Now we use Equation (10.14) to compute this enthalpy of reaction: kJ ⎞ kJ kJ ⎞ ⎛ ⎛ ΔH ! = ΣΔH !f , products − ΣΔH !f , reactants = ⎜ −2323.7 ⎟ − ⎜ −103.8 ⎟ = −2219.9 ⎝ mol ⎠ ⎝ mol ⎠ mol This is the same value we calculated before in Example 10.2. Example 10.3 Use Equation (10.14) to calculate the enthalpy of reaction for the following reaction: SiCl 4 ( l ) + 2H 2O( l ) → SiO2 ( s ) + 4HCl ( g ) Tabulating the enthalpy of formation values from Appendix B, we have:

302

Thermochemistry and Kinetics Reactants

Products

ΔH ! (kJ/mol)

ΔH !f (kJ/mol)

SiCl4(l)

–687.0

SiO2(s)

–910.7

2H2O(l)

2(–285.8) = –571.6

4HCl(g)

4(–92.3) = –369.2

Sum (Σ)

–1258.6

–1279.9

Inserting these two sums into Equation (10.14), we have the following: kJ ⎞ kJ kJ ⎞ ⎛ ⎛ − ⎜ −1258.6 = −21.3 ΔH ! = ΣΔH !f , products − ΣΔH !f , reactants = ⎜ −1279.9 ⎟ ⎟ ⎝ mol ⎠ mol mol ⎠ ⎝

10.3

Free Energy

10.3.1 Brief Review of Enthalpy and Entropy We are now ready to put enthalpy and entropy together into a single law that can predict whether a process spontaneously occurs. As we saw in Chapter 8, the tendency in nature is for processes to go in a direction that decreases potential energy and increases entropy. But these tendencies may or may not point in the same direction, so we need a way to weigh them against each other so we can determine if a reaction occurs or not. The goal of this section is to present the equation that does this. Let’s first summarize what we have seen so far about enthalpy and entropy. For enthalpy, we know the following: Enthalpy Summary 1. For processes at constant pressure, the change in enthalpy, ΔH, is equal to the heat absorbed or released during the process. 2. For exothermic reactions, ΔH is negative; for endothermic reactions, ΔH is positive. 3. Most reactions that occur in nature are exothermic. 4. When an exothermic reaction occurs, heat is released. This means that the potential energy in the molecules of the products is lower than in those of the reactants, so the product molecules are more stable. 5. Nature favors processes in which the enthalpy decreases, but processes in which the enthalpy of a system increases also occur. 6. The First Law of Thermodynamics for constant-pressure processes may be written as ΔH = ΔU + PΔV . And now, a summary of what we have seen about entropy, along with two new pieces of information added for completeness: Entropy Summary 1. The entropy of a system is a measure of the disorder or randomness in the system. 2. For every process, the entropy of the universe increases. If the entropy decreases in a certain system, this can only happen by a larger increase in entropy somewhere else. (This is one way of stating the Second Law of Thermodynamics.) 303

Chapter 10 3. Nature favors processes in which the entropy increases, but processes in which the entropy of a system decreases also occur. 4. (New) Entropy is denoted by S, and a change in entropy by ΔS. 5. (New) The entropy of a pure crystalline substance at absolute zero (0 K) is equal to zero. As temperature increases, entropy does as well.

10.3.2 More on Entropy

entropy (J/mol∙K)

The last point above about entropy deserves comment. I have described entropy as a measure of the disorder in a system. The mathematical way physicists characterize entropy is by assessing the number of different possible states that are available for a system to be in. In this statement, the term “states” refers to every possible combination of position and velocity available to all the atoms in the system. Since we are talking about the states of atoms, the term scientists use for all these states is microstates.3 At absolute zero, molecular motion ceases. A perfectly ordered crystal in which there is no movement at all can only be in one state, so there is no randomness at all—no uncertainty at all about where an atom might be or how fast it might be moving. Thus, the entropy of a pure crystal at 0 K is zero. Now, if the temperature of the crystal is a bit above 0 K, the atoms in the crystal are vibrating. There is not much randomness in this crystal because the atoms are not free to translate, but since there is now a bit of uncertainty associated with the position and velocity of each of the vibrating atoms, there is entropy in the system. In general, the higher the temperature of a system solid gets, the more entropy there is because there is more randomness—more microstates available boiling to the particles in the system. A large increase liquid in entropy occurs when the substance melts to become a liquid because then the particles are free to move around a bit. And when vaporization occurs, an even larger increase in entropy gas melting takes place. In the gaseous state, the velocities of particles are high and the particles are spread out, so the number of microstates available to all 0 the particles in the system is huge. These gen0 temperature (K) eralizations about the entropy of a system with Figure 10.11. Variation of entropy with temperature respect to temperature are depicted graphically and phase transitions in a substance. in Figure 10.11. One more observation we can make about entropy pertains to the numbers of moles indicated by chemical equations. The Haber-Bosch process for the manufacture of ammonia goes according to this equation: N 2 ( g ) + 3H 2 ( g ) → 2NH3 ( g )

(10.15)

3 It was Ludwig Boltzmann’s work in statistical mechanics in the 1870s that led to the equation that quantifies the entropy of a system: S = k logW, where W is the number of microstates available to the particles in a system, and k is the Boltzmann constant, 1.3806 × 10–23 J/K. At Ludwig Boltzmann’s grave in Vienna, Austria, this equation is engraved in the marble above a bust of Boltzmann. Nowadays, the equation is typically written S = kB log Ω. 304

Thermochemistry and Kinetics We have seen this reaction a few times before. Note that for every four moles of reactants, only two moles of products are produced. This reaction entails a reduction in entropy since there are fewer moles of gas in the products than in the reactants. The next equation represents a process called steam reforming, which is used to produce a gas mixture called syngas. CH 4 ( g ) + H 2O( g ) → CO( g ) + 3H 2 ( g )

(10.16)

Syngas is a mixture of carbon monoxide and hydrogen used as a fuel and as a precursor for the manufacture of methanol and a number of other industrial substances. The steam reforming reaction is strongly endothermic, but there is also an increase in entropy, since every two moles of reactants produce four moles of gas products. An increase in the number of moles of gas means more microstates are available to the particles in the gas, and thus indicates an increase in entropy.

10.3.3 Gibbs Free Energy There are three names associated with the foundations of statistical mechanics that you should know. We have already encountered James Clerk Maxwell and Ludwig Boltzmann (Section 6.1.2), who gave us the Maxwell-Boltzmann speed distributions for gas particles. The third is the great American scientist Josiah Willard Gibbs (Figure 10.12). In the 1870s, Gibbs introduced a concept he called “available energy,” a quantity based on both the enthalpy and entropy in a system. Today, this quantity is called the Gibbs free energy or simply free energy. The Gibbs free energy, G (J/mol), in a system is defined as G = H −TS

(10.17)

In this equation, H, T, and S are the molar enthalpy (J/mol), temperature (K), and molar entropy [J/(mol∙K)] of the system. As Figure 10.12. American scientist Josiah Willard Gibbs (1839–1903). with enthalpy, the Gibbs free energy cannot be measured directly, but changes in the free energy can be measured. Thus, we work with the change to the Gibbs free energy, ΔG. For processes at constant temperature involving the standard states of reactants and products, Equation (10.17) may be re-expressed as follows: ΔG ! = ΔH ! −TΔS!

(10.18)

Spontaneous processes in nature favor energy decreases and entropy increases. The Gibbs free energy equation relates them together, and spontaneous processes in nature always go in the direction that reduces the Gibbs free energy of the system. In other words, if ΔG ! for a process is ! negative, the process happens spontaneously. Also, any process in which ΔG increases at the stated conditions does not occur spontaneously. Since the absolute temperature is always positive, the terms in Equation 10.17 that determine if ΔG ! is negative are ΔH ! and ΔS!. Either of these can be positive or negative, and this gives us four possibilities to consider when seeking to determine if a reaction will occur. These are summarized in Table 10.3. First, any exothermic process in which the entropy increases occurs spontaneously. In this case, nature’s preferences for minimizing energy and increasing entropy are both being satisfied, and such a process always occurs. The use of the term “spontaneous” here is the traditional language used to describe a process with a negative value for ΔG !, but the term can be misleading and requires some explanation. 305

Chapter 10 ΔH !

ΔS!

ΔG !

Example

negative (exothermic)

positive (increases disorder)

always negative (spontaneous)

2O3 ( g ) → 3O2 ( g )

negative (exothermic)

negative (decreases disorder)

negative at lower temperatures (spontaneous at lower temperatures)

H 2O( l ) → H 2O( s )

positive (endothermic)

positive negative at higher temperatures (increases disorder) (spontaneous at higher temperatures)

H 2O( s ) → H 2O( l )

positive (endothermic)

negative (decreases disorder)

3O2 ( g ) → 2O3 ( g )

never negative (never spontaneous)

Table 10.3. Four possible combinations of ΔH and ΔS lead to the different possibilities for the sign of ΔG.

The fact that the Gibbs free energy is negative means that a chemical reaction will occur so long as the activation energy is present to initiate the reaction. (We address activation energy in just a ! few pages.) Reactions such as fuel combustion always have a negative value for ΔG , but fuels generally don’t ignite “spontaneously” (for which we are thankful). A spark or flame is required to initiate the reaction; after that it goes by itself. The last row in the table is for the opposite case—enthalpy increases and entropy decreases. Such a process never occurs spontaneously because it goes against the natural tendencies of both the enthalpy and entropy. (Such process can be made to occur only by supplying enough energy to drive the process, which increases entropy even more somewhere else.) The two middle rows deal with the cases in which the enthalpy and entropy pull in opposite directions. The deciding factor in each of these scenarios is the temperature because it multiplies the change in entropy and thus determines how significant the entropy contribution is to the Gibbs free energy. Note that in cases such as the phase transition examples shown in the table, the process does occur literally spontaneously. No activation energy is needed to make an ice cube melt; it just happens. Example 10.4 For the steam reforming reaction to make syngas, shown in Equation (10.16), ΔH ! = 206.1 kJ/mol and ΔS! = 0.215 kJ/(mol∙K). Determine if this reaction will occur at standard temperature, 298 K. We use Equation (10.18) to compute the change in the Gibbs free energy for this process, ΔG !. If it is negative, then the reaction occurs. ΔG ! = ΔH ! −TΔS! ΔG ! = 206.1

kJ kJ kJ − 298 K ⋅0.215 = 142.0 mol mol⋅K mol

At standard temperature, the Gibbs free energy increases for this process. Thus, the reaction does not occur spontaneously.

306

Thermochemistry and Kinetics Example 10.5 The equation for the decomposition of ammonium chloride to ammonia and hydrogen chloride is as follows: NH 4Cl( s ) → NH3 ( g ) + HCl ( g ) Given ΔH ! = 176 kJ/mol and ΔS! = 285 J/(mol∙K), address the following questions about this process: 1. Will the reaction occur spontaneously at standard conditions? 2. What is the effect of increasing the temperature? 3. Is the given value for ΔS! consistent with the reaction equation shown above? To answer the first question, we need to calculate the change in the Gibbs free energy. The temperature for standard conditions is 25°C, or 298 K. Note that the value for ΔS! is given in J/(mol∙K). This must be changed to 0.285 kJ/(mol∙K) to make the energy units consistent. Using Equation (10.18) and the given information, we calculate the change in the Gibbs free energy for this process as follows. ΔG ! = ΔH ! −TΔS! ΔG ! = 176

kJ kJ kJ − 298 K ⋅0.285 = 91 mol mol⋅K mol

The change in the Gibbs free energy is positive, indicating that this reaction does not occur spontaneously at 25°C. A note about the significant digits in the calculation may be in order here. The product of 298 K and 0.285 kJ/(mol∙K) is 84.93 kJ/mol. When this value is subtracted from 176 kJ/mol, the decimal places are all lost because of the addition rule for significant digits. Thus, the result is 91 kJ/mol. Regarding the second question, we note that although this reaction is endothermic, the entropy change is positive. Thus, at a high enough temperature, the product TΔS will be larger than ΔH, making ΔG ! negative and enabling the reaction to occur. Regarding the third question, the reaction equation indicates that one mole of a solid reactant forms two moles of gaseous products, indicating an increase in entropy. Since ΔS! is positive, it is consistent with the equation.

Three more important points remain to be made in this section. First, the mere fact that the Gibbs free energy is negative says nothing about the rate at which a reaction occurs. For the reaction below, ΔG ! = −3.3 J/ ( mol⋅K ) , and thus the reaction is favorable: C ( s, diamond ) → C ( s, graphite ) However, the reaction takes place so slowly it is never observed. Second, make note of this: the general principle embedded in Equation (10.14) may be used to calculate ΔS! for any reaction. Data for S! are tabulated in standard reference sources for thou307

Chapter 10 sands of standard compounds, including Table B.8 in Appendix B. Thus, ΔS! for any reaction may be calculated as ! ! ΔS! = ΣSproducts − ΣSreactants The method for calculating ΔS! is the same as that illustrated in Example 10.3. You simply tabulate standard molar entropy values for products and reactants, multiply each by the number of moles in the equation, determine the two sums, and take the difference of the two sums. Thus, one can easily determine whether the reaction entails an increase or decrease in entropy. Make note, though, that standard entropy values are usually given in J/(mol∙K) rather than kJ/ (mol∙K), so you have to watch out for the need to convert the units. Third, the same idea applies to calculating the change in the Gibbs free energy. Data for the standard free energy of formation, ΔG !f , are tabulated in reference sources for common compounds, including Table B.9 in Appendix B. Just as with the change in enthalpy in Equation (10.14), the change in the Gibbs free energy, ΔG !, for any reaction may be calculated as ΔG ! = ΣΔG !f , products − ΣΔG !f , reactants Of course, ΔG ! may also be calculated according to Equation (10.18), where ΔH ! and ΔS! are calculated from standard enthalpy of formation and entropy data as described above.

10.4

Reaction Kinetics

One of the goals of research in chemistry is to understand more precisely what actually happens during a chemical reaction. Such research seeks answers to questions like these: How exactly does chemical species A get transformed into chemical species B? Does the transformation happen all at once or in a sequence of steps? What factors govern how fast a reaction occurs? What are the physical mechanisms of such factors? The subject of the physical mechanisms of reactions and factors that govern the rates of chemical reactions is known as reaction kinetics. In this section, we take a brief tour of these topics.

10.4.1 Collision Theory Collision theory is a general term that refers to our present understanding of chemical reactions at the molecular level. The essential idea behind collision theory is that chemical reactions occur when molecules of different chemical species collide. However, just because molecules jostle around together does not mean a chemical reaction occurs. You can pour gasoline on the ground but it does not begin reacting with the oxygen in the air (combustion) without a source of ignition. When molecules collide, the clouds of electrons in the orbitals surrounding the atoms repel each other with a great deal of force. In order for a chemical reaction to occur, a molecular collision must involve enough energy to force the molecules close enough together that existing bonds between atoms can be broken and new bonds formed. Not only does this require enough energy; it also requires that the molecules have the right orientation in space at the moment they collide. For example, the steam reforming reaction shown in Equation (10.16) is depicted in Figure 10.13. On the left side of the figure, the CH4 and H2O molecules collide and bounce apart because the orientation of the H2O molecule is not favorable relative to that of the CH4 molecule. On the right side of the figure, the H2O molecule is oriented so that the oxygen atom can disrupt the bonds in both molecules and bond with the carbon atom in CH4 to produce one molecule of CO. 308

Thermochemistry and Kinetics Figure 10.13. If molecular orientation is not favorable, a molecular collision may not result in a reaction (left). A different orientation of molecules may enable a reaction to occur (right). During the reaction, intermediate species form as part of the steps in the reaction mechanism.

intermediates

The right side of Figure 10.13 also indicates that during the reaction, intermediate species form that are not part of the product species. We return to considering intermediates shortly.

10.4.2 Factors Influencing Reaction Rate The reaction rate of a chemical reaction is the rate at which the concentrations of reactants and products are changing. Based on collision theory, there are four important factors we can control that can influence the rate of chemical reactions. These are as follows: 1.

Temperature As you know, the higher the temperature of the reactants, the higher the kinetic energy of the moving particles. At higher speeds, molecules collide more frequently, providing more opportunities for the favorable orientations that lead to the breaking of bonds and formation of new bonds. Note that increasing the temperature only increases the reaction rate if the temperature is high enough for the reaction to commence. We discuss this issue more in Section 10.4.4. On rare occasions, higher temperatures can decrease a reaction rate. This can occur, for example, if the higher temperature increases the reverse reaction rate at the expense of the forward reaction rate. We see an example of this in the next chapter, Section 11.1.4.

Surface Area As a factor in chemical reactions, the surface area of solid reactant applies to heterogeneous reactions—reactions in which there is more than one state of matter represented among the reactants. A solid reactant that is to react with a liquid or gas has a certain surface area exposed to the liquid or gas reactant. According to collision theory, reactions happen through collisions, so the larger the surface area of the solid reactant, the more the atoms in the solid are exposed to collisions from the fluid reactant molecules. A 1-cm cube of solid material has a surface area of 6 cm2. If the same amount of material is ground to small particles with grains that are also cubes but with side lengths of 0.1 mm, the total surface area of the solid is 600 cm2. The finer the grains in the powder, the larger the total surface area of the solid. Sawdust, coal dust, and powderized aluminum are all examples of substances that may not sound all that dangerous but which can explode violently in a so-called dust explosion. In all these cases, the explosion happens because of the enormous surface area provided by a reactant consisting of small particles. If the dust forms a suspension in air, explosive conditions result. Sawdust is essentially powdered carbon, as is coal dust. A spark in a dusty woodworking shop or coal mine has often led to a tragic explosion. Dust from dry grain, flour, powdered milk, and other sources of carbon can produce the same explosive suspen-

309

Chapter 10 sion in air. Both aluminum and titanium metals will also form explosive suspensions in air when in powdered form. 3.

Concentration Since chemical reactions occur by means of molecular collisions, the more particles of reactant species there are available, the faster a reaction occurs. When one or more of the reactants are in a solution of liquid or gas, the higher the concentrations of those reactants are, the more molecules there are in the solution to react. As an example, oxygen accounts for about 21% of the gas molecules in the air. At the same pressure, the concentration of oxygen molecules in pure oxygen gas is nearly five times the concentration in air. As a result, a combustion reaction fueled by pure oxygen occurs much faster than the same reaction in air, releasing heat energy at the same increased rate. We have seen that acetylene is an unstable molecule with a large positive enthalpy of formation (Table 10.2). When this gas is combined with pure, high-pressure oxygen and ignited, we have a combustion reaction that gives off so much heat so rapidly that the flame can be used for cutting steel, as illustrated in Figure 10.14. The oxyacetylene torch shown has two separate gas tubes arriving at the nozzle, one for the C2H2 and one for the O2. The two gases Figure 10.14. The flame from an are stored in separate steel storage tanks. oxyacetylene torch cuts right

The reaction between oxygen and acetylene is an example through steel plate. of a homogeneous reaction—one in which all the reactants are in the same state (gas in this case). 4.

Catalysis A catalyst is a substance that expedites or facilitates a chemical reaction without itself being consumed by the reaction. The process by which a catalyst influences the rate of a chemical reaction is known as catalysis. Metals often act as catalysts because of a phenomenon known as adsorption. Adsorption is the collection of molecules on a surface, attracted there by electrostatic attraction or Van der Waals forces. When reactant molecules adsorb onto the surface of a catalyst, the surface forces cause reactant molecules to separate into individual atoms, at which time they are free to bond to other atoms to form new molecules. The presence of iron as a catalyst is the key that allows the formation of NH3 in the HaberBosch process. After the N2 and H2 molecules adsorb onto the surface of the iron, the diatomic molecules of nitrogen and hydrogen come apart and individual atoms of nitrogen and hydrogen form on the metal surface. After that, hydrogen atoms begin bonding with nitrogen atoms, building up NH3 molecules by adding one hydrogen atom at a time. The catalyst works by providing a completely different energy pathway, significantly reducing the energy required to make the reaction occur. We will look at this energy pathway more closely shortly. A catalyst that is in the same state as the reactants in a chemical system is referred to as a homogeneous catalyst. A catalyst in a different state, such as the metal catalyst used with the gases in the Haber-Bosch process reaction, is called a heterogeneous catalyst. Note that

310

Thermochemistry and Kinetics in reversible reactions such as equilibrium systems, a catalyst increases the reaction rate equally in both directions.

10.4.3 Reaction Mechanisms Experimental evidence now strongly supports the theory that chemical reactions almost never take place in one simple step. Instead, reactions typically require several separate steps in which the bonds of reactant molecules are rearranged into the bonds of the product molecules. The sequence of steps in which a reaction takes place is called the reaction mechanism. A reaction mechanism is a model depicting a sequence of steps and intermediates involved in a chemical reaction. As mentioned above, intermediates are chemical species that form temporarily during the reaction, but which are not part of the net reaction equation. As an example, when hydrogen gas, H2, and iodine gas, I2, are combined in a sealed container, the gases form the following equilibrium: H 2 ( g ) + I2 ( g ) ! 2HI( g ) Hydrogen and hydrogen iodide are both colorless gases; iodine gas is purple. At equilibrium, all three gases are present in the container. By means of some ingenious experimental techniques— which include hitting the molecules with certain wavelengths of light—scientists discovered that the H2 and I2 molecules do not simply come apart and recombine to form HI gas. Instead, various intermediates are involved. Figure 10.15 shows three different reaction mechanisms that have been proposed as models for this reaction. All three begin with the formation of individual atoms of iodine, an intermediate. In the first mechanism, the dissociated iodine atoms interact with H2 to form HI. In the second mechanism, the I2 ! 2I intermediate I atoms form H2I, another intermediate, before finally forming HI. In the third mechanism, the dissociated 2I + H 2 ! 2HI I atoms attach to either side of the H2 molecule to form yet another intermediate, IH2I. The attached iodine atoms then cause the breaking of the H2 bond, leaving separate molecules of HI. As another example, consider again the Haber-Bosch I ! 2I 2 process. The reaction mechanism for this process, as presently understood, is shown in Figure 10.16 on the next page. This diagram is a combination enthalpy diagram, such as we I + H 2 ! H 2I saw in Figures 10.9 and 10.10, and energy pathway diagram, such as those shown in Figure 10.2. The figure ties together many of the concepts we have already studied in this chapH 2I + I ! 2HI ter, as well as a couple of ideas we haven’t even gotten to yet, so let’s take the time to look it over carefully. I2 ! 2I The diagram begins on the left with N2 and H2 gas and ends on the right with NH3. Two different energy pathways 2I + H 2 ! IH 2I are shown. Notice that as Hess’s law predicts, the total enthalpy change for the reaction (ΔH = –46 kJ/mol) is the same regardless of the energy pathway. It just happens that IH 2I ! 2HI one pathway is feasible for an industrial process and the other is not. Figure 10.15. Three proposed reaction The upper path models the way the reaction might pro- mechanisms for the formation of HI gas ceed without the metal catalyst. The first step is a huge en- from H2 and I2. 311

Chapter 10 Figure 10.16. Energy pathways for the formation of NH3 from N2 and H2. The upper pathway indicates the enormous enthalpies that are involved without the iron catalyst. The catalyst provides a different energy pathway which is initially exothermic.

N + 3H 314

NH + 2H

1129 kJ/mol

1400

~960

389

NH2 + H

543 17

/2 N2 + 3 /2 H2

ΔH = 46 kJ/mol

~21

~33

/2 N2

259 106

+ /2 H2

460

Nad + 3Had

NHad + 2Had

~41

NH2ad + Had

NH3 NH3ad

dothermic reaction, in which 1,129 kJ/mol of heat must be supplied to break the N2 triple bond and the H2 bonds. This produces separate atoms of N and H, which are intermediates. Afterwards, a sequence of exothermic steps leads to the net enthalpy change of ΔH = –46 kJ/mol. In these exothermic steps, hydrogen atoms are added one at a time to the nitrogen to form the intermediates NH and NH2, and then finally NH3. Look now at the fascinating energy pathway at the bottom of the figure, which shows the mechanism for the catalyzed process. The first step is the adsorption (ad) of the N2 and H2 molecules on the surface of the metal catalyst. Next, while adsorbed on the metal surface, the diatomic molecules separate and this separation is actually exothermic under these conditions (!). So instead of having to supply the reactants with 1,129 kJ/mol of heat to break apart the N2 and H2 molecules, only about 21 kJ/mol have to be supplied to get over the little energy hump there (more on that soon) and then the molecules release 259 kJ/mol as they dissociate while adsorbed on the catalyst surface. This forms the adsorbed N and H intermediates. From here, several small endothermic steps lead to the intermediates NH and NH2, and then finally to NH3. Notice that all these steps occur while the molecules are still adsorbed on the catalyst. Only after the NH3 molecules are formed is a final boost of heat supplied (50 kJ/mol) to remove the NH3 molecules from the catalyst and form them into NH3 gas. The net enthalpy change for this pathway is the same as for the other energy pathway, and the same intermediate species are formed. But with the intermediates adsorbed onto the catalyst, the energies involved are quite different.

10.4.4 Activation Energy and the Activated Complex Look again at the lower energy pathway in Figure 10.16. Before each energy transition, the reactants need a small energy boost in order for the next step in the reaction to commence. This small energy boost is seen for both exothermic transitions and endothermic transitions. Figure 10.17 shows an enlarged view of generic energy pathways for both exothermic and endothermic reactions. To commence the reaction, an amount of energy labeled Ea or Eaʹ must be supplied. This energy is the activation energy, Ea (or Eaʹ ), roughly defined as the amount of energy required to initiate the chemical reaction. Depending on the specifics of the reaction, the activation energy may be supplied by heating, a spark, a flame, an electrical current, or even by mechanical means such as physical shock or compression (which supplies energy to a gas by 312

Thermochemistry and Kinetics activated complex

products

energy

ΔH (negative)

products

reactants

energy

activated complex

ΔH’ (positive)

Ea’

reactants

exothermic reaction pathway

endothermic reaction pathway

time

Figure 10.17. To initiate the exothermic or endothermic reaction, the activation energy Ea or Ea’ must first be supplied. Ea is the energy required to form the activated complex. Once the activated complex is formed, the reaction commences.

doing work on the gas). However the activation energy is supplied, its effect is to increase the temperature, giving the molecules the kinetic energy they need so that effective molecular collisions occur with enough force to break molecular bonds. Current theory holds that when these effective, energetic collisions occur, a so-called activated complex results. An activated complex is a range of transitional molecular structures of partially formed bonds that results from an effective molecular collision with adequate energy. The activated complex lasts only while bonds are in the process of breaking and reforming. The formation of the activated complex is shown in Figure 10.17 at the peak of the energy curve; once this peak is reached and the activated complex forms, the reaction commences. The formation of the activated complex gives us a more nuanced definition for the activation energy: the activation energy is the minimum energy necessary to transform the reactant molecules into the activated complex for a given reaction. Figure 10.17 displays separate diagrams for exothermic and endothermic reactions. However, these diagrams can also be interpreted as representing the energy pathways for the forward and reverse directions in a single reversible reaction. In the exothermic reaction, the total energy change from the energy of the activated complex to the energy of the reaction products is Ea − ΔH = Ea + ΔH This amount of energy becomes the activation energy, Ea', for the reverse reaction. Ea′ = Ea + ΔH The energy required to form the activated complex is the same in either direction. But the energy that must be supplied to the reactants, Ea or Ea', depends on the energy the reactants are starting with. In some cases, the reactants already have enough energy at room temperature to form an activated complex. Just pour a bit of vinegar into some baking soda and watch what happens— no additional heating is necessary. But in a great many cases, an extra amount of energy must be supplied. As an example, when there is a natural gas leak, why doesn’t the gas spontaneously combust when coming in contact with the air? The reason is that although the reactants (methane and oxygen) are together, the activation energy has not been supplied. An open flame might be enough to do the job. Another example is illustrated in Figure 10.18. I hope you have seen 313

Chapter 10

energy

the bright light of burning magnesium before because the picture doesn’t do it justice. One thing I always point out to my students when performing this demonstration is that the magnesium does not ignite with a match, but it does light with a butane gas-grill lighter. The flame of the burning match is not as hot as the butane flame, so it does not produce enough heat to supply the activation energy. I also like to note that the flame always goes out when it is about 1/4 inch from the pliers I use to hold the strip of magnesium metal. The pliers are metal and conduct heat pretty well, cooling the magnesium metal strip. When the Figure 10.18. Burning magnesium, which lights flame gets close enough to the pliers, the pliers with a butane lighter, but not with a match. conduct heat away from the flame so fast that the energy supplied by the flame falls below the activation energy needed to keep the reaction going. Without the requisite amount of energy, the activated complex cannot form and the reaction does not happen. Now that we have introduced the concepts of reaction mechanism and activation energy, let’s return for a moment to the general topic of how catalysts work. Figure 10.19 compares the energy pathway for a single-step exothermic reaction to the alternative, three-step energy pathway provided by the presence of a catalyst. Instead of the large activation energy, Ea, required to initiate the single-step process, the catalyst lowers the activation energy by providing an alternative reaction mechanism. In the generic case shown in the figure, the alternative mechanism consists of three energy pathway without catalyst separate reaction steps, each with Ea its own activation energy. But since energy pathway none of these activation energies is with catalyst reactants Ei as great as the activation energy in Ea1 Ea2 the pathway without the catalyst, the Ea3 catalyst allows the desired reaction to occur more expeditiously. We saw in the case of the iron catalyst used in products Ef the Haber-Bosch process that surface exothermic reaction pathway attractions between metal atoms and adsorbed particles reduce the energy time required to break apart molecular Figure 10.19. The catalyst provides an alternative energy bonds in the reactants, so much so pathway for the reaction with lower activation energy. The that the reactant dissociation actually alternative path shown entails a three-step reaction. becomes exothermic. Endothermic steps occur after that as the product molecules begin to form, but again energy requirements are substantially lower than the energy required to break apart reactant molecules without the catalyst.

10.4.5 Reaction Rate Laws A few pages back, we saw that the concentration of reactants in solution affects the rate of the reaction. In referring to reaction rate, we are referring to the rate at which the molarity of

314

Thermochemistry and Kinetics the reactants and products are changing per unit time. This means that the reaction rate, R, of a chemical reaction has units of M/s, molarity per second. Consider the reaction between ammonium and nitrite ions to form nitrogen and water: NH +4 ( aq ) + NO−2 ( aq ) → N 2 ( g ) + 2H 2O( l )

(10.19)

This reaction has two reactants. One of the ways to investigate the effect of concentration on the reaction rate is to vary the concentration of one reactant while holding the concentration of the other reactant constant. For the reaction above, experiments have shown that the reaction rate, R, is directly proportional to the NH4+ concentration, [NH4+]. This means that if [NH4+] is doubled or tripled, the rate of the reaction doubles or triples, respectively. We model this relationship algebraically as: R = k ⎡⎣ NH +4 ⎤⎦ As we have seen before, k is a constant of proportionality. Thus, this expression says that the reaction rate, R, is directly proportional to the ammonium ion concentration. It has also been found experimentally that for the reaction in Equation (10.19), the following holds: R = k ⎡⎣ NO−2 ⎤⎦ Or, in words, the reaction rate is directly proportional to the nitrate ion concentration. Combining these two expressions gives R = k ⎡⎣ NH +4 ⎤⎦ ⎡⎣ NO−2 ⎤⎦

(10.20)

Equation (10.20) is an example of a rate law—a mathematical expression that characterizes reaction rates in terms of the concentrations of the reactants. The constant k is known as the rate constant. The units for the rate constant are such as to allow the reaction rate to have units of M/s. (The units for k are illustrated in the next example.) In general, we can consider a generic chemical equation such as the following: aA + bB → cC + dD

(10.21)

The reaction rate law for such a reaction is of the following form: R = k [ A ] [B ] m

(10.22)

In Equation (10.22), the terms in square brackets are the concentrations of the reactants. The exponents m and n, as well as the rate constant k, must be determined experimentally. The values of the exponents m and n are typically 0, 1, or 2. In Equation (10.20), the exponent on both concentrations is 1, so we say that “the reaction rate is first-order in [NH4+]” and “the reaction rate is first-order in [NO2–].” When two first-order terms are multiplied in a rate law, the overall rate law is second order—the orders on the individual terms add together to make the overall rate law order. Note that the rate constant, k, is a constant at a given temperature, but the concentrations and reaction rate are clearly not. From the initial concentrations, these vary exponentially (at least, for first-order reactions), decreasing from their initial concentrations and finally stabilizing at 315

Chapter 10 their equilibrium concentrations. But once we know the rate law, we can calculate the rate for any given set of concentrations. Example 10.6 Use the rate data in Table 10.4 to derive the rate law and rate constant for the reaction of ammonium and nitrate ions in water. Looking at trials 1 and 2, we see that when the initial value of [NH4+] is doubled while holding [NO2–] constant, the initial reaction rate doubles. Trials 1 and 3 show us that when [NH4+] is increased by a factor of four ([NO2–] constant), the initial rate increases by the same factor. From these data, we see that the reaction rate is first order in [NH4+], or R = k ⎡⎣ NH +4 ⎤⎦

Trial

[NH4+], initial (M)

[NO2–], initial (M)

R, Initial (M/s)

0.0050

0.100

1.35 × 10–7

0.0100

0.100

2.70 × 10–7

0.0200

0.100

5.40 × 10–7

0.100

0.0100

2.70 × 10–7

0.100

0.0200

5.40 × 10–7

0.100

0.0300

8.10 × 10–7

Table 10.4. Rate data for the reaction of ammonium and nitrite ions in aqueous solution (25°C).

Looking at trials 4 and 5, we see that when [NO2–] is doubled ([NH4+] constant), the initial rate doubles. Trials 4 and 6 show that when [NO2–] is tripled ([NH4+] constant), the initial rate increases by the same factor. From these data, we see that the reaction rate is first order in [NO2–], or R = k ⎡⎣ NO−2 ⎤⎦ We now know that the rate law is of the form R = k ⎡⎣ NH +4 ⎤⎦ ⎡⎣ NO−2 ⎤⎦ To determine the rate constant, we solve the rate law for k and insert experimental data from one of the trials. Data from any trial gives the same value for k. We will use data from trial 2 so that all values have three significant digits. M 2.70 ×10−7 R s k= = = 2.70 ×10−4 M −1s −1 ⎡⎣ NH +4 ⎤⎦ ⎡⎣ NO−2 ⎤⎦ ( 0.0100 M )( 0.100 M ) The complete rate law is found to be R = 2.70 ×10−4 M −1s −1 ⎡⎣ NH +4 ⎤⎦ ⎡⎣ NO−2 ⎤⎦

10.4.6 Rate Laws and Reaction Mechanisms The form and order of a reaction rate law is determined by the reaction mechanism. For a single-step reaction, the powers m and n on the reactant concentrations are equal to the coefficients on the reactants. Thus, if Equation (10.21) represents a single-step reaction, the rate law is 316

Thermochemistry and Kinetics R = k [ A ] [B ] a

However, if a reaction takes place by means of a multistep mechanism, the steps occur at different rates and one of the steps occurs more slowly than the others. This step is referred to as the rate-determining step. The rate-determining step determines the rate law, and the coefficients on the reactants in the rate-determining step form the powers on the terms in the rate law. This is a powerful clue that helps chemists figure out the actual reaction mechanism. For example, consider the following reaction: CO( g ) + NO2 ( g ) → CO2 ( g ) + NO( g )

(10.23)

It has been found experimentally that the rate law for this reaction is R = k [ NO2 ]

This rate law is second order in [NO2] and doesn’t depend on [CO] at all. The fact that the rate law in this case is second order in [NO2] implies that the reaction in Equation (10.23) is a multistep reaction, and that its rate-determining step begins as follows: 2NO2 → We conclude this because, as mentioned above, the rate law for a single-step reaction is formed by multiplying the reactant concentrations together and raising each one to a power that is the same as the coefficient on that reactant. In fact, the mechanism proposed for the reaction in Equation (10.23) is the following: step 1:

2NO2 → NO3 + NO

(slow)

step 2:

NO3 + CO → NO2 + CO2

(fast)

Example 10.7 Nitrogen dioxide reacts with fluorine as follows: 2NO2 ( g ) + F2 ( g ) → 2NO2F ( g ) A proposed mechanism for the reaction is the following two-step sequence: step 1:

NO2 + F2 → NO2F + F

(slow)

step 2:

F + NO2 → NO2F

(fast)

Identify the rate-determining step and propose a rate law for the reaction of nitrogen dioxide and fluorine. The rate-determining step in a multistep reaction is the slowest step. Thus, the first step in the proposed mechanism is the rate-determining step. From this reaction step, the rate law is R = k [ NO2 ][ F2 ]

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Chapter 10 Example 10.8 Consider the reaction between nitric oxide and hydrogen: 2NO( g ) + 2H 2 ( g ) → N 2 ( g ) + 2H 2O( g ) From experiments, the rate law for this reaction has been found to be R = k [ NO ] [ H 2 ] 2

Address these questions: 1. What is the reaction order for this rate law? 2. If the concentration of H2 is doubled, what is the effect on the reaction rate? 3. If [NO] is tripled, what is the effect on the reaction rate? The rate law is second order in [NO] and first order in [H2]. We add these together to get the overall order of the rate law. This rate law is third order. The rate law is first order in [H2]. Thus, doubling [H2] doubles the reaction rate. The rate law is second order in [NO]. Tripling [NO] increases the reaction rate by a factor of (3)2 = 9.

Chapter 10 Exercises SECTION 10.1

1. Describe how a thermochemical system might do mechanical work. Consider (a) a gas in a cylinder, and (b) reaction products in the open air. 2. Why is ΔH negative for an exothermic reaction? 3. Use the first law of thermodynamics to distinguish between the change in enthalpy and the change in internal energy of a thermochemical system during a reaction. 4. Use the enthalpy diagram shown on the next page to address the following questions. 1 a. Is the reaction CO2 ( g ) + 2H 2O( l ) → CO( g ) + 2H 2O( l ) + O2 ( g ) exothermic or en2 dothermic? b. What is the enthalpy change for CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H 2O( l ) ? c. Write the thermochemical equations, including the enthalpy change, for the three endothermic processes represented in the diagram. 5. Distinguish between enthalpy of combustion and enthalpy of formation. 6. Why is it important to specify the states of the compounds in a thermochemical equation? SECTION 10.2

7. Use the enthalpy diagram shown below as an example to explain Hess’s law. 8. What practical advantages are there to being able to use Hess’s law to compute the enthalpy of reaction for a given process?

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Thermochemistry and Kinetics 9. For each of the following compounds, write the reaction illustrating the formation of the compound from its elements. Show the enthalpy of formation with each equation (see Appendix B, Table B.7 for enthalpy of formation data). a. C2H2(g) b. Li2SO4(s)

CH4(g) + 2O2(g)

c. SO3(g) d. NH3(g) 10. Use the method illustrated in Example 10.2 to determine the enthalpy of reaction for each of the following reactions. In each case, solve by using enthalpy of formation data from Appendix B. 1 a. NO( g ) + O2 ( g ) → NO2 ( g ) 2 b. Fe 2O3 ( s ) + 3CO( g ) → 2Fe( s ) + 3CO2 ( g ) c. Ca ( OH )2 ( s ) → CaO( s ) + H 2O( g )

ΔH2 = 607 kJ ΔH1 = 890 kJ CO(g) + 2H2O(l) + ½O2(g) ΔH3 = 283 kJ CO2(g) + 2H2O(l)

d. CaCO3 ( s ) → CaO( s ) + CO2 ( g )

e. Mg ( OH )2 ( s ) → MgO( s ) + H 2O( l ) 11. Use Equation (10.14) and the method illustrated in Example 10.3 to verify your results for the five reactions in the previous item. 12. Determine the enthalpy of formation for the following compounds. Solve by using enthalpy of combustion data from Table 10.1. (No combustion equation is required for O2.) a. ethylene, C2H4(g) b. formic acid, CH2O2(l) c. benzene, C6H6(l) 13. In a combustion reaction at standard conditions, each of the following compounds produces CO2(g) and H2O(l). Determine the enthalpy of combustion for one mole of each by using enthalpy of formation data. a. methanol, CH3OH b. ethanol, C2H5OH SECTION 10.3

14. For the reactions listed below, inspect the equation and predict whether ΔS! is positive or negative. a. 2SO2 ( g ) + O2 ( g ) → 2SO3 ( g )

b. CO( g ) + 2H 2 ( g ) → CH3OH( l )

c. Ba ( OH )2 ( s ) → BaO( s ) + H 2O( g ) 15. Describe how the entropy of a substance relates to the state (solid, liquid, or gas) and temperature of the substance. 16. Why is the entropy of a pure crystalline substance at absolute zero equal to zero? 17. Consider a reaction for which ΔH ! > 0 and ΔS! < 0. Explain why this reaction does not happen spontaneously in nature.

319

Chapter 10 18. The boiling point for benzene, C6H6, is 80°C. At 60°C, which do you expect to be greater for the vaporization of benzene, ΔH or TΔS? Explain your response. 19. Determine the change in the standard Gibbs free energy for each of the following reactions at 25°C. For each, state whether the reaction is spontaneous at the given conditions. a. N ( g ) + O ( g ) → 2NO( g ) , ΔH ! = 182.6 kJ and ΔS! = 24.8 J/K 2

b. C( s ) + O2 ( g ) → CO2 ( g ) , ΔH ! = –393.5 kJ and ΔS! = 2.70 J/K

20. For the following reactions, ΔH and ΔS are given for the temperature indicated. Calculate ΔG for each and indicate whether the reaction is spontaneous at the given conditions. a. ΔH = 108 kJ/mol, ΔS = 0.0375 kJ/(mol∙K), T = 292 K b. ΔH = –93.1 kJ/mol, ΔS = 0.105 kJ/(mol∙K), T = 135°C c. ΔH = –278 kJ/mol, ΔS = 0.375 kJ/(mol∙K), T = 775 K 21. Using enthalpy and entropy data in Appendix B (Tables B.7 and B.8), determine the change in the standard Gibbs free energy for the Haber-Bosch process, in which ammonia is formed ! from nitrogen and hydrogen gases. First calculate ΔG at 25°C, then calculate ΔG at 500°C, assuming that the values for ΔHf and S do not change. (In fact, they do change a little, but this calculation is a good approximation.) Comment on what these two values tell you about the prospect of running the Haber-Bosch process at a pressure of 1 bar (standard pressure). 22. Using data from Appendix B, calculate ΔG !, ΔH !, and ΔS! for each of the following reactions. Then compare your ΔG ! values to the values obtained from ΔG ! = ΔH ! −TΔS! for each reaction. (You will find the values are virtually the same.) a. 6Cl 2 ( g ) + 2Fe 2O3 ( s ) → 4FeCl 3 ( s ) + 3O2 ( g ) b. H 2 ( g ) + F2 ( g ) → 2HF ( g )

c. NO2 ( g ) + N 2O( g ) → 3NO( g )

d. SO2 ( g ) + 2H 2 ( g ) → S( s ) + 2H 2O( g )

e. MgCl 2 ( s ) + H 2O( l ) → MgO( s ) + 2HCl ( g ) SECTION 10.4

23. Explain the two factors that determine whether a molecular collision will result in the breaking and reforming of chemical bonds. 24. What is a reaction mechanism? 25. In a certain demonstration, a teacher fills a metal can with non-dairy coffee creamer and makes it explode. The demonstration works by placing a small amount of explosive black powder in the bottom of the metal can, covering the black powder with a paper towel, and filling up the can with coffee creamer. The black powder is ignited with a fuse and this blows the coffee creamer out of the can. The instant the coffee creamer is out of the can it explodes. Explain why coffee creamer, which is basically carbon, explodes as described. (Do not try this demonstration without expert supervision and full attention to safety procedures.) 26. The change in Gibbs free energy for the combustion of gasoline is a large negative number. Explain why gasoline doesn’t immediately burst into flames when it is spilled on the ground. 27. A certain reaction is found to occur in a single step, with the following mechanism: A + 2B → AB2

320

Thermochemistry and Kinetics Determine the rate law for this reaction and use it to predict the effect on the reaction rate of tripling [B]. 28. Given the following reaction mechanism: step 1: A 2 + A 2 → C + D3 slow step 2: B + D3 → A 2 + E 2 fast Determine the overall balanced equation for the reaction, identify the rate determining step, identify the intermediates, and propose an appropriate rate law for the reaction. 29. Draw energy diagrams similar to those shown in Figure 10.18 to fit the following reactions: a. ΔH = –20 kJ/mol, Ea = 15 kJ/mol b. ΔH = 30 kJ/mol, Ea = 50 kJ/mol c. ΔH = –5 kJ/mol, Ea = 45 kJ/mol 30. Consider a reversible reaction. Compare the activation energy and enthalpy change in the forward direction to those of the reverse direction. 31. A certain chemical reaction is described by the single-step mechanism 2A + B → C . Rate data are collected for this reaction, indicating the reaction rates for different concentrations of reactants. These data are shown in the chart. a. Determine the rate law for the reaction. b. Determine the value of the rate constant. c. Determine the initial value of the reaction rate if the initial concentrations of A and B are both 0.25 M. 32. Determine the volume in liters of 3.55 g of SO3 gas at STP.

Trial

Initial [A]

Initial [B]

Initial Rate of Forming C

33. Aluminum oxidizes to form Al2O3. Assume 0.046 mol Al is combined with 0.028 mol O2. Determine the limiting reactant and the mass of Al2O3 formed.

0.10 M

0.20 M

1.15 × 10–4 M/s

0.10 M

0.40 M

2.30 × 10–4 M/s

0.20 M

0.40 M

9.20 × 10–4 M/s

GENERAL REVIEW EXERCISES

34. Fluorine gas reacts violently with water to produce HF and O3. What volumes of HF and O3 are produced if 5.500 L of F2 reacts with water? Gas volumes are all at STP. 35. Determine the molarity of a solution of citric acid, C6H8O7, if 132.7 g of citric acid are diluted in water and the resulting solution is diluted to a volume of 2500.0 mL. 36. Determine the pOH, [H3O+], and [OH–] for a solution with a pH of 9.81. 37. Determine the expected boiling and freezing points of brine solution formed by dissolving 175.00 g KBr in 450.0 g H2O. 38. A volume of 17.00 mL of 0.650 M NaOH is required to reach the equivalence point in a titration of vinegar. Determine the molarity of acetic acid in the vinegar if the vinegar volume is 18.50 mL.

321