TEST BANK
TEST BANK
AN INTRODUCTION TO PHYSICAL SCIENCE 15TH EDITION BY JAMES SHIPMAN, JERRY D WILSON, CHARLES A HIGGINS, BO LOU SOLUTIONS MANUAL Chapter 1-24
Chapter 1
MEASUREMENT Chapter 1 is important because all quantitative knowledge about our physical environment is based on measurement. Some chapter sections have been reorganized and rewritten for clarity. The 1.2 Section, ―Scientific Investigation,‖ introduces the student to the procedures for scientific investigation. Major terms such as experiment, law, hypothesis, theory and scientific method are introduced. The idea that physical science deals with quantitative knowledge should be stressed. It is not enough to know that a car is going ―fast‖; it is necessary to know how fast. A good understanding of units is of the utmost importance, particularly with the metricBritish use in the United States today. The metric SI is introduced and explained. Both the metric and the British systems are used in the book in the early chapters for familiarity. The instructor may decide to do examples primarily in the metric system, but the student should get some practice in converting between the systems. This provides knowledge of the comparative size of similar units in the different systems and makes the student feel comfortable using what may be unfamiliar metric units. The Highlight, ―Is Unit Conversion Important? It Sure Is,‖ illustrates the importance of unit conversion. The general theme of the chapter and the textbook is the students’ position in his or her physical world. Show the students that they know about their environment and themselves through measurements. Measurements are involved in the answers to such questions as, How old are you? How much do you weigh? How tall are you? What is the normal body temperature? How much money do you have? These and many other technical questions are resolved or answered by measurements and quantitative analyses.
DEMONSTRATIONS Have a meter stick, a yardstick, a timer, one or more kilogram masses, a one-liter beaker or a liter soda container, a one-quart container, and a balance or scales available on the instructor’s desk. Demonstrate the comparative units. The meter stick can be compared to the yardstick to show the difference between them, along with the subunits of inches and centimeters. The liter and quart also can be compared. Pass the kilogram mass around the classroom so that students can get some
idea of the amount of mass in one kilogram. Mass and weight may be compared on the balance and scales. When discussing Section 1.6, ―Derived Units and Conversion Factors,‖ have class members guess the length of the instructor’s desk in metric and British units. Then have several students independently measure the length with the meter stick and yardstick. Compare the measurements in terms of significant figures and units. Compare the averages of the measurements and estimates. Convert the average metric measurement to British units, and vice versa, to practice conversion factors and to see how the measurements compare. Various metric unit demonstrations are available from commercial sources.
ANSWERS TO MATCHING QUESTIONS a. 15 b. 8 c. 10 d. 2 e. 19 f. 14
g. 21
h. 13
p. 4 q. 23
v. 22
w. 7
r. 17
s. 5 t. 20 u. 16
i. 18
j. 6 k. 11
l. 3 m. 12
n. 1 o. 9
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.c
2. b
3. c 4. b 5. b
6. c 7. d
8. b 9. d 10. c
11. b
12. b 13. a 14. b
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. biological
2. hypothesis
7. longer
8. fundamental
12. mass
13. less
3. scientific method 9. time or second
4. sight, hearing 9
10. one-billion, 10
5. limitations
6. less
11. liter
ANSWERS TO SHORT-ANSWER QUESTIONS 1. An organized body of knowledge about the natural universe by which knowledge is acquired and tested. 2. Physics, chemistry, astronomy, meteorology, and geology. 3. The 5 elements of scientific method are: 1. Observations and Measurements, 2. Hypothesis, 3. Experiments, 4. Theory, and 5. Law. 4. Hypothesis
5. A law is a concise statement about a fundamental relationship of nature. A theory is a welltested explanation of a broad segment of natural phenomena. 6. It illustrates the need to improve the standard of education among the general public and to emphasize the importance of a well-developed scientific method. 7. Sight, hearing, touch, taste, and smell. 8. They have limitations and can be deceived, thus providing false information about our environment. 9. (a) No. (b) Yes. (c) Lower line. 10. A fixed and reproducible value. 11. They are the most basic quantities of which we can think. And they are not dependent on
other physical quantities. 12. A group of standard units and their combinations. 13. mile/hour 14. No, the United States is the only major country that has not gone completely metric. 15. Kilogram, a platinum-iridium cylinder. 16. Mass. Weight varies with gravity. 17. Meter-kilogram-second, International System of Units, and centimeter-gram-second. 18. Base 10 easier to use (factors of 10). 19. kilo- (k), mega- (M), milli- (m), micro- (µ) 20. Mass of a cubic liter of water. 21. kg/cubic meter. 22. Three fundamental quantities generally used are: Length(m), Mass(Kg), and
Time(s). 23. The compactness of matter. 24. It is given a new name. 25. No. An equation must be equal in magnitude and units. 26. Yes. And it could be confused with ―meters‖ instead of ―miles.‖ 27. To express measured numbers properly. 28. The 3 rules for determining significant figures are: 1. Non-zero digits are always significant,
2. Zeros at the beginning of a number are not significant, 3. Internal or end zeros are significant.
For example - 0203.089 have 6 significant figures (2,0,3,0,8,9). 29. Three. 30. One.
ANSWERS TO VISUAL CONNECTION a. meter, b. kilogram, c. second, d. mks, e. foot, f. pound, g. second, h. fps
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Intrinsic properties are invariant. Kilogram cylinder and meterstick are subject to wear, dirt, and change. 2. A liter, because it is larger than a quart. 3. Scientific laws describe; legal laws regulate. Scientific laws are about the nature of things; legal laws concern society. 4. 1 kgf > 1 lbf (force; 1 kgf = 2.2 lbf or 1 kgm = 2.2 lbm); 1 m3 > 1 gal; notable exception is the slug. 5. No, a man did not buy a new rod because the box has dimensions 3 ft × 4 ft so he put his 5 ft rod diagonally. 6. 1 m = 3.28 ft 828 m (3.28 ft/m) = 2.72 ×103 ft; 508 m (3.28 ft/m) = 1.67 × 103 ft Δ = 1.05 × 103 ft
ANSWERS TO EXERCISES 1. 100,000 cm or 105 cm 2. 16000 MB 3. 106 mm3 4. 1 m3 = 103 L. 1 m3 = 102 cm x 102 cm x 102 cm = 106 cm3 (1 L/103 cm3) = 103 L = 1000 L 5. 0.50 L (1 kg/L) = 0.50 kg = 500 g 6. 15 cm x 25 cm x 30 cm = 11250 g and 11.25 kg 7. (a) 0.55 Ms = 0.55 × 106 s (b) 2.8 km = 2.8 103 m (c) 12 mg = 12 10–3 g = 1.2 10–5 kg (d) 100 cm = 1.00 m
8. (a) 32 GB (b) 54.3 mL (c) 0.5421 m (d) 6.21 kilobucks 9. 6 ft 10 in. = 82 in. (2.54 cm/in.) = 208.28 cm = 2.0828 m 10. 6 ft 7 in. 11. Yes, to two significant figures 12. (a) 70 mi/h (1.61 km/mi) = 112.7 km/h (113 km/h); (b) 65 mi/h (1.61 km/mi) = 104.65 km/h (105 km/h) 13. No, 300 L ~ 300 qt (1 gal/4 qt) = 75 gal 14. Yes. That would make the room about 3 m × 4 m, which would be about 10 ft × 13 ft that could be the size of a small dorm room. 15. See AYK # 6, Height of Burj Khalifa - Height of Taipei 101 = 828m – 508m = 320 m =
32000 cm(1/2.54 in./cm) = 12598.43 in.(1/12 ft/in.) = 1049.86 ft = 1050 ft 16. 900 ft (1 m/3.28 ft) = 274.32 m; 1,900 ft = 579 m 17. cm, km 18. 103 kg (2.2 lb/kg) = 2,200 lb. 103 kg heavier by 200 lb 19.
= m/V = 500 g/47 cm3 = 10.64 g/cm3 (the density of the metal)
20. V =
= 2000 g/7.9 g/cm3 = 253.2 cm3
21. (a) 7.7 (b) 0.0030 (c) 9500 (d) 0.00034 22. (a) 4.3 (b) 1.0 (c) 16 (d) 5.5 23. 4.3 24. (a) 55 (b) 0.58 (c) 1870 (d) 14 25. (3.15 m × 1.53 m)/0.560 m = 8.61 m 26. 6.75 (3 sf)
Chapter 2
MOTION This chapter covers the basics of the description of motion. The concepts of position, speed, velocity, and acceleration are defined and physically interpreted, with applications to falling objects, circular motion, and projectiles. A distinction is made between average values and instantaneous values. Scalar and vector quantities are also discussed. Also, an interesting Highlight on Galileo and the Leaning Tower of Pisa discusses the status of the tower. Problem solving is difficult for most students. The authors have found it successful to assign a take-home quiz on several questions and exercises at the end of the chapter that is handed in at the beginning of class. (It may save time and be instructive to have students exchange and grade papers as you go over the quiz.) This may be followed by an in-class quiz on one of the take-home exercise, for which the numerical values have been changed. The procedure provides students with practice and helps them gain confidence.
DEMONSTRATIONS A linear air track may be used to demonstrate both velocity and acceleration. If an air track is not available, a 2-in. 6-in. 12-ft wooden plank may be substituted. It will be necessary to have a V groove cut into one edge of the plank to hold a steel ball of about 1-in. diameter. The ball will roll fairly freely in the V groove. Also, various free-fall demonstrations are commercially available. (General references to teaching aids are given in the Teaching Aids section.)
ANSWERS TO MATCHING QUESTIONS a. 14
b. 2 c. 3 d. 12
o.18
p. 8
e. 16
f. 13
g. 1
h. 6 i. 10
j. 7
q. 9 r. 4
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a
2. c
3. d 4. d 5. d
7. c
8. d
9. d 10. c 11. b
6. a 12. c
k. 17
l. 11
m. 5
n. 15
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. position
2. scalar
7. time, t2 8. gravity
3. vector
4. distance
5. speed
6. constant or uniform
9. m/s2 10. centripetal (center-seeking)
11. 9
12. motion, velocity
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Classical Mechanics. 2. An origin or reference point and a unit measurement scale are needed. 3. Motion is a change in position of an object over time. Hence, the time rate of change of position is the basis of describing motion in terms of speed and velocity (length/time). 4. A scalar has magnitude, and a vector has magnitude and direction. 5. Distance is the actual path length and is a scalar. Displacement is the directed, straight-line distance between two points and is a vector. Speed is distance per unit time, and velocity is displacement per unit time. 6. The statement is correct; displacement is a vector whose length is the shortest distance between the initial and final points, whereas distance may take a different path between the same two points. 7. (a) They are equal. (b) The average speed has a finite value, but the average velocity is zero because the displacement is zero. 8. Either the magnitude or direction of the velocity, or both. An example of both is a child going down a wavy slide at a playground. Another example is a car changing speed and direction in traffic. 9. Yes, both (a) and (b) can affect speed and therefore velocity. 10. An object will slow down if the direction of velocity and acceleration are opposite. 11. Initial speed is zero. Initial acceleration of 9.8 m/s2, which is constant. 12. The object would remain suspended. 13. No, in uniform circular motion, velocity changing direction, centripetal acceleration. 14. Center-seeking. Necessary for uniform circular motion. 15. A tighter curve has a smaller radius, which would result in a higher magnitude of centripetal acceleration than a gentle curve. 16. (a) & (b) Inwardly toward the Earth's axis of rotation. (c) The person himself is spinning hence the direction would be inwards towards center axis of the person.
17. g and vx Where g is an acceleration due to gravity and vx is a constant horizontal velocity. 18. No, it will always fall below a horizontal line because of the downward acceleration due to gravity. 19. Greater range on the Moon, gravity less (slower vertical motion). Range on the Moon will be approximately 6 times the range on the Earth. 20. Initial velocity, projection angle, and air resistance. 21. Both have the same vertical acceleration. 22. Less than 45o because air resistance reduces the velocity, particularly in the horizontal direction.
ANSWERS TO VISUAL CONNECTION a. Speed, velocity, and acceleration increasing, b. Speed is constant. Velocity is changing as direction is changing. Constant centripetal acceleration and zero tangential acceleration, c. Speed, velocity and acceleration decreasing
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. More instantaneous. Think of having your speed measured by a radar. This is an instantaneous measurement, and you get a ticket if you exceed the speed limit. 2. (a) & (b) We feel we are in motion when we see a change in our position with respect to the surrounding. As the Earth revolves with high speed around the Sun we also move with the same speed. We are in the same frame of reference and hence we feel to be stationary. (c) We can easily sense this motion by observing the change in position of the Sun and the Moon. 3. Free fall is any object in motion under the influence of gravity; therefore, an object projected vertically upward is in free fall as it is influenced by gravity. 4. (a) & (b) Inwardly toward the Earth's axis of rotation. (c) The person himself is spinning, hence the direction would be inwards towards the center axis of the person. d ½ gt2 , so t 2d / g 5.
2(11 m) 1.5 s 9.8 m/s2 Balloon lands in front of prof. Student gets
an ―F‖ grade. 6. Skydiver uses ways to increase air resistance and achieve terminal velocity quicker. Hence,(a) During an updraft the upward current of air would help the diver to balances his/her downward weight allowing to reach the terminal velocity quicker. (b) During a
downdraft, the downward current of air would add up to the downward weight of the diver which would oppose the achieving of terminal velocity. 7. The ball would still take 1.50 s to hit the ground. 8. ac = v2/r = (4 m/s)2/2 m = 8 m/s2
ANSWERS TO EXERCISES 1. The gardener walked 7 m with a magnitude displacement of 5 m. 2. 𝑣̅ = d/t = 7 m/12 s = 0.58 m/s for the average speed with the magnitude of the average velocity of 0.42 m/s. 3. 𝑣̅ = d/t = 100 m/11.5 s = 8.7 m/s 4. 𝑣̅ = d/t = 2𝜋r/t = (𝜋 × 300 m)/10 minutes (10 × 60 = 600 m/s) = (3.14)(300 m)/600 m/s = 1.57 m/s; the magnitude of the jogger’s average velocity would be zero because the jogger started and stopped at the same place on the track which makes the displacement 0. 5. t = d/v = 750 mi/(55.0 mi/h) = 13.6 h 6. t = d/v = (7.86 × 107 km)/(3.00 × 108 m/s) = 2.62 × l02 s. Speed of light (constant). 7. (a) d = v t = (52 mi/h)(1.5 h) = 78 mi (b) v = d/t = 22 mi/0.50 h = 44 mi/h (c) v = d/t = 100 mi/2.0 h = 50 mi/h 8. (a) d/150 s (b) d/192 s (c) d/342 s. Omission.d inadvertently left out. Assuming 100 m, (a) 100 m/150 s = 0.667 m/s (b) 100 m/192 s = 0.521 m/s (c) 200 m/342 s = 0.585 m/s 9. (a) v = d/t = 300 km/2.0 h = 150 km/h, east (b) Same, since constant 10. (a) v = d/t = 750 m/20.0 s = 37.5 m/s, north (b) Zero, since displacement is zero 11. a = (vf – vo )/t = (12 m/s – 0)/6.0 s = 2.0 m/s2 12. (a) a = (vf – vo )/t = (0 – 8.3 m/s)/1200 s = –6.9 × 10–3 m/s2 (b) v = d/t = (5.0 × 103 m)/(1.2 × 103 s) = 4.2 m/s (Needs to start slowing in plenty of time.) 13. (a) a = (vf – vo )/t = (8.0 m/s – 0)/10 s = 0.8 m/s2 in the direction of motion. (b) v = 8 m/s + 0.8 m/s2 × 5 s = 12 m/s 14. (a) 44 ft/s/5.0 s = 8.8 ft/s2, in the direction of motion (b) a = (88 ft/s – 44 ft/s)/4.0 s = 11 ft/s2 in the direction of motion (c) (66 ft/s – 88 ft/s)/3.0 s = –7.3 ft/s2 in the opposite direction of motion (d) a = (66 ft/s – 0)/12 s = 5.5 ft/s2 in the direction of motion 15. 𝑡 = √
2𝑑 g
2(0.156 𝑚)
= √ 9.80 𝑚/𝑠 2 = 0.18 𝑠
16. v = vo + gt = 0 + 9.80 m/s2 × 0.18 𝑠 = 1.75 m/s
17. d = ½ gt2, t = sq.root [2(274 m)/9.80 m/s2] =7.5 s 18. d = ½ gt2. t as in exercise 17. 4.3 s – 2.5 s = 1.8 s 19. 90.0 km/h = 25.0 m/s. ac = v2/r = (25.0 m/s)2/500 m = 1.25 m/s2 20. (a) ac = v2/r = (10 m/s)2/70 m = 1.4 m/s2 (b) ac = 14% of g; Yes, you would be able to sense the car’s acceleration. 21. 𝑡 =
2𝑑
2(2 𝑚)
√ g = √ 9.80 𝑚/𝑠2
= 0.64 𝑠
22. 45o – 42o = 3o, so 45o + 3o = 48o. Note: air resistance can usually be disregarded because the golf ball has a smooth surface, a solid core, and is small.
Chapter 3
FORCE AND MOTION This chapter is one of the most important in the textbook because it deals with Newton’s laws of motion and gravitation, as well as the concepts of linear and angular momentum. Also, involving a force, buoyancy and Archimedes’ principle is discussed in this chapter. The material naturally follows that of Chapter 2. With the foundations of kinematics established, the agents that produce motion are considered. This branch of mechanics is known as dynamics. Sufficient time should be spent on this material to be sure that students have a firm understanding of these concepts. Force and net force are discussed in an initial chapter section because of the importance of understanding these concepts. It is suggested that students be required to make complete statements of Newton’s laws and to give examples. When stating Newton’s second law of motion, stress that the force is the unbalanced force acting on the total mass, and that the mass is the total mass being accelerated. Also that the acceleration is in the direction of the unbalanced, or net, force. Acceleration (or deceleration) is evidence of the action of an unbalanced force. The Highlight on automobile airbags includes side airbags and ―depowering‖ features. Also new to the thirteenth edition is the Highlight: Surface Tension, Water Striders, and Soap Bubbles.
DEMONSTRATIONS A linear air track can be used to illustrate Newton’s laws of motion and the concept of linear momentum. An Atwood machine provides an excellent demonstration for illustrating Newton’s second law of motion. Best results can be obtained if the student is led through the demonstration (see the Laboratory Guide) by questions rather than having the instructor merely perform the experiment.
An apple may be brought to class to illustrate the idea of one newton of weight. Be sure the apple weighs about 3.6 ounces. Also, a spring balance calibrated in newtons can be displayed supporting a 1-kg mass. Free-fall can be demonstrated with a feather and a coin in a glass tube from which the air can be evacuated. Let your students handle the glass tube for the best results. Newton’s third law of motion can be demonstrated by using a toy rocket that holds water under pressure-equal and opposite forces are demonstrated as the rocket accelerates along a string. Releasing a blown balloon also illustrates the law. The law of conservation of angular momentum is demonstrated dramatically using a turntable or rotating stool and two masses (for example, 1 kg each) held in the hands. While rotating, the masses are brought closer to the body (reduced moment of inertia), and the rate of rotation increases. When beginning the demonstration, point out that you can't get started by yourself. You must have an external force or torque, which can be supplied by a student. Students often will ask to try the demonstration. Permission should be granted with caution. The rotation can make a person quite dizzy. (General references to teaching aids are given in the Teaching Aids section.)
ANSWERS TO MATCHING QUESTIONS a. 2 b. 9 c. 15 d. 6 i. 7 j. 4
k. 17
e. 14
l. 13
f. 5
m. 19
g. 11
h. 16
n. 10 o. 1
p. 12 q. 3
r. 8 s. 18
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.d
2. d 3. c 4. d
5. a
6. d 7. b
8. d
9. c
10. a
11. a
12. d
13. c 14. d
15. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. vector
2. capable
3. could
4. unbalanced force
5. inertia 6. inversely
7. kg ∙ m/s2 8. static, kinetic or sliding 9. equal, opposite, different
10. 19.6 11. less
more 13. net or unbalanced 14. torque
ANSWERS TO SHORT-ANSWER QUESTIONS 1. A force is a push or pull, and a net force is a combination of all forces acting on an object.
12.
2. No, the force may be balanced, meaning that the forces canceled each other out, and the net force zero. 3. If we pull a tablecloth quickly without disturbing dishes and glasses, they will remain in place because of inertia. 4. (a) goes slightly forward (b) goes slightly backward 5. The inertia of the hammerhead continues motion and tightens on the handle. 6. A quick jerk gives inertia for tearing. Larger rolls have more inertia and resistance to motion. 7. If a is zero in F = ma, then F = 0, and an object is at rest or moving with a constant velocity (first law). 8. Yes, if the sum of the forces is zero. 9. Yes, the body is in motion already as per the first law of motion. 10. Zero, since velocity is constant. 11. It requires less force to keep moving than to initiate motion. Because initially when the body is at rest to set the body in motion it has to overcome its inertia and static friction. When it is in motion it only wants to overcome kinetic friction. 12. (a) Ten times the force, but also ten times the inertia, or mass; therefore, it falls at the same rate. (b) Similar, except that the acceleration of the rocks would be less—that is, g/6. 13. The two forces of the force pair will act on different objects. For example - Jet propulsion. The rocket and exhaust gas exert equal and opposite forces on each other and hence, accelerate in opposite directions. 14. No, the pad is there only to hold the rocket. The expanding combustion gases exert a force on the rocket, and the rocket exerts a force on the gases. Consider firing a rocket in space— there is nothing to ―push against.‖ 15. There are equal and opposite forces for all forces, and the net force is zero. Force due to wall on block and force due to hand on block are same and opposite. Gravitational pull is balanced by frictional force between hand and block, block and wall. 16. 9.8 N. Holding one end of the string before the pulley, the scale would measure only one mass. 17. Weight zero (gravity essentially zero). Same mass, 70 kg. 18. Because F
then F approaches zero only as r approaches infinity. So, there will be a
force of attraction between any two bodies in the Universe. 19. According to Newton’s Law of Gravitation, if the distance between the Earth and the Moon is doubled then the force will become 0.25 (1/4) times that of what it was before.
20. It is less because the Moon has a smaller mass and size (radius) than the Earth. 21. Yes, value of g = 0 is possible. At the center of the Earth value of g is zero because the mass of Earth at center is zero. Gravity does exist everywhere in the universe. However, deep into space far from any celestial body, the acceleration due to gravity can be almost zero. 22. The density of helium is much less than the density of air. Hence, there will be a buoyant force on a balloon.23. The material selected for the manufacturing of a life jacket should have much less density. Hence it will displace enough volume of water. 24. Zero. Just another liter of water. 25. As we lower the iron piece the support will decrease due to the increase in viscous force and at certain point iron piece will achieve terminal velocity where the support will be zero. 26. The glass bulb floats in liquid because of buoyant force. Higher the density of liquid higher the buoyant force and hence glass bulb float higher. 27. The Pine wood block will have more volume above the water surface since the density of the oak block is greater than that of pine. The oak block would sink quicker than pine. 28. kg · m/s 29. In the absence of an unbalanced external force, an object or system has a constant velocity and hence a constant momentum. 30. (a) Gravity (the weight force) and the normal upward force of the surface on the blocks. (b) These forces cancel each other, and there is no net external force on the block. 31. Through the conservation of angular momentum. Tucking when diving reduces the r of the mass distribution and the rotational speed increases.
ANSWERS TO VISUAL CONNECTION a. inertia, b. mass, c. constant velocity, d. net force, e. acceleration, f. m/s2, g. action, h. equal and opposite reaction, i. different objects
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. In the 17th century Galileo performed experiments on motion using a ball rolling down an inclined plane onto a level surface. He concluded that if a very long surface could be made perfectly smooth, there would be nothing to stop the ball, so it would continue to slide in the absence of friction indefinitely or until something stopped it. Thus, contrary to Aristotle, Galileo concluded that objects could naturally remain in motion rather than coming to rest. 2. Opposite reaction in both cases affects weight. (Downward force of arms, upward force on scale, and more weight.)
3. Set mg equal to Eq. 3.4. Moon has smaller M and R. 4. (a) To increase the amount of water displaced and increase the buoyancy. (b) Saltwater is denser than regular water, and the average density of a person is less than that of saltwater. 5. The adhesion between the water and the clothes is not sufficient to provide the necessary centripetal force for the water to rotate with the clothes, and hence the water becomes separated. 6. According to the third law of motion, for every action, there is an equal and opposite reaction. When a gun is fired, the forward momentum of the bullet is balanced by the backward momentum of the gun to maintain zero momentum. Also, before the gun is fired the total momentum is zero as neither object is moving. 7. Longer lever arm, more torque.
ANSWERS TO EXERCISES 1. (a) 1.0 N in direction of 9.0 N force. (b) 17.0 N in direction of forces. 2. 350 N to equal fmax. 3. F = ma = (4.0 kg)(5.0 m/s2) = 20 N 4. a = F/m = 2.1 N/(7.0 × 10–3 kg) = 300 m/s2 5. a = F/m = 950 N/1000 kg = 0.95 m/s2 6. a = F/m = 350 N/20 kg = 17.5 m/s2 7. w = mg = (6.0 kg)(9.8 m/s2) = 59 N 8. w = mg = (4.0 kg)(9.8 m/s2) = 39 N 9. (a) (120 1b)(4.45 N/lb) = 534 N (b) Personal 10. (a) w = mg = (75 kg)(9.8 m/s2) = 735 N. (b) Same, zero acceleration. 11. (a) F = Gm1m2/r 2= (6.67 × 10–11 N ∙ m2/kg2)(103 kg)(103 kg)/(25 m)2 = 1.1 × 10–7 N (b) Much, much smaller; w = mg = (103 kg)(9.8 m/s2) = 9.8 × 103 N 12. F = Gm1m2/r2 = (6.67 × 10–11 N ∙ m2 /kg2)(3.0 kg)(3.0 kg)/(0.15 m)2 = 2.7 × 10–8 N 13. (a) r2 = 2r1, and F2/F1 = (r1/2r1)2 = (1/2)2 = ¼; force will be 1/4 times the initial value of force. (b) r2 = r1/2, and F2/F1 = (2)2 = 4; force will be 4 times the initial value of force. 14. (a) r2 = (2/3)r1, and F2/F1 = (r1/r2)2 = (3/2)2 = 9/4 = 2.25 (b) r2 = 3r1, and F2/F1 = (1/3)2 = 1/9 15. (a) wM = wE /6 = 150 lb/6 = 25 lb (b) Personal. 16. g = F/m = 49 N/125 kg = 0.39 m/s2
17. Float. Density = m/V = 120 g/125 cm3 = 0.96 g/cm3, less than water. 18. Float. Density = 0.28 g/cm3 19. Buoyant force = Density of water × V × g = (1000 kg/m3)(1 m3) (9.8m/s2) = 9800 N Weight of all students = Buoyant force n(75kg)(9.8m/s2) = 9800 N n =13.33 ≈ 13 students 20. Length of ice cube = l = (0.92 cm)/1.414 = 0.65 cm Weight = Buoyant force = Density of water × Volume of water displaced × g = (1 g/cm3)(0.27 cm3)(980 cm/s2) = 269.1 dyne 21. Momentum = p = mv = (1000 kg)(20 m/s) = 2 × 104 kg • m/s eastward. 22. Momentum = p = mv = (900 kg)(30 m/s) = 2.7 × 104 kg ∙ m/s northward. So, car has more momentum than truck. There is no role of direction here. 23. P = momentum = mv Pg = –Pb, or mgvg = –mbvb, and with m = w/g, vg = (-mb/mg)vb = (–wb/wg)vb = (–750 N/550 N)(0.50 m/s) = –0.68 m/s 24. m1v1 = -m2v2 750 𝑁 × 0.50 𝑚/𝑠
Since mass cannot be negative, m2 = 9.8 m/s2× 0.65 𝑚/ = 58.87 kg 25. v2 = rlvl/r2 = (600 × 106 mi)(15000 mi/h)/(100 × 106 mi) = 90000 mi/h the comet’s velocity 26. L = mvr L = 1.29 kg × 33.5 m/s × 50000 m = 2.2 × 106 kg m2/s
Chapter 4
WORK AND ENERGY This chapter should be covered thoroughly in lecture and assignment. The relationship of work and energy is of the greatest importance in understanding many daily activities. Also, the development of the physical environment is closely associated with the control of energy. The law of conservation of energy is one of the most important general laws and has been a key to many of nature's secrets. Thus it is important for the student to know the meanings of work and energy and to be familiar with various forms of energy.
Although this chapter deals primarily with general concepts and mechanical energy, it should be pointed out how easy it is to change other types of energy, such as chemical and electrical energy, to other forms and use them to do work. The textbook tries to get the student to think in terms of symbols, and this is a good chapter to stress this kind of thinking. For example, when referring to kinetic energy, think ½ mv2, and when thinking of gravitational potential energy, think mgh. In addition, the status of energy consumption and resources is covered, including alternative and renewable sources.
DEMONSTRATIONS A simple pendulum can be used to display the transformation of potential energy to kinetic energy, and vice versa. As the pendulum swings back and forth, ask the students at what stages the velocity, acceleration, potential energy, and kinetic energy have their minimum and maximum values. Demonstrate the examples of work as shown in the illustrations in the textbook. A radiometer can be used to show that light can do work. (General references to teaching aids are given in the Teaching Aids section.)
ANSWERS TO MATCHING QUESTIONS a. 9 b. 11, c. 1 d. 3 j. 10
k. 4
l. 15
e. 14 f. 8 g. 13
m. 6
n. 7
h. 5 i. 2
o. 12
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.a
2. d 3. d 4. b
5. c
6. c
7. d
8. a
9. b
10. a 11. c
12. c
13. b 14. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. parallel
2. joule
3. scalar
4. work
7. square 8. constant 9. power, watt
5. kinetic
6. motion, position
10. 0.75 11. electrical energy 12. natural gas 13.
renewable 14. ethanol (alcohol)
ANSWERS TO SHORT-ANSWER QUESTIONS 1. (1) When you apply a force and there is no displacement, then work done by force is zero. (2) When force is perpendicular to displacement.
2. No, there must be motion. No work is done in holding an object stationary, but work is done in lifting. Also, when the force and displacement are perpendicular to each other, the vertical component does not work. Like when a lawn mower is pushed. 3. Friction. Dust reduces friction by introducing particles between the floor and the object, thus changing the coefficient of friction to that between the small particles and the puck and not the floor and the puck. 4. No work while stationary. Work was done on the weights in lifting. 5. No work while stationary. Work was done on the weights in lifting. 6. kinetic energy = 1/2 × m × v2 Hence, if the speed becomes half then the kinetic energy becomes 1/4 of what it was initially. 7. As the speed of an object is doubled, its kinetic energy will become 4 times. So, 4 times work is required to stop. 8. Water towers are tall structures and placed on high elevation because due to high elevation potential energy of water increases. Higher the potential energy longer the range up to water can be transported without doing external work. 9. Both may be correct, depending on the zero reference point chosen. 10. (a) The height depends on the initial kinetic energy, mgh = ½ mv2. (b) By the conservation of energy, it would have the same speed as it had initially. 11. Change in the potential energy between two positions is the same in any case. Even if the reference point is changed, the potential energy would be calculated according to that and their difference would never change. 12. Total energy includes all forms of energy. Mechanical energy is the sum of kinetic and potential energies. 13. 50 J; potential energy converted to kinetic energy. 25 J; the object falls from half the height it will have half kinetic energy before hitting the ground. 14. Total energy: when energy neither enters nor leaves a system and thus has a constant value. Mechanical energy: no energy loss. 15. (a) a and e. (b) c. (c) c. (d) a and e. (e) a and e. (f) c. (g) a, b, d, and e. (h) b, c, and d. (i) b, c, and d. (j) a, b, d, and e.
16. (a) b and c. (b) a. (c) a. (d) b and c. (e) b and c. (f) a. (g) b, c, e, and d when going toward h = 0. (h) a, and e, d, when spring is compressing. (i) a, and e, d, when spring is compressing. (j) b, c, and e, d when going toward h = 0. 17. Because the snowballs have the same initial speed from the same height, both will have the same speed on striking the ground. This is an illustration of the conservation of energy. 18. Yes, kinetic energy is converted into potential energy and the potential energy into kinetic energy. Mechanical energy is conserved.19. P = W/t, person B does more work because of longer time. So, person A, with the shorter time, expends more power. 20. Yes, but motors with different sizes of engines will require different amounts of time. One must run the ¼ hp three times as long, or run a larger motor one-third the time. 21. (a) More work is done in a given time. (b) Doing a given amount of work faster. 22. Energy, kilowatt-hour (kWh). 23. LED consumes 12 J energy per second, while incandescent light bulb consumes 60 J per second. Saved energy = 48 J per second. Percentage saving with respect to 60 W= (48 J/60 J) × 100 = 80%. 24. The average human radiates approximately 350000 J per hour that is 97.2 J per second. 25. Oil is consumed the most in the United States. Natural gas has been the largest source of generation of electricity in the US. 26. Industry 27. Radiant, chemical, nuclear, sound, and heat. 28. Examples of alternative energy sources: Biofuel and Tidal Energy 29. Examples of renewable energy sources: hydropower, wind power, solar power, geothermal, and tides 30. Solar and wind.
ANSWERS TO VISUAL CONNECTION a. When athlete is approaching it is doing work and gaining kinetic energy. b. Athlete doing work, possesses potential energy, kinetic energy transformed to potential energy c. Athlete not doing work, possesses gravitational potential energy, potential energy transformed to gravitational potential energy, potential energy is maximum at this point. d. Athlete not doing work, possesses both kinetic and gravitational potential energy, part of gravitational potential energy transformed to kinetic energy.
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes, this is numerically possible. If there is no motion, the kinetic energy is zero. If the student selects his or her position as the zero reference point, then the potential energy is zero. 2. Same speed. Upward stone has the same speed as the other stone on return to initial height. 3. When a person lands on the trampoline they push the trampoline by bending their legs. Because of this trampoline exerts opposing force which is used to achieve more height. 4. We can feel heat and vibrations, see energetic phenomena and light, and hear sound. However, it is doubtful that you could smell or taste energy. 5. Incandescent bulbs, Fluorescent Light bulbs, and Light Emitting Diode (LED)
bulbs. LED bulbs and Fluorescent Light bulbs are 83% and 75% more efficient than Incandescent bulbs respectively. 6. Turn off lights when not needed, turn off appliances when not being used, keep thermostats properly set, and have good home insulation, use ―smart‖ controls to regulate the temperatures and light, use ―smart‖ chargers that turn off when a device is completely charged.
ANSWERS TO EXERCISES 1. W = Fd = (250 N)(2.0 m) = 500 J 2. F = W/d = 400 J/2.0 m = 200 N 3. W = mgh = (6.0 kg)(9.8 m/s2)(1.5 m) = 88 J 4. W = mgh = (31.75 kg)(9.8 m/s2)(0.5 m) = 155.6 J5. W = (60%) Fd = (0.60)(200 N)(6.0 m) = 7.2 × 102 J 6. W = (40%) Fd = (0.40)(200N)(6.0 m) = 4.8 × 102J 7. W = mgh = (0.150 kg)(9.80 m/s2)(10.0 m) = 14.7 J 8. W = mgh = (0.150 kg)(9.80 m/s2)(7.50 m) = 11.0 J 9. (a) Ek = ½mv2 = ½(1000 kg)(25 m/s)2 = 3.125 × 105 J (b) W = Ek = 3.125 × 105 J 10. Ek= ½mv2, or v = √2𝐸/𝑚 = √2(1.2 × 104 J)/60 kg = 20 m/s [3.6 km/h/(m/s)] = 72 km/h 11. Eka = ½mv2 = ½(2.0 × 10-3 kg)(4.0 × 102 m/s)2 = 1.6 × 102 J. Ekb = ½mv2= ½(6.4 × 107 kg)(10 m/s)2= 3.2 × 109 J, so ocean liner has greater kinetic energy. 12. Ek = ½mv2 = ½(20 kg)(9.0 m/s)2 = 810 J 13. K = W = Fd = (8.0 N)(4.0 m) = 32 J
14. To have a speed of 12 m/s the 0.5 Kg block needs: 1/2 mv2 = 1/2 (0.5)(12 × 12) = 36 J of kinetic energy. Since d = W/F = 36 J/8.0 N = 4.5 m and from Exercise 13 we know that d = 4 m, 0.5 m farther would the force have to act for the block to have a speed of 12 m/s. 15. Ep = mgh = (3.00 kg)(9.80 m/s2)(-10.0 m) = –294 J. Below ground zero point. 16. W = mgh = (5 kg)(9.8 m/s2)(7.5 m) = 367.5 J 17. ½𝐸𝑃̂(lost) = ½mgh = ½mg(12 m), and h = 6.0 m 18. (0.33𝐸𝑃̂) = (0.33)mgh = (0.33)mg(6.0 m), and h = 2.0 m 19. (a) E = mgh = (60 kg)(9.8 m/s2)(12 m) = 7.1 × 103 J. (b) Same. (c) Same. 20. mgh = ½mv2, and v = √2𝑔ℎ (h from top) [[Ed. Square root sign over all]] (a) v = √2𝑔(5.00 𝑚) = 9.90 m/s (b) v = √2𝑔(7.50 𝑚) = 12.1 m/s 21. P = W/t W = P x t = (700 W)(8 s) = 5600 J 22. In Exercise 21, Work (W) = P × t = (700 W) x (8 s) = 5600 J. We know that Work (W) = mgh. Therefore, the height of a hill he can cycle in that 8.0 s with 5600 J of work will be: h = W/mg = (5600 J)/(60 kg)(9.8 m/s2) h = 9.5 m 23. (a) W = Fh = (556 N)(4.0 m) = 2.2 × 103 J (b) P = W/t = 2.2 × 103 J/25 s = 89 W 24. P = W/t = mgh/t = 125 lb(1 kg/2.2 b)(9.8 m/s2)(4.0 m)/5.0 s = 4.5 × 102 W 25. E = Pt = (1.60 kW)(1/6 h) + (1.10 kW)(4/60 h) = 0.34 kWh(18¢/kWh) = 6.12¢ 26. (a) E = Pt = (1.25 kW)(4.0/60 h) = 0.083 kWh (b) 0.083 kWh(12¢/kWh) = 1¢
Chapter 5
TEMPERATURE AND HEAT Chapter 5 is an important chapter because temperature and heat are two of the most common physical concepts that students experience. In general, temperature measurements are given, and we say that heat is a form of energy. Hence it is important that a basic understanding and distinction of temperature and heat is obtained. In large part, the chapter is concerned with the measurement of macroscopic quantities of heat, such as specific heat and latent heat. The general trend is to express these heats in joules (J). However, calculations will be done primarily in kilocalories (kcal) because of the difficulty of
adding numbers expressed in powers of 10, which is necessary when using joules. Calculations are much easier when done in kilocalories, and the results can be converted to joules if so desired. Because heat transfer has many applications in daily life, this is an important and interesting topic that should be covered in some detail. The chapter contains interesting Highlights: Freezing from the Top Down, and Hot Gases: Aerosol Cans and Popcorn. Finally, the basics of thermodynamics are discussed in Section 5.7.
DEMONSTRATIONS A thermometer may be calibrated in class by using boiling water for the steam point and ice water for the ice point. Uncalibrated thermometers are available, and students find it interesting and obtain a grasp for temperature scales when the interval between the ice and steam points is divided into degrees. (How should it be done?) Also, have a calibrated thermometer on hand so that you can check and see how accurate your calibration is. The bulb of one of two thermometers may be painted black and exposed to sunlight or a heat lamp to show the difference in radiation absorption.
ANSWERS TO MATCHING QUESTIONS a. 16 b. 9 n. 11
c. 15
o. 2 p. 17
d. 25 e. 21 q. 12
f. 19
g. 6
r. 23 s. 14
h. 22
i. 7 j. 5
t. 8 u. 10
v. 18
k. 13
l. 20
w. 3
x. 24
10. c 11. c
12. b
m. 4
y. 1
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.c
2. a
3. b 4. c 5. b
6. c 7. a 8. b
9. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. lower 2. temperature 7. coin
8. gas
3. 1000
4. J/ kg C
9. Kelvin (absolute)
10. inversely
5. temperature 6. pressure 11. direction
12. completely
ANSWERS TO SHORT ANSWER QUESTIONS 1. Fahrenheit 2. Celsius or Kelvin 3. Alcohol, low; mercury, high 4. Because of the thermal expansion of the bimetallic coil on which it sits. 5. The volume of water increases and density of water decreases when you cool it from 4°C to 0°C. This is known as the anomalous behavior of water.
6. Cold in, hot out 7. If a body with less internal energy is hotter, then heat will flow from a body with low internal energy to high internal energy. 8. It will relatively take more heat per mass to heat/cool it and it will also relatively hold the heat longer. 9. Water in the filling has high specific heat. 10. No. The amount of heat necessary to change the temperature of a given amount of a substance (in this case water), depends on three factors: the mass, the specific heat, and the temperature change. 11. Yes, it is possible. For example, at boiling point and melting point, heat given to a substance does not increase its temperature but used to change its phase. 12. Specific heat, c = J/kg · °C latent heat, J/kg. The latent heat process occurs at a particular temperature, hence there is no temperature change. 13. Copper provides better thermal conductivity, or heat transfer, and faster cooking. A good conductor has molecules in a lattice (as in the case of metal) which allows for quick and uniform vibration (heating). 14. The tile floor will feel colder because the tile floor has a greater thermal conductivity. 15. Bridges are more exposed to cold air than roads. Cold air underneath causes freezing. Therefore bridge loose heat and get colder quickly. 16. Loose fitting allows for the air to be trapped. Trapped air is a natural insulator and reduces heat loss by conduction and convection. 17. Solids have definite shape and volume. Liquid does not have definite shape but has definite volume. Gas neither has definite shape nor volume. 18. The molecules exert little or no force on one another except when they collide. The distance between molecules in a gas is quite large compared with the size of the molecules on the other hand, solids and liquid molecules are compactly arranged to one other hence they are difficult to compress. 19. (a) Sublimation. (b) Deposition. 20. As the temperature increases, the molecules move at higher speeds, the kinetic energy and pressure will increase. 21. A gas made up of point particles that interact only by collisions. 22. Decrease the volume of the gas. 23. Molecular collisions with the walls of the gas container.
24. Pressure can be increased by giving heat to the system. Also keeping temperature constant and decreasing volume. Example, (1) If you heat a container it bulges, and also pressure of gas inside increases. (2) If gas is compressed its pressure increases. 25. Heat is removed from the system (balloon), and negative work is done as the balloon collapses. 26. First law: Energy is conserved in thermodynamic processes. Second law: The direction of a process and whether or not a process will take place spontaneously. 27. First law (conservation of energy) and second law (entropy increases in every natural process). 28. In general, no, except in the case of compressed gas. Expanding them makes them colder than the cool area, and compressing them makes them hotter than the warm area. By doing external work it is possible to make heat flow from a low temperature to a higher temperature. A refrigerator is an example of extracts heat from lower temperature to higher temperature. 29. It never decreases—the universe is the largest closed system of which we can think. 30. No, according to the third law of thermodynamics. 31. Zero.
ANSWERS TO VISUAL CONNECTION (a) deposition, (b) sublimation, (c) freezing, (d) melting, (e) condensation, (f) vaporization
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. As the air near the campfire is hot, so this hot air is displaced and empty space is created. Cold air’s molecules are more closely packed and they move more quickly. They seek empty space to achieve equilibrium and average the density across the location. Hence one can feel cooler breeze on backs. 2. When steam condenses, latent heat is given up. Also, the temperature of the steam is greater than that of hot water. 3. No, there is conduction (by coolant and through metal) and convection (by fan). 4. Soup contains water, which is one of the highest specific heats and can store more heat energy. Therefore, it takes longer to cool down. Blowing on the bowl of soup helps remove the heat energy from the soup.
5. When the sheet is heated, the metal sheet expands. The Circumference of ring expands in the same ratio as the width of the plate. So, the hole gets larger. 6. Refrigerator extract the heat from the objects placed inside and then it releases that heat outside the refrigerator which is within the room, so opening of refrigerator to cool down room does not make any sense. 7. No. The energy that is released to the surroundings leads to an increase in the entropy of the surroundings. This is greater than the decrease in entropy of the ice. When we consider both systems and surroundings, the change in entropy is positive. The Second Law is preserved. 8. Heat pumps extract heat from around the evaporator and dump heat into the environment around the condenser. Electricity is being used to concentrate and shift heat, not to produce heat directly as in a resistive electric radiator or fan heater.
ANSWERS TO EXERCISES 1. (22 × 9/5) + 32 = 71.6°F 2. (a) 245°C is the higher temperature because 245°F = 118.33°C. (b) 200°C is the higher temperature because 200oC = 392oF. 3. TC = 5/9 (TF – 32) = 5/9(68 – 32) = 20°C 4. (38 × 9/5) + 32 = 100.4°F 5. (a) TC = 5/9(– 40 – 32) = – 40°C (b) TK = – 40 + 273 = 233 K 6. (a) TC = TK – 273 = 3 – 273 = –270°C (b) TF = (9/5)(–270) + 32 = –454°F 7. 250 kJ × 1 kcal/(4.184 × 103 kJ) = 5.975 × 10-2 kcal 8. 100 kcal/h (4186 J/kcal) = 4.2 × 105 J/h = 418.6 kJ/h 9. (0.45 Cal/ h·lb)(8 h)(150 lb) = 540 Cal 10. 3500 Cal/(0.45 Cal/h) = 7.8 × 103 h (approx. 324 days, almost a year). 11. We know that 1 Watt = 1 J/s. Hence, 1 kWh = 3.6 x 106 J and if 1 kWh = 3412 Btu, then 1 Btu = (3.6 × 106)/(3412) J = 1055.1 J. 12. We know that 1 Btu = 1055.1 J; therefore, the power consumption of the heater in Watts will be (5000 × 1055.1) J/3600 sec = 1465.42 J/s or 1465.42 Watts 13. H = mc T = (0.50 kg)(1.0 kcal/kg ∙ oC)(10°C) = 5.0 kcal 14. H = mc T = (1.0 kg)(4186 J/kg ∙ oC )(100°C) = 4.186 105 J 15. (a) H = mc T = (1.0 kg)(1.0 kcal/kg ∙ oC )(80°C) = 80 kcal (b) (80 kcal)(0.00116 kWh/kcal)(18¢/kWh) = 1.67¢ 16. The energy required to heat the water from 25°C to 100°C would be:
(2.5 kg)(4186 J/kg · °C)(100°C – 25°C) = 784875 J. Since the heater supplies 100 J/s, the time taken to raise the temperature of water to its boiling point will be (784875 J)/(100 J/s) = 7848.75 sec = 130.81 min = 2.2 hrs. 17. HL = mc T = (0.500 kg)(0.50 kcal/kg ∙ oC)(10°C) = 2.5 kcal, H2 = mLf = (0.500 kg)(80 kcal/kg) = 40.0 kcal, H3 = (0.500)(1.00 kcal/kg ∙ oC)(20°C) = 10.0 kcal Total= 52.5 kcal 18. H1 = (0.300 kg)(0.50 kcal/kg∙oC)(10°C) = 1.5 kcal, H2 = (0.300 kg)(540 kcal/kg) = 162 kcal, H3 = (0.300 kg)(1.0 kcal/kg∙oC)(100°C) = 30 kcal, H4 = (0.300 kg)(80 kcal/kg) = 24 kcal Total = 217.5 kcal 19. The pressure drops to one-third of the previous pressure. 20. p2 = (T2/T1) p1 = (1.2) p1 21. T1 = 20 + 273 = 293 K, T2 = 40 + 273 = 313 K p2 = (T2/T1) p1 = (313 K/293 K) p2 = (1.07) p1 22. T1 = 20 + 273 = 293 K T2 = (p2/p1) T1 = (1.5)293 K = 440 K; 440 K – 273 = 167°C 23. p2 = (V1/V2) p1 = (0.500 m3/0.150 m3)(200 Pa) = 667 Pa 24. 25°C = 298.15 K; 5°C = 278.15 K; 278.15 K/298.15 K = 0.93 Volume outside = 0.93 × 0.15 m3 = 0.14 m3
Chapter 6
WAVES AND SOUND The concepts of waves, sound, and light are discussed in this chapter. Because most information about our environment comes to us by means of waves (see Section 1.3, ―The Senses‖), the general properties of waves are studied to prepare the student for the many physical concepts that involve waves. The general properties of waves are considered. Light is treated generally as waves, not electromagnetic waves since electric and magnetic fields have not been studied (Chapter 8). Major emphasis is given to sound waves because of their relationship to the environment. This includes the Doppler effect with its many practical applications.
A qualitative discussion of resonance and standing waves without mathematics is presented.
DEMONSTRATIONS There are many demonstrations that illustrate waves, sound, and the Doppler effect. For the concept of transverse wxaves, a long length of rubber hose is very useful. The demonstration of longitudinal waves can be made with a toy Slinky. A number of demonstrations are available for sound and electromagnetic waves. Consult the Teaching Aids section (Appendix) for sources of these. Sound demonstrations are particularly easy and interesting. Illustrate the production of sound by several methods. Perform the ringingbell demonstration in a bell jar from which the air can be evacuated. The Doppler effect can be demonstrated using an ordinary hair dryer. Use an oscilloscope for graphic representation of different sounds, including speech. Standing waves can be demonstrated using a mechanical vibrator and a piece of string with a suspended weight.
ANSWERS TO MATCHING QUESTIONS a. 13 k. 3
b. 9 l. 8
c. 20 m. 1
d. 12 e. 4 f. 7
n. 21 o. 14
p. 6
g. 18
h. 15
i. 16
q. 17
r. 2 s. 19
j. 11 t. 5
u. 10
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a
2. b
3. a 4. a 5. d
6. c 7. c
8. b
9. d 10. c
11. a
12. a
13. d
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. matter
2. perpendicular
3. wavelength
4. wavelength 5. light or 3.00 × 108 m/s
electromagnetic or transverse 7. longitudinal
8. 20 kHz 9. intensity
6.
10. 3 11. higher
12. moving away 13. natural or characteristic
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Energy is propagated by waves, not the matter. 2. No, electromagnetic waves need no medium. Travels in vacuum. 3. (a) parallel to wave velocity direction. (b) perpendicular to wave velocity direction 4. (a) m (b) Hz (or 1/s)
(c) s
(d) m
5. Frequency is inversely proportional to period (f = 1/T). Energy is proportional to the square of its Amplitude (A²)
6. (a) 2 cm (b) f = 1/0.6 s = 1.7 Hz 7. No, light travels at c in a vacuum. 8. Longer wavelength: red end. Higher frequency: blue end. 9. Microwaves have longer wavelengths than visible light but have lower frequencies than visible light. 10. Visible light = 4.3 x 1014 Hz to 7.5 x 1014; yes 11. A compression is a high-pressure region and a rarefaction is a low-pressure region within a sound wave. 12. The energy of a wave is directly proportional to the square of its amplitude. Hence, if amplitude is doubled, the energy becomes 4 times. 1
13. 25, 25 14. (a) Frequency. (b) Intensity or amplitude. 15. No. Sound needs something to travel through to get from one place to another. In space, since there is no air (vacuum), sound cannot travel. So, there are no sounds in the space. 16. Intensity is defined to be the rate of sound energy transfer per unit area carried by wave. Energy is expressed in joules (J). Intensity is expressed in J/s per m2 or W/m2. 17. No. The dB scale is nonlinear. (An increase of 3 dB doubles sound intensity.) 18. Light travels faster than sound. 19. (a) Shortened. (b) Lengthened. 20. (a) Shift to a higher frequency. That is when source and listener relatively move towards each other. (b) Shift to a lower frequency. That is when source and listener relatively move away from each other. 21. (Mach) = (680 m/s)/(340 m/s) = 2.0, 22. Detection involves the reflection of waves from an object. Speed and ranging involve the Doppler shift of the reflected waves. 23. Maximum energy transfer to the system when driven at a resonance frequency. The resonance-driven frequency should be equal to the natural frequency of the system. 24. Node is the point on standing wave whose displacement is minimum. The end point of a guitar string is a fixed point and hence amplitude will be minimum. 25. The tension is applied to a string. Different musical notes (frequencies) on a guitar or violin are obtained by placing a finger on a string, which effectively shortens its length.
ANSWERS TO VISUAL CONNECTION
(a) transverse, (b) electromagnetic, (c) longitudinal, (d) sound, (e) inverse, f = 1/T, (f) wave speed
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Greater speed downwind. Slower speed against the wind. Strong wind only you may hear yourself. 2. No. There is no atmosphere on the Moon. Communication is by radio waves, which are electromagnetic waves that require no medium for propagation. This is so interesting and helps debunk some conspiracy theories. 3. Through sunburns and suntans. UV also can cause cataracts and blindness (not a good detection method). 4. Faster than the speed of sound in water. 5. Longer strings produce sounds with a lower frequency; shorter strings produce sounds with a higher frequency. 6. Open: L = 1/2, 3/2, and 5/2 wavelengths Closed: L = 1/4, 3/4, and 5/4 wavelengths
ANSWERS TO EXERCISES 1. T = 1/f = 1/(0.25 × 103 Hz) = 4.0 × 10–3 s 2. T = 1/f = 1/5.0 Hz = 0.20 s 3. = c/f = (344 m/s)/(3000 Hz) = 0.115 m 4. (a) f = v/ = (2.0 m/s)/(1.5 m) = 1.3 Hz
(b) T = 1/f = 1/(1.33 Hz) = 0.75 s
5. (a) = c/f = (3.00 108 m/s)/(1460 103 Hz) = 205.48 m (b) wavelength = (3.00 108 m/s)/(107 106 Hz) = 2.8 m 6. = (3.00 l08 m/s)/(1018 Hz) = 3.00 10–10 m, or 0.300 nm 7. f = c/λ = (3.00 108 m/s)/(632.8 10–9 m) = 4.741 × 105 GHz 8. f = c/ = (3.00 108 m/s)/6.00 10–6 m = 0.500 1014 Hz 9. d = ct = (3.00 × 105 km/s)(3.16 × 107 s) = 9.48 × 1012 km 10. (a) d = 3000 mi = 4.8 × 106 m, t = d/c = (4.8 × 106 m)/3.0 × 108 m/s = 1.6 x 10-2 s (b) d = 3.50 × 105 m, t = d/c = (3.50 × 105 m)/3.0 × 108 m/s = 1.2 × 10-3 s 11. Audible Range of wavelength is: Lower limit wavelength ( ) = v/f = (344 m/s)/(20 Hz) = 17.2 m Upper limit wavelength ( 2) = 344 m/s/(2.0 104 Hz) = 0.0172 m
12. = v/f = (344 m/s)/(50 × 103 Hz) = 6.9 × 10–3 m 13. vm = 15(344 m/s) = 5.2 × 103 m/s. Wavelength = vm/f = (5.2 × 103 m/s)/20 × 103 Hz = 0.26 m 14. vm = f × wavelength = (0.333 m)(15.0 × 103) = 5.00 × 103 m/s; vm/vs = (5.00 × 103 m/s)/344 m/s = 14.5 times 15. d = vt = (1/3 km/s)(4.5 s) = 1.5 km = (1/5 mi/s)(4.5 s) = 0.90 mi 16. d = (500 m)(2.0) = 1000 m t = d/vs = (1000 m)/(344 m/s) = 2.9 s 17. 100 dB – 90 dB = 20 dB or 2 B, then 102 or 100 times greater. 18. 76 dB
Chapter 7
OPTICS AND WAVE EFFECTS The six major properties of waves (reflection, refraction, dispersion, polarization, diffraction, and interference) are presented and discussed in this chapter. It is convenient to introduce the properties to the students by means of demonstrations. Reflection can be illustrated with a plane mirror. Also, note that everything in the classroom is seen by reflection except the light source. A pencil or a small ruler in a glass of water is a good demonstration of refraction, and a prism can be used to illustrate dispersion. Viewing a candle through the slit between two fingers is good for illustrating diffraction, as is viewing a candle flame through a feather. Interference of light waves is best illustrated by Young’s experiment. If an experiment with a spectroscope is not done in the laboratory, the instrument should be presented and demonstrated in class. Polarized light is easily demonstrated with crossed Polaroids. (Polarizing sunglasses may be used.) Considerable class time should be spent in constructing ray diagrams and locating real and virtual images for different types of spherical mirrors and lenses. To simplify matters, only raw diagrams are used to locate images. The mirror and lens equations have been omitted.
DEMONSTRATIONS There are many demonstrations for illustrating waves, sound, and the basic laws of optics. A few have been mentioned in the introduction above. The ripple tank is an excellent piece of apparatus
for the production and projection of water waves. The apparatus can be used to study reflection, refraction, diffraction, and interference of waves. Distribute replica diffraction gratings to the students in class and let them see the spectrum from an incandescent light source. Demonstrate polarization with linear polarizers and double refraction with an Iceland spar crystal. If available, a laser is a spectacular demonstration in itself, and many optical properties can be demonstrated with commercial laser kits.
ANSWERS TO MATCHING QUESTIONS a. 7
b. 19
c. 20
d. 1
e. 15
f. 3
g. 18
h. 9
i. 10
j. 8
k. 17
l. 2
m. 11
n. 22
o. 5 p. 16
q. 14
r. 4
s. 6 t. 13
u. 21
v. 12
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.d
2. d 3. b 4. c 5. b
6. d 7. b
8. c
9. c 10. a
11. a
12. b
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. geometrical (light) and physical (wave) total internal 11. greater
6. converging
7. can
2. specular, diffuse
8. thinner
3. medium
9. concave or diverging
4. toward
5.
10. transverse
12. destructive interference
ANSWERS TO SHORT-ANSWER QUESTIONS 1. (a) diffuse. (b) specular. 2. Left, seen as right in mirror. (self-portrait) 3. 12 inches long because image size will be same as an object. distance is not a factor. 4. Walk toward you at same speed. In-step but opposite feet (right-left reversal). 5. In mirror with right-left reversal 6. Index of refraction is the ratio of speed of light in vacuum to the speed of light in medium. The speed of light is greatest in vacuum; hence index of refraction is always greater than 1. 7. If the index of refraction of glass is higher than that of water, the speed of light in glass is slower than the speed of light in water. Water has a higher speed of light. cm = c/n; the higher n, the smaller cm 8. Refraction at water-glass and glass-air surfaces. 9. When the angle of incidence increases beyond the critical angle. 10. The stars would not twinkle since that is caused by our atmosphere refracting light.
11. Yes, sunlight refracted over horizon by atmosphere. No atmosphere, no refraction – shorter. 12. Reflecting surface of the convex mirror bulges outside while for the concave mirror it bulges inward. Example of concave mirror mirrors used in automobile headlights and an example of a convex mirror is mirrors used in optical instruments and calling bell. 13. Real images are formed on the object side of the mirror and can be displayed or focused on a screen. Virtual images are formed ―behind‖ or ―inside‖ the mirror and cannot be formed on a screen. 14. (a) Concave: real, Do > f; virtual, Do < f. (b) Convex: always virtual. 15. The light ray that travels parallel to the optic axis of the concave mirror goes through the focal point of the mirror after reflection 16. The ray of light after the reflection becomes parallel to the principal axis. 17. This is done to concentrate light in one direction. A concave mirror is obtained by curving the back surfaces of the automobile headlights and the bulb is placed at the focus of this concave mirror. Mirrors used in cosmetics are concave mirrors which are magnifying. 18. convex mirror 19. At the center 20. (a) Convex: real, Do > f; virtual, Do < f. (b) Concave: always virtual. 21. Convex lenses concentrate the light energy to one spot on the paper so that the heat energy accumulates on that one small spot of paper. As the heat increases, combustion will occur when the spot becomes too hot and the paper will burn. 22. The rods are more sensitive than the cones and are responsible for light and dark ―twilight‖ vision; the cones are responsible for color vision. 23. Concave lenses are used to correct nearsightedness and convex lenses are used to correct farsightedness. 24. Sunglasses and LCDs (for example). 25. No, sound waves are longitudinal and cannot be polarized. 26. A virtually invisible filter can be built into lenses to eliminate the amount of reflecting light that enters the eye. Polarized glasses not only reduce glare, but they also make images appear sharper and clearer, increasing visual clarity and comfort. 27. If the axes of Polaroid filters are perpendicular to each other then they will act as window shade as the light cannot pass through the assembly.
28. Use long wavelength of light compared to the size of an opening or object. Also, use a diffraction grating. 29. For diffraction, the size of the edge or opening should be in the range of wavelength of the incident wave. Audible sound waves have wavelengths on the order of centimeters to meters. The wavelengths of sound are larger than or about the same size, as objects and diffraction readily occurs for sound. 30. AM radio waves because the longer the wavelength compared to the size of the opening or object, the greater the diffraction (bending). 31. a) When two pulses are in phase they form constructive interference. b) When two pulses are out of phase they form destructive interference. 32. Interference.
ANSWERS TO VISUAL CONNECTION (a) object; (b) convex (converging); (c) real, (d) inverted, (e) reduced
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. A spherical convex mirror. 2. Reflections from different surfaces of glass. 3. The convex mirror produces a reduced image, and the smaller image may be interpreted as being more distant than it actually is. 4. The one on the left. 5. The fish would see the 360° above-water panorama in a circular cone defined by the critical angle. 6. Using polarizing sunglasses to see if you can darken the sunglasses with cross-polarization. 7. Thin-film interference; the different wavelengths of light determine the color.
ANSWERS TO EXERCISES 1. θi = θr = 40° 2. θr = 90° – 30° = 60° 3. Refer Fig. 7.4. Applying the law of reflection, it can be shown that the total image
seen in a plane mirror is only half the person’s height. Also, the distance from the mirror is not a factor. 4. 76 in./2 = 38 in., and 62 in./2 = 31 in. So, Δ = 7 in.
5. cm = c/n = (3.00 × 108 m/s)/1.45 = 2.07 × 108 m/s 6. n = c/cm = (3.00 × 108 m/s)/(1.40 × 108 m/s) = 2.14 7. cwater/c = nvacuum/nwater; 1/1.33 = 0.75; Speed of light in water is 75% of the speed of light in vacuum. 8. cm /c = 0.413 = 1/n, and n = 1/0.413 = 2.42, diamond. (See Table 7.1.) 9. Sketch ray diagram. Real, inverted, same size. 10. Sketch ray diagrams. Reduced image becomes larger as objects move toward the mirror approaching the center of curvature, with M = 1 at that point. Continues to become larger as object approaches focal point, and image becomes virtual inside focal point. 11. f = R/2; f = 20 cm; R = 40 cm; Position is behind the center of curvature; real, inverted and magnified. 12. Sketch ray diagram. 5.5 cm, virtual, upright, and smaller. 13. Sketch ray diagram. Real, inverted, same size. 14. Sketch ray diagrams. Reduced image becomes larger as object moves toward lens approaching the 2f position, with M = 1 at that point. Then M > 1 as object approaches the focal point, and image becomes virtual inside the focal point. 15. Sketch ray diagram. f = R/2; f = 20 cm; R = 40 cm; Position is between the focus and the center of curvature; real, inverted, reduced. 16. Sketch ray diagrams. Di = 2f. 17. Sketch ray diagram. f = R/2; f = 20 cm; R = 40 cm; Position is in front of the object; virtual, upright, reduced. 18. Sketch a ray diagram. f = R/2; f = 20 cm; R = 40 cm; No image formed; rays parallel.
Chapter 8
ELECTRICITY AND MAGNETISM This is an important chapter because students should have a basic understanding of electric circuits because electricity plays such an important role in our lives. Also, knowledge of the fundamental concepts of electricity and magnetism is necessary in the study of atomic physics. Electric and magnetic laws and fields should be stressed so as to provide an understanding of some of the underlying concepts in the study of modern physics and chemistry. The discussion of electric fields has been expanded in the Thirteenth Edition, so as to make electromagnetism and waves better understood. This chapter is somewhat long, but the study of electric and magnetic phenomena is well worth the effort.
DEMONSTRATIONS Lecture demonstrations with an electroscope that can be projected on a large screen are very good and create interest. Demonstrate electromagnetic induction. (Commercial apparatuses are available.) Demonstrate the electric generator and electric motor. (May be constructed or commercially available.) Demonstrate the compass and dip needle. Pass around a cheap compass and bar magnets.
ANSWERS TO MATCHING QUESTIONS a. 13
b. 20 c. 25
d. 11
e. 6 f. 16
m. 23
n. 2
p. 5
q. 22
o. 18
g. 8
h. 24
i. 19
j. 14
k. 7
l. 17
r. 4 s. 12
t. 21
u. 3
v. 9
w. 15
x. 10
y. 1
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a
2. c
3. a
4. b 5. a
6. b 7. a 8. b
9. d
10. d
11. c
12. b
13. c 14. b
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. negatively
2. semiconductors
increase 7. decrease 8. current
3. amp-volt or (amp)2ohm 9. smallest
10. Curie
4. charge
11. south
5. ohm
6.
12. secondary
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Protons and neutrons. Protons and electrons. 2. Equal and opposite, Newton’s third law. 3. Static electricity 4. When a charged rubber comb (positive charge) is exposed to the bits of paper, the electric field polarizes the positive and negative charge of the paper, and hence the bits of paper attract to the charged rubber comb. Similarly opposite charges are induced on balloon sticks and ceiling wall. 5. Electrical potential energy arises due to work done against an electric force. Voltage is electrical energy (or work) per unit charge, V = W/q. 6. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. Ohm’s law is the relationship expressed as V = IR or voltage = current × resistance.
7. P = V 2/R; Small resistance, large joule heat, and vice versa. 8. Electric field takes place with the speed of light and the free electrons everywhere in the wire begin drifting almost at once. So, though the electron drift velocity is very small, an auto battery influences the starter as soon as you turn the ignition switch. 9. Direct current has a voltage with constant polarity and flows in only one direction. Alternating current results from a changing polarity, and the current alternates in direction. 10. So that all will have the same voltage and independent paths so they can be operated independently. 11. (a) Opens a circuit by the melting of a metal strip when current is too large. (b) Opens a circuit by magnetic or thermal means when current is too large. (c) Dedicated grounding wire to prevent object being at high voltage. (d) Use of ground wire as a grounding wire. 12. Cars are safe from lightning because of the metal cage surrounding the people inside the vehicle. This may sound counter-intuitive because metal is a good conductor of electricity, but the metal cage of a car directs the lightning charge around the vehicle occupants and safely into the ground. 13. Head-to-tail is series combination; all heads and all tails are connected in a parallel combination. 14. The law of charges states that like charges repel and unlike charges attract. The law of poles state that like poles repel and unlike poles attract. Hence, both are the same. 15. When iron filings are exposed to a magnetic field, the filings are themselves temporarily magnetized along their long axis. Now being magnets, they tend to align in the bar magnet's field. 16. (a) One that is easily magnetized. Iron, nickel, and cobalt. (b) The ferromagnetic material becomes an induced magnet. Above the Curie temperature, ferromagnetic materials lose their magnetism. 17. A current gives rise to a magnetic field in an iron core. 18. (a) The magnetic field of a bar magnet with the bar’s north magnetic pole near the Earth’s geographic south pole. (b) Declination is the horizontal angular distance between the Earth’s magnetic field lines or compass direction (magnetic north) and true north. Maps are based on true north, so it is necessary to know the declination to navigate properly. 19. Based on torque on a current-carrying wire.
20. (a) Nothing. (b) Experiences a force at a right angle to motion, the direction given by the right hand rule. 21. Electromagnetic induction. Used to step-up and step-down voltage and current so as to reduce loss in electrical power transmission. 2
22. Much of the power would be lost through I R losses. 23. Circuit not complete. 24. Human body has a large resistance, so for small voltages, the current is also small. If all you are concerned about is not stopping your heart, then it becomes much more complicated, because your heart and nervous system are very complicated and even small amounts of current can be dangerous. That is why high voltage is written on boards.
ANSWERS TO VISUAL CONNECTION a. 1 A, b. 2 A, c. 2 A, d. 5 A. Combined resistance: 2.4 or 36/15 Ω
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Electrostatic charge, moisten. 2. (a) Midway for electron or proton, equal and opposite forces. (b) No. 3. (a) No (b) No 4. With one hand in the pocket, you are not likely to grasp a high voltage with both hands and provide a path across your chest. 5. Parallel arrangement of one 10 Ω resistor on one side and two 10 Ω resistors on the other. R1 × R2/R1 + R2 = 10 Ω × 20 Ω/10 Ω + 20 Ω = 200 Ω2/30 Ω = 6.7Ω 6. 1/Rp = 1/R1 + 1/R2 = (R2 + R1)/R1R2 and Rp = R1R2/(R1 + R2) 7. Two magnets would be obtained. Creating magnetic monopoles in this manner is not possible. Magnetism is the result of domain alignment. 8. Moving electric charges give rise to magnetic fields (speaker and microphone). A magnetic field may deflect a moving electric charge (electric motor, generator, and transformer).
ANSWERS TO EXERCISES 1. q = ne = (106)(1.6 × 10–19 C) = - 1.6 × 10–13 C 2. n = q/e = 1.00 C/1.60 × 10–19 C/electron = 6.25 × 1018 electrons 3. F = kq1q2/ r = (9.0 109 N · m2/C2)(-1.0 C)(2.0 C)/(3.0 m)2 = -2 10–9 N, mutually attracted 2
4. F = kq1q2/ r2 = k (1.6 × l0–19 C)2/(5.3 × 10–11 m)2 = -8.2 × 10–8 N Gravitational force = F = Gmemp/r2 = (6.67 × 10–11Nm2/kg2)(9.11 × 10–31 kg)(1.67 × 10–27 kg)/(5.3 × 10–11 m)2 = 3.6 × 10-47 5. I = q/t = ne/t = (4.8 1018)(1.6 × 10–19 C)/0.25 s = 3.1 A 6. q = It = (1.50A)(6.5 s) = 9.75 C n = 9.75 C/e = 9.75 C/(1.6 × 10–19) C = 6.1 × 1019 electrons 7. 40 J 8. V = W/q = 40 J/0.25 C = 160 V 9. I = V/ R = 120 V/50 = 2.4 A 10. V = IR = (0.50 A)(30 ) = 15 V 11. (a) I = P/V = 0.50 W/3.0 V = 0.17 A (b) R = V/I = 3.0 V/0.17 A= 18 12. (a) P = I × V = 0.25 A × 12 V = 3 W (b) R = V/I = 12 V/0.25 A = 48 13. W = Pt = (1.5 kW)(0.50 h/day)(30 days) = 22.5 kWh × $0.18/kWh = $4.05 14. E = (1.0 kW)(3.0 h/day)(30 day) = 90 kWh × $0.18/kWh = $16.20 15. P = V / R = (12 V)2/24 = 6 W (6 J/s) 2
16. P = IV and V = IR or I = V/R, then P = (V/R)V = V 2/R 17. (a) R = V/I = 110 V/10 A = 11 (b) P = IV = (10 A)(110 V) = 1100 W 18. (a) I = P/V = 9 W/120 V = 0.075 A
(b) R = V/I = 120 V/0.075 A = 1600
(c) P = E/t = 9 J/s, or 9 J each second 19. (a) I = V/RS = V/(R1 + R2) = 12 V/60 = 0.20 A (b) P = V 2/RS = (12 V)2/60 = 2.4 W 20. RP = R1R2/(R1 + R2) = (25 )(35 )/60 = 14.6 a) I = V/Rp = 12 V/14.6 = 0.82 A b) Power = P = VI = 9.84 W 21. (a) I1 = I2 = V/Rtotal = 24 V/30 Ω = 0.8 A (b) P1 = I2 × R1 = 0.82 × 10 = 6.4 W; P2 = I2 × R2 = 0.82 × 20 = 12.8 W 22. (a) I1 = V/R = 24 V/10 Ω = 2.4 A I2 = 24 V/20 Ω = 1.2 A (b) P1 = V × I1 = 24 V × 2.4 A = 57.6 W P2 = 24 V × 1.2 A = 28.8 W 23. (a) I = V/R = 12 V/120 Ω = 0.1 A (b) V = I × R
V1 = 0.1 A × 20 Ω = 2 V V2 = 0.1 A × 40 Ω = 4 V V3 = 0.1 A × 60 Ω = 6 V 24. (a) I1 = V/R = 12 V/20 Ω = 0.6 A I2 = 12 V/40 Ω = 0.3 A I3 = 12 V/60 Ω = 0.2 A (b) Voltage in parallel arrangement is constant. V1 = V2 = V3 = Vtotal 25. (a) Step-down transformer (b) Ratio of 3 to 1, V2 = 100 V/3 = 33.33 V; V1/V2 = I2/I1 I2 = (100 V/33.33 V) × 0.25 A = 3 × 0.25 A = 0.75 A 26. (a) V2 = 12 V × 3 = 36 V (b) V1/V2 = I2/I1 I2 = (12 V/36 V) × 2.0 A = 0.67 A 27. N2 = (V2 / V1 ) N1 = (220 V/4400 V)(1000) = 50 turns 28. N2 = (40000 V/240000 V)(900) = 150 turns
Chapter 9
ATOMIC PHYSICS The subject matter of atomic physics includes some of the revolutionary concepts and developments of twentieth-century physics. First, for background, we review the models of the atom up to 1911. The dual nature of light is then introduced, accompanied by a Highlight on Einstein and his work. The next section discusses the Bohr theory of the hydrogen atom and includes a Highlight on fluorescence and phosphorescence. Quantum applications—microwave ovens, X-rays, lasers—are discussed. The Heisenberg uncertainty principle and the concept of the wave nature of matter are then explained, followed by a Highlight on electron microscopes. The chapter ends with a treatment of the quantum model of the atom. We have kept the discussions of these difficult topics as simple and straightforward as possible. Sections 9.1, 9.2, and 9.3 are basic material. Instructors should use their judgment as to what additional parts of the chapter to cover and emphasize.
DEMONSTRATIONS Many of the topics in this chapter are difficult to demonstrate. Fluorescence and phosphorescence may be demonstrated with an ultraviolet lamp and appropriate samples from supply houses. Students especially seem to like finding that fluorescent chalk has been used to write their homework assignment on the board. A common laser may be demonstrated, but use appropriate safety precautions. A food item may be heated in a microwave oven to demonstrate that the center remains cold if sufficient time is not allowed for conduction of heat to the center.
ANSWERS TO MATCHING QUESTIONS a. 9 b. 6 c. 1 d. 15
e. 11 f. 14
g. 5
k. 3
o. 17
q. 7
l. 4
m. 2
n. 12
p. 10
h. 16
i. 8
j. 13
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.a
2. c
3. a
4. d 5. c
6. a 7. b
8. d
9. d 10. a
11. b
12. b
13. b
14. D
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. electron 7. lower
2. Rutherford 8. water
3. greater 4. photon
9. unknown
10. uncertainty
5. emission 6. increases 11. wave
12. Probability
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Classical mechanics: macrocosm, motion of large objects. Quantum mechanics: microcosm, behavior of atoms and electrons on the atomic level. 2. Thomson found that electrons were deflected by electrical and magnetic fields in such a way that their electric charge had to be negative. 3. Thomson’s ―plum pudding‖ model pictured the atom as consisting of electrons randomly positioned like raisins in an otherwise homogeneous mass of positively charged ―pudding.‖ Thomson’s model was abandoned when Rutherford discovered that each atom had a tiny core, or nucleus, in which 99.9% of the mass and all the positive charge were concentrated and around which the electrons circulated. 4. The wavelength becomes shorter as the temperature increases. 5. It is because radio waves have very small energy and very big wavelength (more than wavelength of gamma photons, which is described by the corpuscular theory of light). Thus,
the behavior of radio waves is described by the wave theory of light, not the corpuscular theory of light. 6. Polarization of light is the phenomenon which can be explained by wave nature of light and Photoelectric effect can be explained by particle nature of light. 7. A photon is a quantum or ―packet‖ of electromagnetic energy (light), and its energy depends on the frequency, E = hf. 8. Frequency and wavelength are inversely proportional. 9. A photon of red light has less energy, lower frequency, and longer wavelength than a photon of violet light. 10. Albert Einstein. The theory of relativity. 11. Only light above a certain frequency causes electrons to be ejected, and thus the photon must have a certain minimum energy to cause ejection; E = hf, the higher the frequency, the greater the photon energy. 12. No. Excited neon atoms emit photons of specific, discrete energies when the atoms make transition from high energy states to lower energy states. These creates a line spectrum, not a continuous spectrum. 13. The emission spectrum is represented by collection of separate-colored lines, with dark spaces between them while the absorption spectrum is represented by collection of dark lines, with colored spaces between them. For same atom, number of emission lines and absorption lines are same. 14. The principal quantum number determines the distance of an electron from the nucleus and hence the energy level of the electron. 15. If an atom continuously radiates energy, electrons will collapse in the nucleus. Hence, Bohr postulated that an orbiting electron does not radiate energy when in an allowed, discrete orbit but does so only when it makes a quantum jump, or transition, from one allowed orbit to another. 16. The ground state for an electron is the lowest energy state. Energy states above the ground state are called excited states. 17. Four visible lines—red, blue-green, and two violet—make up the line emission spectrum of hydrogen, as shown in Figures 9.9 (b) and 9.13. 18. The energy value is -13.60 eV for the ground state and it would require that much energy input to ionize a hydrogen atom.
19. Bohr explains the line spectrum (discrete) of hydrogen atom by assuming electron moved in circular orbit whose angular momentum is integral multiple of h/2π. Energy of the photon emitted is the energy difference between transition orbits. And hence we get discrete absorption and emission spectra. 20. n = 3 to n = 1 21. Six, 4 to 3, 3 to 2, 2 to 1, 4 to 2, 4 to 1, and 3 to 1. 22. The potato contains water, whereas the ceramic plate does not. 23. Light Amplification by Stimulated Emission of Radiation 24. Laser light is monochromatic, coherent, and has amplified intensity. 25. The intensity of laser light can cause eye damage. 26. X-rays are called braking rays because they are formed when high-speed electrons are stopped as they hit a metal plate. 27. It is impossible to know a particle’s exact position and velocity simultaneously. 28. Heisenberg’s uncertainty principle is applicable for Quantum mechanics (atomic level particles like electrons protons.) 29. A moving particle has a wave associated with it called a matter wave, or de Broglie wave. The associated wavelength is significant only for atoms and subatomic particles. 30. De Broglie realized that if you use the wavelength associated with the electron, and assume that an integral number of wavelengths must fit in the circumference of an orbit, you get the same quantized angular momentum that Bohr did. 31. Davisson and Germer showed that a beam of electrons undergoes diffraction when passed through a slit. 32. The electron microscope. 33. Erwin Schrödinger. 34. Quantum model.
ANSWERS TO VISUAL CONNECTION a. Dalton, billiard ball model; b. Thomson, plum pudding model; c. Rutherford, nuclear model; d. Bohr, planetary model; e. Schrodinger, electron cloud (or quantum) model
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Lyman series have higher energy levels than Balmer series, while Paschen series have lower energy levels. For Lyman series, transitions take place from higher energy level to ground state that is to level n = 1, while in Paschen series, transitions take place from higher energy level to level n = 3. Hence energy difference is higher for Lyman. The differences are a result of the wavelength and the amount of energy electrons need to make transitions.
2. A quantized cruise control might be digital and allow speeds to be set on 5 mph units, such as 50 or 55 mph, but not 51 mph. 3. As Highlight Figure 2 shows, individual atoms can be imaged and even moved around by use of a scanning tunneling microscope. 4. Mercury vapor or sodium vapor lights give off only certain wavelengths of light (those in their line emission spectra). Thus only certain colors are available to be absorbed or reflected from the variously colored cars. The reflected light by which we see the cars thus has a different overall composition (color) from reflected sunlight or reflected light from an incandescent bulb. 5. Heisenberg’s uncertainty principle is used in quantum mechanics which is applied for the atomic level particles not for macroscopic particle. A car is much too large to allow v (or x) to be of significant size.
ANSWERS TO EXERCISES 1. E = hf = (6.63 × 10–34 J · s)(5.00 × 1014 1/s) = 3.32 × 10–19 J 2. E = hf = (6.63 × 10–34 J · s)(5.45 × 1014 1/s) = 3.61 × 10–19 J 3. (a) E = hf, f = E/h = (6.63 × 10–19 J)/(6.63 × 10–34 J · s) = 1.0 × 1015 Hz (b) wavelength = c/f = 3.00 × 108 m/s/ 1015 = 300 × 10-9 m = 300 nm 4. (a) E = hf; f = (1.75 × 10-19 J)/(6.63 × 10-34 J · s) = 2.64 × 1014 Hz (b) Wavelength = (3.00 × 108 m/s)/(2.64 × 1014 1/s) = 1.136 × 10-6 m = 1136 nm 5. rn = (0.053 nm)(22) = (0.053 nm) × 4 = 0.212 nm 6. rn = (0.053 nm) × 42 = (0.053 nm)16 = 0.848 nm 7. En = -13.6/22 = -13.6/4 = -3.4 eV 8. En = -13.6/42 = -13.6 eV/16 = -0.85eV 9. E5 = (-13.6/(52)) = -0.544eV ΔE = -3.40 eV - (-0.544 eV) = 2.856 eV 10. Ephoton = Eni –Enf = –13.60 eV – (–0.85 eV) = –12.75 eV (The minus sign indicates photon absorption.) 11. λ = h = 6.63 × 10-34 J · s/(0.55kg × 45 m/s) = 2.68 × 10-35 m = 2.68 × 10-26 nm mv
12. Answers vary. (10 mi/h = 4.5 m/s, and convert mass, l lb = 0.45 kg.) = (6.63 × 10-34 J · s)/(55 kg. × 6.67 m/s) = 1.81 × 10-36 m = 1.81 × 10-27 nm (The value of the speed should not exceed 10 m/s.)
Chapter 10
NUCLEAR PHYSICS This chapter concludes the physics section of the textbook by discussing the atomic nucleus. Because of the great impact that nuclear energy has had—and will have—on society, this chapter is extremely important. The material in this chapter is so fundamental that we recommend it be covered in its entirety. Among the topics discussed are the names and symbols of elements, the structure and composition of nuclei, atomic mass, radioactive decay, half-life, nuclear reactions and reactors, fission, fusion, and the biological effects of radiation. The small amount of mathematics included is kept simple. Applications of radioactivity are stressed, and three Highlights emphasize the historical aspects of the discovery of radioactivity, the building of the first nuclear bombs, and the problem of nuclear waste disposal. Also in the Thirteenth Edition, a section on elementary particles has been added for those that wish to have an introduction to this topic. The student will need to refer frequently to the periodic table on the inside front cover of the textbook. We recommend that a wall chart of the periodic table be on display at all times in the lecture classroom, including during exams.
DEMONSTRATIONS Students generally find a demonstration of a Geiger counter or other radiation detector to be interesting and helpful. A comparison of the penetrating power of alpha particles, beta particles, and gamma rays is instructive (a solid source set #32852 can be obtained from CENCO or another science supply company). Another demonstration involves charging an electroscope and illustrating its quick discharge with a radioactive source. Of course, use appropriate safety precautions when dealing with radiation. An old X-ray of a human or animal obtained from a doctor or veterinarian may be passed around the class. A display Chart of the Nuclides is available at moderate cost from General Electric Company, 175 Curtner Ave., Mail Code 684, San Jose, CA 95125.
ANSWERS TO MATCHING QUESTIONS a. 5 b. 21 c. 13
d. 9
e. 26
f. 17
g. 25
h. 10
i. 24
j. 6 k. 22
l. 8 m. 2
o.19
r. 1
s. 14
t. 4
u. 11
v. 15
w. 7
x. 18
z. 23
p. 3 q. 16
y. 12
n. 20
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.a
2. a
3. c
4. b 5. d 6. c
7. b
8. b 9. a
10. c
11. d 12. c 13. b 14. b 15. D
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. potassium
2. neutron number
(electron) 7. three 8. two less
3. nucleons 9. mass number
4. isotopes 10. critical
5. half-life
6. beta particle
11. deuteron (or deuterium)
12. 80 13. quarks
ANSWERS TO SHORT-ANSWER QUESTIONS 1. (a) Nitrogen.
(b) Helium.
2. (a) C (b) Cl
(c) Pb
(c) Iron.
3. Because some chemical symbols are derived from their Latin names, for example, silver, ag (argentum) or sodium, Na (natrium). 4. Nucleons. 5. More than 99.97% 6. Atomic radius of hydrogen is 53 pm and atomic radius of Uranium 230 pm. They are not same. 7. Z (atomic number) is the number of protons in the nucleus of each atom of that element. N (neutron number) is the number of neutrons in a nucleus. A (mass number) is the number of protons plus neutrons in the nucleus: the total number of nucleons. 8. Atomic mass number is number of nucleons that is number of protons and neutrons. Hence number of neutrons is N = A - Z. 9. 1H is protium (or just hydrogen); 2H is deuterium (D) 3H is tritium (T) 1
1
1
10. 1H is protium (or just hydrogen); 2H is deuterium (D) 3H is tritium (T) 1
1
1
12
11. All atomic masses are based on the C atom, which is assigned a relative atomic mass of exactly 12 u. 12. All the protons are closely packed in nucleus so there will be huge force due to repulsion of positively charged protons, still nucleons are bound in nucleus by strong nuclear forces. As the nucleons are bound within nucleus hence nuclear forces are stronger than electric forces. 13. The strong nuclear force (or just nuclear force). It drops to zero at distances greater than about 10–14 m. 14. Transformation into different nucleus.
15. Proton number increases by one. 16. Less by mass of 2 protons and 2 neutrons hence decrease in the mass number by 4. 17. (a) Gamma.
(b) Beta.
(c) Alpha.
(d) Gamma.
(e) Alpha.
18. One-sixteenth. 19. Activity becomes too weak and unable to count. 20. a + A B + b 21. Mass numbers and atomic numbers are conserved. 22. Smoke detector has americium 241. As it decays, the alpha particles that are emitted ionize the air inside part of the detector. The ions form a small current that allows a battery (or house voltage) to power a closed circuit. If smoke enters the detector, then the ions become attached to the smoke particles and slow down, causing the current in the circuit to decrease and an alarm to sound. 23. Neutron. Source of energy is the binding energy of the nucleus. 24. The minimum amount of fissionable material necessary to sustain a chain reaction is called the critical mass. A subcritical mass would be less than that, and no chain reaction would be possible. A supercritical mass would be more than that, and an expanding chain reaction would be possible. 25. 0.7%, this must be enriched to about 3% for use in a U.S. nuclear reactor, and 90% or more for use in nuclear weapons. 26. No. Nucleus is ―split‖ into two nuclei. 27. Control rods adjust the number of neutrons available to cause fission. Moderators slow down fast neutrons from fission so that they can more effectively cause other fissions. 28. For a nuclear explosion to occur, the fissionable material must be of much greater purity than the fuel in nuclear reactors. 29. Meltdown is a severe nuclear reactor accident that results in core damage from overheating. A meltdown was once referred to as the China Syndrome because the ―melt‖ is heading downward through the center of the Earth ―toward China.‖ Of course, this description is inaccurate because China is not on the other side of the Earth from the United States; both countries are in the Northern Hemisphere. 30. Plutonium-239 is made from uranium-238. 31. Hydrogen 32. D is a deuterium nucleus or deuteron; T is a tritium nucleus or triton. 33. A plasma is a very hot gas of electrons and protons or other nuclei. In magnetic confinement, electric fields are used to form the plasma, and electric and magnetic fields are
used to confine its charged particles. The second process is inertial confinement using lasers. Inertial confinement, a technique in which simultaneous high-energy laser pulses from all sides cause a fuel pellet containing deuterium and tritium to implode, resulting in compression and high temperatures. If the pellet stays intact for a sufficient time, then fusion is initiated. 34. Fusion has the advantages of low cost and abundance of fuel, fewer nuclear waste disposal problems, and the impossibility of a runaway accident. Its disadvantages are that it has not yet proved practical, and that fusion plants will be more costly to build and operate than fission plants. 35. Albert Einstein. E is energy, m is mass, and c is the speed of light. 36. High temperature needed to start fusion reaction. 37. The decrease in mass in a nuclear reaction is called the mass defect. If a decrease in mass has resulted, then the reaction is exoergic (releases energy) 38. Binding energy per nucleons for Iron is highest hence Iron is most stable. 39. Radiation that can ionize atoms or molecules to form ions. Such radiation can kill living cells. 40. In order of needing more shielding: alpha particles, beta particles, and gamma rays, X-rays, and neutrons. Clothing and our skin are sufficient protection against alpha particles unless the particles are ingested or inhaled. Beta particles can burn the skin and penetrate into body tissue, but not far enough to affect internal organs. Heavy clothing can shield against beta particles. Gamma rays, X-rays, and neutrons are more difficult to stop. 41. Exchange particles are responsible for force interactions. 42. Three quarks (two with +2/3 and one with -1/3 electronic charge).
ANSWERS TO VISUAL CONNECTION (a) 73Li4, (b) 126C6, (c) 16 8O 8, (d) 23 11Na 12
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Charged particles would be deflected in different directions. (See Chapter 8.5.) 2. More intensity, greater concentration of carbon-14. Samples would be older. 3. When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced.
4. Answers might include not smoking, fewer high-altitude airplane flights, living closer to sea level, watching TV from a greater distance, fewer X-ray diagnoses, and radon checks of your crawl space or basement. 5. Exchange or virtual particles interact with particles to produce the effects of attraction or repulsion. They do this by shuttling back and forth between the particles, carrying small packets of energy. For repulsion, the effect is much the same as two basketball players passing a heavy ball between them.
ANSWERS TO EXERCISES 1.
Fill in the nine gaps in this table. Element Symbol
Ga
P
Cl
Protons
31
15
17
Neutrons
39
16
20
Electrons
31
15
17
Mass number
70
31
37
2.
Fill in the nine gaps in this table. Element Symbol
B
F
Ar
Protons
5
9
18
Neutrons
6
10
22
Electrons
5
9
18
Mass number
11
19
40
3. For oxygen there are three isotopes, A = 16, Z = 8, N = 8, A = 17, Z = 8, N = 9, A = 18, Z = 8, N = 10. 16𝑂 , 17𝑂 , 18𝑂 8 8 8 9 8 10
4. Z = atomic numbers, N= number of neutrons.
Phosphorous-31 (Z = 15, N = 16), Bromine-80 (Z = 35, N = 45), Iron-56 (Z = 26, N = 30), Titanium-48 (Z = 22, N = 27), Silver-108 (Z = 47, N = 61), Platinum-179 (Z = 78, N = 101). Number of neutrons are greater than atomic numbers. 5. (a) 147𝑁, beta (b) 42𝐻𝑒, alpha
(c) , gamma decay
6. a) , gamma decay
b) 228 88𝑅𝑎, alpha decay c) −10𝑒, Beta decay 7. (a) 226𝑅𝑎 → 222𝑅𝑛 + 4𝐻𝑒 88
86
60𝐶𝑜 → (b) 27
2
60𝑁𝑖 + 0 28 −1𝑒
8. (a) 225𝐴𝑐 → 221𝐹𝑟 + 4𝐻𝑒 89
(b) 221𝐹𝑟 87
87
2
→ 221𝑅𝑎 + 87
0𝑒
−1
9. (a) 222𝑅𝑛 → 218𝑃̂𝑜 + 4𝐻𝑒 (Alpha particle) 86
84
2
(b) Granddaughter in case of alpha decay is Lead Pb-214 218𝑃̂𝑜 → 214𝑃̂𝑏 + 4𝐻𝑒 84
82
2
Granddaughter in case of beta decay is Astatine At-218 218𝑃̂𝑜 → 218𝑃̂𝑏 + 0𝑒 84
85
−1
10. (a) & (b) 1) Alpha decay, 233 91𝑃̂𝑎 2) Beta decay, 233 92𝑈 3) Alpha decay, 229 90𝑇ℎ 4) Alpha decay, 225 88𝑅𝑎 5) Beta decay, 225 89𝐴𝑐 6) Alpha decay, 221 87𝐹𝑟 7) Alpha decay, 217 85𝐴𝑡 8) Alpha decay,21383𝐵𝑖 Now the double decay mode 9) Beta decay,213𝑃̂𝑜 followed by Alpha decay, 209𝑃̂𝑏 84
82
or Alpha decay, 209𝑇𝑖 followed by Beta decay, 209𝑃̂𝑏 81
10) Beta decay, 209𝐵𝑖 83
11. (a) 249𝐶𝑓 (Z > 83) 98
76𝐴𝑠 (odd-odd)
(b) 33
(c) 158𝑂 (fewer n than p) 12. (a) 179𝐹 (fewer n than p)
82
(b) 226 88𝑅𝑎 (Z > 83) (c) 209𝐹 (odd-odd) 13.
30 h = 5 half-lives 6.0 h/half-life
14. 16 cpm 8 cpm 4 cpm 2 cpm Three arrows are shown, so three half-lives are needed. 3 × 5730 years = 17190 years. The organism died 17190 years ago. 15. 32.4 d/8.1 d/half-life = 4 half-lives 1 1/2 1/4 1/8 1/16 75 h
= 5 half-lives
16. 15 h/half-life
480 cpm 240 cpm 120 cpm 60 cpm 30 cpm 15 cpm will be the activity.
17. 1 1/2 1/4 1/8 1/16 Four arrows mean four half-lives have elapsed. (4 half-lives)(12.3 y/half-life) = 49.2 y 18. 2000 cpm 1000 cpm 500 cpm 250 cpm 125 cpm Four arrows mean four half-lives. 28 min/4 half-lives = 7.0 min/half-life 19. 22 days 20. 8 days 21. 66 days 22. 30 days 24𝑀𝑔 23. (a) 12 28𝐴𝐼 (b) 13
(c) 11𝐻 (d) 254 102𝑁𝑜 24. (a) 11𝐻 (b) 10𝑛 (c) 2𝐷 Or 2𝐻 (either use D or H for symbol of Deuterium) 1
93𝑆𝑟 (d) 38
25. 106 43𝑇𝑐 26. 3 10𝑛
1
27. 3 × 4.00260 u = 12.00780 u on left – 12.00000 u on right = 0.00780 u mass loss (0.00780 u)(931 MeV/u) = 7.26 MeV or energy produced 28. 2.0140 u + 2.0140 u u = 4.0280 u showing on the left. 3.01601 u + 1.0078 u = 4.0239 u showing on the right. The difference is 0.0041 u more on the left, so mass is lost and (0.0041 u)(931 MeV/u) = 3.8 MeV of energy produced.
Chapter 11
THE CHEMICAL ELEMENTS Chapter 11 begins the four-chapter chemistry section by discussing the chemical classification of matter, the discovery and occurrences of the elements, the periodic table, and the basics of compound nomenclature. A chapter Highlight discusses Berzelius and How New Elements Are Named. Conceptual questions and answers are presented on the uniqueness of compounds as well as on why a ―periodic table of compounds‖ is impractical. Much of the material discussed in this chapter is necessary for dealing successfully with the next three chapters. The student already should have learned the names and symbols of the 46 elements in Table 10.2 of the textbook and will now need to learn the names and formulas of the eight polyatomic ions listed in Table 11.6 and the eleven common compounds shown in Table 11.3. Probably the most efficient and effective way of doing this is to make and use flashcards.
DEMONSTRATIONS When discussing the differences between compounds and mixtures, it is useful to pass around two Erlenmeyer flasks, one containing the compound zinc sulfide and the other containing a heterogeneous mixture of powdered sulfur and mossy zinc. Solubility can easily be demonstrated with a beaker, salt, and water. Try changing the temperature of the water to see how much more salt can be dissolved. Elemental abundance in Earth’s crust can be demonstrated by looking at the chemical formulas for common igneous, sedimentary, and metamorphic rocks in Chapter 22, and specifically in Table 22.8.
ANSWERS TO MATCHING QUESTIONS a. 19 b. 5 c. 6 d. 15 e. 20 f. 17 g. 9 h. 11 i. 24 j. 22 k. 4 l. 8 m. 16 n. 3 o. 23 p. 25 q. 18 r. 21 s. 13 t. 14 u. 2 v. 10 w. 7 x. 1 y. 12
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. c 3. a 4. b 5. d 6. c 7. c 8. b 9. b 10. d 11. a 12. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Organic chemistry 2. solvent 3. element 4. technetium (Tc) 5. atoms 6. aluminum 7. diamond 8. Mendeleev 9. four 10. increases 11. acetate 12. sodium hydroxide
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Homogeneous Mixture. 2. Physical, analytical, organic, inorganic, and biochemistry. 3. Illustrations b and d show different atoms and/or molecules and thus represent mixtures. Illustration e shows identical molecules composed of the same two elements and so represents a compound. Illustrations a and c show identical atoms and identical diatomic molecules of atoms of the same element, respectively, and thus represent elements. Illustration c shows diatomic molecules of an element, whereas e shows diatomic molecules of a compound. 4. A homogeneous mixture, featuring a type of solid solution, results when copper and zinc are mixed to form brass. 5. Mixtures can be separated into, and prepared from, pure substances by physical processes. 6. In the example of common salt, common salt (compound) is made up of Sodium and Chlorine. Sodium has metallic properties but salt does not. Chlorine is gas but salt is solid at room temperature. Melting point and boiling point of salt and sodium and chlorine changes. Density of sodium and salt is changed.
7. Compounds can be prepared from, and decomposed into, elements by chemical processes only. 8. An appropriate set of examples would be: bronze, salt water, air. 9. Carbonated beverages are bottled and capped tightly to maintain storage under high pressure. Once a carbonated beverage is opened, the pressure inside the bottle is reduced to normal atmospheric pressure over time, and the CO2 starts escaping from the liquid, eventually resulting in the flat taste. 10. Any six of these: gold, silver, lead, copper, tin, iron, carbon, sulfur, antimony, arsenic, bismuth, or mercury. Most of these are metals. The first gases discovered were hydrogen, oxygen, nitrogen, and chlorine. 11. The first synthetic element, technetium (Tc) was produced by nuclear bombardment of molybdenum with deuterons. 12. At present, about 118 elements are known, 98 of which occur naturally on Earth. 13. (a) Oxygen and silicon. (b) Iron and nickel. (c) Oxygen and carbon. (d) Nitrogen and oxygen. 14. The three noted allotropes of carbon are diamond, graphite, and fullerenes. In diamond, each carbon atom is in the middle of a geometric structure called a regular tetrahedron, where the four bonds of each carbon atom point toward the four corners of the tetrahedron, resulting in a three-dimensional network that helps make diamond the hardest substance known. Graphite consists of carbon atoms that are bonded in a network of flat hexagons, where each carbon atom is bonded to three other carbon atoms that lie in the same plane, forming sheets of flat, interlocking hexagons. Fullerenes consist of soccer-ball -like arrangements of interlocking hexagons and pentagons formed by carbon atoms. 15. Dmitri Mendeleev developed the periodic table in 1869. 16. Atomic number is now used. The periodic trends in the table are based on atomic number, which correlates more exactly with electron configuration than does atomic mass. 17. Periodic characteristics of elements include: 1) atomic number, 2) electron configuration, 3) valence electrons, 4) atomic sizes, and 5) ionization energy. One can also include metallic character in this list.
18. (a) Periods. (b) Groups. 19. The valence electrons are the ones that form chemical bonds. 20. (a) Metals usually have 1 to 3 valence electrons; nonmetals usually have 4 to 8. (b) Metals are good conductors of heat and electricity; nonmetals are not. (c) Metals are solid at room temperature (except for Hg); many nonmetals are gaseous or liquid. 21. Metallic character increases down a group and decreases from left to right across a period. (a) Be, Mg, Ca, Sr, Ba, Ra. (b) Ne, F, O, N, C 22. Semimetals are those elements that display both metallic and nonmetallic characteristics (i.e., they only conduct electricity under certain conditions). Examples are boron, silicon, and germanium; they are typically used in the semiconductor industry for electrical circuits. 23. Mercury and bromine are liquid. Hydrogen, oxygen, nitrogen, fluorine, and chlorine, (plus six noble gases) are gases. 24. (a) Ba, Sr, Ca, Mg, and Be, and (b) C, N, O, F and Ne. Ionization energy increases from left to right across a period. 25. Methane, nitrous oxide, nitric oxide, and sulfuric acid, respectively. 26. (a) An atom is the smallest particle of an element; a molecule is an electrically neutral particle composed of two or more atoms chemically bound together. (b) An atom is electrically neutral; an ion has an electrical charge. (c) A molecule is an electrically neutral particle composed of two or more atoms; a polyatomic ion is an electrically charged particle composed of two or more atoms. 27. The Greek prefixes designate the number of atoms of the element that occur in the molecule. 28. Because they have the same number of valence electrons. 29. Noble gases have interesting properties in that they are not very reactive. That is, they are monotonic, and it does not bond easily with other elements. Their valence electron shells are completely full, and thus are stable. They have very low melting and boiling points. Noble gases are used in lighting in two ways: they are used in incandescent bulbs to insulate the hot tungsten filament from other reactive gases, and they can be used in ―neon‖ signs giving an emission spectrum.
30. Fluorine and chlorine are gases, bromine is liquid, and iodine is solid. Fluorine is the most reactive element. 31. Sodium iodide is added to prevent thyroid problems due to iodine deficiency. 32. K2CO3, Na2CO3, NaOH, NaHCO3, and Mg(OH)2, respectively. 33. Rare Earth metals are the 15 lanthanoids, plus yttrium and scandium. They are used in electronic devices like cell phones. The military uses them in night-vision goggles, GPS and guidance devices as well as for communications. 34. Hydrogen is considered a nonmetal, not a metal. However, hydrogen appears in Group 1A because it usually reacts similar to other alkali metals.
ANSWERS TO VISUAL CONNECTION a. pure element, b. mixture of an element and a compound, c. mixture of an element and a compound, d. pure compound, e. pure element, f. mixture of two elements
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Element 113 (Nihonium) would belong to Group 13 and Period 7 – it would resemble the Representative (Main group), boron group or post-transition metals. Element 115 (Moscovium) would belong to Group 15 and Period 7 – it would resemble the Representative (Main group), pnictogens or post-transition metals. Element 117 (Tennessine) would belong to Group 17 and Period 7 – it would resemble the halogens. Element 118 (Oganesson) would belong to Group 18 and Period 7 – it would resemble the noble gases. 2. The letter J. 3. Although homogenized milk looks uniform to the unaided eye, a microscope shows fat globules dispersed throughout the water. Therefore, it is not a true solution, because it is not mixed on the atomic or molecular level. 4. Graphite makes a good lubricant because of its slipperiness, so the key should turn easily. 5. Being ―snobs‖ refers to the noble gases’ lack of chemical reactivity, or ―interaction,‖ with other elements.
6. Fluorine is the most reactive element of all and thus is so dangerous that it should be handled only by experts. 7. For this hypothetical case, Ca-20 would have 20 protons and 10 electrons. No, neutral hydrogen needs 1 proton and 1 electron, each with the same charge. 8. HgS
ANSWERS TO EXERCISES 1. (a) Homogeneous mixture. (b) Compound. (c) Element. (d) Heterogeneous mixture. 2. (a) Element, (b) Compound, (c) Homogeneous mixture, (d) Heterogeneous mixture 3. About 100 g/100 g H2O. 4. About 2½ times (interpolated from graph). 5. Unsaturated; about 240 g of sugar are soluble per 100 g H2O at 40°C. 6. Saturated; interpolation from graph shows that only about 130 g of NaNO3 are soluble in 100 g H2O at 60°C. 7. (a) S (b) Na (c) Ts 8. (a) Pt, (b) Po, (c) Ra. 9. (a) Nitrogen. (b) Potassium. (c) Zinc. 10. (a) Antimony, (b) Arsenic, (c) Mercury 11. a) Period 5, Group 4B b) Period 6, Group 8B c) Period 2, Group 8A 12. (a) Group 7A and Period 6, (b) Group 2A and Period 5, (c) Group 3A and Period 5. 13. (a) Representative, nonmetal, gas. (b) Transition, metal, solid. (c) Inner transition, metal, solid. 14. (a) Representative /metal/solid, (b) Transition/metal/solid, (c) Representative/nonmetal/gas. 15. a) For Li total electrons = 3, number of valence electron = 1, number of shells containing electrons = 2 b) For Cl total electrons = 17, number of valence electron = 7, number of shells containing electrons = 3 16. (a) 4, 2, 2 (b) 16, 6, 3
17. (a) 6.94 u, 3, 3, 3 (b) 197.0 u, 79, 79, 79 18. (a) 39.9 u, 18, 18, 18 (b) 87.6 u, 38, 38, 38 19. (a) 2, 5 (b) 2, 8, 5 20. (a) 2, 8, 1 (b) 2, 8, 8 21. (a) Ca, Mn, Se (b) Po, Se, O 22. (a) Na, P, Cl (b) Br, Cl, F 23. (a) Sr, Sn, Xe (b) Ar, Ne, He 24. (a) Cs, K, Na (b) Ca, As, Br 25. (a) Ca, Br, Kr (b) Cs, Rb, Li 26. (a) Be, C, Ne (b) Pb, Ge, Si 27. (a) Sulfuric acid. (b) Nitric acid. (c) Hydrochloric acid. 28. (a) Phosphoric acid. (b) Acetic acid. (c) Carbonic acid. 29. (a) Calcium bromide. (b) Dinitrogen pentasulfide. (c) Zinc sulfate. (d) Potassium hydroxide. (e) Silver nitrate. (f) Iodine heptafluoride. (g) Ammonium phosphate. (h) Sodium phosphide. 30. (a) Aluminum carbonate. (b) Ammonium sulfate. (c) Lithium sulfide. (d) Sulfur trioxide. (e) Barium nitride. (f) Barium nitrate. (g) Silicon tetrafluoride. (h) Disulfur dichloride. 31. Li2S, Li3N, LiHCO3 32. Ba(NO3)2, BaCl2, BaSO4, and Ba3(PO4)2 and Al(NO3)3, AlCl3, Al2(SO4)3, and AlPO4
Chapter 12
CHEMICAL BONDING This chapter outlines the early history of chemistry and discusses the law of conservation of mass, the law of definite proportions, and Dalton’s atomic theory. The chapter’s first Highlight focuses on Lavoisier, the father of chemistry. The chapter’s second Highlight discusses sunglasses and photochromic lenses. The basic ideas of ionic, covalent, polar covalent, and hydrogen bonding are discussed. Review electric charges
and electric forces if necessary (Chapter 8.1). After the student is able to determine ionic charges from chemical formulas, the Stock system of nomenclature is introduced. Conceptual questions and answers are giving to help understand covalent, ionic, and hydrogen bonding. The material in this chapter should be covered in its entirety.
For success in Chapters 12 through 14, the student must understand the trends of ionic charge and number of covalent bonds formed for atoms of the representative elements, as given in Figure 12.7 and Table 12.4, and must develop the ability to write the formulas of ionic and covalent compounds using the ideas presented in Sections 12.4 and 12.5.
DEMONSTRATIONS Large models of NaCl, CaCO3, and solid CO2 are helpful when discussing the structure of ionic and molecular solids. These are available from sources such as Klinger. Have the students build various models of the atom (see Chapter 9.1) and discuss the properties of each. Remind students the scale of the atom compared with the size of the nucleus (Chapter 10.2). Have the students calculate the size of a hydrogen atom using a peanut as the nucleus.
ANSWERS TO MATCHING QUESTIONS a. 13
b. 3
c. 15 d. 8
e. 7
f. 16 g. 2 h. 9
i. 12 j. 17
k. 4 l. 14 m. 11 n. 19 o. 1 p. 5 q. 6 r. 18 s. 10
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. a 3. d 4. c 5. d 6. a 7. c 8. d 9. c 10. c 11. c 12. d 13. d 14. a 15. c 16. d 17. b 18.c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. same 2. excess 3. atoms 4. Dalton's atomic 5. FeO 6. anion 7. two 8. M2X 9. ionic 10. double 11. decreases 12. nitrogen 13. higher, 14. lower, 15. greater
ANSWERS TO SHORT-ANSWER QUESTIONS 1. The law of conservation of mass states that no detectable change in total mass occurs during a chemical reaction. For example, if a sealed vessel containing copper and oxygen has a mass of 245.00 g and is heated so that copper oxide is formed, the vessel and its content will still have a mass of 245.00 g after reaction. 2. Lavoisier discovered the law of conservation of mass, explained the role of oxygen in combustion, established the principles for naming chemicals, wrote the first modern chemistry textbook, and introduced quantitative methods into chemistry. 3. Lavoisier performed the first quantitative experiments that helped him prove the conservation of mass using the most precise balance of that time. 4. The isotope is 12C and its assigned mass 12 u. 5. Dry ice is solid carbon dioxide, which is very cold (–78°C) and sublimes (passes directly from the solid phase to the gaseous). Dry ice is colorless and odorless. 6. The reactant used completely is the limiting reactant (B in this example). A is the excess reactant. 7. Dalton hypothesized (1) that each element is composed of small indivisible particles called atoms, which are identical for that element but different from atoms of other elements, (2) that chemical combination is the bonding of a definite number of atoms of each of the combining elements to make one molecule of the formed compound, and (3) While all atoms of an element were identical, different elements had atoms of differing size and mass. 8. Carbon and Oxygen combine different molecules when they are combined in different proportions. In case oxides of carbon that is carbon monoxide (CO) and carbon dioxide (CO2). A fixed mass of carbon say 100 grams, may react with 133 grams of oxygen to form CO, or with 266 grams of oxygen to form CO2 then the ratio of the masses of oxygen that can react with 100 grams of carbon is 266:133 = 2:1. 9. In forming compounds, atoms tend to gain, lose, or share valence electrons to achieve the electron configurations of the noble gases. That is, they tend to get eight electrons (an octet) in the outer shell (except for H, which tends to get two electrons in the outer shell, like the configuration of He). Hydrogen is an exception because it must use the first shell, which can hold only two electrons.
10. The valence electrons are those in the outermost shell of the atom. Valence electrons can participate in the formation of chemical bonds. Individual atoms can achieve a noble gas electron configuration in two ways: by transferring electrons or by sharing electrons. 11. The net electric charge on an ion is the number of protons minus the number of electrons. 12. Sodium bonds with chlorine via an ionic bond. Sodium will lose its valence electron to form a sodium ion of charge +1 that has a stable shell (octet) of electrons. This extra electron is used by a chlorine atom to form a stable chloride ion with charge -1. Electrical forces attract these ions together to form a strong, ionic bond. The summary reaction is this: Na+ + Cl- → NaCl 13. Two K+ ions and one S2– ion. 14. The number of dots is the same as the atom’s number of valence electrons. 15. A Lewis symbol shows the valence electrons of a single atom or ion. Lewis structures show the valence electrons in the molecules and ion combinations that make up compounds. 16. Metals form positive ions, whereas nonmetals form negative ions. Cations are positive ions; anions are negative. 17. Isoelectronic means having the same electron configuration (in this case, two electrons in the first shell and eight in the second). 18. The total electric charge must be zero, and all atoms must have noble gas configurations. Additional rule, when writing formula for ionic compounds the cation comes first, followed by anion, both with a numeric subscript to indicate the number of atoms of each. 19. If it conducts electricity when molten, it is ionic. Ionic compounds have high melting point and are hard and brittle. 20. In the solid phase, ions are held in place and thus cannot move, thereby unable to conduct an electric current. 21. The Stock system is preferred whenever a metal that can form more than one type of ion is part of the compound. 22. Covalent compounds are formed by the sharing of pairs of electrons.
23. A single bond is one shared pair of electrons between two atoms, whereas double and triple bonds result from the sharing of two and three pairs, respectively. 24. Electronegativity increases across (left to right) in a period and decreases down a group. 25. Fluorine has the highest electronegativity, and oxygen has the second highest. 26. A polar covalent bond. 27. No; because there is only one polar bond, the polarity could not be offset by an opposing polar bond. 28. The carbon and oxygen atoms within the polyatomic ion are covalently bonded, while the whole CO32- aggregation behaves like an ion (due to the overall -2 charge) when forming ionic compounds. 29. No. Carbon tetrachloride is a compound of two nonmetals; therefore, it uses covalent bonds. Materials with covalent bonds are poor conductors of electricity in all forms. 30. Covalent. With a boiling point of –10°C, it is a gas at room temperature, and ionic compounds are not gaseous at room temperature. 31. The lone pair of electrons on the nitrogen central atom results in a trigonal pyramidal geometry with an overall molecular dipole according to VSEPR. 32. Like dissolves like, which means that polar liquids tend to dissolve polar and ionic compounds but not covalent compounds, whereas covalent liquids tend to dissolve other covalent compounds but not polar and ionic substances. 33. When water freezes, the alignment of the hydrogen bonds between the water molecules in ice produces a more open structure, thereby increasing its volume and making its density less than liquid water.
ANSWERS TO VISUAL CONNECTION 1 (a) ionic (b) covalent 2 (a) ionic (b) covalent (c) double 3 (a) double (b) covalent (c) ionic
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS
1. The lowest common denominator of 12 and 8 is 24. To get 24 buns you would require the purchase of two packages; to get 24 wieners, you would purchase three packages. This problem is similar to finding the correct ratio of positive and negative ions needed to give net neutrality and thus to find the correct formula for an ionic compound. 2. The law of conservation of mass states that matter cannot be created or destroyed. Although pollutants may be burned and changed into other substances, no total mass is lost, so waste products are still present. Dissolving in water may only disperse the original molecules and not change them at all. Disposing of the pollutants in water will also affect marine life which is detrimental to the ecological balance. 3. To completely use the 200 bodies would require 800 wheels. You have 900 wheels on hand, so 200 cars can be produced, and 100 wheels will be left over (in excess). This problem is similar to having an excess of one reactant in a chemical reaction. 4. Vinegar is polar, and oil is nonpolar, so a heterogeneous mixture results, with the less dense oil on top. Shaking causing a dispersion (intermixing) so that some of each component pours out if you act fast! 5. Batteries just don't work as well in extremely cold temperatures. Cold temperatures affect the fluid inside the battery, and it becomes sluggish. 6. There are a few exceptions to the trend that electronegativity decreases down a group. Some are gallium, indium, and thallium in Group 3A. Others are germanium, tin, and lead in Group 4A. These elements have smaller atomic size than is expected based on the trends in their individual Groups and Periods (see Fig. 11.23). The smaller size allows the nucleus to draw bonding electrons to it. The nuclei in largersized atoms are more shielded and thus have lower electronegativity values. Notice that some of these exceptions are seen in the ionization energies (Fig 11.24).
ANSWERS TO EXERCISES 1. 68 tons + 96 tons – 36 tons = 128 tons 2. 0.942 g + 56.413 g – 57.136 g = 0.219 g 3. (a) 12.0 u + (2 × 16.0 u) = 44.0 u (b) 12.0 u + (4 × 1.00 u) = 16.0 u (c) (3 × 23.0 u) + 31.0 u + (4 × 16.0 u) = 164.0 u
4.
(a) 180.2 u, (b) 342.3 u, (c) 749.0 u
5.
100.0% – 25.5% = 74.5%
6.
100.0% – 72.2% = 27.8% Nitrogen (N)
7.
(a) (23.0 u/58.5 u) × 100% = 39.3% Na; (35.5 u/58.5 u) × 100% = 60.7% Cl (b) 42.1% C, 6.4% H, 51.5% O (FM = 342.0 u)
8. (a) 40.1% Ca, 12.0% C, 48.0% O (FM = 100.1 u) (b) 41.7% Mg, 54.9% 0, 3.4% H (FM = 58.3 u) 9. (a) (6.1 g/14.1 g) × 100% = 43% Mg (b) 14.1 g – 6.1 g = 8.0 g S, law of conservation of mass. 10. (a) (7.75 g/67.68 g) × 100% = 11.5% P (b) 67.68 g – 7.75 g = 59.93 g Br, law of conservation of mass. 11. Still 14.1 g of MgS, because 6.1 g of magnesium will react with only 8.0 g of sulfur, leaving 2.0 g of sulfur in excess; law of definite proportions. 12. Still 67.68 g of PBr3, because 59.93 g of bromine will react with only 7.75 g of phosphorus, leaving 2.25 g of phosphorus in excess; law of definite proportions. 13. (a) 2– (b) 1+ (c) 1– (d) 3– (e) 2+ (f) 0 (g) 4+ (h) 3+ 14. (a) 1–, (b) 1+, (c) 2–, (d) 2+, (e) 3+, (f) 3–, (g) 0, (h) 4+
15.
16.
17. (a) CsI (b) BaF2 (c) Al(NO3)3 (d) Li2S (e) BeO (f) (NH4)2SO4 18. (a) RbI, (b) Cs3P, (c) Li2SO4, (d) Ag2CO3, (e) Zn3(PO4)2, (f) Al(NO3)3 19. (a) Iron (II) hydroxide. (b) Copper (II) chloride. (c) Gold (III) iodide. 20. (a) Zinc nitrate. (b) Chromium (II) sulfide. (c) Copper (I) oxide. 21. (a) 2 (b) 0 (c) 1 (d) 3 (e) 4 22. (a) 1 (b) 0 (c) 1 (d) 4 (e) 3
23.
24.
25. (a) HCI (b) NCI3 (c) SCI2 (d) CCI4 26. (a) CH4 (b) CBr4 (c) CS2 (d) No compound formed. 27. (a) Covalent, two nonmetals. (b) Ionic, Group 1A metal and nonmetal. (c) Ionic, metal and polyatomic ion. (d) Covalent, two nonmetals. (e) Covalent, all nonmetals. (f) Ionic, two
polyatomic ions.
28. (a)
Covalent, two
nonmetals. metal
(b) Ionic, Group 1A and nonmetal. (c) Covalent, two nonmetals. (d) Ionic, Group 2A metal and nonmetal. (e) Ionic, metal and polyatomic
ion. (f) Covalent, three nonmetals. 29. (a) S and Cl are in the same period, but Cl is farther right than S, so Cl is the more electronegative, and the arrows would point to the two Cl atoms.(b) C and O are in the same period, but O is farther right and so is the more electronegative. Thus, the arrows would point to the two oxygen atoms. 30. (a) Cl is higher than Br and in the same group, and so Cl is the more electronegative atom, and the dipole arrow will point toward the Cl. (b) N is higher than I, but I is farther right. Figure 12.12 shows that N has EN = 3.0, whereas I has EN = 2.5. So N is the more electronegative atom, and the dipole arrow in each bond will point toward the N. 31. (a) linear, polar; (b) trigonal pyramidal, polar; (c) angular, polar; (d) trigonal planar, nonpolar 32. (a) linear, polar; (b) tetrahedral, nonpolar; (c) trigonal planar, nonpolar; (d) trigonal pyramidal, polar
Chapter 13
CHEMICAL REACTIONS This chapter discusses the differences in chemical and physical changes, how and why equations are balanced, the role of energy in chemical reactions, factors that influence the rate of reactions and the basic types of chemical reactions (combination, decomposition, hydrocarbon combustion, single-replacement, and double-replacement). The students are made aware of chemical reactions that occur in their everyday lives. Acids, bases, and pH are discussed, and the concepts of oxidation and reduction are explained. The chapter’s final topic is a discussion of Avogadro’s number, the mole concept and examples of mole to gram conversions. Chapter Highlights discuss the chemistry of tooth enamel and the chemistry of airbags. Conceptual questions and answers are given to help understand the rates of reaction and acids and bases. We recommend that all sections of this chapter be covered. The completing and balancing of equations that illustrate a few basic reaction types is the most important skill developed by the student in this chapter. Information and skills attained in Chapters 11 and 12 will be used extensively.
DEMONSTRATIONS Any demonstrations of chemical reactions will be valuable aids to student understanding and appreciation. Examples include the combination reaction when magnesium burns, a combustion reaction involving burning ethanol, an acid-carbonate reaction such as CaCO3 and HCl, a single replacement reaction such as placing a zinc strip in aqueous CuSO4, or (carefully) demonstrating the reaction of a small piece of Zn with dilute hydrochloric acid. Another good demonstration would be the MnO2-catalyzed decomposition of KClO3, but take appropriate safety precautions, including a safety shield. An excellent demonstration of an exothermic reaction is to place KMnO4 crystals to a height of about one inch in a 16 × 150 mm test tube, place the test tube in an Erlenmeyer
flask, and add about seven drops of glycerol. After a few seconds, smoke will start to rise, and very shortly a flame will shoot up in the tube. It is advisable to use a safety shield. The difference between a physical change and a chemical change can be demonstrated by mixing powdered sulfur and iron filings. That this is only a physical change is shown by using a magnet to separate the iron filings from the sulfur. Each substance keeps its own characteristic properties, and no chemical combination has occurred. If the mixture is placed in a test tube and heated, the iron and sulfur combine to give FeS. Now a magnet will not separate the iron from the sulfur.
ANSWERS TO MATCHING QUESTIONS a. 12 b. 7 c. 2 d. 5 e. 26 f. 24 g. 18 h. 15 i. 8 j. 16 k. 10 l.25 m. 11 n. 23 o. 3 p. 21 q. 19 r. 22
s. 14 t. 1 u. 17
v. 6 w. 20 x. 4 y. 9 z. 13
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. d 3. b 4. b 5. d 6. a 7. c 8. a 9. c 10. b 11. a 12. d
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. physical 2. valance 3. AB 4. A + B 5. lower 6. enzymes 7. higher 8. catalyst 9. hydroxide or OHˉ 10. neutral 11. salt 12. redox 13. will not 14. two
ANSWERS TO SHORT-ANSWER QUESTIONS 1. The reactants are carbon dioxide and water. The products are oxygen and glucose, the catalyst is chlorophyll, and the energy source is the Sun. 2. That it is blue-black, crystalline, solid, and that it sublimes are all physical properties. That it reacts with aluminum and many other metals is a chemical property. 3. (a) Physical. (b) Physical. (c) Chemical. (d) Chemical. 4. Reactants disappear, new substances (products) appear, and energy is released or absorbed. 5. In a chemical reaction the reactants are diminished (used up partially or completely) to form the products. These products have different chemical properties than the
reactants. For the reaction to take place energy in the form of heat, light, electricity, or sound is either given to the system (absorbed) or released from the system. 6. In a chemical reaction the coefficients represent the relative number of substances in the reaction. These numbers can be changed to correctly balance the equation. The subscripts, however, represent the absolute number of atoms or molecules in each substance. These subscript numbers are fixed and cannot be changed. 7. First would be carbon, second would be hydrogen, third would be oxygen. 8. (a) Fractional coefficients are generally not used. (b) The smallest set of whole numbers has not been used; 4, 2, 4 should be changed to 2, 1, 2. (c) Hydrogen should be written H2, not H, on the product side. (The correct set of coefficients would then become 2, 2, 2, 1.) (d) Helium is nonreactive. 𝑀𝑛02
9. Solid (s), aqueous or water solution(aq), catalyst −−−→,, liquid (l), yield →, gas (g), equilibrium reaction rate
.
10. (a) 2 H2O → 2 H2 + O2 - this is a decomposition reaction (b) 2 Ca + O2 → 2 CaO - this is a combination reaction (c) 2 NaN3(s) → 2 Na(s) + 3 N2 (g) - this is a decomposition reaction 11. The reaction vessel feels warm for an exothermic reaction because energy is being liberated (released) during the reaction. An endothermic reaction, however, requires energy input to the system therefore, the reaction vessel will feel cold. 12. A chemical reaction can either cause energy to be absorbed to break bonds, or release energy to form new bonds. 13. (a) Reactants. (b) Products. (c) Activation energy. (d) Heat of reaction. (e) Exothermic. 14. Carbon dioxide and water are the products, and oxygen is the other reactant. If O2 is in short supply, carbon monoxide and carbon are also products. 15. For reaction to occur, the collision must occur at the proper places in the molecules (proper orientation) and must have at least enough energy to break bonds (the activation energy). 16. Temperature, concentration, surface area, and catalyst.
17. (a) An increase in temperature increases the frequency of collisions and their average energy. (b) An increase in reactant concentration means that collisions will be more frequent. 18. Powdered zinc and powdered sulfur are in finer states of subdivision than the lumps, and so more reactant surface is available for collisions. 19. A catalyst speeds up the rate of the reaction by providing a new pathway with lower activation energy. 20. A catalytic converter changes the poisonous carbon monoxide (CO) and nitric oxide (NO) gases coming from a car’s engine into less harmful carbon dioxide (CO2) and harmless nitrogen (N2). 21. 7; 10000 times. 22. Blue, red. 23. pH meter; the milk of magnesia, at the right. 24. Acids: conduct electricity, turn blue litmus red, taste sour, react with bases to neutralize their properties, react with active metals to liberate H2. Bases: conduct electricity, change litmus from red to blue, react with acids to neutralize their properties. 25. Arrhenius acids are substances that give hydrogen ions, H+ (or hydronium ions, H3O+), in water. Arrhenius bases give hydroxide ions, OH–, in water. 26. (a) Hydrochloric acid, HCl. (b) Magnesium hydroxide, Mg(OH)2. (c) Sodium hydroxide, NaOH. (d) Sodium bicarbonate (sodium hydrogen carbonate), NaHCO3. (e) 5% acetic acid, HC2H3O2. 27. Water and salt. The most common mistake is writing the wrong formula for the salt. 28. Hydrates. 29. Water, carbon dioxide, and salt. 30. Sodium hydrogen carbonate eliminates odors by neutralizing or removing the odor compound. 31. Particles of solid form in the mixture of the two clear solutions as the positive ions of one solution combine with the negative ions of the other to form an insoluble salt.
32. Oxidation is a gain of oxygen or a loss of electrons. Reduction is a loss of oxygen or a gain of electrons. 33. The relative activity of a metal indicates its tendency to lose electrons to ions of another metal or to hydrogen ions. 34. Hydrogen gas and salt. This is a single-replacement reaction. 35. Gold and silver are not very active. Sodium and magnesium are active metals, high in the activity series, and so form compounds readily. 36. 1017. The Avogadro constant, usually denoted by NA or L is the factor that, multiplied by the amount of substance in a sample, measured in moles, gives the number of constituent particles in that sample. 37. The molarity is the number of moles of solute per liter of solution. 38. Molarity is defined to be the ratio of the number of moles of solute per volume of solution. To calculate the number of moles of salt in a solution, multiply the molarity and the volume.
ANSWERS TO VISUAL CONNECTION a. combination b. AB → A + B c. single replacement d. AB + CD → AD + CB e. combustion
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Water acts as an acid when it reacts with sulfur trioxide to form sulfuric acid, or acid rain. Water can act as a base when combined with lemons to make lemonade. 2. Acidic 3. Charcoal burned indoors would liberate a dangerous amount of carbon monoxide gas. 4. The citric acid dissolves and ionizes, and the hydrogen ions react with the sodium hydrogen carbonate to liberate CO2 gas—the bubbles. 5. In the body, enzymes catalyze the reactions; thus, activation energy requirements are lower. 6. Sticks have more surface area than the same mass of logs, so they react more rapidly when heated in the presence of oxygen.
7. Baking soda is an excellent absorber of gases that can cause unpleasant odors. 8. The baking of bread. 9. The water comes from the combustion of the gasoline. Hydrocarbon burning gives CO2 and H2O. 10. Strong acids ionize readily; that is, they give hydrogen ions (H+) in solution. This means that the valence electrons of the compound are easily removed, and therefore, the ionization energy is very small. The valence electrons ―live‖ higher in the potential energy well (see Fig 9.11) and it takes less energy to remove them from the well. Weak acids have their valence electron deeper in the well and require more energy to become ionized. Thus they do not become ionized as readily in solutions.
ANSWERS TO EXERCISES 1. (a) 1, 1, 1, 2 (b) 4, 3, 2 (c) 2, 3, 1, 3 (d) 1, 2, 1, 1 (e) 2, 2, 1 (f) 2, 15, 12, 6 2. (a) 2, 1, 2 (b) 4, 3, 2, 6 (c) 2, 13, 8, 10 (d) 2, 2, 4, 1 (e) 8, 3, 4, 9 (f) 1, 2, 1, 1 3. (b) Combination. (e) Decomposition. 4. (a) Combination. (d) Decomposition. 5. (a) N2 + 3 H2 → 2 NH3 (b) 2 KCl → 2 K + Cl2 6. (a) Ca + Cl2 → CaCl2 (b) Na2CO3 → Na2O + CO2 7. C5H12 + 8 O2 → 6 H2O + 5 CO2 8. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 9. (a) HNO3 + KOH → H2O + KNO3 (b) 2 HC2H3O2 + K2CO3 → H2O + CO2 + 2 KC2H3O2 (c) H3PO4 + 3 NaOH → 3 H2O + Na3PO4 (d) H2SO4 + CaCO3 → H2O + CO2 + CaSO4 10. (a) 2 HCl + Ba(OH)2 → 2 H2O + BaCl2 (b) 6 HCl + Al2(CO3)3 → 3 H2O + 3 CO2 + 2 A1C13 (c) H3PO4 + 3 LiHCO3 → 3 H2O + 3 CO2 + Li3PO4 (d) 2 Al(OH)3 + 3 H2SO4 → 6 H2O + A12(SO4)3 11. (a) AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) (b) BaC12(aq) + K2CO3(aq) → BaCO3(s) + 2 KCl(aq) 12. (a) K2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2 KNO3(aq)
(b) 2 K3PO4(aq) + 3 CaBr2(aq) → 6 KBr(aq) + Ca3(PO4)2 13. (a) Will not be possible. (b) Will be possible. (c) Will be possible. 14. (a) Will (Zn is above Fe). (b) Will not (Pb is below Fe). (c) Will not (Ag is below H). 15. (a) Ni + Pt(NO3)2(aq) → Pt + Ni(N03)2(aq) (b) Zn + H2SO4(aq) → H2 + ZnSO4(aq) 16. (a) Mg + 2 HCl → H2 + MgC12 (b) Al + 3 AgNO3(aq) → 3 Ag + Al(NO3)3(aq)) 17. 12.04 × 1023 molecules 18. 24.08 × 1023 molecules 19. 6.02 × 1023 atoms, 23.0 g; 3.00 mol, 69.0 g; 2.00 mol, 12.04 × 1023 atoms. 20. 18.06 × 1023 molecules, 132 g; 1.00 mol, 6.02 × 1023 molecules; 0.50 mol, 22.0 g. 21. 185.25 g 22. 0.094 mol 23. 2.7 × 1021 molecules 24. 9.40 × 1024 molecules 25. 721.3 grams 26. 1.15 × 103 g 27. Molarity = Number of moles of solute/Volume of solution in litre. Molarity = 0.5/2 = 0.25 mole/litre 28. Molarity = Number of moles of solute/Volume of solution in litre. Molarity = 1/0.25 = 4 moles/litre
Chapter 14 ORGANIC CHEMISTRY This chapter deals with the division of chemistry that probably most affects our daily lives. Organic compounds are all around us and, in fact, basically are us. The chapter begins with a discussion of the bonding rules for organic molecules. Aromatic and aliphatic hydrocarbons are then treated, followed by a section on derivatives of hydrocarbons and synthetic polymers. The last section is an introduction to biochemistry including brief discussions of carbohydrates, fats, and proteins. The two chapter Highlights discuss some relatively new and interesting products in organic chemistry. The conceptual questions
and answers illustrate one use of polymers and the structure of some biochemical molecules. For success in handling this chapter, the student must learn at once the basic bonding rules in Table 14.1 (these also were covered in Chapter 12) and the names and molecular formulas of the first eight alkanes (Table 14.2). In order for the student to understand simple isomers, the instructor must continually emphasize the tetrahedral geometry of the four single bonds to carbon and that no difference exists in drawing a bond to carbon ―up,‖ ―down,‖ or ―sideways‖ on the two-dimensional blackboard. Some instructors may not feel comfortable teaching organic chemistry, but every effort has been made to make the presentation of the chapter material clear, consistent, and coherent. If time is short, the instructor can stop at any point in the chapter but should not skip a section to go to one further on, because each section builds on the preceding section.
DEMONSTRATIONS The use of ball-and-stick models to show bonding is recommended, especially for the aromatic hydrocarbon compounds. Also, the naming of organic compounds and the structure of isomers are most easily discussed with the aid of such models.
ANSWERS TO MATCHING QUESTIONS a. 16 b. 15 c. 20 d. 3 e. 7 f. 17 g. 13 h. 23 i. 11 n. 26 o. 9 p. 18 q. 22 r. 4 s. 10 t. 12
j. 14 k. 25 l. 5 m. 24
u. 19 v. 8 w. 1 x. 6 y. 21 z. 2
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. c 3. d 4. c 5. a 6. b 7. a 8. d 9. a 10. c 11. c 12. b
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Biochemistry 2. three 3. Covalent 4. aliphatic 5. carcinogen 6. cycloalkane 7. 109.5° 8. alcohols 9. amide 10. Polyethylene 11.
nylon 12. carbohydrate
ANSWERS TO SHORT-ANSWER QUESTIONS
1. Organic chemistry studies the compounds that contain carbon. Biochemistry studies the chemical compounds and reactions that occur in living cells. 2. Organic compounds are formed by covalent bond among the constituent atoms. 3. Carbon (4 bonds), hydrogen (1 bond), oxygen (2 bonds), sulfur (2 bonds), nitrogen (3 bonds), a halogen (1 bond). 4. Benzene is an aromatic hydrocarbon; it is a clear, colorless liquid, and as the name indicates, it has a very strong odor. The molecular formula is C6H6. 5. Kekulé,
preferred
. 6. An aromatic hydrocarbon contains one or more benzene rings. 7. Alkanes, CnH2n + 2, CH3CH3, etc. Cycloalkanes, CnH2n, C3H6, etc. Alkenes, CnH2n, CH2 = CH2, etc. Alkynes, CnH2n - 2, C2H2 ≡C2H2, etc. 8. (i) Methane, CH4 (the principal component of natural gas) (ii) Ethane, C2H6 (iii) Propane (iv) Butane (v) Pentane (vi) Hexane (vii) Heptane (viii) Octane Properties of alkanes: Alkanes are colorless. Alkanes are less dense than water (alkanes float on top of water). Alkanes are non-polar molecules, so they are more soluble in non-polar solvents than they are in polar solvents. Alkanes are insoluble in water. The melting and boiling points of the shorter chain alkanes are low, but the melting and boiling point of alkanes increase as the number of carbon atoms in the carbon chain increases.
Petroleum is made up chiefly of alkanes but also contains other classes of hydrocarbons. Gasoline contains the alkanes from pentane to decane (n = 5 to n =10). The alkanes with higher values of n make up other products, such as diesel fuel, fuel oil, petroleum jelly, paraffin wax, and lubricating oil. The largest alkanes make up asphalt. Alkanes are also used as starting materials for products such as paints, plastics, drugs, detergents, insecticides, and cosmetics. 9. The bonds point toward the corners of a regular tetrahedron. They form angles of 109.5° to one another. 10. Constitutional isomers are compounds that have the same molecular formula but different structural formulas. 11. For organic compounds, the basic rules of nomenclature are: 1) the longest chain of carbon is found, and the compound is named as a derivative of the alkane, 2) the positions and names of the substituents replacing hydrogen on the longest chain are added, and 3) the carbon atoms are numbered starting at the end of the chain closest to the substituents. 12. The full and condensed structural formulas for methane, methyl group, ethane, and ethyl groups are: H
H
|
|
H–C–H
H–C–
|
|
H
H
Methane, CH4
methyl group, CH3 –
H H
H H
|
|
|
H–C–C–H
|
H–C–C–
| |
|
H H
H H
Ethane, CH3CH3
| ethyl group, CH3CH2 –
13. Benzene and Cyclohexane are structurally different. Benzene is an aromatic hydrocarbon with 3 double bonded carbons in a ring (condensed formula, C6H6), while cyclohexane is a saturated hydrocarbon ring with only single-bonded carbon atoms (condensed formula, C6H12). 14. Cyclohexane, C6 H12,
15. Ethylene, H2C = CH2. Acetylene, HC ≡CH. 16. Saturated hydrocarbons are those in which the hydrogen content is at a maximum. Unsaturated hydrocarbons have double or triple bonds, which allow the addition of more hydrogen atoms. Saturated hydrocarbons have a relatively low chemical reactivity. Unsaturated hydrocarbons are more reactive than their saturated counterparts. Saturated hydrocarbons generally burn with a blue flame. Unsaturated hydrocarbons generally burn with a sooty flame. 17. An alkene can add a molecule of H2 to the double bond. An alkyne can add two molecules of H2 to the triple bond. Alkanes cannot undergo addition reactions because to do so would make five bonds to a carbon atom, which can form only four.
18. Chlorofluorocarbon. Used in air conditioners, refrigerators, and heat pumps. Upon reaching the stratosphere, CFCs provide chlorine atoms that can destroy ozone molecules that protect life by absorbing UV radiation. Simple alkyl halides are generally colorless, odorless liquids at room temperature and are insoluble in water. They are polar in nature. 19. CH3OH 20. Alcohols have the general formula R-OH, contain one or more hydroxyl groups (-OH), and have names ending in -ol.
21. Amines have the general formula R-NH2, contain one or more amino groups (NH2).They have unpleasant odors like the odor of a raw fish which comes from the amines it contains. Amines have many applications as medicinal and as starting materials for synthetic fibers. 22. The general formula for a carboxylic acid is RCOOH. It contains a carboxyl group (COOH).
23. Esters are popular because they have pleasant odors. They are found in oranges and other ripe fruits, and in the fragrances of flowers. The general formula for an ester is RCOOR'.
24. Amides are nitrogen-containing compounds and have the general formula RCONH-R’. They are used as solvents and plasticizers. 25. Addition polymers are formed when alkene monomers add to one another, so a carboncarbon double bond is necessary. 26. Condensation polymers are constructed from molecules that have two or more reactive groups. Each molecule attaches to two others by ester or amide linkages. 27. Dacron is a polyester, and nylon is a polyamide. Common addition polymers include Polyethylene, Teflon, Styrofoam, and Polyvinyl chloride.
28. Amino acid is monomer of proteins. Simplest monomer of amino acid is glycine.
29. Humans get energy from carbohydrates by breaking bonds in digestion to form simple sugars which can be used in cells to produce energy. The three main types of carbohydrates are sugars, starches, and cellulose, although humans cannot digest cellulose. Plants make carbohydrates by using energy from the Sun in photosynthesis. 30. Glucose and fructose combine to form sucrose. Glucose. 31. Herbivores have bacteria in their digestive tracts that have the enzymes necessary to break the glucose linkages in cellulose. 32. Differences between fats and oil are: 1) Fats remains solid at room temperature whereas oils are liquid at room temperature. 2) Fats have more hydrogen percentage than oils. 3) Fats have single bond between carbons whereas oils have double bond.
33. Saturated fats are chemically different from unsaturated fats. Saturated fats contain all single bond carbon atoms that are saturated with hydrogen, whereas unsaturated fats have double bonds between some of the carbon atoms (i.e., are not saturated with single carbonhydrogen bonds). Physically, saturated fats are in solid form at room temperature while unsaturated fats are liquid at room temperature (also called oils). The saturated fats are dangerous in large quantities because they can lead to cholesterol buildup in arteries. 34. Soap is composed of the sodium salts of the fatty acids produced when fats are treated with sodium hydroxide. 35. The alcohol vapor and potassium dichromate are the reactants that produce carboxylic acid and chromium sulfate. The alcohol vapor is oxidized to carboxylic acid and the potassium dichromate is reduced to chromium sulfate.
ANSWERS TO VISUAL CONNECTION a. Benzyl group
b. ester
c. Amino group
d. amide
e. carboxylic acid
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The element silicon, Si, is in the same group (the carbon group) as that of carbon in the periodic table. It is a metalloid i.e. it exhibits nonmetal-like properties (like carbon) and also metal-like properties. However, its bonding capabilities do not come close to those of carbon. 2. ―Cat cracking‖ is short for catalytic cracking, the production of smaller hydrocarbons from larger ones by the use of a catalyst. Many oil refineries are located in Louisiana.
3. Aspirin is an aromatic hydrocarbon (specifically, a carboxylic acid). It contains a carboxyl group (COOH), an ester group (R-CO-O-R'), and a benzene ring. Acetaminophen is also an aromatic hydrocarbon (an amide). It contains an amide group (R-CO-NH-R'), a hydroxyl group (R-OH), and a benzene ring. 4. The four groups are (A) adenine, (C) cytosine, (G) guanine, and (T) thymine; A pairs with T and C pairs with G. A contains amine functional group, while C, G, and T contain amine and amide, functional groups. 5. Peel the onion under the surface of the water in a bowl.
6. Water and soap are used to clean a spot of grease off of clothing. Water is a polar solvent and soap is a long-chain sodium salt of fatty acids; soap is polar at one end of the long molecule and nonpolar at the other end. Water dissolves the polar end of the
soap molecule, and the grease (a nonpolar solvent) dissolves the other end of the soap molecule. The grease is suspended in the water-soap mixture (emulsified) and is swept away by rinsing. The alcohol would dissolve the soap due to which the production of foam would be very less (near to 0). Hence water and soap is a better option for cleaning clothes. 7. Hydrogen bonds exist between some of the H atoms and O or N atoms in other parts of the same protein molecule that have twisted around into close proximity. The result is that proteins twist themselves into long curls and then fold onto themselves.
ANSWERS TO EXERCISES 1. (a) Is valid. (b) Is incorrect because Cl should have one bond, not two. 2. Formula (a) is incorrect because the C has only three bonds when it should have four. Formula (b) is correct. 3.
4.
5. (a) Alkene. (b) Cycloalkane. (c) Alkyne. (d) Alkane. (e) Aromatic. 6. (a) Aromatic. (b) Alkane. (c) Cycloalkane. (d) Alkene. (e) Alkyne. 7. (a) Alkane. (b) Alkyne. (c) Cycloalkane. (d) Alkene. (e) Aromatic. 8. (a) Alkene. (b) Cycloalkane. (c) Alkane. (d) Alkyne. (e) Aromatic. 9. (a) Same compound. (b) Neither. (c) Constitutional Isomers. 10. (a) Neither. (b) Same compound. (c) Constitutional Isomers.
11. (a)
(b)
12 (a)
(b)
13.
14. Three isomers: 1-heptene, CH2CHCH2CH2CH2CH2CH3 2- heptene, CH3CH = CHCH2CH2CH2CH3 3- heptene, CH3CH2CH = CHCH2CH2CH3 15. (a) Amine. (b) Amide. (c) Ester. (d) Carboxylic acid. (e) Alkyl halide. (f) Alcohol. 16. (a) Carboxylic acid. (b) Ester. (c) Alcohol. (d) Amine. (e) Alkyl halide. (f) Amide. 17. (a) CH3CH2-OH and CH3-O-CH3 (b) CH3CH2CH2Cl and CH3CH(Cl)CH3 18. (a) CH3CH2NH2 and CH3NHCH3 (b) CH3CH2CH2OH
CH3CH(OH)CH3 CH3OCH2CH3 19.
20.
21.
22.
23.
24.
25.
26.
Chapter 15 PLACE AND TIME This chapter is very important for the prospective elementary teacher, as well as other students. The authors have found that the topics of this chapter, which are so common environmentally, are of strong interest to students. In this chapter, we examine how events are located, or their positions designated, in space and time. Cartesian coordinates are introduced, and the concept of a grid expanded to locations on the spherical Earth in
terms of latitude and longitude. There is a Highlight on global positioning, daylight saving time, and the calendar. How time is measured is of major importance, and the measurement units of days, hours, standard time zones, daylight savings time, and so on, are discussed. The year is related to the seasons, and the chapter ends with the historical development of our calendar.
DEMONSTRATIONS With colored markers, draw the equator, small circles parallel to the equator, the prime meridian, and one or two other meridians on a 16-in.-diameter sphere. Show the sphere axis tilted 23° down from the vertical. Place a light source representing the Sun at the center of the lecture desk; then move the slate sphere around the light, illustrating how the declination of the Sun changes as Earth revolves around the Sun. A model solar system is very good for showing this, but the system is too small for a large class to see it clearly. Demonstrate the difference between the solar day and the sidereal day by having one of your students sit on a stool in front of the class to play the part of the Sun. You are the Earth, and the other students are stars. Stand in front of the student on the stool facing the class and explain to the others the part they are playing. Then revolve about the student on the stool, keeping your back to the student at all times. After one revolution has been completed, ask the person on the stool how many rotations you made. The answer will be ―None,‖ because your back was in view at all times. Next ask the class (playing stars) how many rotations you made during one revolution. They will answer ―One.‖ Repeat the demonstrations but make one rotation with respect to the student on the stool. This will be two rotations with respect to the class. They will get the point that there is always one more rotation to anyone outside of the rotating and revolving system. Demonstrate how the solar day varies in length by having the lamp on the lecture desk play the part of the Sun; you play the part of the Earth. With a meter stick in your hand (this is more effective than just the extended arm), stand at the side of the desk facing the lamp (Sun). Hold the meter stick out toward the lamp so that, sighting over the meter stick, you observe the Sun. Next rotate counterclockwise and revolve
counterclockwise (take a couple of steps) around the lecture desk and lamp. Rotate 360° with respect to your original position. Note that you cannot observe the sun by sighting across the meter stick but must rotate slightly more than 360° to see the Sun on your meridian. Repeat the demonstration. This time take three steps in revolving. Have the student’s note that you must rotate even more than last time because you have to move around farther in orbit. Thus the length of a solar day depends on the orbital velocity of Earth. The greater the orbital velocity, the longer the solar day, provided that the rotation time remains constant.
ANSWERS TO MATCHING QUESTIONS a. 3 b. 20 c. 15 d. 21 e. 10 f. 23 g. 22 h. 8 i. 16 j. 11 k. 5 l. 18 m. 1 n. 14 o. 9 p. 4 q. 12 r. 6 s. 19 t. 7 u. 13 v. 17 w. 2
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. d 3. c 4. b 5. b 6. b 7. d 8. c 9. d 10. b 11. c 12. b 13. c 14. b 15. b 16. c 17. b 18. c 19. d
20. d
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. rectangular 2. origin 3. parallels 4. great circle 5. equator 6. prime meridian7. Coordinated Universal Time (UTC) 8. Beijing 9. four 10. westward 11. altitude 12. noon altitude 13. 30°E 14. summer solstice 15. Cancer 16. seasons 17. zodiac 18. Sun 19. Common Era 20. top
ANSWERS TO SHORT-ANSWER QUESTIONS 1. The origin in a Cartesian coordinate system is defined as the intersection of the two perpendicular numbered lines that make up the coordinate system. 2. Longitude and latitude. 3. Answer is relative to the student’s location and reference point. 4. Latitude: 0° to 90°N and 0° to 90°S. Longitude: 0° to 180°E and 0° to 180°W. 5. The meridian running through Greenwich, England, which is taken as 0° longitude, and the equator, which is taken as 0° latitude.
6. (a) Meridians. (b) Parallels 7. The equator is defined as the reference point for latitude; therefore, the equator is 0° latitude, by definition. Since the equator is a parallel it does not have a longitude value (or you can say that it corresponds to all longitudes). 8. The center of the Earth. 9. Meridians are lines drawn from the geographic North Pole, perpendicular to the equator, to the geographic South Pole. All meridians are half circles, portions of a great circle, one whose plane passes through the center of the Earth. 10. All parallels are small circles except the equator, which is a great circle. 11. 75° for Eastern Time Zone, 90° for Central, 105° for Mountain, and 120° Pacific. 12. Ante meridiem and post meridiem. Greenwich Mean Time and Coordinated Universal Time. 13. The year is generally defined as the elapsed time needed for the Earth to make one complete orbit around the Sun. However, the year can be defined in different ways: The tropical year, the year of seasons, is the time between successive vernal equinoxes. The sidereal year is defined as the time it takes Earth to complete one orbit around the Sun (i.e. the orbit period of the Earth) with respect to a distant star. 14. To have more hours of sunlight when most people are awake, which also conserves energy. DST encourages more outdoor activities and allows better and safer travel in the evening. All states observe DST except Hawaii and Arizona. The territories of American Samoa, Guam, the Northern Mariana Islands, and Puerto Rico also do not observe DST. 15. A solar day is the elapsed time between two successive crossings of the same meridian by the Sun. A sidereal day is for a star other than the Sun. The solar day is longer than a sidereal day by about 4 minutes. 16. When one crosses the IDL traveling westward, the date is advanced to the next day; when one crosses the IDL traveling eastward, one day is subtracted from the present date. 17. Earth completes its rotation in 24 h i.e., 360°/24h = 15°/ h 18. A ship's longitude can be measured by a marine chronometer, an accurate clock.
19. The point directly over your head. The altitude angle is the angle above the horizon, between the horizon and "the line" while the zenith angle is angle between "the line" and zenith. The altitude and zenith angle make up the angle between the horizon and the zenith i.e., 90°. 20. Estimating the position by dead reckoning. 21. (a) Between 23.5°N and the equator. (b) Between 23.5°S and the equator. 22. The Babylonian calendar is based on the motions of the Moon. The Gregorian calendar is an improved Julian calendar, both of which are based on the position of the Sun. 23. About June 21 and December 22. 24. (a) 23½°N; (b) 23½°S; (c) and (d) 0°. 25. The time for the Moon to go through its cycle of phases. 26. Julian calendar: adopted by Julius Caesar in 46 B.C. with introduction of leap year. Gregorian calendar: adopted by Pope Gregory XIII in 1582 to correct a 10-day discrepancy, along with the addition of a century leap year correction. 27. Every four years, except century years not divisible evenly by 400; that is, there are 97 leap years in every 400 years. 28. The lighted and dark portions of the Earth are equal in each hemisphere during the spring and fall equinoxes. During northern hemisphere summer, the boundary of the circle of illumination moves such that the North Pole region receives 24 hours of daylight. The opposite happens in winter. 29. 2400 A.D. 30. The pole star changes over the precession period. 31. Earth’s precession is caused by gravitational forces (torques) from both the Sun and the Moon acting on the Earth. This occurs because Earth is not a perfect sphere. 32. 25800 years
ANSWERS TO VISUAL CONNECTION (a) North Pole, +90°; (b) South Pole, -90°; (c) Arctic circle, +66.5°; (d) Antarctic Circle, -66.5°; (e) Parallel, +45°; (f) Tropic of Capricorn, -23.5°; (g) Tropic of Cancer, +23.5°; (h) Equator, 0°; (i) Meridian, 90° W
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Sundials measure apparent solar time. The sundial tells the time of day when there is sunlight by the apparent position of the Sun in the sky. 2. The altitude of the North Star equals the latitude. 3. Yes, it is possible. 1° latitude = 111 km = 60 nautical miles which is constant everywhere but 1° longitude = 60 nautical miles near the equator it reduces to zero at poles. 4. It is hotter in Texas because the sunlight is more direct (more intense). 5. June 21, the summer solstice – the Sun travels the longest path through the sky. In the Northern Hemisphere, the North Pole is tilted 23.4 degrees toward the Sun. The shadow will move towards north side. 6. (a) Yes. Parallels are complete circles. (b) No. Meridians are only half circles. 7. Born in a leap year, only a birthday every 4 years, and even less for a century year not divisible by 4. 8. Precession of the Earth’s axis.
ANSWERS TO EXERCISES 1. Place A: 50°N, 90°W 90° – 50° = 40° • N.P. 90° + 60° = 150° Place B: 60°S, 90°E 150° + 40° = 190° 2. Place A: 20°N, 75°W 20° – 0° = 20° Place B: 30°S, 75°W 30° – 0° = 30° 20° + 30° = 50° 3. 190° × (60 n mi/1°) = 11400 n mi 4. 50° × (60 n mi/1°) = 3000 n mi
5. Opposite coordinate of 40°N = 40°S. The same for 116°E = 180 - 116 = 64°W. Hence, the opposite coordinates are 40°S, 64°W.
6. 24°N, 133°E 7. Taking the central meridians of the time zones: 120° – 75° = 45°, or 3 hours, so 3 P.M., Oct. 8. 8. Taking the central meridians of the time zones: 105° – 75° = 30°, or 2 hours earlier, so 11 P.M., Oct 15 (crossed midnight). 9. Work the same way as you did Problem 7. Tokyo is in the 135°E time zone. The answer is 4 A.M. the next day—November 27. 10. Work the same way as you did Problem 7. The same figure can be used again (as Fig. 15.9). The answer is 2 A.M. the next day—February 23. 11. 5 a.m., MST 12. 10 p.m.
13. Drawing a time circle and counting from 105°W (Utah on MST) to 15°E, the time and date would be 3 A.M., Feb. 23. 14. Drawing a time circle and counting from 0° to 15°E, the time and date would be 7 A.M., July 1. 15. Since its June 21st, Latitude (L) = Zenith (Z) + Sun degrees (S). This gives, Z = L – S = 49° - 23.5° = 25.5°. Now altitude (A) = 90° - 25.5° = 64.5°. 16. Since its December 22nd, Latitude (L) = Zenith (Z) - Sun degrees (S). This gives, Z = L + S = 49° + 23.5° = 72.5°. Now altitude (A) = 90° - 72.5° = 17.5°. 17. June 21 the Sun’s declination is 23.5°N. Altitude = 71.5° so Zenith angle = 90° – 71.5° = 18.5°N. Hence, latitude = 23.5° + 18.5° = 42°N. 18. December 22 the Sun’s declination is 23.5°S. Altitude = 65.5° so Zenith angle = 90° – 65.5° = 24.5°N. Hence, latitude = 24.5° - 23.5° = 1ºN. 19. On or about June 21, 74.5ºN .
θ = 39° – 23.5° = 15.5° 90° – 15.5° = 74.5ºN (Maximum altitude) 20. On or about December 22, 27.5°N 21. The Sun’s altitude in the sky at noon changes by an average of 0.26°/day. Sun is overhead at 0° on March 21. From March 21 to May 10 there are 50 days. So, the latitude on May 10 would be: (0.26°/day)(50 days) = 13°N. 22. 18° S 23. On December 21, the sun is overhead at 23.5°S. Zenith angle = 90º - 86.5º = 3.5°. Latitude = 3.5º - 23.5º = 20°S. Actual time of sailing = 12 - 7 = 5 h. The movement of sun is 15°/h towards the west; 5 h = 75°. Hence, the location of the ship is 20°S and 75°E (In the Indian Ocean). 24. On June 21, the sun is overhead at 23.5°N. Zenith angle = 90º - 66.5º = 23.5°. Latitude = 23.5º + 23.5º = 47°N. Actual time of sailing = 12 - 2 = 10 h. The movement of sun is 15°/h towards the west; 2 h = 150°. Hence, the location of the ship is 47°N and 150°E (In the North Pacific Ocean).
Chapter 16 THE SOLAR SYSTEM In this chapter, we essentially begin our study of astronomy as we move out into space. The chapter begins with a definition of astronomy followed by some historical considerations and Kepler’s laws. This sets the stage for comparing and contrasting the properties of the planets. Tables 16.1 and 16.2 are presented to summarize the properties of and comparisons between the terrestrial and Jovian planets. Dwarf planets are also defined and the few that
have been discovered are highlighted. Also, the origin of the solar system is considered, along with the latest reported information concerning newly discovered planetary systems beyond our solar system. The two highlights explain how spacecraft travel to the outer solar system and how astronomers discover exoplanets. The conceptual questions and answers discuss how the Foucault pendulum works and how the rotations of gaseous planets are determined.
DEMONSTRATIONS Make a model solar system that shows motion of planets. Build a Foucault pendulum. Take care to use a long wire attached to a ceiling with a pivot. Use a large sphere to illustrate rotation and revolution of the Earth. Also use a large sphere to illustrate Eratosthenes' calculation. Show a class demonstration of parallax (finger of each hand held together, then one held close to the face and the other at arm's length).
ANSWERS TO MATCHING QUESTIONS a. 15
b. 4
c. 21
d. 10
e. 22
f. 6
g. 24
h. 8
i. 2
j. 11
m. 20
n. 7
o. 19
p. 13
q. 14
r. 18 s. 3
t. 12
u. 17
v. 1
k. 16 l. 5 w. 23 x. 9 y.
25
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. a 3. c 4. b 5. d 6. b 7. b 8. c 9. c 10. a 11. b 12. b 13. c 14. c 15. b 16. c 17. d 18. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Astronomy 2. geocentric 3. major axis 4. orbital period 5. inferior 6. prograde 7. Conjunction 8. 0.31 9. Foucault 10. Earth 11. greenhouse effect 12. iron oxide 13. ice giants 14. Ceres 15. Eris 16. condensation 17. Doppler effect
ANSWERS TO SHORT ANSWER QUESTIONS 1. Planets, moons, asteroids, meteoroids, dwarf planets, comets, and dust.
2. A geocentric model is Earth-centered; a heliocentric model is Sun-centered 3. When farthest from the Sun— conservation of angular momentum. 4. (orbital period)2 α (semimajor axis)3 5. The major planets orbit in a flat disk with elliptical, yet nearly circular, orbits around the Sun. All the planets revolve around the Sun counterclockwise. 6. Rocky planets near the Sun are placed in the terrestrial group, while the gaseous giant planets are in the Jovian group. 7. No it’s a superior planet. 8. No, it is an oblate spheroid—bulging at the equator. 9. The Earth is about 4 times larger than the Moon. 10. Rotation: Foucault pendulum. Revolution: Parallax and aberration of starlight 11. Mercury, Venus, the Earth, and Mars all have rocky surfaces, high densities, orbits close to the Sun, metallic cores. These characteristics resemble those of the Earth. 12. The Venusian surface is really hot (467°C or 873°F) due the large amount of carbon dioxide in the atmosphere which produces a large ―greenhouse effect‖. 13. Earth’s Grand Canyon is 277 mi long and was carved from flowing water. Mars’ Valles Marineris is a system of canyons about 2500 mi long crack in the surface. 14. Terrestrial—solid (more dense) and small. Jovian—gaseous and large. Jupiter, Saturn, Uranus, and Neptune are collectively known as the Jovian planets. They possess strong magnetic fields, have many moons and rings, and are very distant from the Sun, with orbits far apart from one another. 15. The rings of Saturn are clearly larger and more evident than the rings of the other Jovian planets. Saturn’s rings are composed of ice and rock material which makes it much more reflective and easier to see in the sunlight. The rings of Jupiter, Uranus, and Neptune are much fainter due to the smaller amount of material they contain as well as the composition being dark, dusty material. Each set of rings show individual bands and gaps due to the influence of gravity and the presence of shepherd moons. 16. Uranus’ axis lies almost in its orbital plane. 17. Methane in the atmosphere absorbs red light, reflecting blue light. 18. Dwarf planets are spherical objects orbiting the Sun that have not fully cleared their orbit path.
19. Pluto has not cleared its orbit (it crosses the orbit of Neptune). Pluto is small, cold and has a noncircular and highly tilted orbit. 20. A flattened region of dwarf planets and comets beyond the orbit of Neptune 21. The early cloud of dust and gas, out of which the Sun and the planets formed. Liquids and solid material condensed out of this early cloud to form the planets. 22. Interstellar dust was key in allowing the condensation process to work. 23. Astrometry is the branch of astronomy that deals with the measurement of positions and motions of celestial objects, to measure their wobble. 24. The Doppler Effect and transits are methods to detect exoplanets.
ANSWERS TO VISUAL CONNECTION (a) Jupiter, (b) Mercury, (c) Sun, (d) asteroid belt, (e) Mars, (f) Saturn, (g) Uranus, (h) Earth, (i) Neptune, (j) Venus, (k) Pluto
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The advance knowledge of rocket fuels, computers, communication, and the initiative of political leaders and the public to fund such research, has helped in expanding our knowledge of the cosmos including the solar system. Many developed space probes and ground based telescopes have majorly helped in understanding the solar system like the Juno spacecraft revealing surprising truths about Jupiter's clouds.
2. This Foucault pendulum is swinging in a plane that is parallel to the Earth’s spin axis. The pendulum also swings in a plane fixed with a distant star, Polaris, the North Star in this case. As the Earth rotates the pendulum remains in the same plane, fixed with the north-south plane. Therefore, the pendulum does not rotate. 3. A complete set of phases are observed for Mercury and Venus because the angle between the planet-Sun-Earth viewing angle changes significantly, between 0° and 180° which allows us to look at full range of phases of these planets. However, for the outer planets, the viewing angle does not change significantly and we see the outer planets in a near-full phase all the time.
4. Astronomers used hypotheses and tests in order to justify the changes to a planet’s definition. Reclassifying allows scientists to better compare properties of similar objects, learning more in the process. 5. The solar nebula theory suggests that the nebula contracted under the influence of its own gravity, began rotating, and then flattened into a swirling disk of gas and dust. Through the process of condensation, the nebula evolved into the system we observe today. The stars like the Sun spin from the angular momentum that was there in the solar nebula from which it formed. All orbital motion of the planets (including the spin) is due to this original angular momentum. 6. One method uses astrometry (measuring the wobble) - As the star moves through space (galaxies and stars move), the gravitational tugging shows up as tiny deviations from the straight-line path that a star without a planet(s) would follow, and the star is observed to wobble slightly. A second method of detection is to measure the Doppler shift of the starlight as the star moves towards and away from us due to the planet's and the star's revolution about their center of mass (a gravitational balance point). A third technique using gravitational wobble is to measure time. Pulsars are ultradense, rapidly spinning stars that emit radio waves at very regular time intervals. By accurately measuring the time of arrival of these pulses, planets can be discovered from variations in these radio signals. The most successful method is the transit method. Light from a star is dimmed if a planet passes in front of the star (transit) and causes an eclipse. The amount of the light dimmed and the length of time of the eclipse can give clues to the size, mass, and atmosphere of the orbiting planet.
ANSWERS TO EXERCISES 1. T2 = 𝑘𝑎3 = ( 2. T2 = 𝑘𝑎3 = (
1 𝑦2
) × (10 𝐴𝑈)3 = 1000 𝑦2; 𝑇 = √1000 𝑦2 = 31.6 y
(𝐴𝑈)3
1 𝑦2 (𝐴𝑈)3
) × (518 𝐴𝑈)3 = 138991832 𝑦2; 𝑇 = √138991832 𝑦2 =
11789.5 y 3. T2 = 𝑘𝑎3; 𝑎3 =
𝑇2 𝑘
4. R = 30.1 AU
5. T = 2.8 y
3
3 = (0.62 𝑦)2 ( (𝐴𝑈) ) ; 𝑎 = √0.3𝐵 𝐴𝑈 3 = 0.724 AU
1𝑦2
6. T = 8 y 7. 260 million miles 8. From Table 16.1, for Jupiter, 𝑅𝐽 = 5.2 AU, and for Mars, 𝑅𝑀 = 1.5 AU. So, from Exercise 7 answer, the asteroid belt is closer to Mars. 9. T2 = kR3. Therefore, for smaller R, T would also be smaller. 10. T2 = kR3, so the smaller R means a smaller T. Smaller orbital periods would lead to greater orbital speeds because orbital speed is inversely proportional to time. 11. Mercury, Venus, Earth, Mars. 12. Jupiter, Saturn, Uranus, Neptune.
Chapter 17 MOONS AND SMALL-SOLAR SYSTEM BODIES Students should be given an outside assignment to observe the Moon for one month. Have them determine the time of meridian crossing, and record the rising and setting times of the Moon. The instructor should point out how the frame of reference is formed for the surface of the Moon. Many astronomy texts show the Moon reversed left to right, because many telescope pictures are reversed. All photos in this text are oriented to show what someone on the Earth would see if looking at the Moon with the unaided eye.
DEMONSTRATIONS Use a model solar system with a Moon that shows the motion of the Earth and the Moon and also a large Earth sphere and an 8-in.-diameter Moon that the instructor can move manually around the Earth sphere. Let the classroom wall be the Sun. Explain to the class the position of the Moon at new, first quarter, full, and last quarter phases. With the Moon at any of these positions, rotate the Earth to show the rising and setting times of the Sun and the Moon. Demonstrate with a model sphere of the Earth and the Moon, plus an incandescent lamp for the Sun, the positions necessary for solar and lunar eclipses.
ANSWERS TO MATCHING QUESTIONS
a. 24
b. 22 c. 17 d. 5 e. 12 f. 2
g. 19 h. 23 i. 14
j. 6 k. 8 l. 25 m. 4
n. 1
o. 13 p. 18 q. 15 r. 9 s. 21 t. 11 u. 10 v. 20 w. 7 x. 16 y. 3
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.b 2. c 3. b 4. b 5. a 6. c 7. a 8. b 9. d 10. a 11. a 12. c 13. b 14. c 15. a 16. d 17. b 18. d 19. d 20. c 21. c 22. a 23. a 24. b 25. c 26. e
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. maria 2. rays 3. later 4. 6 A.M. 5. first quarter 6. waning 7. solar 8. Earth’s 9. spring 10. twelve 11. zero 12. asteroids 13. Ganymede 14. Io 15. Titan 16. Charon 17. Water ice 18. asteroids 19. meteorite 20. nucleus
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Highlands - They are the light-colored rock surface easily seen on the Moon, Maria - They are large, flat areas believed to be craters formed by the impacts of huge objects from space and later filled with lava, Craters - They are numerous and are a well-known feature on the Moon’s surface. The floors of larger craters are flattened and form basins. 2. The lunar mare are lower elevation regions on the Moon, probably formed by huge impacts early in the Moon’s formation. These lowlands were later filled with lava which is darker-colored rock. 3. (July 20) 1969. There were five other moon landings. 4. 3.1 – 4.4 billion years. 5. There are two periods involved: First is sidereal month which is little over 27.3 days and the second is synodic month which is about 29.5 days. 6. Because the Moon revolves around the Earth (360°) in 27.3 days, it must travel 360°/27.3 days = 13.2°/day. So, the Earth must turn an extra 13.2° for the Moon to be in the same position in the sky as the previous night. And it takes around 52.8 minutes for the Earth to turn 13.2°. 7. Waxing: illuminated portion of the Moon growing or increasing.
Waning: illuminated portion of the Moon decreasing. 8. Crescent moon: less than one half of the observed surface illuminated. Gibbous moon: more than one half of the observed surface illuminated. 9. An orbit in which the rotation period equals the revolution period. 10. Umbra: region of total darkness of the Moon’s shadow. Penumbra: semidark region of the Moon’s shadow. 11. First quarter, Sun to west of Moon. Last quarter, Sun to east of Moon. 12. A partial eclipse of the Sun occurs when the Moon moves in between the Earth and the Sun, but alignment is not perfect. The penumbral (semidark) shadow is cast on the observer and only part of the Sun is covered. 13. The observer is in the Penumbra region. 14. Lunar eclipse: Sun, Earth, and Moon. Solar eclipse: Sun, Moon, and Earth. 15. Differential gravitational attraction on opposite sides. The Earth is ―pulled away" from water on the opposite side. 16. Spring tides: at new or full phases, greatest variation. Neap tides: at first or lastquarter phases, minimum variation. 17. Three; the Earth has 1 while Mars has 2 moons.18. Two: Phobos and Deimos. 19. Four. These four largest moons on the planet Jupiter were discovered by Galileo. 20. Jupiter’s Io. This is due to timed gravitational tugs between Jupiter and the Moons Europa and Ganymede which causes stresses that continually flex the Moon’s interior. 21. Ganymede, Jupiter. 22. Saturn’s large Moon Titan is unique because it has a thick hydrocarbon atmosphere, mostly methane gases. The hydrocarbon smog may have some interesting chemistry going on. The surface of Titan also shows channels and lakes of liquid methane. Evidence also suggests it might have a liquid subsurface ocean. 23. Neptune’s moon Triton is interesting because it has geologic activity in the form of geysers spewing out of its icy surface. The surface shows complex landforms and poles of frozen nitrogen. Triton also orbits retrograde (backward) and thus is believed to be a gravitationally captured moon. 24. Titan, Saturn; Titania, Uranus; Triton, Neptune.
25. Rotation and revolution are the same (like the Moon and the Earth). 26. Hydra 27. Eight times smaller 28. Dwarf planets are probably composed primarily of water ice mixed with some rocky material. Because of their great distance from the Sun, the composition is similar to comets. 29. Most asteroids are between the orbits of Mars and Jupiter. 30. A meteor when it enters the Earth’s atmosphere, and a meteorite if it strikes the Earth’s surface. 31. Comet tails are composed of two different materials: gas and dust. The gases are charged particles of hydrogen, oxygen, nitrogen, and carbon. The dust particles are very small grains of carbon-type particles. 32. Short-period comets have orbital periods of less than 200 years while long-period comets have orbits that lie within the inner solar system.
33. 76 years. 34. Zodiacal light is a faint band of light along the zodiac whereas the gegenschein (German for ―counterglow‖) is also due to sunlight reflected from dust particles. This faint glow is observed exactly opposite the Sun. Appearing as a diffuse, oval spot, it is more difficult to observe than zodiacal light.
ANSWERS TO VISUAL CONNECTION (a) new, (b) waning crescent, (c) 3rd quarter, (d) waning gibbous, (e) full, (f) waxing gibbous, (g) 1st quarter, (h) waxing crescent
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes. You will observe phases of Earth similar to the phases we observe of the Moon with the new phase beginning when the Moon, Earth, and the Sun are in the same plane with Earth between the Sun and the Moon. 2. The phases of the Moon would essentially occur in the opposite order if the Moon orbited backward. Following the new Moon, would be the crescent, then the 3rd quarter, gibbous, full, gibbous, 1st quarter, crescent, then back to the new Moon. But
the 3rd quarter moon would have to be renamed to 1st quarter since it comes first. The appearance of the phases would not change, but the direction the Moon moves in the sky would be westward instead of eastward. 3. The crescent Moon is pointing in the wrong direction. It appears the sunlight is coming from above. 4. Kirkwood gaps are caused by gravitational interaction between Jupiter and asteroids with simple fractions (e.g. 1/2, 1/3, etc) of Jupiter's orbital period. The effect of these mean-motion resonances is a change in the asteroid's orbital elements (particularly semimajor axis) sufficient to create these gaps in semimajor axis space. 5. Mountains and canyons on Pluto’s moon Charon are evidence of cryovolcanism, icy surface features melted and shifted due to internal stresses from gravitational energy.
ANSWERS TO EXERCISES 1. Approximately 133 N. 2. Approximately 27 1b. 3. Approximately 354 days (29.5 × 12). 4. Approximately 330 days (27.5 × 12). 5. 6:50 A.M. 6. The Moon rises about 50 minutes later each successive day. After seven days the Moon will rise about 5 hours 50 min. later. So approximately, 11:50 P.M. 7. (a) The right side or west side is bright, because the Sun has just set in the west. (b) It would be the same phase in Australia and the west side would be bright, but this is the left side. 8. (a) The eastern side or the left side. (b) The same phase (last quarter), and the east or right side would be bright. 9. Halfway through waxing crescent phase is 1/8 cycle. Last-quarter phase is 3/4, or 6/8, cycle. So, 6/8 – 1/8 = 5/8, and 29.5 days (5/8) = 18 days. 10. From the start of a waning gibbous phase to the start of the waning crescent phase, is 1/4 cycle. Then, 29.5 (1/4) = 7.38 days. 11.
12.
13. (a) High tides are approximately 180° apart. 180° from 84°W is 96°E. (b) Longitude 90° east and west of 96°E is 6°E and 174°W. 14. (a) 180° from 95°W is 85°E. (b) Longitude 90° east and west of 85°E is 5°E and 175°W.
Chapter 18 THE UNIVERSE The universe as viewed from planet Earth is composed of stars in the Milky Way galaxy and other galaxies. The galaxies, in turn, form the myriad of galaxy clusters that are observed throughout the depths of space. This chapter discusses these basic building blocks (stars and galaxies) and presents information on the structure, shape, form, age, and extent of the known universe.
The origin of the universe is slowly being understood, and the basic ideas behind our understanding are presented in this chapter. The beginning of the chapter is devoted to the concept of the celestial sphere and celestial coordinates. After a thorough overview of our star (the Sun), other stars are classified and compared. The birth, life, and death of stars are fairly well understood, and the discussion includes low and high mass stars and their fates (white dwarfs, neutron stars, and black holes).The Hertzsprung-Russell diagram should be explained and made familiar to the student. A take-home open-book quiz is an excellent way for the student to learn how the diagram is plotted and what it explains. This chapter also presents information on the Milky Way galaxy, other galaxies, the expansion of the universe, dark energy, and dark matter. The latest information on the modification of the standard model of the Big Bang is also discussed. Highlights of how astronomers determine distances and the age of the universe are included. Three conceptual questions and answers help students overcome misconceptions in astronomy.
DEMONSTRATIONS Use a transparent globe to show star clusters and the motion of the Earth and the Sun with respect to the stars. Color slides showing the spectrum of a nearby star and a star moving rapidly away from the Earth are helpful in illustrating the Doppler shift. A variety of pictures of celestial objects is available on the Internet. Models of the galaxy can also be drawn and cut out on cardboard; calculations of the overall size and scale are a good way to give perspective. A good demonstration of the expansion of the universe is made by grid lines drawn onto a transparency, then enlarge this by 20% on a photocopy machine. Make another enlargement of 40% onto another transparency. Overlay the transparencies to demonstrate how the spatial grid (the universe) expands.
ANSWERS TO MATCHING QUESTIONS a. 18 b. 25 c. 6
d. 15 e. 3 f. 21 g. 1 h. 13
n. 17 o. 12 p. 5 q. 7 aa. 9
r. 24 s. 14 t. 10
i. 8 j. 22
u. 16
v. 20
k. 27
l. 4 m. 11
w. 23
x. 2 y. 26
z. 19
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.a 2. a 3. d 4. b 5. d 6. a 7. a 8. b 9. d 10. d 11. c 12. a 13. b 14. b 15. a 16. d 17. b 18. d 19. c 20 a
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. parallax 2. Polaris 3. right ascension 4. declination 5. corona 6. helium 7. Flares 8. constellations 9. blue 10. spectrum 11. brightness 12. main sequence 13. Red giants 14. Planetary nebula spiral
15. Nebula 16. nova 17. neutron star 18. supernova 19. barred
20. 13.8
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Cosmology is the study of the structure and evolution of the universe. 2. Declination, right ascension, distance. 3. The ecliptic and celestial equator are tilted by 23.5°. The reason for this is the 23.5° tilt of the Earth’s axis. 4. The vernal equinox is the point of intersection of the celestial equator and the ecliptic, where the Sun is moving northward. The celestial prime meridian is the great circle passing from the North Celestial Pole to the South Celestial Pole, perpendicular to the celestial equator and intersecting the celestial equator at the vernal equinox. 5. The parsec is larger. One parsec equals 3.26 1y. 6. Stellar parallax is caused by the motion of the Earth about the Sun. In order to calculate distance to a star, astronomers measure the parallax angle of the star, and, using the known Earth-Sun distance, triangulation is used to solve the distance to the star. 7. 5800 K. 8. The estimated year of the next solar maximum is 2025.
9. The Sun’s corona is visible during a total solar eclipse.
10. The diameter of Sun is approximately 100 times bigger than Earth. The Sun has a volume = M/D = 1.43 × 1027 m3. Also, Earth has a volume of 1.1 × 1021 m3. Hence, Sun is so large that it could fit = 1.33 million Earths 11. The Core, the Radiation and Convection zones, the Photosphere (―surface‖), the Chromosphere, the Transition region, and the Corona. 12. 41𝐻 → 4𝐻𝑒 + 2( 0𝑒 + 𝑣̅ + ) 1
2
+1
13. A star is a giant ball of gas which is stable because of the balance of two forces. Pressure pushes the gas outward, but gravity pulls the gas inward. 14. Apparent magnitude is a measure of the brightness of a star as observed from the Earth. Absolute magnitude is a measure of the brightness a star would have if it were positioned 10 parsecs from the Earth. 15. Over one day, the Moon, the stars, and the constellations move across the sky from East to West. Because Earth rotates eastward, the objects in the heavens will move westward, rising in the East and setting in the West. 16. Proxima Centauri is 4.3 ly distant. 17. O, B, A, F, G, K, M. 18. A main sequence star is Sirius, a red giant is Arcturus, a supergiant is Betelgeuse, and a white dwarf is Procyon B. 19. An emission nebula is a bright nebula in which the emission comes from hydrogen, oxygen, and other atoms ionized by the energy from nearby stars. A reflection nebula is a bright nebula in which the energy of the nearby stars is not sufficient to ionize these gases, and the dust just reflects and scatters the starlight. 20. The Sun has the fuel supply of a solar-mass star; it is expected to become a red giant in about 5 billion years. The Sun will become so large that it will engulf the orbits of Mercury, Venus, and probably the Earth. 21. Nucleosynthesis is the creation of nuclei inside stars. 22. The mass of a star determines its lifetime. The mass determines how big and hot the star will get during its life. The temperature is very important because the temperature determines how quickly the nuclear reactions take place in the core of the star. The faster these nuclear reactions take place, the shorter the lifetime of the star.
23. A brown dwarf is a failed star, one unable to sustain fusion long enough to move onto the main sequence. About 100 brown dwarfs have been identified. 24. Protostar, main-sequence star, red giant, planetary nebula, white dwarf. 25. A nova is an explosion from the surface of a white-dwarf star in a binary star system. A supernova is a violent stellar explosion that can shine as brightly as an entire galaxy of billions of normal stars. A supernova is a much more powerful explosion than a nova. 26. A typical white dwarf is half as massive as the Sun, yet only slightly bigger than Earth. The White dwarf is 200,000 times as dense as the Earth. A neutron star has a mass of about 1.4 times that of our Sun and is about 20 km in diameter. This means that a neutron star is so dense that on Earth, one teaspoonful would weigh a billion tons. 27. A pulsar is a rapidly rotating neutron star. It emits energy along its magnetic poles that are spinning around in space. When that pole is pointing directly towards Earth, we receive a burst of emission; the emission turns on and off as the pulsar rotates. This is similar to a lighthouse because the light in a lighthouse turns ―on‖ and ―off‖ as it rotates around. 28. (The sketch should resemble Fig. 18.22.) 29. A black hole does not emit energy so it cannot be ―seen‖. However, astronomers can detect the presence of a black hole by its gravitational influence on nearby objects. When a black hole is in a binary system with a star, captured hot gases form an accretion disk just outside the event horizon, and emit X-rays as they orbit and spiral into the black hole. Another method is by using gravitational waves (ripples in space-time) which can be detected when two black holes collide. 30. Spiral, elliptical, irregular. 31. Quasars are the cores of galaxies that were forming when the universe was young. They are the most distant objects observed. 32. The two types of spiral galaxy are the normal and barred type galaxies. The normal types have a round nuclear region with spiral arms that curve out from the nucleus. The barred spiral galaxies have an elongated nucleus (shaped like a rectangular bar) with the spiral arms curving out from the ends of the nuclear bar.
33. Dark matter is unobserved matter hypothesized by astronomers to explain why a cluster of galaxies exists as a gravitationally bound system. Dark energy seems to be accelerating the expansion of the universe. 34. The Milky Way is a barred spiral galaxy about 100000 ly in diameter. Its disk is about 2000 ly thick, and its nuclear bulge is about 13000 ly thick. 35. In a halo around the Milky Way lie about 200 globular clusters, which are spherical collections of a few tens of thousands of stars. Their distribution led to the conclusion that our solar system was located in a spiral arm of the Galaxy, not at its center. 36. Dark matter is one of the greatest mysteries in astronomy. The composition of dark matter is unknown at this time, but it is believed to exist in two forms: ordinary matter and exotic matter. The ordinary matter is believed to be small, dense objects that we cannot see because they are too dim. The exotic matter is believed to be subatomic particles that we have and have not discovered. 37. Astronomers classify the Milky Way as a barred spiral galaxy, type b (SBb). 38. The Hubble-Lemaître law states that the greater the recessional velocity of a galaxy, the farther away the galaxy. This indicates that the universe is expanding. Astronomers can determine the age of the universe by using this law. It is calculated to be 13.8 billion years. 39. Galaxies are not evenly distributed in space but congregate in huge superclusters and long filaments that are separated from each other by huge voids containing very few galaxies. 40. The cosmological redshift, the cosmic microwave background, and the H/He ratio. 41. The most recent modification to the Big Bang model is called the inflation model. It differs from the standard model in that it includes a tremendous release of energy at 10-35 seconds after the big bang. This tremendous release of energy caused the universe to expand (inflate) very rapidly and thus flatten space-time. The release of energy is thought to be caused by the breakup of a unified force as the universe cooled slightly from the beginning. 42. The universe is about 71.4% dark energy, 24% dark matter, and 4.6% ordinary matter.
ANSWERS TO VISUAL CONNECTION (a) Luminosity (LSun); (b) Temperature (K); (c) White Dwarfs (hotter, dim, smallest, blue-white color); (d) Main Sequence (upper part: hotter, blue, bright, massive; middle part: Sun-like; lower part: cooler, red, dim, least massive); (e) Giants (cooler, bright, red); (f) Supergiants (cooler, bright, red, largest-size)
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Astronomers took a long time to accurately measure the distances to stars because the distances are so far away. It took more modern technology and telescopes to provide precise enough measurements. Note however that the basic method to measure stellar distances was known very well by the Greeks over 2000 years ago. 2. If Sirius exploded into a supernova it would be a cataclysmic event for anything within about 50-100 light-years distant. The Crab supernova released so much energy from about 6500 light-years from Earth that it was brighter than the full Moon for two weeks. Sirius is only 8.7 light-years from Earth; it would release so much energy in the form of gamma rays and x-rays that it would probably destroy all life on the Earth. 3. Star A is blue, star B is yellowish, and star C is reddish. Because of the StefanBoltzmann law (σT4), higher temperature stars emit more light. Star A is the brightest star. 4. Gravity waves, or distortions in space created by supernova energy events, travel at the speed of light, which allows scientists to determine the distance to far-away galaxies. The property of supernovae which is very useful in this method is the extreme brightness. 5. Composition of white dwarf is determined by the X- ray spectroscopy and ultraviolet spectroscopy. 6. It will be several billion years before the Sun reaches the red giant phase. 7. The future of the universe depends on the overall energy density. If the density is over a critical value, then the universe will stop expanding and collapse. If the density is the critical value, then the universe will continue to expand forever, but take an infinite amount of time to do so. If the density is less than the critical value,
then the universe expands forever. With the recent discovery of the dark energy accelerating the expansion of the universe, the estimated future of the universe is a faster expansion over time.
ANSWERS TO EXERCISES 1. d = 1/p = 1/0.20 arcsec = 5.0 pc 2. d = 1/p = 1/0.38 arcsec = 2.6 pc (= 8.6 ly) 3. (365 d/y)(24 h/d)(60 min/h)(60 s/min) = 3.15 × 107 s/y 4. (1.86 × 105 mi/s)(3.15 × 107 s/y) = 5.86 × 1012 mi/ly 5. 196 pc 6. (140 pc)(3.26 ly/pc) = 456.4 ly 7. –3 is 10 magnitudes brighter than +7, and so it is 100 × 100 = 10000 times brighter. 8. Venus is 5 magnitudes brighter than the star, so it is 100 times brighter 9. 211 Mpc Use Hubble’s Law Vr = Hd, where H = 71 km/s/Mpc. The distance to the galaxy cluster can be determined from d = Vr/H = 15000/71 = 211 Mpc (mega-parsecs). 10. For the Hydra supercluster having a distance of 800 Mpc, the recessional velocity is calculated from Vr = Hd, Vr = 71 × 800 = 56800 km/s. 11. The age of the universe is calculated from Hubble’s constant, the rate of expansion of the universe. If H = 100 km/s/Mpc then, Age = (9.78 × 1011)/H = (9.78 × 1011)/100 = 9.8 × 109 years = 9.8 billion years. If H = 50 km/s/Mpc then, Age = (9.78 × 1011)/50 = 0.1956 x 1011 years = 19.6 billion years. 12. If H = 75 km/s/Mpc, the age of the universe is Age = (9.78 × 1011)/75 = 0.1304 x 1011 years = 13 billion years
Chapter 19
THE ATMOSPHERE For students to understand and appreciate weather principles and phenomena, it is essential that they have a knowledge of the fundamental physical properties of the atmosphere. The current
emphasis on the environment offers a timely introduction to the study of the atmosphere. Everyone is aware of atmospheric pollution problems (Chapter 20); however, to really understand, students should know the normal conditions of our atmosphere that are being jeopardized. These conditions are presented in this chapter. The chapter discusses the composition of the atmosphere and how the relative percentages of the major constituents are maintained. Attention is given to the physical properties that distinguish one part of the atmosphere from another, particularly the divisions based on temperature. The section on the energy content of the atmosphere gives the student an understanding of what is responsible for the dynamics of the atmosphere. As pointed out in Chapter 1, measurement is necessary to describe conditions, and this also applies to the atmosphere. Common atmospheric measurements, with which the student should be familiar, are discussed. The concept of relative humidity and the dew point temperature should be covered thoroughly. The last two sections of the chapter are concerned with the movements of the gases of the atmosphere-winds and air currents-and the associated formation of clouds. Convection cycles and their relationship to local winds and world circulation patterns have important effects to which the student can relate. This is what makes the study of meteorology so interesting. Clouds are one of the most common sights in a student’s environment. This promotes interest in the study of various cloud types and characteristics. A knowledge of cloud formation is prerequisite to understanding precipitation processes, mechanisms, and types, which will be considered in the next chapter.
DEMONSTRATIONS 1. Atmospheric Pressure A dramatic demonstration of atmospheric pressure is the crushing of a metal can. There are two variations of this demonstration. One way is to use a gallon metal can (e.g., a well-rinsed paint thinner can). Place a small amount of water in the can and heat the open can sitting on a ring stand with a Bunsen burner. When the water is boiling vigorously, as evidenced by condensed water vapor coming from the can (not steam, which is invisible), turn off the burner and place a rubber stopper securely in the can opening. (A rubber stopper is recommended over a screw, gasketed cap. The rubber stopper is safer should you forget to turn off the burner.) As the can cools and the steam condenses, a partial vacuum forms and the pressure difference causes the can to slowly collapse. After the can is crushed, slowly remove the stopper. An audible rush of air illustrates the partial vacuum in the can (like that in a vacuum-packed coffee can).
A quicker, and perhaps more convenient, way to demonstrate atmospheric pressure is to use an aluminum soft drink can. (Large metal cans are sometimes difficult to obtain.) Place a small amount of water in the rinsed can, and hold the can over the flame of a Bunsen burner using a pair of tongs. Upon hearing vigorous boiling, quickly invert the open top of the can into a container of water. The can is crushed immediately. When the can is raised from the water, water will run from the can, having been forced into the partial vacuum. (Have an extra can available, because students will want to see the demonstration repeated.) The hole of a small can may also be stopped with a piece of clay. 2. Dew Point and Atmospheric Pressure For a classroom demonstration of dew point and atmospheric pressure, an iced soft drink in a glass with a straw can be used. After studying the section on humidity, the student should be able to explain why condensation occurs on the glass. This demonstrates that the air in the vicinity of the glass has been cooled to its dew point, and because the air is saturated (100% relative humidity), the water vapor condenses on the glass. Drinking through a straw demonstrates atmospheric pressure. Aspiration through the straw reduces the pressure within, and liquid is forced up the straw by the pressure difference relative to the outside atmospheric pressure. This can be effectively shown by punching a small hole in the side of the straw. This equalizes the pressure and makes the straw useless. 3. Humidity A simple psychrometer can be easily constructed. The only essential equipment is two thermometers, a water reservoir, and a cloth wick to keep one of the bulbs wet. Care should be taken not to place the wet bulb too near the water reservoir, as this will impair free evaporation. One method is to use a paper quart milk container as a reservoir and attach the thermometers to the side using garbage bag ties. A hole can be made for the wick to extend into the water. The relative humidity and the dewpoint temperature can be determined from the difference in the thermometer readings by using the tables given in the text Appendix VIII. This demonstration can be set up during one class period and a reading taken the next, to allow time for the evaporation rate to come to equilibrium. The students themselves can construct a psychrometer as a project, taking readings in their dormitory for a period of days and plotting the results to show the humidity variation over time. 4. Weather Instruments
Weather-measuring instruments, such as barometers and anemometers, can be used to demonstrate how measurements are obtained. See the companies listed in the Teaching Aids section. 5. Convection Cycles A commercially available product called a Lava-Lite works well in demonstrating convection cycles. The Lava-Lite consists of a lamp base with an upper portion containing liquids of different densities. One liquid is colored and rises in the other as a result of heat from the lamp. As the globules of liquid rise and cool, they fall back to the bottom of the container. The LavaLite should be turned on an hour or more before the scheduled demonstration, as time is required to initiate the action.
ANSWERS TO MATCHING QUESTIONS a. 3 b. 29 c. 13 n. 20
d. 24
o. 2 p. 30
4 ab. 23
ac. 19
e. 7 f. 14
q. 28
r. 10
g. 18
s. 27
h. 22
i. 15
t. 9 u. 17
j. 6
v. 21
k. 25
w. 1
l. 16
x. 8
m. 11
y. 26
z. 12
aa.
5. 14.7
6. 76,
ad. 5
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.c
2. c
3. b 4. b 5. a
6. a 7. d 8. b 9. b
10. d 11. d 12. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Meteorology 2. stratosphere 3. thermosphere 30 7. maximum
8. opposite
9. isobar
4. carbon dioxide, methane
10. thermal circulation
11. counterclockwise
12. vertical development
ANSWERS TO SHORT-ANSWER QUESTIONS 1. 78% nitrogen, 21% oxygen, and 0.9% argon. 2. No, photosynthesis by plants replenishes the oxygen. 3. Chlorophyll. 4. (a) Decreases with increase in altitude (b) Increases with increase in altitude (c) Decreases with increase in altitude (d) Increases with increase in altitude. 5. It absorbs and protects us from harmful ultraviolet radiation in sunlight. 6. Solar disturbances, and the recombination of ionized gas molecules and electrons in the Earth’s upper atmosphere.
7. Earth has a larger albedo as it reflects 33% of incident sunlight than the Moon. The Moon has an albedo of 7%. 8. The atmosphere receives most of its direct heat from the Sun. The overall heating is accomplished in three main ways - (1) Absorption of terrestrial radiation, (2) latent heat of condensation, (3) conduction from Earth’s surface. 9. Extreme scattering through dense atmosphere mixes colors-- white 10. Least scattering, and so can be seen for the greatest distance. 11. The absorption and transmission properties of regular glass are similar to those of the atmospheric ―greenhouse gases‖ in general, visible radiation is transmitted and infrared radiation is absorbed. 12. A change in temperature shifts the wavelength of the radiation such that it is transmitted or absorbed, and the Earth cools and heats periodically. 13. Temperature, thermometer; Pressure, barometer; Humidity, psychometer; Wind direction and speed, anemometer. (Also, Precipitation, rain gauge.) 14. Atmospheric pressure determines the force applied to the liquid in a tube causing it to rise. The normal height of a mercury barometer is 76 cm (30 in.). Since water is less dense than mercury, it would need to be much taller than the mercury tower. 15. The temperature of the air in the vicinity of the glass is lowered to the dew point. 16. The heat index is the temperature we feel. This depends on how humid the atmosphere is, which may interfere with perspiration or our cooling down process. 17. In the direction from which the wind is coming, because of the fins. 18. The wind fills the pivoted sock, and the tail of the sock points in the direction toward which the wind is blowing. The angle of the sock relative to the horizontal is an indication of the wind speed. (With a high wind speed, it stands "straight out.") 19. Doppler radar can give wind speed and direction, whereas conventional radar cannot. 20. A thermal cycle set up by localized heating. Near a body of water, land breezes and sea breezes are set up during the night and day, respectively. 21. As viewed from above, in the Northern Hemisphere, cyclones rotate counterclockwise and anticyclones rotate clockwise. The rotations are opposite in the Southern Hemisphere. 22. Because of being in the westerlies wind zone. 23. West, upwind. 24. The windward side may need extra insulation.
25. (a) Low.
(b) High.
26. (a) Cirrocumulus.
(c) Middle.
(d) Low. (e) Vertical development.
(b) Cirrostratus.
(c) Cumulonimbus.
27. The rate of temperature decrease with altitude. Condensation takes place when the dew point of rising air is reached. Greater the lapse rate, the greater the cooling. 28. Most clouds appear white. This is because the water droplets scatter all the visible wavelengths (colors) of the sunlight. Together the colors make up white light. Clouds may be dark or gray when either a cloud is in another’s shadow or the top of a cloud casts a shadow on its own base. Nimbus clouds with dense, larger droplets appear dark.
ANSWERS TO VISUAL CONNECTION a. cumulonimbus g. altocumulus
b. cumulus h. stratus
c. cirrus
d. cirrocumulus
i. stratocumulus
j. nimbostratus
e. cirrostratus
f. altostratus
k. advection
l. radiation
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. (a) Specific heat capacity of water is higher than that of the earth. Lower the specific heat, lesser heat to lose or quicker heat loss. (b) Very dry and little water vapor or clouds to absorb terrestrial radiation and insulate from heat loss. 2. Without greenhouse gases in the atmosphere, Earth's average temperature would become 0º F (-18º C) and be covered in ice making the planet uninhabitable. This is because the greenhouse gases trap radiations from the Sun and act like a thermal blanket around the planet. 3. During rain, the cloud is in a saturated state. When subjected to ground state the precipitation is greater than the humidity state and hence the psychrometer reads below 100%. 4. The projectile would land to the east of its due north path. This variation would occur because the projectile was moving eastward faster at the Equator than was its target farther north. 5. If we place the palm of our hand tightly over the mouth of the cup, the pressure at the top of the water would increase and eventually exceed the pressure of the water leaking from the hole. This would stop the leaking. 6. The straw allows a reduction of the atmospheric pressure on the liquid, which pushes the liquid up the straw. When the finger is released, the atmospheric pressure on the liquid increases and pushes the liquid back out of the straw.
ANSWERS TO EXERCISES 1. (a) (50 – 16 km)/16 km = 2.1
(b) (80 – 50 km)/16 km = 1.9
(c) (200 – 80 km)/16 km =
7.5 2. The following unit conversions can be drawn or compared with fig. 10.4. (a) 14000 ft = 4.3 km = 2.7 mi. (b) 29000 ft = 8.8 km = 5.5 mi. (c) 35000 ft = 10.7 km = 6.6 mi. (d) 65000 ft = 19.8 km = 12.3 mi. (e) 400 mi = 643.7 km. (f) E layer 100 km = 62.1 mi; F layer 320 km = 198.8 mi. (g) 400 - 550 km = 249 - 342 mi. (h) 23000 mi = 37014.9 km. 3. (a) T = To – Rh = 20°F – (3.5°F/1000 ft)(29028.87 ft) = -81.6°F (where R is rate) (b) T = To – Rh = 20°F – (3.5°F/1000 ft)(39370.079 ft) = -117.8°F (where R is rate) 4. (a) T = To – Rh = 20°C – (6.5°C/km)(8.848 km) = -37.5°C (b) T = To – Rh = 20°C – (6.5°C/km)(12 km) = – 58° C 5. From tables H.1 and H.2: (a) 46%,
(b) 62°F,
(c) 12.7 gr/ft3, (d) AC = MC × RH =
(12.7 gr/ft3v)(0.46) = 5.8 gr/ft3 6. From tables H.1 and H.2: (a) 89%, (b) 91°F, (c) 17.1 gr/ft3, (d) AC = (17.1 gr/ft3)(0.89) = 15.2 gr/ft3. 7. (a) AC = MC × RH = (23.4 gr/ft3)(0.83) = 19.4 gr/ft3 (b) DP = 99°F (table H.2), and 105°F – 99°F = 6°F 8. (a) AC = (2.4 gr/ft3)(0.45) = 1.1 gr/ft3
(b) No. Coolness is due to water evaporating (latent
heat removed from bulb). Air temperature is above freezing, and water is replaced from reservoir (also at 35°F). 9. (a) AC = MC × RH = (9.4 gr/ft3)(0.58) = 5.5 gr/ft3
(b) DP = 69°F (from table H.1 and H.2)
10. dT = 5°F (from Table H.1), and DP = 94°F (from Table H.2), so 95°F – 94°F = 1°F. 11. No. The DP of given temperature is 65°F and the temperature of wet bulb is 67°F. 12. According to table H.1, dT = 7°F indicating the wet bulb temperature to be 63°F. The relative humidity would be 63% if dT = 8°F. The decrease in temperature would be 1°F. So, now the wet bulb temperature would be 62°F.
Chapter 20
ATMOSPHERIC EFFECTS Having covered the fundamentals in the preceding chapter, we now turn our attention to the processes and dynamics of weather phenomena. Since we have studied cloud formation, we can now look at precipitation processes and types, storms, and other weather phenomena. In general, the movement of large air masses across the country affects our weather changes. A person who knows about the movements and characteristics of these air masses is better prepared to understand and predict changes in the weather. In this chapter, students will encounter terms familiar to them from weather forecasts, such as fronts and air masses. The material is easily related to daily observations, and particularly at this point, students should be encouraged to understand how meteorological conditions affect the environment. A section is devoted to sensational weather phenomena—storms. These atmospheric disturbances sometimes give rise to property damage and even death. The properties and characteristics of some of the most powerful storms are discussed. Also, safety procedures are emphasized, particularly for tornadoes and hurricanes. A Highlight discusses El Niño and La Niña, whose oceanic effects on the atmosphere have received a great deal of attention in the past decade. The environmental and climatic effects of air pollution are major concerns today. Discussions of these effects commonly appear in daily newspapers and can be heard on radio and TV. Having studied the atmosphere, the student can now better understand and appreciate these problems. A brief history of air pollution is given, and the chief pollutants are identified, along with their sources. Discussions such as the one on acid rain point out how pollution affects our environment. Also, the ramifications of ozone depletion and the ozone hole over the South Pole are discussed. The long-term climatic effects of atmospheric pollution, for which we are responsible, can only be speculative. However, there is much concern over the emission of greenhouse gases and global warming
DEMONSTRATIONS Chemical reactions of some air pollutants, SO2 for example, can be demonstrated in the classroom. However, the utmost care should be taken, and this type of demonstration is not advised. A safer demonstration is the collection of particulate matter by using filter paper. A damp filter paper can be left exposed in the classroom, or large volumes of air can be filtered for
particulates by using a vacuum cleaner. The particulate matter on the filter may then examined with a scanning microscope. Film slides may be made if such equipment is available. As a class project, students can collect samples at different outside locations. Commercial air-sampling kits are also commercially available.
ANSWERS TO MATCHING QUESTIONS a. 4 b. 20 c. 12 n. 7
o. 25
d. 3 e. 9
p. 11
f. 23
g. 10
q. 15 r. 8 s. 18
h. 17
i. 19
t. 5 u. 6
j. 1
v. 24
k. 14
l. 22
m. 16
w. 21
x. 13
y. 2
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.b
2. a
3. d 4. d
5. c
6. c 7. b 8. d
9. c
10. b 11. d 12. c
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Bergeron process occlusion
7. tree
2. rain
3. temperature
8. warning 9. 24
4. Continental tropical
5. warm
6.
10. subsidence 11. photochemical 12. carbon
dioxide
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Bergeron process essentials: (1) ice crystals, (2) supercooled water vapor, and (3) mixing. Silver iodide crystals are substituted for ice crystals, and dry ice is used to cool vapor and form ice crystals. 2. (a) No. Frost is the deposition of water vapor. (b) The successive vertical ascents of ice pellets into supercooled air and regions of condensation produce large, layered ―stones‖ of ice. 3. Classified according to temperature and moisture content. The surface of the source region gives an indication of the moisture content of an air mass. Whereas the general latitude of a source region gives an indication of the temperature of an air mass. 4. See Table 20.1 and Figure 20.4. 5. Warm fronts occur when warm air mass advances over a colder surface. Cold fronts occur when cold air mass advances over a warmer surface. 6. The sharpness of a front's vertical boundary gives an indication of the rate of change of the weather. In general, cold fronts have sharper vertical boundaries than warm fronts, and hence lead to more sudden weather changes. As a result, cold fronts are accompanied by more
violent or sudden changes in weather. The sudden decrease in temperature is often described as a ―cold snap.‖ Dark altocumulus clouds often mark a cold front’s approach. The sudden cooling and the rising warm air may set off rainstorm or snowstorm activity along the front. A warm front may be characterized by precipitation and storms. The more gradual slope of a warm front is usually heralded by a period of lowering clouds. Cirrus and mackerel scale (cirrocumulus) clouds drift ahead of the front, followed by alto clouds, and as the front approaches, cumulus or cumulonimbus clouds resulting from the rising air produce precipitation and storms. Most precipitation occurs before the front passes. 7. (a) Within a cloud (intracloud discharge), between clouds (cloud to cloud discharge), and cloud to ground (ground discharge), between a cloud and the surrounding air (air discharges). (b) Heat lightning: lightning below horizon or behind clouds. Bolt from the blue: lightning in clear skies. 8. Resuscitation and keep warm (shock). 9. Warm front - the boundary of a warm air mass advancing over a cold surface. If the temperature of the cold air and Earth’s surface is below freezing, precipitation falling may cool and freeze on contact, producing an ice storm. 10. The tornado. Although the hurricane has more energy, the energy of a tornado is concentrated in a small region, giving a greater energy density. The larger hurricane, however, is usually more destructive on making landfall. 11. Seek Shelter fast. The basement of a home or building but avoid chimneys and windows. You can also get under a sturdy piece of furniture and cover your head. If you are in a mobile home, evacuate it and seek shelter elsewhere. 12. Latent heat. When the wind speed reaches 74 mi/h (119 km/h). 13. A tornado watch is issued when conditions are favorable for a tornado. A tornado warning is issued when a tornado is seen or is radar indicated. A hurricane watch is issued for coastal areas when there is a threat of a hurricane within 24 to 36 hours. A hurricane warning indicates that hurricane conditions are expected within 24 hours. 14. (a) August, September, October. (b) April, May, June. 15. Any atypical contribution to the atmosphere resulting from human activities. 16. Air pollution results primarily from the products of combustion and industrial processes that are released into the atmosphere. Smoke and soot from the burning of coal plagued England more than 700 years ago. London recorded air pollution problems in the late 1200s. 17. Radiation and subsidence temperature inversions. With a temperature inversion, emitted gases and smoke cannot rise and are held near the ground. Continued combustion causes the
air to become polluted, creating particularly hazardous conditions for people with heart and lung ailments. 18. Complete—CO2 and H2O; incomplete—CO, hydrocarbons, and soot (carbon). 19. The high temperatures of complete combustion cause a reaction between the nitrogen and oxygen of the air. Contributes to acid rain and photochemical smog. 20. Classical smog is smoke-fog. Photochemical smog results from photochemical reactions of hydrocarbons and other pollutants with oxygen in the presence of sunlight. Ozone is the prime indicator of photochemical smog. 21. Rising air meets colder air and spreads out 22. China and the United States. 23. Sulfur dioxide (and nitrogen oxides) combine with water to form acids, which fall as rain (or other types of precipitation). The acid raises the pH of bodies of water. The problem is most acute in the northeastern United States and eastern Canada because of major industrial areas to the west, but acid precipitation may now be found almost everywhere. 24. (a) Transportation. (b) Stationary sources (for example, electrical generating plants). (c) Industry.
(d) Transportation and Stationary sources.
(e) Photochemical smog.
25. Long term, average weather conditions of a region. 26. (a) Increase in the greenhouse effect and global warming and can affect the Earth’s albedo, (b) The Earth’s thermal balance by decreasing transparency of the atmosphere to isolation, (c) Particulate matter and gaseous chemical reactions in the stratosphere. 27.
It was discovered that CFCs cause the ozone depletion. Industrialized nations agreed
to reduce the production of CFCs and replace them with environmentally friendly products. 28. The interaction of CFCs with the ozone layer could deplete the ozone and allow more UV to reach Earth, thereby increasing Earth's temperature. In the extreme, an increase in the average temperature could affect the environment (e.g., lengthen the growing season) and melt the polar ice caps. 29. There are no natural mechanisms in the stratosphere to remove pollutants, and the pollutants could give rise to changes in climate. 30. Melting of polar ice caps, rise of oceans, change of agriculture, etc.
ANSWERS TO VISUAL CONNECTION (a) rain
(b) snow
(c) sleet
(d) hail
(e) dew
(f) frost
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. In general, fair weather is associated with high pressure, since the pressure reduces the formation of clouds and hence precipitation. Bad weather is associated with low pressure, which allows cloud formation and precipitation. 2. By shifting towards renewable energy sources like solar energy, wind energy and tidal energy and alternative energy sources that do not produce CO2, such as nuclear generating stations. 3. Chantal 4. Lightning, about five times. 5. Decrease our use of products with harmful chlorofluorocarbons, driving less, recycling more, turning off electronic devices when not using them, planting a tree.
ANSWERS TO EXERCISES 1. (a) Greenland, (b) Northern Atlantic and Pacific Oceans, (c) Northern Mexico, Southwestern United States, (d) Caribbean Sea, Gulf of Mexico, and Pacific Ocean, (e) Alaska and Canada. 2. (a) cT, (b) mP, (c) cA, (d) mT, (e) cP. 3. dc = vct = (35 km/h)(24 h) = 840 km (522mi), dw = vwt = (20 km/h)(24 h) = 480 km (298 mi) 4. It depends on state to state. If d is the thickness of your state, then, t = d/v would give you the time it takes for a cold and a warm front to travel from west to east across your home state. Here, average v from Exercise 3 may be used. 5. d = vt = (1/3 km/s)(10 s) = 3.3 km; d = vt = (1/5 mi/s)(10 s) = 2.0 mi 6. d = vt = (1/5 mi/s)(6 s) = 1.2 mi
Chapter 21 STRUCTURAL GEOLOGY AND PLATE TECTONICS Our present conception of the Earth’s structural geology began when geologists directed their studies to the oceans in the 1950s. The measurement of ocean depths and the charting of the topography of the ocean floor revealed a system of mid-oceanic ridges that extended through various oceans. In 1952, Harry H. Hess of Princeton University introduced the concept of convection currents and production of the ridges by rising
magma from the Earth’s deep interior, which in turn cause the continents to move. This new theory was called seafloor spreading. Five years later, Jason Morgan of Princeton presented to the American Geological Union the concept that the surface of the Earth consists of rigid plates that move with respect to each other. Verification of Alfred Wegener’s early theory (1915) of continental drift was finally concluded. Our discussion of structural geology begins with the overview of the Earth’s interior structure, then and introduction to the concepts of continental drift and seafloor spreading. This is followed by plate tectonics as the primary mover and shaker of the Earth’s outer shell, the lithosphere. Crustal motion on other planets and moons is discussed in a highlight. The large majority of volcanoes and earthquakes are associated with the movement of lithospheric plates; therefore, they are our next topic for discussion. Section 21.5 presents the causes for earthquakes and safety measures to be taken before, during, and after an earthquake. Conceptual questions and answers emphasize earthquakes, and highlights focus on earthquake risks and the deadly byproducts of earthquakes: tsunamis. The forces that build up in the vicinity of plate boundaries result in crustal deformation and mountain building. The chapter ends with a discussion to these two topics.
DEMONSTRATIONS The concept of isostasy can be demonstrated using large and small ice cubes in a glass. Other sources of teaching aids such as demonstration equipment and supplies, audiovisual material, and publications of use to instructors are listed in the Teaching Aids section at the end of this guide.
ANSWERS TO MATCHING QUESTIONS a. 12 b. 16
c. 6
d. 3
e. 1
f. 19
g. 25
h. 11
i. 8
j. 15
k. 18
l. 26
m. 2 n. 28
w. 17
x. 10
y. 4
z. 20
o. 7 p. 9
q. 27
r. 14
s. 5
t. 24
u. 22
v. 13
aa. 23 ab. 21
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.b 2. d 3. c 4. d 5. a 6. d 7.c 8. c 9. c 10. c 11. a 12. b
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Geology 2. Compressional or P 3. asthenosphere 4. outer core 5. deep-sea trenches 6. mid-ocean 7. isostasy 8. transform 9. ring of fire 10. Richter scale 11. volcanic 12. strike-slip
ANSWERS TO SHORT-ANSWER QUESTIONS 1. 6400 km, 4000 miles. 2. Crust, mantle, outer core, inner core. 3. The inner core is about 6000°C (10800°F). 4. The asthenosphere is relatively plastic and able to move, allowing plates above it to shift. 5. The mechanism behind continental drift is seafloor spreading most noticeable at the mid-ocean ridges. 6. The Mid-Atlantic Ridge is a ridge that runs along the center of the Atlantic Ocean between the continents. 7. Continuity of geologic features, biological evidence, glacial evidence. 8. Remnent magnetism gives the long-term record of the magnetism of the Earth. It can indicate the direction of the Earth’s magnetic field, show that Earth’s magnetic field undergoes reversals, and supports the mechanism of seafloor spreading. 9. The lithosphere is now viewed not as one solid rock but as a series of solid sections or segments called plates that are constantly interacting with one another in very slow motion. 10. The lithosphere is the solid outer shell of the Earth; it includes the crust and some of the upper mantle. The asthenosphere is the region below the lithosphere that is hot enough to be easily deformed and is capable of internal flow.
11. The concept that the Earth’s crustal material ―floats‖ in gravitational equilibrium on a ―fluid‖ substratum. 12. Divergent (Red Sea); convergent (Himalayan Mountains); transform (San Andreas fault). 13. The place, or zone, where the plate descends into the asthenosphere is called a subduction zone. 14. The primary force that moves plates is gravity. 15. The oceanic plate is subducted. Oceanic crust has a higher density (3.0 g/cm3) than does continental crust (2.7 g/cm3). 16. Volcanoes are produced primarily at plate boundaries because of the stresses caused as plates shift. 17. There are very few volcanoes in Africa; the African continent is surrounded mostly by divergent plate boundaries. 18. 1) Mount Fuji in Japan, 2) The Andean Mountains of western South America, 3) The Cascade Mountains of western North America. 19. Earthquakes generally occur at plate boundaries. 20. The Pacific plate is moving northward relative to the North American plate. 21. The focus of an earthquake is the point or region of the initial energy release or slippage. The epicenter is the location on Earth’s surface directly above the focus. 22. Deadly tsunamis can occur because of undersea earthquakes which can travel thousands of miles. 23. (a) The Richter scale gives an absolute measure of the energy released by calculating the energy of seismic waves at a standard distance from the epicenter. (b) See Table 21.1 for the effects of each whole-number magnitude. 24. Quake magnitude and duration, location of the focus and epicenter, and environment of the region around the epicenter. 25. A sketch like Fig. 21.24. An anticline is a rock fold with downward sloping on both sides of a common crest. A syncline is a rock fold with upward-sloping on both sides of a common trough.
26. (a) Normal fault: occurs as the result of expansive forces that cause its overlying side to move downward relative to the side beneath it. (b) Reverse fault: occurs as the result of compressional stress forces that cause the overlying side of the fault to move upward relative to the side beneath it. (c) Transform, or strike-slip fault: occurs when the stresses are parallel to the fault boundary such that the fault slip is horizontal. 27. A fold mountain is characterized by folded rock strata; they also show external evidence of faulting and internal evidence of high temperature and pressure changes. They have thick sedimentary strata indicating that they were once at the bottom of an ocean basin. 28. (a) Fault-block. (b) Fold. (c) Volcanic. (d) Fold.
ANSWERS TO VISUAL CONNECTION (a) mantle, rock; (b) crust, rock; (c) outer core, molten iron and nickel; (d) inner core, solid iron
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Both use the process of investigation to obtain and identify clues to discover the unknown. The geologist investigates for clues to understand the Earth’s past and present. The detective investigates for clues to solve a crime. 2. Seismologists know how fast the two types of waves travel through the Earth and calculate the location by comparing the times when shaking was reported at different stations that record this information. 3. Tectonic plate motions are accurately measured using space-based methods. That is, scientists use satellites to carefully measure the motions of each plate relative to a frame of reference. The global positioning system (GPS) measures the movement of ground stations and has allowed scientists to make vast improvements in the accuracy of the plate motions.
4. Oil and gas deposits are likely found at or near subduction zones as the movement of the plates, generating high pressure and temperature, lead to their production. 5. The answer depends upon your location. By looking at figure 1, in highlight 21.2, area(s) of North America having highest seismic risk can be located. 6. The two continents on Earth that have the fewest earthquakes and volcanoes are Antarctica and Australia respectively. On which of these two continents would you live? It would be relatively easier to live in Australia. 7. Tectonic forces caused by gravity, pressure and friction forces between rocks and plates are the main forces involved in mountain building. Other factors include the density, composition, and temperature of the rock: certain rocks under high stress will deform by faulting (fracturing), but other rocks will buckle and fold (folding) under high stresses.
Chapter 22 MINERALS, ROCKS, AND VOLCANOES The fundamental principle underlying most geology is simply that the processes occurring at present on the Earth have occurred throughout geologic time. Thus ancient rocks can be interpreted with respect to present processes. The composition of rocks is expressed in terms of minerals, which are the building blocks of rocks. Our study of geology begins with the study of minerals. Mineral classification is based on physical and chemical properties. The physical properties distinguish different forms of minerals with the same composition. In addition, physical properties provide a convenient means of mineral identification. Although the number of common minerals is small—fewer than 20 make up over 95% of the rocks in the Earth’s crust—the student may find the number (and names) overwhelming. As a result, only a few minerals are examined in detail. Among these is silicon dioxide (SiO2), which occurs abundantly as quartz, sand, and many other minerals. Following the basic information concerning minerals and igneous rocks, we consider the igneous activity that produces the majority of rocks that form the Earth’s crust. The
chapter concludes with sections on sedimentary rocks and metamorphic rocks. The chapter Highlight discusses how volcanoes can produce gold in rocks. The conceptual questions and answers focus on some interesting applications and the broad context of geologic processes.
DEMONSTRATIONS The best demonstrations for this chapter are representative mineral and rock collections that students can examine and analyze. Also, use rock charts that illustrate types, characteristics, identifying features, and the interrelationships of various types of rock. A chart showing the rock cycle should be a continuous exhibit in the classroom and laboratory during the teaching of geology. The chart on the chemistry of rocks is a good demonstration exhibit.
ANSWERS TO MATCHING QUESTIONS a. 17 b. 9
c. 2
d. 18
e. 24
f. 15
g. 16
h. 21
i. 7
j. 20
k. 22
n. 6
p. 12
q. 8
r. 11
s. 13
t. 4
u. 14
v. 5
w. 23
x. 3 y. 25
o. 10
l. 1
m. 19
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. b 3. b 4. c 5. b 6. b 7. b 8. a 9. b 10. b 11. c 12. a 13. d 14. c 15. d 16. c 17. a 18. a 19. d 20. b
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. silicon dioxide or silica 2. plagioclase
3. metallic luster 4. gem 5. igneous 6.
uniformitarianism 7. cementation
8. basalt 9. magma 10. steam
11. cinder cone 12. concordant 13.
hot spot 14. pluton 15. gas 16. coal 17. stalactites 18. slate 19. Foliation 20. hydrothermal
ANSWERS TO SHORT-ANSWER QUESTIONS 1. A mineral is a naturally occurring, crystalline, inorganic substance (element or compound) that possesses a fairly definite chemical composition and a distinctive set of physical properties. The study of minerals is called mineralogy.
2. The silicon-oxygen tetrahedron is a covalently bonded structure composed of one silicon atom surrounded by four oxygen atoms. The structure is the basic building block of silicate materials. See Figure 22.2(a) for a sketch of its structure. 3. The oxygen-to-silicon ratio in silica is 2-to-1. In other silicate rocks this ratio is greater than 2-to-1 and it can vary significantly. 4. Nonsilicate materials include pure elements like gold and silver, gemstones like diamonds and sapphires, and ores like iron, copper, and nickel. 5. The Mohs scale has limits of 1 (soft) - Talc to 10 (hard) - Diamond. 6. Diamonds are initially cut by strikes from steel blades along their weakest structural planes, the tetrahedral planes. For fine cuts and polishing, diamond saws and polishing wheels coated with diamond dust are used. 7. Luster is the appearance of a mineral’s surfaces in reflected light. Streak refers to the color of the powder of a mineral. 8. Mica has one perfect cleavage plane forming sheets. 9. A rock is defined as a solid, cohesive, natural aggregate of one or more minerals. The study of rocks is called petrology. 10. Igneous rocks are formed by the solidification of magma at or below the Earth’s surface. Sedimentary rocks are formed at the Earth’s surface by compaction and cementation of layers of sediment. Metamorphic rocks are formed below Earth’s surface by the alteration of preexisting rock in response to the effects of pressure, temperature, or the gain or loss of chemical components. 11. This concept, known as uniformitarianism, is the very foundation on which the science of geology rests; that is, the present is the key to the past. The principle of uniformitarianism, the concept of the rock cycle and the recognition of the Earth’s great age remain central themes of modern geology. 12. Sedimentary rocks must undergo heat and pressure to form metamorphic rock first, then they can be melted and cooled to become igneous rock. 13. The molten material is known as magma as long as it is beneath the Earth’s surface but becomes lava if it flows on the surface.
14. Intrusive rock is igneous rock that formed below Earth’s surface and has a coarsegrained texture. Extrusive rock is igneous rock formed outside Earth’s crust and has a fine-grained texture. 15. Igneous rocks have three textures, coarse, fine, and glassy. Examples of coarse texture igneous rocks are granite, fine texture is basalt, and glassy texture is obsidian. 16. Gabbro has a larger grain size, darker color, and lower silica content than does rhyolite. 17. Plutons are solidified magma that forms below the surface while lava is the molten rock that flows on the surface. When lava cools it will form solidified igneous rock. Plutons are classified according to the size and shape of the intrusive bodies and according to their relationship to the surrounding rock they penetrate. 18. The three products of a volcano are gas, lava, and pyroclastics (solids). 19. Temperature and silica content 20. The Hawaiian Islands formed from the Pacific plate moving over a hot spot. 21. Volcanoes can erupt either peacefully or explosively. The type of eruption largely depends on the viscosity and temperature of the magma inside it. Low viscosity, high temperature magma flows easily and peacefully. High viscosity, low temperature magma, however, can get stuck in vents and chambers which allow pressures to build up and potentially explode. 22. A shield volcano has a gently sloping, low profile. A stratovolcano has a steeply sloping, layered composite cone composed of both lava and tephra. A cinder cone volcano is steeply sloping and composed primarily of pyroclastics (tephra). 23. Caldera is a term applied to a roughly circular, steep-walled depression near the summit of a volcano. It may be up to several kilometers in diameter. Calderas result primarily from the collapse of the chamber at the volcano’s summit from which lava and ash were emitted. Crater Lake on top of Mount Mazama in Oregon occupies a caldera. 24. Coal is an organic chemical sedimentary rock. While minerals are not transported in solution to make the rock, coal is classified as sedimentary because it is the lithified remains of plant matter.
25. Stalactites are formed from calcium carbonate precipitating from dripping water from the ceiling of caves. Stalagmites are built up from the floor from the dripping water. You can remember stalactites extend down from the ceiling (c for ceiling) and stalagmites are built up from the ground (g for ground). 26. Physical characteristics of sedimentary rock include color, rounding, sorting, bedding, fossil content, ripple marks, mud cracks, footprints, and raindrop prints. 27. Sedimentary rocks often contain fossils. 28. Temperature and Pressure both increase with depth into the Earth. Temperature increases at a rate of about 30 C°/km, while pressure in the crust increases about 267 atmospheres per kilometer. At these high temperatures and pressures inside the Earth metamorphic rock is formed. 29. Contact metamorphism is change brought about primarily by heat, with very little pressure involved. Shear metamorphism changes rock more by pressure than temperature. Hydrothermal metamorphism changes rock by the circulation of chemically active solutions. 30. Foliated means the parallel arrangement of certain mineral grains that gives the rock a striped appearance. Foliation forms when pressure squeezes the flat or elongate minerals within a rock so they become aligned. 31. Verde Antique is a valuable hydrothermally metamorphosed rock because it has a beautiful, dark green color with whitish accents. It is also durable, resists staining, and keeps its shine.
ANSWERS TO VISUAL CONNECTION (a) heat/pressure, (b) melting, (c) cooling, (d) weathering/erosion, (e) compaction/cementation
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. It is uncommon to find fossils in igneous rock because that type of rock was directly formed from lava and magma. Therefore any fossils would have been destroyed in the making of the rock. Most fossils are found in sedimentary rock.
2. Flood basalt structures are landforms created from fissure eruptions where lava flows onto the Earth's surface. Examples include the Columbia Plateau in northwestern United States and in Iceland. The gentle sloped Shield volcanoes, such as Mauna Loa in Hawaii is another example of the geologic structures formed from flood basalts. The Central Atlantic magmatic province (CAMP) is the Earth's largest continental large igneous province, which erupted about 201 million years ago. It is composed mainly of basalt. 3. In order to identify the materials, you can see which mineral scratches the other and rank them: Calcite crystal (3.0), window glass (5-6), quartz, (7.0), zircon (7.5), and diamond (10). 4. Hematite. 5. There are many good websites that give tests and characteristics of meteorites. The best method to identify meteorites is to have eye witnesses verify their fall through the atmosphere. If this is not possible, then several tests can be done to prove it is of extraterrestrial origin. Meteorites differ from Earth rock in that they have a fusion crust, they often contain smooth depressions in their surfaces called regmaglypts, are usually magnetic, contain some high density metals like iron and nickel, and contain small, round, mineral features called chondrules. Many times an expert is needed to discern whether an unknown rock is truly a meteorite. 6. Mount Fuji is located in Japan and is classified as a stratovolcano. Mount Kilimanjaro, located in Africa (specifically Tanzania), and Mount Vesuvius, located in Italy, are both also classified as stratovolcanoes. 7. The igneous formation shown is a dike.
Chapter 23
SURFACE PROCESSES Some internal processes, such as volcanism and mountain building, build up the Earth’s surface features. Many surface processes work in the opposite direction, wearing away and leveling surface features. The weathering and erosion of surface materials are important geologic processes, as is mass wasting, which is the downward movement of
weathered materials under the influence of gravity. Evidence of these processes is commonplace. However, some processes take place very slowly and are often overlooked. Erosion is carried out by the so-called agents of erosion—running water, ice, wind, and waves—which are the physical phenomena that supply the energy for the removal and transportation of rock debris. Two of these agents, running water (streams and rivers) and ice (glaciers), will be examined in some detail. Not only is water an important factor in geologic processes, it is also necessary to sustain life and is a major environmental concern. In this chapter, the Earth’s water supply is studied. This supply is often thought to be inexhaustible, because about 70% of the Earth’s surface is covered by water. However, only about 2% of this is fresh water, and most of this 2% is locked in glacial ice sheets. The Earth’s water supply is a reusable resource that is constantly being redistributed. In general, atmospheric processes move moisture from large oceanic reservoirs. The water eventually flows back to the seas, eroding as it goes. Groundwater is an important aspect of this gigantic hydrologic cycle, particularly for domestic water supplies. The geologic features of wells and springs, as well as the quality of water with respect to dissolved minerals, are considered. Highlights on Earth’s largest crystals and Earth’s highest tides show the power of water and time. Because oceans cover about 70% of the Earth’s surface, they are important factors in surface processes. The movement of seawater affects regional climates and is an agent in the erosion of coastlines. The final topic of the chapter deals with the topology of the seafloor. Recent investigations have greatly expanded our knowledge of the surface features of this major unseen portion of the Earth’s surface.
DEMONSTRATIONS Geology models show geologic landforms in three dimensions. Stream tables use flowing water and rainfall to show the effects of water on land-formation processes such as erosion and sedimentation. Topographic relief maps can be used to show landform features, and contour models illustrate the fundamentals of contour mapping. These
models are also very useful in the laboratory when the student is asked to draw contour maps.
ANSWERS TO MATCHING QUESTIONS a. 18 b. 9
c. 23
d. 13
e. 17
n. 3
p. 4
q. 2
r. 22
o. 21
f. 14 s. 12
g. 20 t. 5
h. 6
u. 16
i. 24 v. 7
j. 8
k. 1
w. 15
x. 19
l. 10
m. 11
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a 2. c 3. c 4. c 5. d 6. b 7. b 8. c 9. d 10. b 11. b 12. a
ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. mechanical 2. permafrost 3. Dissolution 4. stream 5. Runoff
6. continental glaciers
7. mudflows 8. moraines 9. Subsidence 10. aquifer 11. abyssal plains
12. continental
slopes
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Mechanical weathering involves the physical disintegration of rock (frost wedging, crystal growth, root penetration, and others), and chemical weathering involves chemical changes in rock composition (dissolution, oxidation, and hydrolysis). 2. Chemical weathering is most prevalent in hot, moist climates. 3. Burrowing animals loosen soil and bring it to the surface. Plant root systems fracture rocks. 4. Permafrost is a layer of subsurface soil that remains frozen permanently. 5. Limestone. 6. A depression resulting from the collapse of a cavern that has been formed by chemical weathering. 7. The natural influence of erosion is gravity (mass wasting). The agents of erosion are wind, streams, waves, and glaciers. 8. The dissolved load consists of water-soluble minerals that are carried along by a stream in solution. Fine particles not heavy enough to sink to the bottom are carried along in suspension. This suspended load is quite evident when the stream appears
muddy after a heavy rain. Coarse particles and rocks along or close to the bed of the stream constitute its bed load. These particles and rocks are rolled and bounced along by the current. 9. A loop-like bend in a stream channel is a meander caused by gravity and the rotating Earth. Oxbow lakes. 10. Floodplains contain very rich soil for agriculture; however they flood regularly. 11. Continental glacial ice sheets cover large areas and flow outward. Valley glaciers form in valleys and flow down the valleys. Cirque glaciers are small glaciers that form in depressions along mountains. 12. Rock material that is transported and deposited by ice is called glacial drift. 13. Earth’s largest deserts, surprisingly, are found in Antarctica and the Arctic, near the poles. 14. (a) Rockslide is a type of landslide occurring in mountain areas when large quantities of rock break off and move rapidly down the steep slopes. Fast. (b) Creep is a type of slow mass wasting involving the slow, particle-by-particle movement of weathered debris down a slope, which takes place year after year. Slow. (c) Slump is a type of landslide involving the downslope movement of an unbroken block of overburden that leaves a curved depression on the slope. The overburden does not travel very fast or very far. Slow. 15. Evaporation. Precipitation. 16. Permeability is a measure of a material’s capacity to carry fluid. Porosity is the percentage volume of unoccupied space in the total volume of a substance. 17. Water wells should be drilled into the zone of saturation. 18. A body of permeable rock through which groundwater moves. Aquifers exist under more than half of the conterminous United States. 19. Groundwater extraction can lead to depletion, subsidence, saltwater, and other contamination. 20. There are many ways in which groundwater can become contaminated. Three ways are: 1) saltwater contamination of wells caused by freshwater extraction, 2) chemical
contamination from industries and homes, like phosphates, pesticides, and organic compounds, and 3) acid rain from air pollution. 21. Figure 23.24 shows three features of coastal deposition. Pocket beaches form in the low-energy wave environment between headlands. Barrier islands extend more or less parallel to the mainland. Between a barrier island and the mainland is a protected body of water called a lagoon. Spits are narrow, curved projections of beach that extend into the sea, elongating the shoreline. 22. A cork in the ocean bobs up-and-down in a circular pattern as the waves pass beneath it. However, when the wave gets close to the shore, the circular wave motion is disrupted by the ocean floor and the paths are elliptical in shape. This causes the waves to break and form surf. If a cork is close enough to the shore then it will be thrown up onto the shore in the surf. 23. The tides are the periodic rise and fall of the water level along the shores of large bodies of water. Both the gravitational influence of the Moon and the Sun contribute to the tides, but the Moon has more influence. 24. Continental shelves are gently sloping, relatively shallowly submerged borders of the continents. Continental slopes are the true edges of the continents, where the continental land masses slope downward to the floors of the ocean basins.
ANSWERS TO VISUAL CONNECTION (a) evaporation, (b) condensation, (c) evapotranspiration, (d) precipitation, (e) runoff
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes. The Moon has mass, and gravity can produce downslope movement of soil and rock. 2. The presence of roads and buildings do not allow rainwater to soak into the ground where it falls. The roads move the water quickly to low lying places where the water can quickly rise to flood these regions. Other subsurface regions are robbed of their water and the groundwater can become depleted. Ground water is also affected because it can easily become contaminated by human, agricultural and industrial wastes.
3. Weathering caused the rapid degradation of Cleopatra’s needle. For thousands of years it stood in the dry desert of Egypt. In only 100 years in New York City, it weathered greatly from the rain, humidity, freezing, and acid rain breakdown of the rock. 4. Earth’s fresh water supply will not change significantly due to the melting of the glaciers, because most of the melting occurs in the salt water ocean. The overall liquid water supply will increase and ocean levels will rise significantly. 5. Different ways in which we can protect our drinking water include reducing or removing pesticides and fertilizers used on lawns, reducing the use of organic compounds and phosphates that can help keep groundwater clean. Finally, avoiding improper dumping of garbage; recycle as much as possible. 6. The boundary between a continental crust and oceanic crust is the Continental Shelf. Erosion from the continental slope defines what the continental shelf will look like. Ocean waves, currents, tides, and wind further erode the shorelines. These elements vary depending on the depth and temperature of the ocean creating different features within the continental shelf. The nature of the plate boundaries makes a difference as well. For example, off the coast of Southern California, a transform plate boundary results in active faulting and many underwater avalanches resulting in turbidites.
Chapter 24 GEOLOGIC TIME This chapter ends by focusing on a topic of interest to most students—geologic time. The chapter begins with a discussion of what fossils are, how they are formed (in a Highlight), and the names of some common ones. The principles used to establish relative geologic time are then treated, and the eon, era, and period time units are explained. The next topic is radiometric dating and how it is used to establish the age of the Earth and absolute geologic time. The last section of the chapter tells how relative and
absolute geologic times are combined into the geologic time scale. Another chapter Highlight discusses the evidence for an asteroid strike causing the demise of the dinosaurs. Few students come into a physical science class having anything but the haziest notions of geologic time, so we recommend that the entire chapter be covered. Two conceptual questions and answers are given to help understand fossilization and the geologic time scale.
DEMONSTRATIONS Fossil sets can be used to demonstrate the major types of fossils, their identification, and the geologic time periods they represent. Fossils actually found in nearby localities are always of interest. Students can be asked to bring in any fossils they may have found. Posters, slides, sets, kits, models, and pale plaques of fossils and Earth history are readily available from scientific suppliers. Figure 24.20 is an excellent demonstration of the scale of geologic time. Your students can try making another one by correlating a distance scale with a time scale (1 mi or 1 km equals one million years of Earth’s history)
ANSWERS TO MATCHING QUESTIONS a. 8 b. 9 n. 19
o. 6
c. 2
d. 17
p. 23
e. 15
q. 18
r. 3
f. 24 s. 11
g. 22
h. 21
i. 7
j. 5
t. 14
u. 13
v. 16
k. 4 w. 1
l. 10
m. 26
x. 20
y. 25
z.
12
ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. c 3. c 4. b 5. a 6. b 7. d 8. b 9. a 10. d 11. d 12. a 13. c 14. c 15. d 16. a 17. c 18. a 19. b 20.c ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. amber 2. replacement fossil 3. algae 4. Mesozoic 5. unconformity 6. cross-cutting relationships 7. relative time 8. Correlation 9. epochs
10. trilobite 11. superposition
12. carbon-14 13. larger 14. primordial 15. assumptions 16. meteorites 17. Rodinia 18. K-Pg Event 19. explosion 20. 100,000
ANSWERS TO SHORT-ANSWER QUESTIONS 1. Geologic time is the time span since the formation of planet Earth. 2. Worms do not have a hard skeleton and thus decompose quickly before a fossil can form. 3. The fossil record shows that as time passed, larger and more complex life forms developed. 4. Examples of replaced remains include petrified wood and carbonization of plants. 5. Arizona is famous for dinosaur's fossil and trace fossils. 6. Coal is formed from the carbonization of plants. This process involves the decomposition of plant remains by bacteria under anaerobic conditions. 7. When an embedded shell or bone is dissolved completely out of a rock, it leaves a hollow depression called a mold. If new mineral material fills the mold and hardens, it forms a cast of the original shell or bone. 8. Fossils of blue-green algae, or cyanobacteria, date back to about 3.5 billion years. 9. Layers of rock that contain certain microfossils, such as some species of Foraminifera, indicate the presence of nearby oil deposits. 10. Three examples of trace fossils are tracks, borings, and burrows. 11. In a sequence of undisturbed sedimentary rocks, lava, or ash, each layer is younger than the layer beneath it. Example: If lava A covers rock layer B, A is younger than B. If the rock layers are disturbed by folding or faulting then the principle may be violated. 12. No, non-horizontal rock layers were originally horizontal and were altered later by other forces. 13. Faults are used in relative aging because a fault must be younger than any of the rocks it has affected. 14. An unconformity is a gap, or break, in the rock record at a given locality due to nondeposition or erosion and usually it is impossible to determine how much time is represented by an unconformity. 15. Correlation is the process of matching rock layers in different localities by use of fossils or other means.
16. Index fossils aid correlation because index fossils are those that are widespread, numerous, easily identified, and typical of a particularly limited time segment of the Earth’s history. 17. Typical of a particular limited time segment, widespread, numerous, easily identified. 18. Events in your daily life can be an analogy to relative geologic time. In relative time, you know the order of events, but you do not know the actual time. Examples from daily life include waking, eating breakfast, going to a meeting at work, picking up children from school, and watching the news on TV. 19. Paleozoic, Mesozoic, Cenozoic. 20. Seven, Permian. 21. Triassic 22. Tertiary (Paleogene and Neogene) and Quaternary. 23. Carboniferous period. 24. Precambrian time is the huge span of time before the Cambrian period of the Paleozoic era, that is, the time before the Phanerozoic eon. 25. The oldest rocks are dated by rubidium-strontium radiometric dating. 26. Daughter products that are gaseous can leak out of the rocks being dated. 27. Carbon-12. 28. No addition or subtraction of the parent or daughter has occurred over the lifetime of the rock other than that caused by radioactive decay. The age of the rock does not differ too much from the half-life of the parent isotope. None of the daughter element was present in the rock when it was formed, or if it was, its initial abundance is known. 29. Lead that comes from radioactive decay is called radiogenic lead, whereas lead that does not is called primordial lead. 30. Carbon-14 dating is used to date objects within about 7000 years old. These include the ages of organic material from bones, charcoal from ancient fires, and the wood beams in the pyramids. 31. Meteorites from the asteroid belt date back to 4.56 billion years, the oldest Moon rocks date back 4.45 billion years, and the oldest known Earth rocks date back 4.4 billion
years. All these parts of the solar system are thought to have formed about the same time. The age of the sun is 4.6 billion years.
32. The oldest known Earth rocks date back to 4.4 billion years; the oldest Moon rocks date back to 4.45 billion years. 33. By dating igneous rocks that intrude or cover sedimentary layers and then using the law of superposition. 34. The Hadean eon starts at Earth’s formation; the Archean eon starts at earliest known Earth rocks; the Proterozoic eon starts at the formation of the North American continental core. 35. Cenozoic (65 million years). 36. Cenozoic era, Quaternary period, Holocene epoch. 37. Asteroids, supernova explosions, greenhouse warming, or ice-age cooling. 38. The end of the Ice Ages (last retreat of glaciers from North America and Europe). 39. Earliest fossils of the genus Homo. 40. The Cambrian explosion. 41. The K-Pg extinction event. It has been hypothesized that a massive meteorite, estimated to be 10 km (6 mi) across, struck the Earth about 66 million years ago, causing widespread fires and driving dust, ash, and other debris (about 100 trillion tons!) high into the stratosphere. 42. Unusually high concentrations of iridium, shocked quartz, spherules, and soot in layers of rock from 66 mya. Indicate an asteroid impact. A large impact crater is found at Chicxulub that dates from 66 mya. (Any two pieces of evidence are a sufficient answer.) 43. The K-T event has evidence that points to an astronomical origin. Evidence includes the rich amount of iridium found in the Earth’s soil at the 66 million-yearold boundary. Iridium is prevalent in meteoritic rock but not generally found in Earth rock. Also shocked quartz rock, spherules, and soot were found at the K-T boundary. This evidence points to a catastrophic event at 65 million years old followed by worldwide fires, dust, ash, and debris. Core drilling also provided evidence of pieces of the carbon-iron-rich meteorite itself. 44. Coal forming forests were alive in the Paleozoic era and Pennsylvanian period.
45. Dinosaurs lived about 14 days on the geologic time calendar; this corresponds to about 175 million years of time.
ANSWERS TO VISUAL CONNECTION (a) Parent 28.3, Daughter 11.7; (b) Parent 20, Daughter 20; (c) Parent 10, Daughter 30; (d) Parent 5, Daughter 35; (e) Parent 2.5, Daughter 37.5
ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The radiometric dating technique is useful because it can be applied to a wide array of rocks over a large range of time. Many different isotopes can be used to crosscheck, calibrate, and validate results, thus reducing errors. One weakness of the method is that the daughter products must be of high enough quantity to make an accurate age estimate. Certain ages of rock require radioactive elements with similar half-lives. Another weakness is that it is sometimes difficult to know whether a rock has any daughter element that was present when it formed. 2. All three objects lived in the Paleozoic era which spans from about 550-250 million years old. Crinoids come from the Mississippian period; they are about 340 million years old. Since all three were found in the same rock, a good estimate of the age is 340 million years old. 3. The last of the dinosaurs died out 65 mya., whereas the oldest Homo fossils known only date to about 2 mya. So a huge 63-million-year gap seems to exist between the last dinosaurs and the earliest humans. 4. The Cambrian explosion occurred approximately 541 million years ago in the Cambrian period. Major life types began to appear in the rock record as fossils. It lasted 13 – 25 million years. Before the Cambrian explosion, most organisms were relatively simple, composed of individual cells, or small multicellular organisms, occasionally organized into colonies. As the rate of diversification subsequently accelerated, the variety of life became much more complex, and began to resemble that of today. The Cambrian follows the Ediacaran Period, during which time the continents had been joined in a single supercontinent called Rodinia. As the
Cambrian began, Rodinia began to fragment into smaller continents, which did not always correspond to the ones we see today. 5. Dinosaurs lived from about 250 to 65 MYA. The half-life of carbon-14 is about 5700 years and it can only date materials younger than about 70000 years. Perhaps carbon dating was confused with another type of radiometric dating.
ANSWERS TO EXERCISES 1. (a) Mississippian, Pennsylvanian, Permian. (b) Silurian. (c) Platycrinites and Lingula. (d) Elrathia. 2. (a) Ordovician and Silurian. (b) Devonian. (c) Phacops, Zygospira, and Lingula. (d) Neither lived during that period. (e) Lingula. 3. (a) A is the younger; superposition. (b) E is the younger; cross-cutting relationships. 4. (a) D is the younger; cross-cutting relationships. (b) E is the younger; cross-cutting relationships. 5. If only 25% of its original potassium-40 remains, then the age of the rock must be twice of its half-life. Hence, the minimum age is 2.6 billion years. 6. 3.9 billion years. 7. Older than 210 Ma but younger than 245 Ma. 8. The rock must be between 440 Ma and 350 Ma old, the time period during which both fossils lived.