Heat and Thermodynamics

PHY130 – FUNDAMANTAL OF PHYSICS

CHAPTER 9 Heat and Thermodynamics

9.1 Temperature, thermal expansion and the ideal gas law A. Temperature • • • •

It is a measure of how hot or cold an object is.

Many properties of matter change with temperature.

Thermometers is an instruments designed to measure temperature.

Most common thermometers rely on the expansion of a material with an increase in temperature.

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The numerical scale to measure temperature is Celsius, Fahrenheit and Kelvin.

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The conversion between Celsius, Fahrenheit and Kelvin: 9 đ?‘‡đ??š = đ?‘‡đ??ś + 32 5

đ?‘‡đ??ž = đ?‘‡đ??ś + 273.15

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B. Thermal Expansion i. Linear Expansion

Figure 1 Linear expansion

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A rod of a material has an initial length đ??żđ?‘œ at some initial temperature đ?‘‡đ?‘œ.

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When the temperature increases by an amount ∆đ?‘‡, the length increases by ∆đ??ż.

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If ∆đ?‘‡ is not too large, ∆đ??ż is directly proportional to ∆đ?‘‡ and also proportional to đ??żđ?‘œ.

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This relationship can be express in an equation: Δđ??ż = đ?›źđ??żđ?‘œ Δđ?‘‡

• �, is the proportionality constant, called the coefficient of linear expansion.

• Unit of �, is (°C)-1.

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If a body has a initial length, đ??żđ?‘œ at temperature đ?‘‡đ?‘œ, then its length đ??ż at a temperature đ?‘‡ is given by: đ??ż = đ??żđ?‘œ + Δđ??ż

•

đ??ż = đ??żđ?‘œ 1 + đ?›źÎ”đ?‘‡

∆đ??ż is referring to the increase of length at ∆đ?‘‡.

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EXAMPLE 1 A copper bar is 80 cm long at 15°C. What is the increase in length when it is heated to 35°C? The linear expansion coefficient for copper is 1.7 x 10-5 °C-1. đ??żđ?‘œ = 0.8 đ?‘š, đ?‘‡đ?‘œ = 15°đ??ś, đ?‘‡1 = 35°đ??ś, ď Ąđ?‘?đ?‘œđ?‘?đ?‘?đ?‘’đ?‘&#x; = 1.7 đ?‘Ľ 10 − 5 °đ??ś

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EXAMPLE 2 A surveyor uses a steel measuring tape that is exactly 50 m long at a temperature of 20°C. What is its length on a hot summer day when the temperature is 35°C? đ??żđ?‘œ = 50 đ?‘š, đ?‘‡đ?‘œ = 20°đ??ś, đ?‘‡1 = 35°đ??ś, ď Ąđ?‘ đ?‘Ąđ?‘’đ?‘’đ?‘™ = 12 Ă— 10−6 ℃

B. Thermal Expansion ii. Area Expansion •

If an area expands to đ??´ = đ??´đ?‘œ + Δđ??´ when the temperature rise is ∆đ?‘‡, then the increase in area, ∆đ??´, is given by

∆đ??´ = đ?›žđ??´đ?‘œ ∆đ?‘‡ •

� , is the coefficient of area expansion.

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Unit of đ?›ž , is ℃

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For isotropic solids:

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. � = 2� 9

EXAMPLE 3 A hole of cross sectional area 100 cm2 is cut in a piece of steel at 20oC. What is the area of the hole if the steel is heated from 20oC to 100oC?

ď Ą steel = 11 x 10-6 oC-1. đ??´đ?‘œ = 100 đ?‘?đ?‘š2 = 0.01 đ?‘š2, đ?‘‡đ?‘œ = 20°đ??ś, đ?‘‡1 = 100°đ??ś, ď Ąđ?‘ đ?‘Ąđ?‘’đ?‘’đ?‘™ = 11 Ă— 10−6 ℃−1

B. Thermal Expansion iii. Volume Expansion •

For both solid and liquid materials, increasing temperature usually causes increases in volume.

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If ∆đ?‘‡ is not too large, the increase in volume ď „đ?‘‰ is proportional to ď „đ?‘‡, and also proportional to the initial volume, đ?‘‰đ?‘œ.

• The relationship can be expressed as follows:

∆đ?‘‰ = đ?›˝đ?‘‰đ?‘œ ∆đ?‘‡ • đ?›˝, is the coefficient of volume expansion.

• Unit of �, is (°C)-1.

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• If a body has a initial volume, Vo at temperature To, then its volume, V at a temperature T is given by:

đ?‘‰ = đ?‘‰đ?‘œ + Δđ?‘‰

đ?‘‰ = đ?‘‰đ?‘œ 1 + đ?›˝Î”đ?‘‡

• For isotropic solids:

� = 3�

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C. Gas Law i. Boyle’s Law • Named after the Irish physicist Robert Boyle (1627-1691).

•

Boyle’s law states that the volume of a given mass of gas at a constant temperature is inversely proportional to its pressure.

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C. Gas Law i. Boyle’s Law •

When temperature, đ?‘ť is constant:

đ?‘ƒđ?‘– đ?‘‰đ?‘– = đ?‘ƒđ?‘“ đ?‘‰đ?‘“

Figure 2 Pressure vs. volume of a gas at a constant temperature, showing the inverse relationship as given by Boyle's law: as the pressure increases, the volume decreases.

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C. Gas Law ii. Charles’s Law • Named after the French scientist J.A.C. Charles (1746-1823). •

Charles’s law states that the volume of a fixed mass of a gas at constant pressure is directly proportional to its temperature.

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When pressure, � is constant:

�� �� = �� ��

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Figure 3 Volume of a gas function of (a) Celsius temperature, and (b) Kelvin temperature, when the pressure is kept constant.

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C. Gas Law iii. Gay-Lussac’s Law • Named after the Joseph Gay-Lussac (1778-1850). •

Gay-Lussac’s law states that at constant volume, the pressure of a gas is directly proportional to the absolute temperature.

•

When pressure, đ?‘˝ is constant:

đ?‘ƒđ?‘– đ?‘ƒđ?‘“ = đ?‘‡đ?‘– đ?‘‡đ?‘“

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C. Gas Law iv. Ideal Gas Law • The ideal equation is the combined equation of the three gas laws, which states that for a fixed mass of ideal gas,

đ?‘ƒđ?‘– đ?‘‰đ?‘– đ?‘ƒđ?‘“ đ?‘‰đ?‘“ = đ?‘‡đ?‘– đ?‘‡đ?‘“ • The complete ideal gas law equation:

đ?‘ƒđ?‘‰ = nRT

Where : n = Number of moles R = Universal gas constant

https://youtu.be/robEY-idcLU

= 8.315 J/mol•K

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• Total number of molecules, N : đ?‘ = đ?‘›đ?‘ đ??´ • Ideal Gas Law in terms of molecules:

đ?‘ƒđ?‘‰ = đ?‘ đ?‘˜đ?‘‡

• đ?‘˜, is called the Boltzmann’s constant.

EXAMPLE 4

An ideal gas at 20ď‚°đ??ś and a pressure of 1.50 Ă— 105 Pa is in a container having a volume of 1.0 đ??ż. a) Determine the number of moles of gas in the container. b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmosphere pressure. Find the final temperature. đ?‘‡đ?‘– = 20°đ??ś + 273.15 đ??ž = 293.15 đ??ž đ?‘ƒđ?‘– = 1.50 Ă— 105 đ?‘ƒđ?‘Ž đ?‘‰đ?‘– = 1.0 đ??ż = 1. 0 Ă— 10−3 đ?‘š3

𝑃𝑉 = nRT

𝑉𝑓 = 2.0 𝐿 = 2. 0 × 10−3 𝑚3 𝑃𝑓 = 1.013 × 105 𝑃𝑎

𝑃𝑖 𝑉𝑖 𝑃𝑓 𝑉𝑓 = 𝑇𝑖 𝑇𝑓

EXAMPLE 5

Compute the number of molecules in a gas contained in a volume of 10 cm3 at a pressure 1.013 x 105N/m2 and a temperature of 30 K. đ?‘‰ = 10 đ?‘?đ?‘š3 = 1 Ă— 10−5 đ?‘š3 đ?‘ƒ = 1.013 Ă— 105 đ?‘ /đ?‘š2,

đ?‘ƒđ?‘‰ = nRT

đ?‘‡ = 30đ??ž

đ?‘ = đ?‘›đ?‘ đ??´

9.2 Heat A. Heat •

Heat is the energy that is transferred from one body to another because of a difference in temperature.

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Unit : Joule (J)

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The equation used to measure the quantity of heat, ∆đ?‘„, required to increase the temperature of a mass, đ?‘š, of a certain material by an amount, ∆đ?œƒ

∆đ?‘„ = đ?‘šđ?‘?∆đ?œƒ ∆đ?‘„đ?‘”đ?‘Žđ?‘–đ?‘› = đ?‘šđ?‘? đ?œƒđ?‘“ − đ?œƒđ?‘–

@

∆đ?‘„đ?‘™đ?‘œđ?‘ đ?‘Ą = đ?‘šđ?‘? đ?œƒđ?‘– − đ?œƒđ?‘“

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B. Calorimetry •

A method to measure heat energy transfer.

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It follows the Thermal Equilibrium Principle ďƒź It states that in a close system, temperature equilibrium can

be achieve from net heat transfer from higher temperature body to a lower temperature body. ďƒź HEAT LOST FORM A HOT BODY = HEAT GAIN BY COLD BODY

∆đ?‘„đ?‘™đ?‘œđ?‘ đ?‘Ą = ∆đ?‘„đ?‘”đ?‘Žđ?‘–đ?‘›

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• To make measurements, a calorimeter is used.

• One important use of the calorimeter is to determine the specific heat of a substance.  By using the “method of mixture” technique, a sample of the substance is heated to a high temperature and then quickly placed in a cool water of the calorimeter.  By measuring the final temperature of the mixture, the specific heat can be calculated.

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EXAMPLE 6 A copper block with mass of 0.5 kg at temperature 77ď‚°C was heated until its temperature reaches 100ď‚°C. How much heat energy is needed to increase the temperature of the copper block?

Answer : 4485 J

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C. Specific Heat Capacity, c •

is the quantity required to raise 1 kg of material by 1°C.

•

Unit : J/kg•°C @ J/kg•K

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EXAMPLE 7 An engineer wishes to determine the specific heat of a new alloy. A 0.15 kg sample of the alloy is heated to 540C. It is then quickly placed in 400 g of water at 10C, which is contained in a 200 g aluminum calorimeter cup. The final temperature of the mixture is 30.5C. Calculate the specific heat of the alloy.

Answer : 497.42 J/kgC

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D. Heat Capacity, C •

is the quantity of heat required to raise material temperature by 1ď‚°C

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Unit : J/°C @ J/K đ??ś = đ?‘šđ?‘?

Where : m = Mass of material c = Specific heat capacity of material https://youtu.be/yhNHJ7WdT8A 33

E. Latent Heat Phase change •

refers to specific state of mater.

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3 types of phase : solid, liquid & gas.

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When a material changes from one phase to another it is called phase change.

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At any given pressure, a phase change occurs without any change in temperature.

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For example, a phase change in the melting of ice.

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E. Latent Heat Latent Heat •

is the heat required by a 1 kg material to change phase without changing its temperature.

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Unit : J/kg

• There are two types of latent heat: a) Latent heat of fusion, Lf

 The heat required to change 1 kg of material from the solid to the liquid phase. b) Latent heat of vaporization, Lv

 The heat required to change 1 kg of material from the liquid to the vapor phase.

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E. Latent Heat •

The heat involved in a phase change depends not only on the latent heat, but also on the total mass of the substance.

∆đ?‘„ = đ?‘šđ??ż Where : m = Mass of substance L = Latent heat of the particular process and substances

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Temperature as a function of the heat added to bring 1.0 kg of ice at -40oC to steam above 100oC Q1 = ∆Q = mc∆ Q2 = ∆Q = mL

Q2

Q1

https://youtu.be/XWoPJN6zF2k Q1

Q2

https://youtu.be/m8GhhF-_ih4 https://youtu.be/EWrXALn6sUQ

Q1

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EXAMPLE 8 How much energy does a refrigerator have to remove from 1.5 kg of water at 20ď‚°C to make ice at -12ď‚°C? Answer : 662880 J

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9.3 The First Law of Thermodynamics A. The Internal Energy, đ?‘ˆ • The internal energy of a system (U) is the total energy content of the system. • It is the sum of the kinetic, potential, chemical, electrical, nuclear, and all other forms of energy possessed by the atoms and molecules of the systems.

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9.3 The First Law of Thermodynamics B. The Work Done By A System, ď „đ?‘Š • If the system loses energy to its surrounding, ď „đ?‘Š will be positive. • When the surrounding do work on the system (give energy), ď „đ?‘Š will be negative.

• In a small expansion ď „đ?‘‰, a fluid at constant pressure p does work by: ∆đ?‘Š = đ?‘?∆đ?‘‰

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The First Law of Thermodynamics It states that if an amount of heat energy ď „đ?‘„ flows into a system, then this energy must appear as increased internal energy ď „đ?‘ˆ for the system and work done ď „đ?‘Š done by the system on its surrounding. ∆đ?‘„ = ∆đ?‘ˆ + ∆đ?‘Š

https://youtu.be/O7HwhkYt6YU

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9.4 Thermodynamics Processes • A process where a system change from one state to another.

• There are four types of themodynamics process.

A. An Isobaric Process • Process carried out at constant pressure.

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9.4 Thermodynamics Processes B. An Isovolumetric Process • Process carried out at constant volume. • When a gas undergoes such a process, ∆đ?‘Š = đ?‘?∆đ?‘ˆ = 0 • And so the first law of thermodynamics becomes

∆đ?‘„ = ∆đ?‘ˆ • Any heat that flows into the system appears as increased internal energy of the system.

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9.4 Thermodynamics Processes C. An Isothermal Process • Process carried out at constant temperature, ď „đ?‘‡ = 0 • In the case of and an ideal gas, ď „đ?‘ˆ = 0 in an isothermal process. • So the first law becomes ∆đ?‘„ = ∆đ?‘Š

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9.4 Thermodynamics Processes D. An Adiabatic Process • In this process, no heat is transferred to or from the system, ď „đ?‘„ = 0.

• So the first law becomes 0 = ∆đ?‘ˆ + ∆đ?‘Š

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• For processes where the pressure varies, the work done is the area under the đ?‘ƒđ?‘‰ curve. • Figure 4 referring to work done by an ideal gas in isothermal process.

• The work done equals the area under the đ?‘ƒđ?‘‰ curve.

Figure 4

• Shaded are equals the work done by the gas when it expands from đ?‘‰đ??´ to đ?‘‰đ??ľ.

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Figure 5 Process ADB consists of an isovolumetric (AD), and an isobaric (DB) process.

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EXAMPLE 9

In each of the following situations, find the change in internal system. a) A system absorbs 2092 J of heat and at the same time does 400 J of work. b) A system absorbs 1255 J and at the same time 420 J of work is done on it.

c) 5020.8 J is removed from the gas held at constant volume.

Answer : 1692 J, 1675 J, -5020.8 J

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EXAMPLE 10 For each of the following adiabatic processes, find the change in internal energy.

a) A gas of 5 J of work while expanding adiabatically. b) During an adiabatic compression, 80 J of work is done on a gas.

Answer : -5 J, 80 J

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