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Chapter 10 - Thermodynamics



ThermodynamicsChapter 10

Table of Contents

Section 1 Relationships Between Heat and Work

Section 2 The First Law of Thermodynamics

Section 3 The Second Law of Thermodynamics



Section 1 Relationships

Between Heat and WorkChapter 10

Objectives

• Recognize that a system can absorb or release energy as heat in order for work to be done on or by the system and that work done on or by a system can result in the transfer of energy as heat.

• Compute the amount of work done during a thermodynamic process.

• Distinguish between isovolumetric, isothermal, and adiabatic thermodynamic processes.



Chapter 10

Heat, Work, and Internal Energy

• Heat and work are energy transferred to or from a system. An object never has “heat” or “work” in it; it has only internal energy.


Between Heat and Work



( )

work = pressure volume change

A FW Fd Fd Ad P V

A A

W P V

Chapter 10

Heat, Work, and Internal Energy, continued

• In thermodynamic systems, work is defined in terms of pressure and volume change.





Chapter 10


• If the gas expands, as shown in the figure, V is positive, and the work done by the gas on the piston is positive.

• If the gas is compressed, V is negative, and the work done by the gas on the piston is negative. (In other words, the piston does work on the gas.)





Chapter 10


• When the gas volume remains constant, there is no displacement and no work is done on or by the system.

• Although the pressure can change during a process, work is done only if the volume changes.





Chapter 10

Thermodynamic Processes

• An isovolumetric process is a thermodynamic process that takes place at constant volume so that no work is done on or by the system.

• An isothermal process is a thermodynamic process that takes place at constant temperature.

• An adiabatic process is a thermodynamic process during which no energy is transferred to or from the system as heat.





Thermodynamic Processes

Chapter 10Section 1 Relationships




Section 2 The First Law of ThermodynamicsChapter 10

Objectives

• Illustrate how the first law of thermodynamics is a statement of energy conservation.

• Calculate heat, work, and the change in internal energy by applying the first law of thermodynamics.

• Apply the first law of thermodynamics to describe cyclic processes.



Chapter 10

Energy Conservation

• If friction is taken into account, mechanical energy is not conserved.

• Consider the example of a roller coaster:– A steady decrease in the car’s total mechanical energy

occurs because of work being done against the friction between the car’s axles and its bearings and between the car’s wheels and the coaster track.

– If the internal energy for the roller coaster (the system) and the energy dissipated to the surrounding air (the environment) are taken into account, then the total energy will be constant.




Energy Conservation

Chapter 10Section 2 The First Law of Thermodynamics



Chapter 10

Energy Conservation




Chapter 10

Energy Conservation, continued

• The principle of energy conservation that takes into account a system’s internal energy as well as work and heat is called the first law of thermodynamics.

• The first law of thermodynamics can be expressed mathematically as follows:

U = Q – W

Change in system’s internal energy = energy transferred to or from system as heat – energy

transferred to or from system as work




Chapter 10

Signs of Q and W for a system




Chapter 10

Sample Problem

The First Law of Thermodynamics

A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat? Has energy been added to or removed from the refrigerant as heat?




Chapter 10

Sample Problem, continued

1. Define

Given:

W = –135 J

U = 114 J


Tip: Work is done on the gas, so work (W) has a negative value. The internal energy increases during the process, so the change in internal energy (U) has a positive value.

Diagram:

Unknown:

Q = ?



Chapter 10


2. Plan

Choose an equation or situation:

Apply the first law of thermodynamics using the values for U and W in order to find the value for Q.

U = Q – W


Rearrange the equation to isolate the unknown:

Q = U + W



Chapter 10


3. Calculate

Substitute the values into the equation and solve:

Q = 114 J + (–135 J)

Q = –21 J


Tip: The sign for the value of Q is negative. This indicates that energy is transferred as heat from the refrigerant.



Chapter 10


4. Evaluate

Although the internal energy of the refrigerant increases under compression, more energy is added as work than can be accounted for by the increase in the internal energy. This energy is removed from the gas as heat, as indicated by the minus sign preceding the value for Q.




First Law of Thermodynamics for Special Processes




Chapter 10

Cyclic Processes

• A cyclic process is a thermodynamic process in which a system returns to the same conditions under which it started.

• Examples include heat engines and refrigerators.

• In a cyclic process, the final and initial values of internal energy are the same, and the change in internal energy is zero.

Unet = 0 and Qnet = Wnet




Chapter 10

Cyclic Processes, continued

• A heat engine uses heat to do mechanical work.

• A heat engine is able to do work (b) by transferring energy from a high-temperature substance (the boiler) at Th (a) to a substance at a lower temperature (the air around the engine) at Tc (c).


• The internal-combustion engine found in most vehicles is an example of a heat engine.



Combustion Engines




Chapter 10

The Steps of a Gasoline Engine Cycle




Refrigeration




Chapter 10

The Steps of a Refrigeration Cycle




Section 3 The Second Law of ThermodynamicsChapter 10

Objectives

• Recognize why the second law of thermodynamics requires two bodies at different temperatures for work to be done.

• Calculate the efficiency of a heat engine.

• Relate the disorder of a system to its ability to do work or transfer energy as heat.



Chapter 10

Efficiency of Heat Engines

• The second law of thermodynamics can be stated as follows:

No cyclic process that converts heat entirely into work is possible.

• As seen in the last section, Wnet = Qnet = Qh – Qc.

– According to the second law of thermodynamics, W can never be equal to Qh in a cyclic process.

– In other words, some energy must always be transferred as heat to the system’s surroundings (Qc > 0).




Chapter 10

Efficiency of Heat Engines, continued

• A measure of how well an engine operates is given by the engine’s efficiency (eff ).


• Because of the second law of thermodynamics, the efficiency of a real engine is always less than 1.

eff Wnet

QhQh –QcQh

1 QcQh



Chapter 10

Sample Problem

Heat-Engine Efficiency

Find the efficiency of a gasoline engine that, during one cycle, receives 204 J of energy from combustion and loses 153 J as heat to the exhaust.


1. Define

Given: Diagram:

Qh = 204 J

Qc = 153 J

Unknown

eff = ?



Chapter 10


2. Plan

Choose an equation or situation: The efficiency of a heat engine is the ratio of the work done by the engine to the energy transferred to it as heat.


eff Wnet

Qh1

QcQh



Chapter 10


3. Calculate

Substitute the values into the equation and solve:


eff 1 QcQh

1 153 J

204 J

eff 0.250

4. Evaluate

Only 25 percent of the energy added as heat is used by the engine to do work. As expected, the efficiency is less than 1.0.



Chapter 10

Entropy

• In thermodynamics, a system left to itself tends to go from a state with a very ordered set of energies to one in which there is less order.

• The measure of a system’s disorder or randomness is called the entropy of the system. The greater the entropy of a system is, the greater the system’s disorder.

• The entropy of a system tends to increase.




Chapter 10

Entropy, continued

• The second law of thermodynamics can also be expressed in terms of entropy change:

The entropy of the universe increases in all natural processes.




Chapter 10

Energy Changes Produced by a Refrigerator Freezing Water


Because of the refrigerator’s less-than-perfect efficiency, the entropy of the outside air molecules increases more than the entropy of the freezing water decreases.



Entropy of the Universe

Chapter 10Section 3 The Second Law of Thermodynamics



Multiple Choice

1. If there is no change in the internal energy of a gas, even though energy is transferred to the gas as heat and work, what is the thermodynamic process that the gas undergoes called?

A. adiabatic

B. isothermal

C. isovolumetric

D. isobaric

Standardized Test PrepChapter 10



Multiple Choice

1. If there is no change in the internal energy of a gas, even though energy is transferred to the gas as heat and work, what is the thermodynamic process that the gas undergoes called?

A. adiabatic

B. isothermal

C. isovolumetric

D. isobaric




Multiple Choice, continued

2. To calculate the efficiency of a heat engine, which thermodynamic property does not need to be known?

F. the energy transferred as heat to the engine

G. the energy transferred as heat from the engine

H. the change in the internal energy of the engine

J. the work done by the engine





2. To calculate the efficiency of a heat engine, which thermodynamic property does not need to be known?

F. the energy transferred as heat to the engine

G. the energy transferred as heat from the engine

H. the change in the internal energy of the engine

J. the work done by the engine




3. In which of the following processes is no work done?

A. Water is boiled in a pressure cooker.

B. A refrigerator is used to freeze water.

C. An automobile engine operates for several minutes.

D. A tire is inflated with an air pump.





3. In which of the following processes is no work done?

A. Water is boiled in a pressure cooker.

B. A refrigerator is used to freeze water.

C. An automobile engine operates for several minutes.

D. A tire is inflated with an air pump.





4. A thermodynamic process occurs in which the entropy of a system decreases. From the second law of thermodynamics, what can you conclude about the entropy change of the environment?

F. The entropy of the environment decreases.

G. The entropy of the environment increases.

H. The entropy of the environment remains unchanged.

J. There is not enough information to state what happens to the environment’s entropy.





4. A thermodynamic process occurs in which the entropy of a system decreases. From the second law of thermodynamics, what can you conclude about the entropy change of the environment?

F. The entropy of the environment decreases.

G. The entropy of the environment increases.

H. The entropy of the environment remains unchanged.

J. There is not enough information to state what happens to the environment’s entropy.





Use the passage and diagrams to answer questions 5–8.

A system consists of steam within the confines of a steam engine, whose cylinder and piston are shown in the figures below.


5. Which of the figures describes a situation in which U < 0, Q < 0, and W = 0?

A. (a)

B. (b)

C. (c)

D. (d)







5. Which of the figures describes a situation in which U < 0, Q < 0, and W = 0?

A. (a)

B. (b)

C. (c)

D. (d)







6. Which of the figures describes a situation in which U > 0, Q = 0, and W < 0?

F. (a)

G. (b)

H. (c)

J. (d)







6. Which of the figures describes a situation in which U > 0, Q = 0, and W < 0?

F. (a)

G. (b)

H. (c)

J. (d)







7. Which of the figures describes a situation in which U < 0, Q = 0, and W > 0?

A. (a)

B. (b)

C. (c)

D. (d)







7. Which of the figures describes a situation in which U < 0, Q = 0, and W > 0?

A. (a)

B. (b)

C. (c)

D. (d)







8. Which of the figures describes a situation in which U > 0, Q > 0, and W = 0?

F. (a)

G. (b)

H. (c)

J. (d)







8. Which of the figures describes a situation in which U > 0, Q > 0, and W = 0?

F. (a)

G. (b)

H. (c)

J. (d)




9. A power plant has a power output of 1055 MW and operates with an efficiency of 0.330. Excess energy is carried away as heat from the plant to a nearby river. How much energy is transferred away from the power plant as heat?

A. 0.348 109 J/s

B. 0.520 109 J/s

C. 0.707 109 J/s

D. 2.14 109 J/s





9. A power plant has a power output of 1055 MW and operates with an efficiency of 0.330. Excess energy is carried away as heat from the plant to a nearby river. How much energy is transferred away from the power plant as heat?

A. 0.348 109 J/s

B. 0.520 109 J/s

C. 0.707 109 J/s

D. 2.14 109 J/s





10. How much work must be done by air pumped into a tire if the tire’s volume increases from 0.031 m3 to 0.041 m3 and the net, constant pressure of the air is 300.0 kPa?

F. 3.0 102 J

G. 3.0 103 J

H. 3.0 104 J

J. 3.0 105 J





10. How much work must be done by air pumped into a tire if the tire’s volume increases from 0.031 m3 to 0.041 m3 and the net, constant pressure of the air is 300.0 kPa?

F. 3.0 102 J

G. 3.0 103 J

H. 3.0 104 J

J. 3.0 105 J





Use the passage below to answer questions 11–12.

An air conditioner is left running on a table in the middle of the room, so none of the air that passes through the air conditioner is transferred to outside the room.

11. Does passing air through the air conditioner affect

the temperature of the room? (Ignore the thermal

effects of the motor running the compressor.)


Short Response





11. Does passing air through the air conditioner affect

the temperature of the room? (Ignore the thermal

effects of the motor running the compressor.)

Answer: No, because the energy removed from the cooled air is returned to the room.


Short Response





12. Taking the compressor motor into account, what

would happen to the temperature of the room?


Short Response, continued





12. Taking the compressor motor into account, what

would happen to the temperature of the room?

Answer: The temperature increases.





13. If 1600 J of energy are transferred as heat to an

engine and 1200 J are transferred as heat away

from the engine to the surrounding air, what is the

efficiency of the engine?





13. If 1600 J of energy are transferred as heat to an

engine and 1200 J are transferred as heat away

from the engine to the surrounding air, what is the

efficiency of the engine?

Answer: 0.25





14. How do the temperature of combustion and the

temperatures of coolant and exhaust affect the

efficiency of automobile engines?


Extended Response



14. How do the temperature of combustion and the

temperatures of coolant and exhaust affect the

efficiency of automobile engines?

Answer: The greater the temperature difference is, the greater is the amount of energy transferred as heat. For efficiency to increase, the heat transferred between the combustion reaction and the engine (Qh) should be made to increase, whereas the energy given up as waste heat to the coolant and exhaust (Qc) should be made to decrease.


Extended Response



Use the information below to answer questions 15–18. In each problem, show all of your work.

A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m. The steam shovel’s engine provides 2.00 105 J of energy as heat for the steam shovel to lift the dirt.


Extended Response, continued

15. How much work is done by the steam shovel in lifting the dirt?







15. How much work is done by the steam shovel in lifting the dirt?

Answer: 3.8 104 J







16. What is the efficiency of the steam shovel?







16. What is the efficiency of the steam shovel?

Answer: 0.19







17. Assuming there is no change in the internal energy of the steam shovel’s engine, how much energy is given up by the shovel as waste heat?







17. Assuming there is no change in the internal energy of the steam shovel’s engine, how much energy is given up by the shovel as waste heat?

Answer: 1.62 105 J







18. Suppose the internal energy of the steam shovel’s engine increases by 5.0 103 J.

How much energy is given up now as waste heat?







18. Suppose the internal energy of the steam shovel’s engine increases by 5.0 103 J.

How much energy is given up now as waste heat?

Answer: 1.57 105 J



19. One way to look at heat and work is to think of

energy transferred as heat as a “disorganized” form

of energy and energy transferred as work as an

“organized” form. Use this interpretation to show

that the increased order obtained by freezing

water is less than the total disorder that results

from the freezer used to form the ice.





19. One way to look at heat and work is to think of

energy transferred as heat as a “disorganized” form

of energy and energy transferred as work as an

“organized” form. Use this interpretation to show

that the increased order obtained by freezing

water is less than the total disorder that results

from the freezer used to form the ice.



Answer: Disorganized energy is removed from water to form ice, but a greater amount of organized energy must become disorganized to operate the freezer.



Chapter 10

Entropy




Chapter 10

Energy Changes Produced by a Refrigerator Freezing Water


Technology

Chapter 10 Powerpoint