1. Chapter 10 Preview Objectives Heat, Work, and Internal Energy
Section 1 Relationships Between Heat and Work
Chapter 10
Preview
Objectives
Heat, Work, and Internal Energy
Thermodynamic Processes

2. Section 1 Relationships Between Heat and Work
Chapter 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.

3. Heat, Work, and Internal Energy
Section 1 Relationships Between Heat and Work
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.
A system is a set of particles or interacting components considered to be a distinct physical entity for the purpose of study.
The environment the combination of conditions and influences outside a system that affect the behavior of the system.

4. Heat, Work, and Internal Energy, continued
Section 1 Relationships Between Heat and Work
Chapter 10
Heat, Work, and Internal Energy, continued
In thermodynamic systems, work is defined in terms of pressure and volume change.
This definition assumes that P is constant.

5. Heat, Work, and Internal Energy, continued
Section 1 Relationships Between Heat and Work
Chapter 10
Heat, Work, and Internal Energy, continued
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.)

6. Heat, Work, and Internal Energy, continued
Section 1 Relationships Between Heat and Work
Chapter 10
Heat, Work, and Internal Energy, continued
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.
A situation in which pressure increases and volume remains constant is comparable to one in which a force does not displace a mass even as the force is increased. Work is not done in either situation.

7. Thermodynamic Processes
Section 1 Relationships Between Heat and Work
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.

8. Thermodynamic Processes
Section 1 Relationships Between Heat and Work
Chapter 10
Thermodynamic Processes
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Visual Concept

9. Chapter 10 Preview Objectives Energy Conservation Sample Problem
Section 2 The First Law of Thermodynamics
Chapter 10
Preview
Objectives
Energy Conservation
Sample Problem
Cyclic Processes

10. Section 2 The First Law of Thermodynamics
Chapter 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.

11. Chapter 10 Energy Conservation
Section 2 The First Law of Thermodynamics
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.

12. Chapter 10 Energy Conservation
Section 2 The First Law of Thermodynamics
Chapter 10
Energy Conservation
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Visual Concept

13. Chapter 10 Energy Conservation
Section 2 The First Law of Thermodynamics
Chapter 10
Energy Conservation

14. Energy Conservation, continued
Section 2 The First Law of Thermodynamics
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

15. Signs of Q and W for a system
Section 2 The First Law of Thermodynamics
Chapter 10
Signs of Q and W for a system

16. Chapter 10 Sample Problem The First Law of Thermodynamics
Section 2 The First Law of Thermodynamics
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?

17. Sample Problem, continued
Section 2 The First Law of Thermodynamics
Chapter 10
Sample Problem, continued
1. Define
Given:
W = –135 J
U = 114 J
Diagram:
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.
Unknown:
Q = ?

18. Sample Problem, continued
Section 2 The First Law of Thermodynamics
Chapter 10
Sample Problem, continued
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

19. Sample Problem, continued
Section 2 The First Law of Thermodynamics
Chapter 10
Sample Problem, continued
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.

20. Sample Problem, continued
Section 2 The First Law of Thermodynamics
Chapter 10
Sample Problem, continued
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.

21. First Law of Thermodynamics for Special Processes
Section 2 The First Law of Thermodynamics
Chapter 10
First Law of Thermodynamics for Special Processes
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Visual Concept

22. Chapter 10 Cyclic Processes
Section 2 The First Law of Thermodynamics
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

23. Cyclic Processes, continued
Section 2 The First Law of Thermodynamics
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.

24. Chapter 10 Combustion Engines
Section 2 The First Law of Thermodynamics
Chapter 10
Combustion Engines
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Visual Concept

25. The Steps of a Gasoline Engine Cycle
Section 2 The First Law of Thermodynamics
Chapter 10
The Steps of a Gasoline Engine Cycle

26. Chapter 10 Refrigeration Section 2 The First Law of Thermodynamics
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Visual Concept

27. The Steps of a Refrigeration Cycle
Section 2 The First Law of Thermodynamics
Chapter 10
The Steps of a Refrigeration Cycle

28. Thermodynamics of a Refrigerator
Section 2 The First Law of Thermodynamics
Chapter 10
Thermodynamics of a Refrigerator

29. Chapter 10 Preview Objectives Efficiency of Heat Engines
Section 3 The Second Law of Thermodynamics
Chapter 10
Preview
Objectives
Efficiency of Heat Engines
Sample Problem
Entropy

30. Section 3 The Second Law of Thermodynamics
Chapter 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.

31. Efficiency of Heat Engines
Section 3 The Second Law of Thermodynamics
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).

32. Efficiency of Heat Engines, continued
Section 3 The Second Law of Thermodynamics
Chapter 10
Efficiency of Heat Engines, continued
A measure of how well an engine operates is given by the engine’s efficiency (eff ).
In general, efficiency is a measure of the useful energy taken out of a process relative to the total energy that is put into the process.
Note that efficiency is a unitless quantity.
Because of the second law of thermodynamics, the efficiency of a real engine is always less than 1.

33. Chapter 10 Sample Problem Heat-Engine Efficiency
Section 3 The Second Law of Thermodynamics
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 = ?

34. Sample Problem, continued
Section 3 The Second Law of Thermodynamics
Chapter 10
Sample Problem, continued
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.

35. Sample Problem, continued
Section 3 The Second Law of Thermodynamics
Chapter 10
Sample Problem, continued
3. Calculate
Substitute the values into the equation and solve:
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.

36. Section 3 The Second Law of Thermodynamics
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 greater probability of a disordered arrangement indicates that an ordered system is likely to become disordered. Put another way, the entropy of a system tends to increase.

37. Chapter 10 Entropy, continued
Section 3 The Second Law of Thermodynamics
Chapter 10
Entropy, continued
Greater disorder means there is less energy to do work.
If all gas particles moved toward the piston, all of the internal energy could be used to do work. This extremely well ordered system is highly improbable.

38. Chapter 10 Entropy, continued
Section 3 The Second Law of Thermodynamics
Chapter 10
Entropy, continued
Because of the connection between a system’s entropy, its ability to do work, and the direction of energy transfer, the second law of thermodynamics can also be expressed in terms of entropy change:
The entropy of the universe increases in all natural processes.
Entropy can decrease for parts of systems, provided this decrease is offset by a greater increase in entropy elsewhere in the universe.

39. Energy Changes Produced by a Refrigerator Freezing Water
Section 3 The Second Law of Thermodynamics
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.