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

2. 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. 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. Section 1 Relationships Between Heat and Work

4. Work and Pressure

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

6. 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.) Section 1 Relationships Between Heat and Work

7. 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. Section 1 Relationships Between Heat and Work

8. Practice A: Page An engine cylinder has a cross-sectional area of 0.010m 2. How much work can be done by a gas in the cylinder if the gas exerts a constant pressure of 7.5 x 10 5 Pa on the piston and moves the piston a distance of 0.040m? Given: P= 7.5 x 10 5 N/m 2, A= 0.010 m 2, d= 0.040m Asked: W=? Units Joules Formula: W= P * A * d =

9. Chapter 10 Assignments Practice A Page 338: 1,a,b,2,3(4)

10. Concept Review Work- transfer of energy by motion of an object W= F*d Heat- transfer of energy from a hot object to a cold object. Involves a change in temperature. Q (lost) = -Q (gained) Q = mΔTc p W= PΔV

11. 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. Section 1 Relationships Between Heat and Work

12. Processes isovolumetric process Δ V = 0, W= 0 isothermal process Δ T = 0, ΔU = 0 adiabatic process Q = 0

13. Click below to watch the Visual Concept. Visual Concept Chapter 10 Section 1 Relationships Between Heat and Work Thermodynamic Processes

14. Chapter 10 Assignments Practice A Page 338: 1,a,b,2,3(4) Section Review Page 341: 1,2(2),3,4(3)(7)

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

16. 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.

17. 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. Section 2 The First Law of Thermodynamics

18. Click below to watch the Visual Concept. Visual Concept Chapter 10 Section 2 The First Law of Thermodynamics Energy Conservation

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

20. 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 Section 2 The First Law of Thermodynamics

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

22. 1 st Law of Thermodynamics  U = Q – W

23. 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? Section 2 The First Law of Thermodynamics

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

25. 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 Section 2 The First Law of Thermodynamics Rearrange the equation to isolate the unknown: Q =  U + W

26. Chapter 10 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: Q = 114 J + (–135 J) Q = –21 J Section 2 The First Law of Thermodynamics Tip: The sign for the value of Q is negative. This indicates that energy is transferred as heat from the refrigerant.

27. 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. Section 2 The First Law of Thermodynamics

28. Chapter 10 Assignments Practice A Page 338: 1,a,b,2,3(4) Section Review Page 341: 1,2(2),3,4(3)(7) Practice B Page 346: 1,2(2),3,4(2)(6)

29. Click below to watch the Visual Concept. Visual Concept Chapter 10 Section 2 The First Law of Thermodynamics First Law of Thermodynamics for Special Processes

30. 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.  U net = 0 and Q net = W net Section 2 The First Law of Thermodynamics

31. 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 T h (a) to a substance at a lower temperature (the air around the engine) at T c (c). Section 2 The First Law of Thermodynamics The internal-combustion engine found in most vehicles is an example of a heat engine.

32. Click below to watch the Visual Concept. Visual Concept Chapter 10 Section 2 The First Law of Thermodynamics Combustion Engines

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

34. Click below to watch the Visual Concept. Visual Concept Chapter 10 Section 2 The First Law of Thermodynamics Refrigeration

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

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

37. Chapter 10 Assignments Practice A Page 338: 1,a,b,2,3(4) Section Review Page 341: 1,2(2),3,4(3)(7) Practice B Page 346: 1,2(2),3,4(2)(6) Section Review Page 349: 4(3), 5, 6(4),7,8(10)

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

39. 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.

40. 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, W net = Q net = Q h – Q c. –According to the second law of thermodynamics, W can never be equal to Q h in a cyclic process. –In other words, some energy must always be transferred as heat to the system’s surroundings (Q c > 0). Section 3 The Second Law of Thermodynamics

41. Heat Engine – cyclical process ΔU = O

42. 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. Section 3 The Second Law of Thermodynamics 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.

43. 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. Section 3 The Second Law of Thermodynamics 1.Define Given:Diagram: Q h = 204 J Q c = 153 J Unknown eff = ?

44. Chapter 10 Assignments Practice A Page 338: 1,a,b,2,3(4) Section Review Page 341: 1,2(2),3,4(3)(7) Practice B Page 346: 1,2(2),3,4(2)(6) Section Review Page 349: 4(3), 5, 6(4),7,8(10) Practice C Page 355: 1,2,3a,b,4(5) Chapter Review Page 360: 9(2), 10, 16, 27,29(2)(10)

45. 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. Section 3 The Second Law of Thermodynamics

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

47. 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. Section 3 The Second Law of Thermodynamics

48. Chapter 10 Entropy, continued 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. Section 3 The Second Law of Thermodynamics Greater disorder means there is less energy to do work.

49. 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. Section 3 The Second Law of Thermodynamics

50. Chapter 10 Energy Changes Produced by a Refrigerator Freezing Water Section 3 The Second Law of Thermodynamics 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.

51. Click below to watch the Visual Concept. Visual Concept Chapter 10 Section 3 The Second Law of Thermodynamics Entropy of the Universe

52. Chapter 10 Assignments Practice A Page 338: 1,a,b,2,3(4) Section Review Page 341: 1,2(2),3,4(3)(7) Practice B Page 346: 1,2(2),3,4(2)(6) Section Review Page 349: 4(3), 5, 6(4),7,8(10) Practice C Page 355: 1,2,3a,b,4(5) Section Review Page 357: 2a,b,c,3, 5(5) Chapter Review Page 360: 9(2), 10, 16, 17(3),27,29(2)(10)