Chapter 24 Entropy and the second law of thermodynamics

1. Chapter 24 Entropy and the second law of thermodynamics

2. 24-1 Two kinds of processes and entropy • Irreversible processes（不可逆过程） (a) “All naturally occuring processes proceed in one direction only. They never, of their own accord, proceed in the opposite direction.” Such spontaneous one-way processes are “irreversible”. See动画库\力学夹\5-08过程的方向性 (b) Although the “wrong-way” events do not occur, none of them would violate the law of conservation of energy.

3. 2. Reversible process（可逆过程） In reversible process, we make a small change in a system and its environment; by reversing that change, the system and environment will return to their original conditions. In a truly reversible process, there would be no friction, turbulence(紊乱), or other dissipative effects, which will cause non-compensatory losses of energy. See动画库\力学夹\5-09可逆过程

4. 3. Entropy (S) Entropy is a physical quantity that controlsthe direction of irreversible processes. It is a property of the state of a system; like T, P, V, Eint. Entropy principle: “If an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases.”

5. If the process is isothermal, (24-2) , , the entropy of that system increases.( if Q<0, ). 24-2 Entropy change for reversible processes 1.The definition of entropy change for a reversible process: (reversible) (24-1) Here dQ is the increment of heat energy that is transferred into (or out) of the closed system at temperatureT.

6. 2. Entropy as a state property ※ What’s the entropy change for ideal gases? Divided by T, we obtain from the first law P=nRT/V Integrate between an arbitrary initial state i and an final state f: Only related to initial and final states

7. Sampleproblem 24-1 A vessel containing 1.8kg of water is placed on a hot plate. Both the water and hot plate being initially at . The temperature of the hot plate is raised very slowly to , at which point the water begins to boil. Find of the water during this process? Solution:

8. 24-3 Entropy change for irreversible processes • Due to unavoidablenon-compensatory losses of energy in an irreversible process, it is difficult to calculate directly for it. How to obtain for irreversible processes? • Two steps: 1.Design a reversible process that has the same initial and final states as those in the irreversible process. 2.Use Eq(24-1) to calculate for this reversible process. The obtained is actually the one for the original irreversible process since entropy change is as a state property.

9. An example: reversible process Irreversible process m m Insulation Insulation h h water water water water Thermal reservoir Insulating slab (a) Initial state (b) Final state (a) Initial state (b) Final state Fig 24-1 Fig 24-2

10. Sample problem 24-2 As Fig 24-1, , , , (a) What is the temperature rise of the system water + stone? (b) Find the entropy change of this system. (c) What would be the for the reverse process—that is, for the system to cool down, transferring its energy to the stone in kinetic form, allowing it to leap 2.5m into the air. The specific heat of water is and that of the stone material is .

11. Solution: (a) In the equivalentreversible process (24-4) In the original irreversible process, we have (internal energy is a state quantity) From (b) (c) In the reverse process, , , it never happen.

12. Fig 24-3 shows a paper cup containing a mass m=0.57kg ofhot water and a similar cup containing an equal mass of cold water. The initial temperature of the hot water is ; that of the cold water is . Sample problem 24-3 Insulation cold hot H C Fig 24-3 Insulation C H Fig 24-4

13. When the insulating shutter separating the two enclosure is removed, the hot water and the cold water eventually come to thermal equilibrium at a temperature of . What is the entropy change of the system for this irreversible process? ( , the heat capacity of the paper cups is negligible)

14. Sample problem 24-4 stopcock closed insulating lead shot vacuum insulating T=293k n=0.55mol of an ideal gas T=293k, V1=V2 Thermal reservoir What is the ? Fig 24-6 Fig 24-5

15. 24-4 The second law of thermodynamics • The three examples in above 24-3 show that the systems for the irreversible processes are closed. And the corresponding . • What’s the behavior of for reversible processes in a closed system? . Reconsider Problem 24-4, it can be known that the gas and the reservoir form a closed system. The entropy change of this closed system is found to be zero: . For expansion process insulating gas Thermal reservoir

16. The second law of thermodynamics: “When changes occur within a closed system, itsentropy either increases (for irreversible processes) or remains constant (forreversible processes). It never decreases.” In equation form this statement becomes (for a closed system) (24-5) The “greater than” sign applies to irreversible processes and the “equals” sign to reversible processes.

17. 24-5 Entropy and the performance of engines • A heat engine is a device that extracts energyfrom its environment in the form of heat and does useful work. • At the heart of every engine is a working substance (gas). • For an engine to do work on a sustained (持续的) basis, the working substance must operate in a cycle. See动画库\力学夹\5-06循环过程 1

18. A Carnot engine（卡诺热机） • Fig 24-7 shows schematically the • operation of a carnot engine. reservoir We assume that all thermodynamic processes are reversible. W reservoir Fig 24-7

19. A P B Q1 D Q2 C 0 V 2.The Carnot cycle（卡诺循环） P-V plot of the Carnot cycle AB: isothermal processes Fig 24-8 BC: adiabatic processes CD: isothermal processes DA: adiabatic processes From the first law (for the cycle): (24-8)

20. 3.The T-S diagram Fig 24-9 shows T-S diagram. Fig 24-8 A T B A B D D C C 0 S QH isothermal adiabatic QL isothermal adiabatic Fig 24-9 Since S is a state quanlity, (24-6) Since is also a state quanlity,

21. 4. Efficiency of a Carnot engine The purpose of an engine is to transform as much of the extracted heat into work as possible. We define “thermal efficiency ” as (24-7) From (24-8):

22. Since we have (Carnot efficiency) (24-12) Since , we usually have . Eq(24-12) gives the efficiency of all Carnot engines working between the same two fixed temperatures (24-11)

23. Fig 24-10 5. Search for a “perfect engine” Inventors continually try to improve engine efficiency by reducing that is “thrown away” during each cycle. In Fig 24-10, is reduced to zero and is converted completely into work, then . or for . That is impossible!

24. (Heat) (Work) The second version of the second law of thermodynamics: “No series of processes is possible whose sole result is the absorption of heat from one thermal reservoir and the complete conversion of thisenergy to work.” ---Kelvin version Or ? Equivalent ---Original version During a cycle: 0 Inconsistent with

25. 永 动 机 的 设 想 图

26. Fig 24-11 p A B Q Q D C v 6. Real engine The thermal efficiency given by Eq(24-12) applies only to idealCarnot engines. Real engine in which the processes are not reversible, have lower efficiency. 7. Other reversible engines Fig 24-11 shows the operating cycle of an ideal (that is reversible) Stirling engine. A->B and C->D are isothermal processes; B->C and D->A are const. volume processes.

27. 1.The Carnot refrigerator Fig 24-12 shows the basic elements of a refrigerator. We call it a Carnot refrigerator. W A B D C 0 24-6 Entropy and the performance of refrigerators Fig 24-12 are satisfied.

28. We define the “coefficient of performance” (24-13) (24-14) for Carnot refrigerator (24-15)

29. 2. Search for a “perfect” refrigerator Fig24-13 shows another “inventor’s dream”, a perfect refrigerator, for which , so that . However, it would violate the second law. (for a cycle) Q Q Fig 24-13

30. ? Kelvin version This result leads to a third version of the second law of thermodynamics: “No process is possible whose sole result is the transfer of heat from a reservoir at one temperature to another reservoir at a higher temperature.” In short, “ there are no perfect refrigerators”. ---Clausius version

31. Sample problem 24-6 A household refrigerator, K=4.5, extracts heat from the food chamber at the rate of . • how much work per cycle is required to operate the refrigerator? (b) how much heat per cycle is discharged to the room? Solution: (a) (b)

32. *24-7 The efficiencies of real engine “No real engine can have an efficiency greater than that of a Carnot engine working between the same two temperatures”. Q c w x Q (a) (b)

33. 24-8 The second law revisited So far we have presented three statements of the second law of thermodynamics. 1. The entropy of such system never decreases, . 2. You can not change heat energy into work with 100% efficiency. That is, there are no perfect engines. 3. You can not transfer heat energy from a low- temperature reservoir to a higher temperature reservoir without doing work. That is, there are no perfect refrigerators. ---Kelvin version ---Clausius version

34. ? Clausius version Kelvin version If statement 2 were false and we could build a perfect engine, connected with a refrigerator as shown in Fig 24-15a. W perfect engine (a) (b) Fig 24-15

35. Thus, a violation of statement 2 implies a violation of statement 3. Therefore, all the three versions of the second law of thermodynamics are equivalent.

36. 24-9 A statistical view of entropy 1. Entropy and multiplicity(多重度,微观状态数) See动画库\力学夹\5-10热二律的统计解释 (24-20) W is the multiplicity associated with the configuration whose entropy we wish to calculate. Eq (24-20) is called Boltzmann’s entropy equation 2. Entropy and disorder See动画库\力学夹\5-12都要说不！

37. About Final Examination • Content: Ch. 17-24 • Time: 1/17(Thurs.) 8:30-10:30 am • Classroom: 3-308 • Question Time(答疑时间): 1/ 10 (Thurs.) 2:00-4:00 pm 1/16 (Wed.) 9:00-11:00 am • Question Place(答疑地点): Rm. 2421, Guanghua Buil. (East) * Content of Reexam.(补考): Ch. (1~24)