Short Version : 19. 2 nd Law of Thermodynamics

1. Short Version : 19. 2nd Law of Thermodynamics

2. 19.1. Reversibility & Irreversibility Block slowed down by friction: irreversible Bouncing ball: reversible Examples of irreversible processes: • Beating an egg, blending yolk & white • Cups of cold & hot water in contact Spontaneous process: order  disorder ( statistically more probable )

3. 19.2. The 2nd Law of Thermodynamics Heat engine extracts work from heat reservoirs. • gasoline & diesel engines • fossil-fueled & nuclear power plants • jet engines Perfect heat engine: coverts heat to work directly. 2nd law of thermodynamics ( Kelvin-Planck version ): There is no perfect heat engine. No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Heat dumped

4. Hero Engine Stirling Engine (any engine) Efficiency (any cycle) (Simple engine)

5. Carnot Engine Ideal gas: AB: Heat abs. B  C: Work done C  D: Heat rejected: D  A: Work done Adiabatic processes: • isothermal expansion: T = Th , W1 = Qh > 0 • Adiabatic expansion: Th Tc, W2 > 0 • isothermal compression: T = Tc , W3 =  Qc < 0 Adiabatic compression : Tc Th , W4 =  W2 < 0 

6. Engines, Refrigerators, & the 2nd Law • Carnot’s theorem: • All Carnot engines operating between temperatures Th & Tc have the same efficiency. • No other engine operating between Th & Tc can have a greater efficiency. Refrigerator: extracts heat from cool reservoir into a hot one. work required

7. perfect refrigerator: moves heat from cool to hot reservoir without work being done on it. 2nd law of thermodynamics ( Clausius version ): There is no perfect refrigerator. No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.

8. Perfect refrigerator  Perfect heat engine Clausius  Kelvin-Planck

9. Hypothetical engine, e = 70% Carnot refrigerator, e = 60%  Carnot engine is most efficient eCarnot = thermodynamic efficiency eCarnot erev > eirrev

10. 19.3. Applications of the 2nd Law Power plant Turbine Generator Electricity Boiler fossil-fuel : Th = 650 K Nuclear : Th = 570 K Tc = 310 K Condenser Heat source Cooling water Waste water Actual values: efossil ~ 40 % enuclear~ 34 % ecar ~ 20 % Prob 54 & 55

11. Application: Combined-Cycle Power Plant Turbine engines: high Th ( 1000K  2000K ) & Tc ( 800 K ) … not efficient. Steam engines : Tc ~ ambient 300K. Combined-cycle : Th ( 1000K  2000K ) & Tc ( 300 K ) … e ~ 60%

12. Example 19.2. Combined-Cycle Power Plant The gas turbine in a combined-cycle power plant operates at 1450 C. Its waste heat at 500 C is the input for a conventional steam cycle, with its condenser at 8 C. Find e of the combined-cycle, & compare it with those of the individual components.

13. Refrigerators Coefficient of performance (COP) for refrigerators : 1st law Max. theoretical value (Carnot) COP is high if Th Tc . W = 0 ( COP =  ) for moving Q when Th= Tc .

14. Example 19.3. Home Freezer A typical home freezer operates between Tc =  18C to Th = 30 C. What’s its maximum possible COP? With this COP, how much electrical energy would it take to freeze 500 g of water initially at 0 C? Table 17.1 2nd law: only a fraction of Q can become W in heat engines. a little W can move a lot of Q in refrigerators.

15. Heat Pumps Heat pump: moves heat from Tc to Th . Heat pump as AC : Heat pump as heater : Ground temp ~ 10C year round (US)

16. 19.4. Entropy & Energy Quality 2nd law: Energy of higher quality can be converted completely into lower quality form. But not vice versa. Energy quality Q measures the versatility of different energy forms.

17. Entropy Carnot cycle (reversible processes): Qh = heat absorbed Qc = heat rejected lukewarm: can’t do W, Q  Qh , Qc = heat absorbed C = any closed path Irreversible processes can’t be represented by a path. S = entropy [ S ] = J / K

18. Entropy Carnot cycle (reversible processes): Qh = heat absorbed Qc = heat rejected  Qh , Qc = heat absorbed lukewarm: can’t do W, Q C = Carnot cycle C = any closed path Irreversible processes can’t be represented by a path. S = entropy [ S ] = J / K Contour = sum of Carnot cycles.

19. S = 0 over any closed path  S21+S12= 0  S21=S21 Entropy change is path-independent. ( S is a thermodynamic variable )

20. Entropy in Carnot Cycle Ideal gas： Heat absorbed: Heat rejected: Adiabatic processes： 

21. Irreversible Heat Transfer Cold & hot water can be mixed reversibly using extra heat baths. T1 = some medium T. reversible processes T2 = some medium T. Actual mixing, irreversible processes

22. Adiabatic Free Expansion Adiabatic  Qad.exp. = 0    S can be calculated by any reversible process between the same states. isothermal p = const. Can’t do work Q degraded.

23. Entropy & Availability of Work Before adiabatic expansion, gas can do work isothermally After adiabatic expansion, gas cannot do work, while its entropy increases by  In a general irreversible process Coolest T in system

24. Example 19.4. Loss of Q A 2.0 L cylinder contains 5.0 mol of compressed gas at 300 K. If the cylinder is discharged into a 150 L vacuum chamber & its temperature remains at 300 K, how much energy becomes unavailable to do work?

25. A Statistical Interpretation of Entropy Gas of 2 distinguishable molecules occupying 2 sides of a box 1/4 2 ¼ = ½ 1/4

26. Gas of 4 distinguishable molecules occupying 2 sides of a box 1/16 = 0.06 4 1/16 = ¼ =0.25 6 1/16 = 3/8 = 0.38 4 1/16 = ¼ =0.25 1/16 = 0.06

27. Gas of 1023 molecules Gas of 100 molecules Equal distribution of molecules Statistical definition of entropy : •  # of micro states

28. Entropy & the 2nd Law of Thermodynamics 2nd Law of Thermodynamics : in any closed system S can decrease in an open system by outside work on it. However, S  0 for combined system.  S  0 in the universe Universe tends to disorder Life ?