Short Version : nd Law of Thermodynamics

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 )

19.2. The 2 nd 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. Heat dumped 2 nd 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.

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

Carnot Engine 1.isothermal expansion: T = T h, W 1 = Q h > 0 2.Adiabatic expansion: T h  T c, W 2 > 0 3.isothermal compression: T = T c, W 3 =  Q c < 0 Adiabatic compression : T c  T h, W 4 =  W 2 < 0 Ideal gas: Adiabatic processes:  A  B: Heat abs. C  D: Heat rejected: B  C: Work done D  A:  Work done

Engines, Refrigerators, & the 2 nd Law Carnot’s theorem: 1.All Carnot engines operating between temperatures T h & T c have the same efficiency. 2.No other engine operating between T h & T c can have a greater efficiency. Refrigerator: extracts heat from cool reservoir into a hot one. work required

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

Perfect refrigerator  Perfect heat engine Clausius  Kelvin-Planck

 Carnot engine is most efficient e Carnot = thermodynamic efficiency e Carnot  e rev > e irrev Carnot refrigerator, e = 60% Hypothetical engine, e = 70%

19.3. Applications of the 2 nd Law Power plant fossil-fuel : T h = 650 K Nuclear : T h = 570 K T c = 310 K Actual values: e fossil ~ 40 %e nuclear ~ 34 %e car ~ 20 % Prob 54 & 55 Heat source Boiler Turbine Generator Electricity Condenser Waste water Cooling water

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

Example 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.

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

Example Home Freezer A typical home freezer operates between T c =  18  C to T h = 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 nd law: only a fraction of Q can become W in heat engines. a little W can move a lot of Q in refrigerators.

Heat Pumps Heat pump as AC : Heat pump as heater : Ground temp ~ 10  C year round (US) Heat pump: moves heat from T c to T h.

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

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

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

Entropy change is path-independent. ( S is a thermodynamic variable )  S = 0 over any closed path   S 21 +  S 12 = 0   S 21 =  S 21

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

Irreversible Heat Transfer Cold & hot water can be mixed reversibly using extra heat baths. Actual mixing, irreversible processes reversible processes T 1 = some medium T. T 2 = some medium T.

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

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

Example 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?

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

Gas of 4 distinguishable molecules occupying 2 sides of a box MicrostatesMacrostatesprobability of macrostate 1/16 =  1/16 = ¼ =0.25 1/16 =  1/16 = 3/8 = 0.38

Gas of 100 molecules Gas of molecules Equal distribution of molecules Statistical definition of entropy :  # of micro states

Entropy & the 2 nd Law of Thermodynamics 2 nd 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 ?