Fa06Soln4 - Physics 7B Fall, 2006 Homework 4 Solutions...

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Physics 7B Fall, 2006 Homework 4 Solutions 20.30 ( a ) For a Carnot refrigerator, we have Q H = nRT H ln( V b / V a ) = nRT H ln( V b / V a ); Q L = nRT L ln( V d / V c ) = nRT L ln( V c / V d ) = nRT L ln( V b / V a ), because V c / V d = V b / V a ; W = Q H Q L . The coefficient of performance is CP = Q L / W = Q L /( Q H Q L ) = nRT L ln( V b / V a )/[ nRT H ln( V b / V a ) – nRT L ln( V b / V a )] = T L /( T H T L ) . ( b ) The efficiency of the reversible heat engine is e = 1 – ( T L / T H ), so T L / T H = 1 – e . Thus we have CP = T L /( T H T L ) = ( T L / T H )/[1 – ( T L / T H )] = (1 – e )/ e . ( c ) The coefficient of performance is CP = T L /( T H T L ) = (257 K)/(295 K – 257 K) = 6.8 . 20.31 The power input is P = W / t = Q L /(CP) t = mL /(CP) t = ρ VL /(CP) t ; 1000 W = (1.00 × 10 3 kg/m 3 ) V (3.33 × 10 5 J/kg)/(7.0)(3600 s), which gives V = 7.6 × 10 –2 m 3 = 76 L . 20.36 ( a ) The change in entropy of the water is Δ S water = Q / T = mL / T = – (1.00 kg)(539 kcal/kg)/(373 K) = + 1.45 kcal/K . ( b ) Because the heat to vaporize the water comes from the surroundings, we have Δ S surr = – Q / T = – 1.45 kcal/K . ( c ) For the universe we have Δ S universe = Δ S water + Δ S surr = + 1.45 kcal/K + (– 1.45 kcal/K) = 0 .
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This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at University of California, Berkeley.

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Fa06Soln4 - Physics 7B Fall, 2006 Homework 4 Solutions...

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