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Unformatted text preview: of part (a), we have 0:2 kg of ice left in our mix. The heat needed to melt that ice is Q = miceLf = 200g 80cal=g = 16000cal. At three calories per second that takes a time t = 16000=3s = 5350s = 89:2min = 1:49h. (c) 5 points] What was the entropy change in part (b) and in part (a)? The entropy change in part (b) is very easy to calculate. For the ice water: S = Q=T = 16000cal=273K = 58:6 cal=K 1 and for the surroundings S = Q=T = ;16000cal=293K = ;54:6 cal=K Then the net result is S = 4:0 cal=K: The entropy change in part (a) is much more complicated to calculate. In general, dS = R dQ=T or S = dQ=T . For the water S= S Z 273K 293K mw cw dT = mw cw ln(273=293) = 3000g 1cal=gK ln(273=293) = ;212 cal=K T For the ice = micecice ln(Tf =Tice) + (mice ; 200g)Lf =Tf = 844 1cal=K ln(273=263) + (844 ; 200) 80cal=273K = 31:5 + 189 cal=K = 220 cal=K The total is then S = 8 cal=K. Note in both cases that S > 0 which is absolutely required since we are performing an irreversible process: letting heat travel from a hot object to a cold one. 2 Physics 7B Fall, 2006 Homework 4 Solutions 20.30 (a) For a Carnot refrigerator, we have QH = nRTH ln(Vb /Va) = nRTH ln(Vb/Va); QL = nRTL ln(Vd/Vc) = nRTL ln(Vc/Vd) = nRTL ln(Vb /Va), because Vc/Vd = Vb/Va; W = Q H QL. The coefficient of performance is CP = QL/W = QL/(QH QL) = nRTL ln(Vb/Va)/[nRTH ln(Vb/Va) nRTL ln(Vb/Va)] = TL/(TH TL). (b) The efficiency of the reversible heat engine is e = 1 (TL /TH), so TL /TH = 1 e. Thus we have CP = TL /(TH TL) = (TL/TH)/[1 (TL /TH)] = (1 e)/e. (c) The coefficient of performance is CP = TL /(TH TL) = (257 K)/(295 K 257 K) = 6.8. 20.31 The power input is P = W/t = QL/(CP)t = mL /(CP)t = VL/(CP)t; 1000 W = (1.00 103 kg/m3)V(3.33 105 J/kg)/(7.0)(3600 s), which gives V = 7.6 102 m3 = 76 L. 20.36 (a) The change in entropy of the water is Swater = 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 Ssurr = Q/T = 1.45 kcal/K. (c) For the un...
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