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MT1-cheatsheet

# 529qh ql qh w 1 eqh 1 0529 qh 0471 qh the

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Unformatted text preview: J/s. We find the rate at which coal must be burned from m/t = (3.25 109 J/s)/(2.80 107 J/kg) = 116 kg/s. e1 20.21 (a) We find the pressures from the ideal gas equation: PV = nRT; Pa(6.0 103 m3) = (0.50)(8.315 J/mol K)(743 K), which gives Pa = 5.15 105 Pa. Pb(15.0 103 m3) = (0.50)(8.315 J/mol K)(743 K), which gives Pb = 2.06 105 Pa. (b) For an adiabatic process, PV = nRTV 1 = constant; THVb 1 = TL Vc 1; (743 K)(15.0 L)1.4 1 = (563 K)Vc1.4 1, which gives Vc = 30.0 L. THVa 1 = TLVd 1; (743 K)(6.0 L)1.4 1 = (563 K)Vd1.4 1, which gives Vd = 12.0 L. p a TH b d TL c V (c) The work done during the isothermal process is Wab = nRTH ln(Vb/Va) = (0.50 mol)(8.315 J/mol K)(743 K) ln(15.0 L/6.0 L) = 2.83 103 J. Note that this is also QH. (d) The heat transfer during the isothermal process is Qcd = QL = Wcd = nRTL ln(Vd/Vc) = (0.50 mol)(8.315 J/mol K)(563 K) ln(12.0 L/30.0 L) = 2.14 103 J. (e) For the adiabatic processes we have W = U = nCV T; Wbc = nCV (Tc Tb) = nCV (TL TH); Wda = nCV (Ta Td) = nCV (TH TL) = Wbc . Thus the net work done for the cycle is W = Wab + Wcd = 2.83 103 J 2.14 103 J = 0.69 103 J. (f) The efficiency is e = W/QH = (0.69 103 J)/(2.83 103 J) = 0.24 = 24%. From the temperatures we have e = 1 (TL /TH) = 1 (563 k/743 K) = 0.24. 20.29 (a) The coefficient of performance for the refrigerator is CP = QL /W = TL/(TH TL) = (256 K)/(298 K 256 K) = 6.09. The heat that is to be removed is QL = m(cwater Twater + L + cice Tice) = (0.50 kg){(4186 J/kg C)(25C 0C) + 3.33 105 J/kg + (2100 J/kg C)[0C ( 17C)]} = 2.37 105 J. We find the work from CP = QL /W; 6.09 = (2.37 105 J)/W , which gives W = 3.9 104 J. (b) The power output is P = W/t = QL/(CP)t = m(cwater Twater + L )/(CP)t 200 W = (0.50 kg)[(4186 J/kg C)(25C 0C) + 3.33 105 J/kg ]/(6.09)t, which gives t = 1.8 102 s = 3.0 min. 20.60 First we check to see if energy is conserved: QL + W = 1.50 MW + 1.50 MW = 3.00 MW = QH , so energy is conserved. The efficiency of the engine is e = W/QH = (1.50 MW)/(3.00 MW) = 0.500 = 50.0%. The maxim...
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