Unformatted text preview: eal engine. TL S = (400 K)(0.618) = 247 J, which agrees within significant figures. Physics 7B Fall, 2006 Homework 3 Solutions
The rate of thermal energy flow is the same for the brick and Tint the insulation: Q/t = A(T1 T2)/Reff = A(T1 Tint)/R2 = A(Tint T2)/R1 , T2 where Tint is the temperature at the brick-insulation interface. By equating the first term to each of the others, we have Reff(T1 Tint) = R2(T1 T2), and Reff(Tint T2) = R1(T1 T2). If we add these two equations, we get Reff(T1 T2) = R1(T1 T2) + R2(T1 T2), which gives Reff = R1 + R2 . k1 This shows the usefulness of the R-value. For the R-value of the brick we have R1 = L1/k1 L1 = [(4.0 in)/(12 in/ft)](3.28 ft/m)/(0.84 J/s m C)(1 Btu/1055 J)( 3600 s /1 h)(5 C/9 F) = 0.69 ft2 h F/Btu. Thus the total R-value of the wall is Reff = R1 + R2 = 0.69 ft2 h F/Btu + 19 ft2 h F/Btu = 19.7 ft2 h F/Btu. The rate of heat loss is Q/t = A(T1 T2)/Reff = [(240 ft2)(10 F)/(19.7 ft2 h F/Btu)](1055 J/Btu)/(3600 s/h) = 36 W. T1 ?Q ?t <= L2 19.90
(a) We find the rate of heat flow through the clothing from Q/t = kA(T/L) = (0.025 J/s m C)(1.9 m2)[34C ( 20C)]/(0.035 m) = 73 W. (b) When the clothing is wet, we have Q/t = kA(T/L) = (0.56 J/s m C)(1.9 m2)[34C ( 20C)]/(0.0050 m) = 1.1 104 W. 19.101
. The heat of fusion from the ice forming at the bottom of the ice must be conducted through the ice to the air. When the ice has a thickness x, we have dQ/dt = kA T/x = (A dx/dt)Lf . We separate the variables and integrate: 2 t D k !T t k !T dt = x dx ; whichgives = D . "Lf 2 0 "Lf 0 For the time to form a sheet 25 cm thick, we have (2.0 J/s m C)[0C (15C)]t/(0.917 103 kg/m3)(3.33 105 J/kg) = (0.25 m)2/2, which gives t = 3.2 105 s = 3.7 d. 20.4
(a) We find the rate at which work is done from P = W/t = (180 J/cycle/cyl)(4 cyl)(25 cycles/s) = 1.8 104 J/s. (b) We find the heat input rate from e = W/QH = (W/t)/(QH/t); 0.25 = (1.8 104 J/s)/(QH/t), which gives QH/t = 7.2 104 J/s. (c) We find the time to use one gallon from t = E/(QH/t) = (130 10...
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