Question 6.8: For the previous example, to illustrate the insulating abili...
For the previous example, to illustrate the insulating ability of a vacuum bottle, how long will it take for the coffee to cool from 368 to 322 K if the only heat loss is by radiation?
The heat capacity of the coffee is ρ_mV_cT_1. Assuming the coffee is always mixed well enough that it is at uniform temperature, the cooling rate is equal to the instantaneous radiation loss. The energy loss by radiation at any time t, given by Equation 6.23, is related to the change of internal energy of the coffee by
Q_1=-Q_2=\frac{A_1\sigma (T_1^4-T_2^4)}{[1/\epsilon _1(T_1)]+[1/\epsilon _2(T_2)]-1} (6.23)
-\rho _MV_C\frac{dT_1}{dt} =\frac{A_1\sigma [T_1^4(t)-T_2^4]}{1/\epsilon _1+1/\epsilon _2-1}
where it is assumed that surface 1 is at the coffee temperature and surface 2 is at the outside environment temperature. Then
-\int_{T_1=TI}^{T_1=T_F}{\frac{dT_1}{T_1^4-T_2^4} }=\frac{A_1\sigma}{\rho _MV_c\left(1/\epsilon _1+1/\epsilon _2-1\right) } \int_{0}^{\tau }{dt}
where T_1 and T_F are the initial and final temperatures of the coffee, and ϵ_1 and ϵ_2 are assumed independent of temperature. Carrying out the integration and solving for t gives the cooling time from T_1 to T_F as
t=\frac{\rho _MV_C\left(1/\epsilon _1+1/\epsilon _2-1\right)}{A_1\sigma} \left[\frac{1}{4T_2^3} \ln \left|\frac{(T_F+T_2)/(T_F+T_2)}{(T_I+T_2)/(T_I+T_2)} \right|+\frac{1}{2T_2^3} \left\lgroup\tan ^{-1}\frac{T_F}{T_2}-\tan ^{-1}\frac{T_I}{T_2} \right\rgroup \right]
Substituting ρ_M = 975 kg/m^3, V =1/6π(0.15)^3m^3, c = 4195 J/(kg · K), \epsilon _1 = \epsilon _2 = 0.02, A_1 = π(0.15)^2 m^2, σ = 5.6704 × 10^{−8} W/(m^2 · K^4), T_2 = 294 K, T_1 = 368 K, and T_F = 322K gives t = 374.7 h. The coffee would require about 16 days to cool if heat losses were only by radiation. Conduction losses through the glass wall of the bottle neck usually cause the cooling to be considerably faster.