Question 2.52: Two identical bodies of constant heat capacity are at the sa...
Two identical bodies of constant heat capacity are at the same initial temperature T_i . A refrigerator operates between these two bodies until one body is cooled to temperature T_2 . If the bodies remain at constant pressure and undergo no change of phase, show that the minimum amount of work needed to do this is :
W_{\min }=c_p\left[\frac{T_i{ }^2}{T_2}+T_2-2 T_i\right]
Let \quad T_i= initial temperature of both the bodies A and B(Fig 2.57)
T_2= temperature of body A after cooling
T_2^{\prime}= final temperature of body B.
Heat removed from body A, Q=c_p\left(T_i-T_2\right)
Heat discharged to body B, Q+W=c_p\left(T_2^{\prime}-T_i\right)
Work input,
\begin{aligned}W &=c_p\left(T_2^{\prime}-T_i\right)-c_p\left(T_i-T_2\right) \\&=c_p\left(T_2^{\prime}+T_2-2 T_i\right)\end{aligned}
Now
\begin{aligned}\Delta S_A &=c_p \ln \left\lgroup \frac{T_2}{T_i}\right\rgroup \\\Delta \mathrm{S}_B &=c_p \ln \left\lgroup \frac{T_2^{\prime}}{T_i}\right\rgroup\\\Delta S &=c_p \ln \left\lgroup \frac{T_2}{T_i}\right\rgroup+c_p \ln \left\lgroup \frac{T_2^{\prime}}{T_i}\right\rgroup=c_p \ln \left\lgroup \frac{T_2 T_2^{\prime}}{T_i^2}\right\rgroup\end{aligned}
For W to be minimum, T_2^{\prime} is to be minimum.
\begin{aligned}\therefore \quad c_p \ln \left\lgroup \frac{T_2 T_2^{\prime}}{T_i^2}\right\rgroup &=0=\ln 1 \\T_2^{\prime} &=\frac{T_i^2}{T_2} \\W_{\min } &=c_p\left[\frac{T_i^2}{T_2}+T_2-2 T_1\right]\end{aligned}
