Question

Final temperature obtained for minimum and maximum work obtainable when two finite bodies at constant heat capacity having temperatures T1 and T2 exchanging heat is respectively given

Answer 1

Let $dQ_1$ is the heat transferred to hot reservoir.
e.g., $dQ_1=CdT_1$

$dQ_2$ is the heat transferred to cold reservoir.
e.g., $dQ_2=CdT_2$

we know, for the sum of the entropy changes of the system and surrounding to be zero.
we must have,
$dS_1+dS_2=0$

$\frac{dQ_1}{T_1}+\frac{dQ_2}{T_2}=0$

$C\frac{dT_1}{T_1}+C\frac{dT_2}{T_2}=0$

$Pln(T_1)+Pln(T_2)=0$

$Pln(T_1T_2)=0$

Let initial temperature is $T_0$, $T_1T_2=T_{10}T_{20}$

we must have, $T_1=T_2=T_f$
where $T_f$ is the final temperature of both reservoir.

so, $\bf{T_f=\sqrt{T_{10}T_{20}}}$