Thermodynamics and Thermochemistry 1 Question 14
14. Two blocks of the same metal having same mass and at temperature $T_{1}$ and $T_{2}$ respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $\Delta S$, for this process is
(2019 Main, 11 Jan I)
(a) ${ }^{2} C_{p} \ln \left[\frac{\left(T_{1}+T_{2}\right)^{1 / 2}}{T_{1} T_{2}}\right]$
(b) ${ }^{2} C_{p} \ln \left[\frac{T_{1}+T_{2}}{4 T_{1} T_{2}}\right]$
(c) $C_{p} \ln \left[\frac{\left(T_{1}+T_{2}\right)^{2}}{4 T_{1} T_{2}}\right]$
(d) ${ }^{2} C_{p} \ln \left[\frac{T_{1}+T_{2}}{2 T_{1} T_{2}}\right]$
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Solution:
- At the thermal equilibrium,
final temperature $T_{f}=\frac{T_{1}+T_{2}}{2}$ $\Rightarrow$ for the 1st block, $\Delta S_{\mathrm{I}}=C_{p} \ln \frac{T_{f}}{T_{1}}$
$\Rightarrow$ for the 2nd block, $\Delta S_{\text {II }}=C_{p} \ln \frac{T_{f}}{T_{2}}$
When brought in contact with each other,
$$ \begin{aligned} \Delta S & =\Delta S_{\text {I }}+\Delta S_{\text {II }}=C_{p} \ln \frac{T_{f}}{T_{1}}+C_{p} \ln \frac{T_{f}}{T_{2}} \ & =C_{p} \ln \left(\frac{T_{f}}{T_{1}} \times \frac{T_{f}}{T_{2}}\right)=C_{p} \ln \left[\frac{T_{f}^{2}}{T_{1} T_{2}}\right] \ & =C_{p} \ln \left[\frac{\left(\frac{T_{1}+T_{2}}{2}\right)^{2}}{T_{1} T_{2}}\right]=C_{p} \ln \left[\frac{\left(T_{1}+T_{2}\right)^{2}}{4 T_{1} T_{2}}\right] \end{aligned} $$
