Thermodynamics and Thermochemistry - Result 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) ${ }^{2} C _p \ln \left[\frac{T _1+T _2}{2 T _1 T _2}\right]$
Show Answer
Solution:
- At the thermal equilibrium,
final temperature $T _f=\frac{T _1+T _2}{2}$ $\Rightarrow$ for the 1st block, $\Delta S _I=C _p \ln \frac{T _f}{T _1}$
$\Rightarrow$ for the 2nd block, $\Delta S _{\text {II }}=C _p \ln \frac{T _f}{T _2}$
When brought in contact with each other,
$$ \begin{aligned} \Delta S & =\Delta S _{\text {I }}+\Delta S _{\text {II }}=C _p \ln \frac{T _f}{T _1}+C _p \ln \frac{T _f}{T _2} \ & =C _p \ln \left(\frac{T _f}{T _1} \times \frac{T _f}{T _2}\right)=C _p \ln \left[\frac{T _f^{2}}{T _1 T _2}\right] \ & =C _p \ln \left[\frac{\left(\frac{T _1+T _2}{2}\right)^{2}}{T _1 T _2}\right]=C _p \ln \left[\frac{\left(T _1+T _2\right)^{2}}{4 T _1 T _2}\right] \end{aligned} $$
