Question

Two blocks of the same metal having same mass and at temperature $T_1$ and $T_2$ respectively, are bought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $\Delta S$ for this process is

• A. $C_{P}ln\left[\frac{(T_1+T_2{)}^{\frac{1}{2}}}{T_1{T}_2}\right]$
• B. $C_{P}ln\left[\frac{(T_1+T_2)}{4T_1{T}_2}\right]$
• C. $C_{P}ln\left[\frac{(T_1+T_2{)}^2}{4T_1{T}_2}\right]$
• D. $C_{P}ln\left[\frac{(T_1+T_2)}{2T_1{T}_2}\right]$

Answer 1

The correct option is C $C_{P}ln\left[\frac{(T_1+T_2{)}^2}{4T_1{T}_2}\right]$

For the first block, $\Delta S_1=C_{p}ln\frac{T_f}{T_1}$
For the second block, $\Delta S_2=C_{p}ln\frac{T_f}{T_2}$
When both the blocks are bought in contact with each other,
$\Delta S=\Delta S_1+\Delta S_2=C_{p}ln\frac{T_f}{T_1}+C_{p}ln\frac{T_f}{T_2}=C_{p}ln\left(\frac{T_f^2}{T_1\times T_2}\right)$
$=C_{p}ln\left[\frac{{\left(\frac{T_1+T_2}{2}\right)}^2}{T_1\times T_2}\right]$
$=C_{P}ln\left[\frac{(T_1+T_2{)}^2}{4T_1{T}_2}\right]$