Question

Two blocks of the same metal having the 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 :

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

Answer 1

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

When two blocks are kept in contact with each other, the final temperature will be given as:

$T_f=\frac{T_1+T_2}{2}$

Now, we have $\Delta S_{sys}=\int \frac{d{q}_{rev}}{T}=n{C}_p\int \frac{dT}{T}$

For the first block of metal the entropy will be given as:

$\Delta S_1=n{C}_p{\int }_{T_1}^{T_f}\frac{dT}{T}=n{C}_{p}ln\frac{T_f}{T_1}$

Similarily, $\Delta S_2=n{C}_{p}ln\frac{T_f}{T_2}$

Now, total change in entropy

$\Delta S_1+\Delta S_2=n{C}_p\frac{{T_f}^2}{T_1{T}_2}$

But , $T_f=\frac{T_1+T_2}{2}$

$\therefore$ Final entropy will be: $n{C}_{p}ln\left[\frac{(T_1+T_2{)}^2}{4T_1{T}_2}\right]$