Question

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, $△S$ for this process is

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

Answer 1

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

As the two boxes are identical, So $C_p$ will be same for both and
$T_{final}=\frac{T_1+T_2}{2}$

Entropy change in the first box will be given by
$△S_1=C_{p}ln\frac{T_f}{T_i}since,T_i=T_1△S_1=C_{p}ln\frac{T_f}{T_1}$

Entropy change in the second box will be given by
$△S_2=C_{p}ln\frac{T_f}{T_i}since,T_i=T_2△S_2=C_{p}ln\frac{T_f}{T_2}$

Total entropy change will be given by addition of above two.
$△S=△S_1+△S_2=C_{p}ln\frac{T_f^2}{T_1{T}_2}$
putting the value of $T_f$
$△S=C_{p}ln\left[\frac{(T_1+T_2{)}^2}{4T_1{T}_2}\right]$