Question

Find the change in the entropy in the following process 100 gm of ice at $0^{\circ }C$ melts when dropped in a bucket of water at ${50}^{\circ }C$ (Assume temperature of water does not change)

• A. – 4.5 cal/K
• B. + 4.5 cal/K
• C. +5.4 cal/K
• D. – 5.4 cal/K

Answer 1

The correct option is B + 4.5 cal/K

Gain of entropy of ice
$S_1=\frac{△Q}{T}=\frac{mL}{T}=\frac{80\times 100}{(0+273)}=\frac{8\times {10}^3}{273}cal/K$
Loss of entropy of water =$S_2=-\frac{△Q}{T}=-\frac{mL}{T}$
$=\frac{80\times 100}{(273+50)}=\frac{8\times {10}^3}{323}cal/K$
Total change of entropy
$S_1+S_2=\frac{8\times {10}^3}{273}-\frac{8\times {10}^3}{323}=\pm 4.5cal/K$