Q. Find the change in the entropy in the following process $100\ g$ of ice at $0℃$ melts when dropped in a bucket of water at $50℃$ (Assume temperature of water does not change)
A) $- 4.5\ cal/K$
B) $+ 4.5\ cal/K$
C) $+ 5.4\ cal/K$
D) $- 5.4\ cal/K$
Explanation
(b) Gain of entropy of ice
$S_{1} = \frac{\mathrm{\Delta}Q}{T} = \frac{mL}{T} = \frac{80 \times 100}{(0 + 273)} = \frac{8 \times 10^{3}}{273}cal/K$
Loss of entropy of water $= S_{2} = - \frac{\mathrm{\Delta}Q}{T} = - \frac{mL}{T}$
$= \frac{80 \times 100}{(273 + 50)} = \frac{8 \times 10^{3}}{323}cal/K$
Total change of entropy
$S_{1} + S_{2} = \frac{8 \times 10^{3}}{273} - \frac{8 \times 10^{3}}{323} = + 4.5\ cal/K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.