Q. When $1\ kg$ of ice at $0℃$ melts to water at $0℃$, the resulting change in its entropy, taking latent heat of ice to be $80\ cal/℃$ is
A) $293\ cal/K$
B) $273\ cal/K$
C) $8 \times 10^{4}\ cal/K$
D) $80\ cal/K$
Explanation
(a) $\mathrm{\Delta}S = \frac{\mathrm{\Delta}Q}{T} = \frac{80 \times 1000}{273} = 293\ cal/K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.