Question

When $1\text{kg}$ of ice at $0^{\circ }\text{C}$ melts to water at $0^{\circ }\text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80\text{Cal}{/}^{\circ }\text{C}$, is :

• A. $293\text{Cal/K}$
• B. $80\text{Cal/K}$
• C. $273\text{Cal/K}$
• D. $8\times {10}^4\text{Cal/K}$

Answer 1

The correct option is A $293\text{Cal/K}$

Given:
mass $=1\text{Kg}$
latent heat of ice $=80\text{Cal}{/}^{\circ }\text{C}$
Entropy change is given as
$\Delta s=\frac{\Delta Q}{T}=\frac{m{L}_f}{273}$
$\Delta s=\frac{1000\times 80}{273}=293\text{Cal/K}$.