Question

The latent heat of fusion of ice at $0^{\circ} C$ is $80\, cal / g$. Entropy change $(\Delta S)$ accompanying the melting of $1\, g$ of ice at $0^{\circ} C$ would be (units: $cal / g / K$ )

• A. 273
• B. 8
• C. 0
• D. 0.293

Answer 1

Answer: (D)

0.293

Answer 2

$\Delta S_{m e l}=\frac{\Delta H_{m e l}}{T}=\frac{80}{273}$
$=0.293\, cal / g / K$