Question

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

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

Answer 1

Answer: (D)

$293Cal/K$

Given,

$m=1kg=1000g$

$L=80cal/g$

$T=0^{0}C=273K$

Entropy for solid and liquid when heat is given to substance to change its state at constant temperature.

$\Delta S=\frac{\Delta Q}{T}=\frac{mL}{T}$

$\Delta S=\frac{1000\times 80}{273}$

$\Delta S=293cal/K$

The correct option is D.