Question

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

• A. 273 cal/K
• B. $8 \times 10^4\, cal/K $
• C. 80 cal/K
• D. 293 cal/K

Answer 1

Answer: (D)

Here
$ L =80\, cal / k$
$ m =1 \, kg =1000 \, g $
$ T =273 \, K$
Heat absorbed in change of phase $Q=m L$
$\Delta Q =1000 \times 80=8 \times 10^{4} \, cal$
$S =\frac{\Delta Q}{T}$
$=\frac{8 \times 10^{4}}{273} $
$=293 \, cal / k$