1. Tardigrade
  2. Question
  3. Physics
  4. The latent heat of ice is 80 Cal/g. The change in entropy when 10 gram of ice at 0 °C is converted into water of same temperature is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The latent heat of ice is 80 Cal/g. The change in entropy when 10 gram of ice at $0 {^{\circ}C}$ is converted into water of same temperature is

UPSEEUPSEE 2019

Solution:

The change in entropy of a system is given as
$\Delta S = \frac{\Delta Q}{T} $
where, $T = $ temperature and $\Delta Q =$ heat absorbed by the system.
Here, latent heat of ice, $L = 80$ cal/g, mass of water, $m= 10 \,g$ and temperature,
$T = 0^{\circ}C = 273\,K$
Heat given to $10 \,g$ ice,
$E = mL = 10 \times 80$
$\Rightarrow E = 800 \,cal$
So, the entropy change
$\Delta s =\frac{E}{T} = \frac{800}{273}$
$\Rightarrow s = 2.93 \, cal/K$