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

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 (A) 0.293 Cal/K (B) 2.93 Cal/K (C) 80 Cal/K (D) 8 Cal/K

Tardigrade Admin · Accepted Answer

The change in entropy of a system is given as

Δ S = \frac{Δ Q}{T}

where,

T =

temperature and

Δ 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 = m L = 10 \times 80

\Rightarrow E = 800 c a l

So, the entropy change

Δ s = \frac{E}{T} = \frac{800}{273}

\Rightarrow s = 2.93 c a l / K

Q. 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

0.293 Cal/K

2.93 Cal/K

80 Cal/K

8 Cal/K

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