Question

The latent heat of ice is 80 Cal/gm. The
change in entropy when 10 gram of ice
at 0°C is converted into water of same
temperature is​

Answer 1

Given :

Latent heat of ice is given = 80 Cal/gm

Mass of the ice given = 10 gm

Temperature given  =0°C

To Find :

Change in entropy when this ice is converted into water of same temperature = ?

Solution :

We know that change in entropy $\Delta$S is given as :

$\Delta S = \frac{\Delta Q}{T}$   -(1)

Here  ,$\Delta Q$ is change in heat and T is the temperature .

Since from the given latent heat we have :

For 1 gm ice 80 Cal heat is needed and so for 10 gm 800 Cal heat will be needed .

So , from equation 1 we have:

$\Delta S = \frac{800Cal}{273K}$

      =2.93 Cal/K

Hence, change in entropy for the given conversion of 10 gm ice into water is 2.93 Cal/K .