Question 16.

**Solve the following problems:**

**A thermally insulated pot has 150 g ice at temperature 0oC. How much steam of 100oC has to be mixed to it, so that water of temperature 50oC will be obtained? (Given: latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, the specific heat of water = 1 cal/g °C)**

Answer: 1 g of ice needs 80 cal to become water at 0 C.

1 g of water needs 50 cal to rise to 50 C from 0 C.

So total the heat needed is 150×80 + 150×50 = 19500 cal.

1 g of steam will supply 540 c to become water at 100 C.

1 g of water at 100 C will supply 50 cal to become water at 50 C.

Hence x gram of steam at 100 C will supply 590 cal while cooling to 50 C.

Hence 19500 = 590 x.

Solving, x = 33.05 gram.