Question

1 g of steam is sent into 1 g of ice. At thermal equilibrium, the resultant temperature of mixture is 

(a) 270°C

(b) 230°C

(c) 100°C

(d) 120°C

Answer 1

Option should obviously not greater than 100°C the temperature of the steam.So it is (c)

Explanation

The heat lost due to conversion of 1g steam at 100°C to water at 100°C is 540cal

Melting of 1g ice to 1g water at 0°C requires 80cal heat gained.

Heat required to raise this 1g water at 0°C to 100°C is 100cal.

So this total 180cal heat can easily be available from steam.And for this purpose only 180/540=1/3g of steam has to lose its latent heat in converting it to 1/3g water at 100°C. And after that the mixture will attain thermal equilibrium.

Answer 2

Correct option (c) 100°C

Explanation :

Heat required to melt 1 g of ice at 0°C to water at 0°C = 1 x 80 cal.

Heat required to raise temperature of 1 g of water from 0°C to 100°C = 1 x 1 x 100 = 100 cal.

Total heat required for maximum temperature of 100°C = 80 + 100 = 180 cal.

As one gram of steam gives 540 cal of heat when it is converted to water at 100°C, therefore, temperature of the mixture = 100°C.