Question

60 g of ice at 0°C is mixed with 60 g of steam
at 100°C. At thermal eqilibrium, the mixture
contains
​

Answer 1

Explanation:

Two physical systems are in thermal equilibrium if there is no net flow of thermal energy between them when they are connected by a path permeable to heat

Heat energy required to melt 60g of ice =60×80=4800Cal.

Heat energy required to raise the temprature of water at 0 degree centigrade to 100 degree centigrade =100×1×60=6000Cal

Total energy required =4800+6000=10800Cal

This energy is given by the steam by getting converted into water.

Amount of steam getting converted into water is

540

10800

=20g

Thus, finally total 80gm of water will be there and 40gm of steam.