Question

One gram of ice is mixed with one gram of steam. After thermal equilibrium, the temperature of the mixture is

• A. $0^{\circ }$ C
• B. ${100}^{\circ }$ C
• C. ${55}^{\circ }$ C
• D. ${80}^{\circ }$ C

Answer 1

The correct option is B ${100}^{\circ }$ C

Heat required to melt 1gof ice at $0^{\circ }$ C is 80 cal.
Heat required to convert 1g of water at $0^{\circ }$ C into 1g of water at ${100}^{\circ }C=1\times 1\times 100=100=100cal.$
Heat required condensing 1g of steam = 1 $\times$ 540 cal = 540 cal
Clearly, whole of steam is not condensed. So, temperature of the mixture is${100}^{\circ }$ C.