Question

One gram of ice is mixed with one gram of steam. At thermal equilibrium the temperature of 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 1g of ice at $0^{\circ }C$ to water at $0^{\circ }C=1\times 80cal.$
Heat required to raise temperature of 1 g of water from $0^{\circ }Cto{100}^{\circ }C=1\times 1\times 100=100cal$
Total heat required for maximum temperature of ${100}^{\circ }C$ = 80 + 100 = 180 cal
As one gram of steam gives 540 cal of heat when it is converted to water at ${100}^{\circ }C$,
therefore, temperature of the mixture would be ${100}^{\circ }C$.