Question

One gram of ice at $0^{\circ }$C is mixed with one gram of steam at ${100}^{\circ }$C . At thermal equilibrium the temperature of mixture is

• A. 0°C
• B. 100°C
• C. 55°C
• D. 80°C

Answer 1

The correct option is B 100°C

Heat required to melt 1 g of ice of $0^{\circ }$C to water at $0^{\circ }$C = 1 $\times$ 80 cal.

Heat required to raise temperature of 1 g of water from $0^{\circ }$C to ${100}^{\circ }$C = 1 $\times$ 1 $\times$ 100 = 100 cal

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°C, therefore, temperature of the mixture would be ${100}^{\circ }$C.