Question

1 gram of ice is mixed with 1 gram of steam. At thermal equilibrium, the temperature of the mixture is

• A. ${50}^{\circ }C$
• B. $0^{\circ }C$
• C. ${55}^{\circ }C$
• D. ${100}^{\circ }C$

Answer 1

The correct option is D

${100}^{\circ }C$

Latent heat of ice $=334J/g^{\circ }C$

Latent heat of water $=2230J/g^{\circ }C$

Heat required to take ice from $0^{\circ }Cto{100}^{\circ }C=334+(1\times 4.2\times 100)=754J$

Hence only some of the steam will be converted back to water and the mixture will remain at ${100}^{\circ }C.$