Question

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

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

Answer 1

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

Total heat gained by ice is equal to the total heat lost by steam.

For ice to completely convert into water, heat required is $m_1{L}_f=1\times 80=80cal$

For steam to completely convert into water, heat released is $m_2{L}_v=1\times 540=540cal$

Hence, first 80 calories will not be enough for the steam to condense completely.

Now, to convert melted water to ${100}^{o}C$ from $0^{o}C$, heat required is $m_{1}s(100-0)=1\times 1\times 100=100cal$

So, total energy required to heat ice to water ${100}^{o}C$ is $100+80=180cal$.

Hence, even this amount of energy is not enough for the steam to condense completely. Hence, the final temperature of the mixture will be ${100}^{o}C$.

Note- finally the mixture will consist of both steam and water at ${100}^{o}C$.