  Question

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

A
0C  B
100C  C
50C  D
55C  Solution

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 $$m_1 L_f = 1 \times 80 = 80 cal$$For steam to completely convert into water, heat released is $$m_2L_v = 1 \times 540 = 540\ cal$$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 $$m_1s(100-0)= 1 \times 1 \times 100 = 100 cal$$So, total energy required to heat ice to water $$100^o C$$ is $$100+80=180\ cal$$. 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 $$100^o C$$.Physics

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