Question

one gram of ice at zero degree celsius is mixed with one gram of steam at hundred degree Celsius after thermal equilibrium the temperature of the mixture is
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Answer 1

Answer:

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Answer 2

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

For ice to completely convert into water, heat required is m1 Lf =1×80=80cal

For steam to completely convert into water, heat released is m2 Lv =1×540=540 cal

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

Now, to convert melted water to 100° C from 0° C, heat required is m1s(100−0)=1×1×100=100cal

So, total energy required to heat ice to water 100° 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° C.

Note- finally the mixture will consist of both steam and water at 100° C.

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