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Question

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


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