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Question

One gram of ice is mixed with one gram of steam. After thermal equilibrium is reached, the temperature of the mixture is
(Latent heat of fusion of ice; Lice=80 cal/g, latent heat of condensation of vapour; LV=540 cal/g)


A
100C
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B
55C 
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C
50C 
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D
0C 
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Solution

The correct option is A 100C
We know,
Latent heat of fusion of ice; Lice=80 cal/g
Latent heat of condensation; LV=540 cal/g of vapour.
Specific heat capacity of water; c=1 cal/gC
Heat supplied by steam if it is to condense totally and convert into water at 100C.
Q1=1×LV=540 cal
Heat required by ice to convert totally into water at 0C 
Q1=1×Lice=1×80=80 cal
Heat required by ice to convert totally into water at 100C
Q3=1×Lice+1×C×ΔT=1×80+1×1×(1000)=80+100=180 cal
Since Q1>Q3, entire steam will not condense and the final temperature of mixture of water and steam will be 100C.

Physics

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