1 kg of ice at 0 ºC is mixed with 1 kg of steam at 100 ºC. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10 5 J/kg and latent heat of vaporization of water = 2.26 × 10 6 J/kg.
Text Solution
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Sol. Given:-
Amount of ice at
Amount of steam at
Latent heat of fusion of ice
Latent heat of vapourisation of water
We can observe that the latent heat of fusion of ice
is smaller that latent heat of vapouisation of water
. Therefore, ice will first change into water as less heat is required for this and there will be equilibrium between steam and water.
Heat absorbed by the ice when it changes into water
Heat absorbed by the water formed to change its temperature from
to
Total heat absorbed by the ice to raise the temperature to

The heat required to change ice into water at
is supplied by the steam. This heat will be released by the steam and will then change into water.
If all the steam gets converted into water, heat released by steam,
Amount of heat released is more than that required by the ice to get converted into water at
. Thus,
Extra heat 

Let the mass of steam that is condensed into water be m . Thus,

Total amount of water at
Steam left 
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