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Question
1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 103 J kg−1 and latent heat of vaporization of water = 2.26 × 106 J kg−1.
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Solution
Given:-
Amount of ice at 0oC = 1 kg
Amount of steam at 100oC = 1 kg
Latent heat of fusion of ice = 3.36 × 103 J kg−1
Latent heat of vapourisation of water = 2.26 × 106 J kg−1
We can observe that the latent heat of fusion of ice (3.36 × 105 J kg−1) is smaller that latent heat of vapouisation of water (2.26 × 106 ). 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 (Q1) = 1×(3.36 × 105) J
Heat absorbed by the water formed to change its temperature from 0oC to 100oC (Q2) = 1 × 4200 × 100 = 4.2 × 105 J
Total heat absorbed by the ice to raise the temperature to 100°C, Q = Q1+Q2 = 3.36 × 105+ 4.2 × 105 = (3.36 + 4.2) × 105 = 7.56 × 105 J
The heat required to change ice into water at 100oC 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, Q' = 1 ×( 2.26 × 106) J = 2.26 × 106 J
Amount of heat released is more than that required by the ice to get converted into water at 100oC. Thus,
Extra heat = Q − Q'
= (2.26 − 0.756) × 106
= 1.506 × 106
Let the mass of steam that is condensed into water be m. Thus,
`m=(7.56xx10^5)/(2.26xx10^6)=0.335kg=335gm`
Total amount of water at 100°C = 1000 + 335 = 1335 g =1.335 g
Steam left = 1− 0.335 = 0.665 kg = 665 gm
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