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Question

10 g of ice at 0C is mixed with 5g of steam at 1000C. If latent heat of fusion of ice is 80 cal/g and latent heat of vaporization of 540 cal/g. Then at thermal equilibrium

A
Temperature of the mixture is 0C
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B
Temperature of mixture is 100C
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C
Mixture contains 13.3g of water and 1.67 g of steam
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D
Mixture contains 5.3g of ice and 9.7g of water
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Solution

The correct options are
B Temperature of mixture is 100C
C Mixture contains 13.3g of water and 1.67 g of steam
Heat gain by ice to melt =mLf=10×80=800 cal
Heat gain by 10 g water to raise its temperature from 0C to 100C=10×1×100=1000cal
Mass of steam converted into water
1800=m×540m=3.33g
Equilibrium temperature 100C
Amount of water 10 + 3.33 = 13.33 g
Amount of steam = 5 – 3.33 = 1.67 g


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