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Q.

The heat required in calories to change 10 g  ice at 0°C  to steam at 100°C? Latent heat of fusion and vaporization for H2O are 80 cal/g and 540 cal/g respectively. Specific heat of water is  1 cal/g.

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answer is 7200.

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Detailed Solution

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10 g ice0°C→ 10 g water0°C →10 g steam100°C

Total heat absorbed =ΔHfusion+ΔH temp rise+ΔHVap =10×80+10×1×100+10×540=7200           [Ans=7200 cal]

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