If minimum possible work is done by a re...

If minimum possible work is done by a refrigenator in converting 100 grams of water at `0^@ C` to ice, how much heat (in calories) is released to the surroundings at temperature `27^@C` (latent heat of ice 80 Cal/gram) to the nearest integer ?

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A refrigerator converts 100 g of water at 25^(@)C into ice at -10^(@)C in one hour and 50 minutes. The quantity of heat removed per minute is (specific heat of ice = 0.5 "cal"//g^(@)C , latent heat of fusion = 80 "cal"//g )

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -

Find the heat required to convert 20 g of iceat 0^(@)C into water at the same temperature. ( Specific latent heat of fusion of ice =80 cal//g )

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Calculate the amount of heat required to convert 80g of ice at 0^(@)C into water at the same temperature . ( Specific latent heat of fusion of ice = 80 cal //g )

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