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A refrigeratior converts 50 gram of wate...

A refrigeratior converts `50` gram of water at `15^(@)C` intoice at `0^(@)C` in one hour. Calculate the quantity of heat removed per minute. Take specific heat of water `=1` cal `g^(-1) .^(@)C^(-1)` and latent heat of ice `=80` cal `g^(-1)`

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The correct Answer is:
79.2 cal `min.^(-1)`
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100 gm of water at 20^@C is converted into ice at 0^@C by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = 1 cal g^(-1) C^(-1) and latent heat of ice = 80 cal g^(-1)

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 )

How many grams of ice at -14 .^(@)C are needed to cool 200 gram of water form 25 .^(@)C to 10 .^(@)C ? Take specific heat of ice =0.5 cal g^(-1) .^(@)C^(-1) and latant heat of ice = 80 cal g^(-1) .

Stream at 100^(@)C is passed into 20 g of water at 10^(@)C . When water acquires a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water =1 cal g^(-1) .^(@)C^(-1) and latent heat of steam =540 cal g^(-1) ]

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C when water acquire a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water = 1 cal g^(-1).^(@) C^(-1) and latent heat of steam = 540 cal g^(-1) ]

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