7500 cal of heat is supplied to 100g of ...

7500 cal of heat is supplied to 100g of ice `0 ""^(@)C`. If the latent heat of fusion of ice is 80 cal `g^(-1)` and latent heat of vaporization of water is 540 cal `g^(-1)`, the final temperature of water is

`100^(@)C`

`90^(@)C`

`0^(@)C`

`20^(@)C`

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How many calories are required to change one gram of 0^(@)C ice to 100^(@)C steam ? The latent heat of fusion is 80 cal // g and the latent heat of vaporization is 540 cal // g. The specific heat of water is 1.00 cal // (g K ) .

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Specific latent of fusion of ice is 80 cal/g. Explain this statement ?

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What is the amount of heat required (in calories) to convert 10 g of ice at -10^(@)C into steam at 100^(@)C ? Given that latent heat of vaporization of water is "540 cal g"^(-1) , latent heat of fusion of ice is "80 cal g"^(-1) , the specific heat capacity of water and ice are "1 cal g"^(-1).^(@)C^(-1) and "0.5 cal g"^(-1).^(@)C^(-1) respectively.

What does the statement, 'latent heat of fusion for ice is 80 cal/g' mean?

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