22320 cal of heat is supplied to 100g of ice at 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^(-10, the final amount of water thus obtained and its temperature respectively are

22320 cal of heat is supplied to 100g of ice at 0^(@)C. If the latent

1.	22320 cal of heat is supplied to 100g of ice at 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^(-10, the final amount of water thus obtained and its temperature respectively are
Answer» `8G, 100^(@)C` `100g, 90^(@)C` `92g, 100^(@)C` `8g, 100^(@)C` Solution :HEAT required to convert 100g of ice at `0^(@)C` to water at `100^(@)C` `=(100g) (80 "cal" g^(-1)) + (100g) (1 "cal" g^(-1) ""^(@)C^(-1)) (100^(@)C)` =8000 cal + 10000 cal = 18000 cal But 22320 cal of heat is supplied, so remaining amount of heat =22320 cal - 18000 cal = 4320 cal Let the amount of water evaporated by remaining heat be m. Then `m(540 "cal" g^(-1)) = 4320` cal or `m= (4320"cal")/(540 "cal" g^(-1)) = 8g` Thus the final amount of water OBTAINED at `100^(@)C = 100g- 8g` = 92g