10g of ice of 0^(@)Cis mixed with 100 g of waterat 50^(@)C in a calorimeter.The final temperatureof the mixtureis [Specificheat of water= 1cal g^(-1).^(@)C^(-1),letent offusionof ice= 80 cal g^(-1)]

10g of ice of 0^(@)Cis mixed with 100 g of waterat 50^(@)C in a calori

1.	10g of ice of 0^(@)Cis mixed with 100 g of waterat 50^(@)C in a calorimeter.The final temperatureof the mixtureis [Specificheat of water= 1cal g^(-1).^(@)C^(-1),letent offusionof ice= 80 cal g^(-1)]
Answer» `31.2^(@)C` `32.8^(@)C` `36.7^(@)C` `38.2^(@)C` Solution :Here, Mass of WATER, `m_(w) = 100 g` Mass of ice, `m_(i) = 100g` Specific heat of water , `s_(w) = 1 CAL g^(-1) .^(@)C^(-1)` Latent heat of fusion of ice, `L_(fl) = 80 cal g^(-1)` Let T be the final temperature of the mixture Amount of heat lost by water `m_(w)s_(w)(DeltaT)_(w) = 100 XX 1 xx (50 - T)` Amount of heat gained by ice `m_(i)L_(fl) + m_(i)s_(w) (DeltaT)_(i) = 10 xx 80 + 10 xx 1 xx (T - 0)` ACCORDING of heat lost by water `m_(i)L_(fl) + m_(i)s_(w) (DeltaT)_(i) = 10 xx 80 + 10 xx 1 xx (T - 0)` According to principle of CALORIMETRY Heat lost = Heat gained `100 xx 1 xx (50 - T) = 100 xx 80 + 10 xx 1 xx (T - 0)` `500 - 10T = 80 +T` `11T = 420` or `T = 38.2 .^(@)C`