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Q.

Calculate the heat of fusion of ice from the following data for ice at 0ºC added to water. Mass of calorimeter = 60 g, mass of calorimeter + water = 460 g, mass of calorimeter + water + ice = 618 g, initial temperature of water = 38°C, temperature of the mixture = 5°C. The specific heat of calorimeter = 0.10 ca lg-1 °C-1.

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answer is 78.35.

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Detailed Solution

Mass of water = 460 -60 = 400 gMass of ice = 618 - 460 =158 gHeat lost by water = Heat gained by ice to melt +Heat gained by (water + calorimeter) to reach 5ºCÞ 400×1×(38-5) =158×L+158×1×5 + 60×0.1×5(where L is the latent heat of fusion of ice)⇒ L = 78.35 calg–1

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