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Question
The amount of heat energy required to convert 1 kg of ice at – 10℃ to water at 100℃ is 7,77,000 J. Calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg-1K-1, Specific heat capacity of water = 4200 J kg-1 K-1.
Solution
Amount of heat energy gained by 1 kg of ice at −10°C to raise its temperature to 0°C = 1 × 2100 × 10 = 21000 J
Amount of heat energy gained by 1 kg of ice at 0°C to convert into water at 0°C = L
Amount of heat energy gained when temperature of 1 kg of water at 0°C rises to 100°C = 1 × 4200 × 100 = 420000 J
Total amount of heat energy gained = 21000 + 420000 + L = 441000 + L.
Given that total amount of heat gained is = 777000 J.
So,
441000 + L = 777000.
L = 777000 − 441000.
L = 336000 J Kg-1
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