. The amount of heat energy required to convert 1 kg of ice at -10oC
The amount of heat energy required to convert 1 kg of ice at -10oC completely into water at 100oC is 777000 J. calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg-1 K-1, Specific heat capacity of water is 4200 J kg-1 K-1.
Best Answer
Amount of heat energy gained by 1 kg of ice at – 100 C to raise its temperature to 00 C = 1 × 2100 × 10
= 2100 J
Amount of heat energy gained by 1 kg of ice at 00 C to convert into water at 00 C = L
Amount of heat energy gained when temperature of 1 kg of water at 00 C rises to 1000 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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