. 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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