Question

calculate the minimum amount of water at 30 degree celsius that would be required to completely melt 40 g of ice the specific heat capacity of liquid water is 4200 joules per kilogram degree celsius and the specific latent heat of fusion of ice is 330 joules per kilogram​

Answer 1

Answer:

Amount of heat energy gained by 1kg of ice at −10°C to raise its temperature to 0°C = 1 × 2100 × 10 = 21000 J

Amount of heat energy gained by 1kg of ice at 0°C to convert into water at 0°C = L

Amount of heat energy gained when temperature of 1kg 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 = 777000J.

So,

441000+L=777000.

L=777000−441000.

L=336000JKg

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