Question

Calculate the amount of heat required to change 120 grams of ice at 0 degree Celsius to steam at 100 degree celsius.given that latent heat of fusion of ice is at 80 cal g power minus 1.specific heat of water is 1 cal g minus 1 degree Celsius minus 1,latent heat of vaporisation of steam is equal to 540 cal g minus1

Answer 1

Given: m=100 g and L

f

=336 J/g

Amount of heat required,Q=mL

f

Q=100×336 J

=33600 J

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