heat gained by 1g of waterwhen heated from 0℃ to 100℃
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Mass of ice=1g
Temp of ice=0 degree C
1 First the ice at 0 degree C is to be converted into water at 0 degree C
Heat required at this stage
= Mass of the ice x Latent heat of fusion of ice
=1x 80 =80 cal
2 Now the temperature of water is to be increased from 0 to 100 degree C
Heat required for this =Mass of water x rise in temperature x specific heat of water
=1 x100 x1= 100 cal
3 Now this water at100 degree C is to be converted into vapour
Heat required for this
=Mass of water x Latent heat
= 1 x 536 =536 cal
Total heat required
=80 +100 +536=716 cal
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