Q.
22320cal of heat is supplied to 100g of ice at 0∘C. If the latent heat of fusion of ice is 80calg−1 and latent heat of vaporization of water is 540calg−1, the final amount of water thus obtained and its temperature respectively are
Heat required to convert 100g of ice at 0∘C to water at 100∘C =(100g)(80calg−1)+(100g)(1calg−1∘C−1)(100∘C) =8000cal+10000cal=18000cal
But 22320cal of heat is supplied, so remaining amount of heat =22320cal−18000cal=4320cal
Let the amount of water evaporated by remaining heat be m. Then m(540calg−1)=4320cal
or m=540calg−14320cal=8g
Thus the final amount of water obtained at 100∘C=100g−8g =92g.