How much heat is required to convert 50 g of ice at Solution

Solution

DH = DH1 + DH2

DH1 = heat required to convert ice at 10 deg to water at 10 deg

DH1 = (5.0/18.0) * 6.00 = 1.67 Kj

DH1 = 5.0 * 1 * (15-10) = 25 KJ

So total heat change = 1.67 + 25 = 26.67 kj

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http://chemistry.about.com/od/workedchemistryproblems/a/Heat-Capacity-Phase-Change-Example-Problem.htm