Posted by Chris on Saturday, April 5, 2008 at 5:40am.
You need to do the problem in three parts: (1) The heating of the ice to 0 C, (2) the melting of the ice, and (3) the heating of the liquid water. First caclulate the amount of heat needed to heat the 10 kg to 0 C, and then the amount needed to heat the ice at 0 C. The remainder of the 4.10*10^6 J (after subtracting the previous two amount) is available to heat the liquid water.
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