What is the final temperature of water in the container?

Question

60 g of ice and 70 g of water are at equilibrium in an insulated container. 42 kJ of heat are delivered to this mixture via an electric heater. What is the final temperature of water in the container? Answer in C.

Answer 1

You have 60 g of ice and 70 g of water all at 0 deg C
Added 10,038 calories (42,000 J / 4.184 cal/ J)
Melt 60 g ice: 4783 calories (79.71 cal/g x 60 g)
5300 calories left to heat 130 g of water
Final temp = 40.8 deg C
(40.8 x 1 cal/g x 130 g = 5304 calories)

Answer 2

First, see if the ice melts. The heat of fusion of water is 333 J/g

60g ice x 333J/1g ice = 20,000 J = 20 kJ

So yes: The ice would melt, and there is 22 kJ heat left over to heat the water. But now there is a total of 130g water at 0C. The heat capacity of water is 4.184J/g-C.

22,000J/130gH2O x 1g-C/4.184J = 40degC

So the water would rise by 40degC from 0 C to 40C.