Posted by
**Dylan** on
.

Two kg of water at a temperature of 50 degrees C have 485.3 kJ of heat transferred from them, resulting in a mixture of ice/water at 0 degrees C. What is the composition of the mixture?

This is what I did and would like verification of my answer.

2*4.183*-50 = -418.3 kJ

I then subtracted that from my total heat energy:

485.3 - 418.3 = 67 kJ

For latent fusion:

67 = 335*m

m = 0.2 kg of ice

My original mass of water was 2 kg so the composition of the final mixture would be:

0.2 kg of ice

1.8 kg of water