Question
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
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