1. The problem statement, all variables and given/known data Ice of mass 2 kg at -20 degrees C and 5 kg water at 20 degrees C are mixed. Find the net amount of water in the container. (Specific heat of water=1cal/g degrees C, specific heat of Ice = 0.5 cal/g degrees C, latent heat of fusion=80cal/g degrees C) 2. Relevant equations Heat lost by water=heat gained by ice Heat=m.C.(t2-t1) Where m is mass of substance, C is specific heat of substance, t1 is initial temperature, t2 is final temperature. Heat=m.L Where, m is mass of substance, and L is latent heat of substance (latent heat of fusion). 3. The attempt at a solution Well, the first thing I did is to equate the heat lost by water to the heat gained by the ice. Putting all the variables in the equation, (2kg)(0.5 cal/g degreesC)(T- (-20 degrees C))=(5kg)(1cal/g degrees C)(T-20 degrees C) Where, I have assumed T to be the final temperature. On solving the equation, I'm getting T=30 degrees C which is obviously not possible since, it is higher than the temperature of water. This probably means that the final temperature is higher than zero. So I add the latent heat equation. (2kg)(0.5 cal/g degrees C)(0-(-20)) + (2kg)(80cal/g degrees C)+(2kg)(0.5 cal/g degrees C)(T-0)=(5kg)(1cal/g degrees C)(T-20) T=70 degrees C (Again, not possible) I'm assuming this means that all the ice has not melted. But, I have no idea as to how to proceed from this point. Since it is asking the amount of water in the container, some amount of ice has melted and some amount has not melted. How do I ascertain how much ice has melted? Help will be appreciated. Thanks.