Expert: Richard J. Raridon - 6/21/2009

Question
Hello, can you please help me with one problem that I have in Physics?

The problem is: 1 gram of steam at 100°C is in thermal contact with 50 grams of ice at 0°C and the system is made to equilibriate. Now the question is what will be the final temperature of the system?

I know that the specific heat of water is 4186 J/kg°C, latent heat of fusion is 3.33x10^5 J/kg and latent heat of vaporization is 2.26x10^6 J/kg. However, everytime I try to solve it (using mL + mcdT + mL + mcdT = 0), I always get around -65°C as the final temperature. I know that the temperature will not drop as heat is introduced to ice with a temperature of 0°C. Can you please help me with this?

Answer
Let's assume the final temperature is t.  
So you have 50(3.33x10^2) +50t(4.186) = 1(2.26x10^3) +1(4.186)(100-t)
so t = -65C
What that tells you is that the heat from 1g of steam is not enough to melt all the ice so the final temperature of the system is 0C.