Posted by Ceres on Thursday, September 24, 2009 at 5:48pm.
Fifty grams of hot water at 80oC is poured into a cavity in a very large block of ice at 0oC. The final temperature of the water in the cavity is then 0oC. Show that the mass of ice that melts is 50 g.
To melt ice to 50g of water, I think that it will take:
50g(334J/g) = 167,000J
When cooling, the hot water should release:
Specific heat capacity of water(ΔT)(grams of water)(conversion to J from calories) = J released by cooling
(.5cal/g*oC)(80oC)(50g)(4.18J/cal) = 8360J
This is only about half of the energy that is needed to convert the ice to water, so I am missing something. Any ideas?

Physics  ice to liquid  DrBob222, Thursday, September 24, 2009 at 6:03pm
I don't know why you converted J to calories. All of that seems unnecessary.
How much heat do you have in the 80 C water if it moves from 80 C to zero (which the problem states as the final T). Therefore,
q = mass water x specific heat water x delta T.
q = 50 x 4.18 x 80 = 16,720 Joules.
How much ice can you melt with that?
16,720/334 = ??
This is almost the same problem we worked on a couple of days ago with the block of ice. 
Physics  ice to liquid  DrBob222, Thursday, September 24, 2009 at 6:15pm
If you want to work it with calories, the specific heat of water is 1.0 calorie/gram and the heat of fusion for ice is 80 cal/g, therefore
mass H2O x 1 cal/g x 80 = 4,000 calories.
4000 calories/80 cal/g = 50 g ice. 
Physics  ice to liquid  Ceres, Thursday, September 24, 2009 at 9:42pm
I am sorry to be a bother  I knew it was like the other problem but I couldn't figure out how. I am taking my class online and am struggling to learn the concepts, basically on my own. While I do well in math, this class has been very difficult for me to get my mind around  it's like learning a new language  with very little help. I now understand what you wrote  my one equation, 50g(334J/g) = 16,700J helped me to see how much energy I needed, I just didn't realize that the book was asking me to go back a chapter to get the equation I used last week. Now I know how to find where that energy comes from, and it even makes sense. Thanks for your patience and help. :)