If you have an insulated glass of water at 25 oC containing 10 fluid oz at 25 oC, how much ice would you have to add to end up with a final temperature of 0 oC and no ice left?
Dan--I answered parts of your question incorrectly because I did not read the question the right way. Here is what I wrote, along with corrections.
Convert 10 oz to grams. There are 28.35 g to an oz. The 283.5 grams you have is correct.
(mass ice x heat fusion ice) + [(mass water x specific heat water x (Tfinal-Tinitial)] = 0
This is the correct formula
Solve for Tfinal, the only unknown.
This last sentence is the incorrect part.
mass ice is the question and that is the unknown for which you solve.
You have used 0.5 for the heat of fusion but that isn't correct. In calories it is about 80 calories/gram and in Joules it is about 334 J/g. Use the appropriate one depending upon your needs. Then Tfinal = 0 from the problem and Tinitial = 25 from the problem. That makes mass ice the only unknown and the solution should be easy enough. I'm logging off so I won't be able to check your work but you shouldn't have any trouble.
So you will have
(mass ice x 80)+[(283.5 x 1 x (0-25)] = 0
Solve for mass ice.
I get something like 88+ grams ice to melt and cool the 25 C water to zero C and have no ice left over. I hope this helps.