chemistry
posted by treasa on .
what is the final temperature of amixture containning 25 grams of ice at 20 degree celciusadded to water at room temperature 22oC mass of water is 50g
followed the steps given by you
but not getting the answer (0degree celsius)
ice at 20 to ice at 0 degree=mcdelta T=25*2.06*20=1030
1ce at 0 degreewater at 0=334*25 =8350
waterat 0 to water at final temp=25*4.18*tf0
heat lost by water at 22 degree celsius=
50*4.18*(Tf22)
1030+8350+104.5Tf=(209Tf4598)( negative since heat lost)
4782=313Tf
tf=15.3


All of your work is ok until here.
50*4.18*(Tf22)
1030+8350+104.5Tf=(209Tf4598)( negative since heat lost)
4782=313Tf
tf=15.3
Should have been
1030 + 8350 + 104.5Tf + 209Tf + 4598 = 0
Then you get xxxTf = some number and that makes Tf = something and you KNOW that can't be right. What it means is that all of the ice didn't melt. If I didn't goof that other response should take care of it. 
i don't understand
can you make it clear 
If you followed your math you ended up with Tf of 15.3. I didn't go through line by line but my explanation was that your work should have produced a negative value for Tf. When that happens we know it can't be right from the data given so we know something is wrong. The wrong part is that I assumed all of the ice would melt and I gave the formula to you for that. Since all of the ice will not melt, that formula I gave you is no good. However, I worked the problem completely. If you can tell me what isn't clear perhaps I can help. But let's not beat a dead horse here. If it is something about how I worked the problem after knowing not all of the ice will melt perhaps I can help. If it's something about the first set up I used I think that is a waste of time since we already know that isn't the way to go.