# chemistry

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 degree-water at 0=334*25 =8350
waterat 0 to water at final temp=25*4.18*tf-0
heat lost by water at 22 degree celsius=
50*4.18*(Tf-22)
1030+8350+104.5Tf=-(209Tf-4598)( negative since heat lost)
4782=313Tf
tf=15.3

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asked by treasa
1. http://www.jiskha.com/display.cgi?id=1399943202

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2. All of your work is ok until here.

50*4.18*(Tf-22)
1030+8350+104.5Tf=-(209Tf-4598)( 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.

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3. i don't understand
can you make it clear

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posted by treasa
4. 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.

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