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Posted by on Monday, May 12, 2014 at 11:49pm.

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
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.
not understood can you make it clear

  • chemistry - , Tuesday, May 13, 2014 at 12:09am

    What specifically do you not understand? Basically I have two explanations. The first one I incorrectly assumed that all of the ice would melt. It doesn't and I showed how to work the problem in an earlier post. That I took your work down to the point where you get a negative number to show that that is how you would know that all of the ice did not melt.

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