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
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

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.