Tuesday
July 7, 2015

Homework Help: chemistry

Posted by treasa on Monday, May 12, 2014 at 10:45pm.

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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