Entropy ice-water

1. Jan 21, 2009

anubis01

1. The problem statement, all variables and given/known data
A 1.30×10-2 Kg cube of ice at an initial temperature of -12.0 C is placed in 0.470 Kg of water at 50.0 C in an insulated container of negligible mass.

mi=1.3x10-2 Kg
mw=0.470 Kg
Ti=-12 C
Tw=50 C
ci=2100
cw=4190
Lf=3.34*10^5

2. Relevant equations
Qi=Qw
S=Q/T

3. The attempt at a solution

okay so first I tried to find Tf
Qi=Qw
mi(Lf+ci(Tf-Ti))=mwcw(Tw-Tf)
1.3x10-2(3.34*105+2100(Tf+12))=0.47*4190(50-Tf)
27.3Tf+327.6+4342=-1969.3Tf+98465
Tf=46.98

for entropy I tried to add up all the stages where ice->0 C + ice->water + ice water->Tf + hot water->Tf

S=[miciln(T2/Ti)]+[miLf/T]+[micw ln(Tf/T)]+mwcw ln(Tf/Tw)
S=[1.3x10-2*2100ln(273.15/261.15)]+[(1.3x10-2*3.34x105/273.15)]+[1.3x10-2*4190 ln(46.98+273.15/273.15)]+[0.470*4190 ln(46.98+273.15/50+273.15)]
=1.2264+15.896+8.644-18.490=7.28

I still get the wrong answer when using this method so if someone could point me in the right direction that would be much appreciated.

2. Jan 21, 2009

Andrew Mason

But once the ice melts, the H2O (at 0 C) then warms up to the final temperature as liquid water, not ice. So you have to use the specific heat of water for that stage.

AM

3. Jan 22, 2009

anubis01

oh I get it now, thanks for the help.