Entropy of Ice-Water: Solving the Puzzle

[anubis01]

Homework Statement

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.

m_i=1.3x10^-2 Kg
m_w=0.470 Kg
T_i=-12 C
T_w=50 C
c_i=2100
c_w=4190
Lf=3.34*10^5

Homework Equations

Q_i=Q_w
S=Q/T

The Attempt at a Solution

okay so first I tried to find T_f
Q_i=Q_w
m_i(Lf+c_i(Tf-Ti))=m_wc_w(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=[m_ic_iln(T₂/T_i)]+[m_iLf/T]+[m_ic_w ln(Tf/T)]+m_wc_w ln(Tf/T_w)
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.

 

[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

 

[anubis01]

oh I get it now, thanks for the help.