- #1

anubis01

- 149

- 1

## 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

_{w}c

_{w}(Tw-Tf)

1.3x10

^{-2}(3.34*10

^{5}+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

_{i}c

_{i}ln(T

_{2}/T

_{i})]+[m

_{i}Lf/T]+[m

_{i}c

_{w}ln(Tf/T)]+m

_{w}c

_{w}ln(Tf/T

_{w})

S=[1.3x10

^{-2}*2100ln(273.15/261.15)]+[(1.3x10

^{-2}*3.34x10

^{5}/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.