Calculate Change in Entropy of Ice Cube-Water System

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Homework Statement

An 8.5 g ice cube at -10°C is put into a
Thermos flask containing 200 cm3
of water at 20°C. By how much has the
entropy of the cube-water system changed when equilibrium
is reached? The specific heat of ice is 2220 J/kg·K.

Cw= 4190
Mi= 8.5
Mw= 200
Lf= 333000
Ci=2220

Homework Equations

Tf=cw*mw(20)-mi(lf+10ci)/mw*cw+mi*cw

Tf= 288.73 K


mi*ci*ln(273.15/263.15)+lf*mi/273.15+mi*cw*ln(288.73/273.15)+mw*cw*ln(288.73/293.15)

The Attempt at a Solution

.70366+10.36+1.975-6.365

I got 6.67 J/k. But I'm not sure that's right.

 

[Andrew Mason]

It is not very clear what you are trying to do here.


The first thing you have to do is figure out whether all the ice melts. If it does, you then have to figure out what the final temperature is.

Then you have to work out what the change in entropy of the ice is in reaching 0C, in changing to water, and in rising to the final temperature. Use the definition of incremental change in entropy: dS = dQ/T.

Then you have to work out the change in entropy of the 200 g of water in cooling to the final temperature.

AM