Posted by **Melissa** on Tuesday, December 7, 2010 at 12:39am.

A 3.90×10^2kg cube of ice at an initial temperature of -18.0degreesC is placed in 0.480kg of water at 41.0degreesC in an insulated container of negligible mass.

Calculate the change in entropy of the system.

I found the equilibrium temperature to be at 304.9K, then used

change in S = mass*c*ln(T2/T1) where T2 is the final temperature and T1 is the initial temperature.

But my answer was wrong.

- Physics -
**drwls**, Tuesday, December 7, 2010 at 2:12am
There are TWO initial temperatures. You can not use the equation you used, except for the cooling of the liquid water. When ice is melting (if that ever happens here), the temperature stays the same and the entropy gain is Qin/273K

Are you sure you copied the numbers correctly? There is nearly 1000 times more ice than liquid water initially, according to the numbers you gave. If that is true, you would heat the ice but not melt any.

Rethink the equilibrium temperature. Then calculate the entropy gain of the ice and subtract the entropy loss of the water.

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