# Find the changes in entropy for the water and for the ice

1. Jan 29, 2005

### Soaring Crane

In a well-insulated calorimeter, 1.0 kg of water at 20 C is mixed with 1.0 g (.001 kg) of ice at 0 C. What is the net change in entropy of the system by the moment the ice completely melts? The heat of fusion of ice is 3.34 x 10^5 J/kg.

I must find the changes in entropy for the water and for the ice, then find the net change.

Net change = /_\S1 + /_\S2

This is what I did so far:
S_ice = (mL)/T, where Q = mL

S_ice = [(3.34 x 10^5 J/kg)*.001 kg]/[273 K] = 1.223 J/K ???

How do I find the change in entropy for H2O??

Thanks.

Last edited by a moderator: Jan 7, 2014
2. Jan 29, 2005

### Andrew Mason

You have found the $\Delta S$ for the ice. The $\Delta Q$ for the water is just the loss of heat to the ice. It takes place at 293K (we can see that there is not a significant temperature reduction - about .1 degree). So:

$$\Delta S_{H_2O} = -334/293 = -1.14 J/K$$

AM