# Entropy - Help! Water -> Ice

1. Nov 18, 2005

Entropy - Help! Water --> Ice

Here is my question:
10 kg of h2o at 20 degrees C is converted to Ice at -10 degrees C by being put in contact with a reservoir at -10 degrees C. The process takes place at constant pressure. The heat capacities at constant pressure of water (Cp_h2o) and Ice (Cp_ice) are
4180 and 2190 (J)/(kg)*(K) ​
respectively. The heat of fusion of water, l_h2o, is
3.35x10^5 (J)/(kg).​
Calculate the change in entropy of the universe.

I know delta S_universe = delta S_system + delta S_surroundings, but i am a little stuck, any pointers? :yuck:

2. Nov 18, 2005

### Dr.Brain

During the fusion process , the heat would be lost to the surroundings which would lead to increase in entropy. The reservoir is at a fixed temperature (-10) and would gain heat ...therefore entropy change for surroundings = Q/T.

For the system , the change in entropy would be due to heat lost to the surroundings As the transformation in from H2o to ice ... the net entropy change would be:

dS = Ca ln (T2/T1) + Cb ln(T2/T3)

BJ