Entropy - Help! Water -> Ice

  • #1
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:
 

Answers and Replies

  • #2
538
2
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
 

Related Threads on Entropy - Help! Water -> Ice

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
20K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
15
Views
2K
Replies
1
Views
6K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
5
Views
20K
Top