# Change in entropy question

1. Apr 6, 2015

### Lisa Marie

1. The problem statement, all variables and given/known data
5 kg of water at 60 degrees are put in contact with 1 kg of ice at 0 degrees and are thermally isolated from everything else. The latent heat of ice is 3.3x105 J/kg

What is the change of entropy of the universe when 100J of energy are transferred from the water to the ice

2. Relevant equations

Q=+/- mL
Q=mcΔT
ΔS=|Q|/Tc-|Q|/TH

3. The attempt at a solution
mcΔT+mL+mcΔT=0
(5)(4200)(Tf-60)+(1)(3.3×105)+2000(1)Tf=0

23000Tf=96000
Tf=41.74 C

ΔS=100/(41.74+273)-100/(273)=-0.0486

I'm not getting the right answer for this one. Help would be greatly appreciated!

2. Apr 6, 2015

### vela

Staff Emeritus
You're assuming that all the ice is going to melt.
What does the last term represent? I figure it has something to do with the ice, but I'm not sure.

This expression also assumes that all of the ice melts. Is this possible if only 100 J of energy is transferred from the warm water to the ice? Can you describe in words what's going to happen?

3. Apr 6, 2015

### Halo CX

Firstly you can not use the relation $$\Delta S= \Delta Q/T$$ to find out entropy change for water, because here T is changing. Instead you should use the general relation $$\Delta S= \int dQ/T$$ to find out the change in entropy. This process should work out.

4. Apr 6, 2015

### Staff: Mentor

There is a bit of ambiguity regarding this problem statement. It is not clear whether the water and ice are in intimate contact, or whether they are separated by a diathermal barrier, and that, once the 100 J are transferred, the diathermal barrier is removed and replaced by an insulation barrier. The version with the barriers is simpler to solve. If 100 J is removed from the 5 kg of water, how much does its temperature drop?

Chet