# Homework Help: How do I find the entropy variation?

1. Dec 13, 2017

### Upupumiau

1. The problem statement, all variables and given/known data
Find the ∆S per mol between liquid water at -5 ºC and ice at -5ºC at 1020hPa
Data:
∆CP,m fusion = 37,3 J K-1 mol-1
∆fusH = 6,01 kJ mol-1
The answer is 21.3 J/K mol

2. Relevant equations
Usually I solve these problems by steps when they are at P=1 atm but since its at P=1020 hPa i don't really know how to proceed.
I tried using the following equation:
∆S= Q/T + nCpm ln T2/T1
Since I am given the Cpm I interpret that the pression is constant throughout the process

3. The attempt at a solution
I first tried to solve it like I normally would, without taking into account that the press ion a was not 1 atm.
I want to try and find the temperature at which water soldifies at 1020 hPa but I don't know how with the data I am given. I thought that since Qp is the enthalpy maybe I could find it using Q = n∆T specific heat of water
but even thought I know what the specific heat of water is (41814 J/K kg) I'm not given this data so I'm unsure if this is what I have to do.
This is what I first tried doing but after I realized my mistake I did this:

I didn't obtain the value for entropy I should have with neither of the methods I tried. Where am I wrong?

2. Dec 13, 2017

### Staff: Mentor

You are correct about thinking of this as a multi-step process.

ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
ice at 0 C and 1 atm to water at 0 C and 1 atm
water at 0 C and 1 atm to water at 0C and 1020 hPa
water at 0C and 1020 hPa to water at 5 C and 1020 hPa

3. Dec 14, 2017

### Upupumiau

I tried doing it this way and it really does make more sense but I can't still obtain the 21.3 J/K I should and I really don't know why. I thought maybe it's because of the Cpm I have been given , but it turns out it doesnt really differ from the one I was given, so I thought I might change the sign since one it's fusion and the other one its for solidification. Still, I haven't obtained the result I need and I don't know why. Could anyone please help me?

4. Dec 14, 2017

### Staff: Mentor

You did this mostly correctly in your first attempt. The first thing to recognize is that 1020 hPa is essentially 1 atm, so there is no need to account for a pressure effect. So the change in entropy in going from 1 mole of ice at -5 C to water at - 5 C is given by:
$$\Delta S=C_i\ln(273/268)+\frac{\Delta H(0)}{273}+C_w\ln(268/273)=(C_W-C_i)\ln(268/273)+22.0$$
We know that $C_w-C_i=37.3\ J/mol.K$

5. Dec 14, 2017

### Upupumiau

I don't quite understand how you're using logarithmic properties here, could you explain it to me please?

6. Dec 14, 2017

### Staff: Mentor

Are you saying that you are not sure how to determine the change in entropy of a solid or liquid when it is heated or cooled from an initial temperature to a final temperature (and why this should involve natural logs of temperatures)?

7. Dec 31, 2017

### Upupumiau

I do, sorry. I had to revisit the basics. Thank you for your help.