# Conceptualizing entropy change of surroundings

1. Oct 3, 2012

### Bipolarity

I asked a question on this forum a few days ago about the entropy change of the surroundings, and am grateful for the insight provided. However, something faulty in my conceptualization is preventing me from solving this problem.

Let's say you have a set processes shown in the following diagram, where each of the arrows indicates a reversible process for which q and w can all be calculated:

A --> B --> C --> D

Suppose that there is a direct path A-->D, but this path is irreversible and the heat of this process is unknown. However, this path is known to be isothermal.

The entropy change (of the system) in going from A-->D is simply the sum of the individual entropy changes (of the system) in each of the above steps (since entropy is a state function and since each step is reversible).

Now comes the confusing part:
The entropy change of the surroundings in going from A-->D is calculated by summing up the heat transferred to the surroundings in each of the individual steps, and then divided by the temperature of the isothermal process A-->D.

This confuses me greatly. I don't understand why this approach is used in calculating the surroundings' entropy change.

Perphaps someone more experienced in thermo can explain?

BiP

2. Oct 4, 2012

### Andrew Mason

You have to be able to define states A and D first. Once you do that, you find a reversible path between states A and D, determine the heatflow for that path (ie the reversible heatflow) and then divide by T to determine the change in entropy.

AM

3. Oct 4, 2012

### DrDu

Maybe I don't quite understand your setup, nevertheless, I'll try to give quite trivial a counter example: Let's say my system consists of a resistor which is well cooled.
The states A, B, C, D of the system are taken to be all the same and the entropy, heat, work etc to go from one to the other are all trivially =0. Nevertheless I can go from A to D irreversibly creating an arbitrary amount of heat by letting flow a current through the resistor for an arbitrary amount of time.

4. Oct 4, 2012

### Bipolarity

OK here is an example of my states:

- All states have 1 atm pressure.
- All states have the same volume.

A: water at 283K
B: water at 273K
C: ice at 273L
D: ice at 283K

As you can see, the freezing of water into ice is only reversible if the process is done at equilibrium conditions, i.e. 273K because that is the freezing point of water, at which the Gibbs energy change is 0. That is, B-->C represents the reversible freezing of water, whereas A--D represents the irreversible freezinong of water.

So to calculate the system's entropy change from A-->D, we would construct the reversible path A-->B-->C-->D, and then sum up the entropy change in each (reversible) step. Right?

Now what about the surroundings? The surrounding's entropy change should just be the heat (transferred to surroundings) of the process divided by 283K. Does "heat of the process" refer to heat in process A + heat in process B + heat in process C?

I seem to have reached some paradox. The entropy change of the universe in A-->D must be >0 because the overall process is irreversible (by second law). Thus, the entropy change of the surroundings in A-->D must be greater in magnitude than that of the system (if the system's entropy change is negative).

But in each process A-->B, B--C, C--D, the entropy change of the surroundings has equal magnitude to that of the system. What's the deal here? Is the entropy change of the surroundings not a state function?

BiP

5. Oct 4, 2012

### DrDu

There won't be ice at 283 K at standard pressure.

6. Oct 4, 2012

### Bipolarity

Yes, it's not a spontaneous process obviously, but what if we were to calculate the entropy change of this process? It's still a process, whether or not it actually occurs....

BiP

7. Oct 4, 2012

### Staff: Mentor

No. The sum of the heats transferred in the individual reversible steps does not necessarily sum to the heat transferred in the irreversible isothermal process. Just take the simple one-step example of a spontaneous isobaric expansion (with heat supplied by a constant temperature reservoir held at the initial temperature) compared to a reversible isothermal expansion at the initial temperature (exchanging heat with a constant temperature reservoir at the same initial temperature). The heats transferred and the works done in these two processes will be different, although, for an ideal gas, the change in internal energy and change in entropy for the system will be the same. But, the entropy change in the surroundings for the irreversible process will be higher. Calculate it in an example and see.

Chet

Last edited: Oct 4, 2012
8. Oct 4, 2012

### Andrew Mason

Did you mean 263K for D?
Correct. What would a reversible path from A to B be? hint: try a combination of adiabatic and isothermal changes (compression or expansions).

AM

9. Oct 4, 2012

### Bipolarity

Oh damn that was a mistake. Andrew it is indeed 263K you are correct.

I will restate the problem:
A: water at 263K
B: water at 273K
C: ice at 273K
D: ice at 263k

Water freezing to ice at 263K (i.e. process A-->D) is not reversible. Its entropy change of system can be calculated by summing up individual entropy changes of the system in each reversible step.

How would you calculate the entropy change of the surroundings in freezing a certain amount of water at 263K?

BiP

10. Oct 4, 2012

### Studiot

Tell us first how you would obtain liquid water at 263°K

11. Oct 4, 2012

### Bipolarity

There are supercooled water droplets below freezing point in the stratosphere...
But I don't see how that has any bearing on the problem.

The question asks to find the entropy change of the surroundings of water freezing to ice at 263K. Easily the Gibbs energy change of this reaction can be calculated, and it must be a spontaneous process.

But I am asking about the entropy change of the surroundings as the water isothermally freezes at this temperature. We can't use the latent heat of fusion because that assumes the water is in equilibrium while freezing (273K and 1atm).

Perhaps it's impossible to calculate the entropy change of the surroundings in this process? Maybe it needs to be empirically determined using a calorimeter and that's the only way?

Ideas? Thanks!

BiP

12. Oct 4, 2012

### Studiot

Supercooled maybe, but have you checked the phase diagram for the conditions under which liquid water is possible at 263°K ?

13. Oct 4, 2012

### Bipolarity

Yes, water doesn't naturally exist at that point. It will turn into ice. And that is the point of my question... when it does turn into ice, what will the entropy change of the surroundings be for this process?

BiP

14. Oct 4, 2012

### Studiot

If it can't exist how can it be there to turn into ice?

15. Oct 4, 2012

### Staff: Mentor

Yes. The entropy change of the surroundings is not unique. Think of the surroundings as a separate system. In one case you can employ a high temperature gradient in the surroundings (holding the temperature at the boundary with your system fixed), and in another case you can employ a low temperature gradient in the surroundings (holding the temperature at the boundary with your system fixed at the same value). The entropy changes in the surroundings in these two cases differ, even though the entropy changes for the system are the same.

16. Oct 4, 2012

### Studiot

If you really want the answer to your question get hold of a Mollier Diagram that goes down to your temperature range.

Mollier diagrams plot (specific) entropy against (specific) enthalpy for different pressure and temperature isolines.

17. Oct 4, 2012

### Andrew Mason

I am still confused. Is the liquid water initially at 263K or 283K? How are you getting water at 263K? Is this under high pressure or super-cooled conditions?

There is a reversible path between two normal states. So, for example, you could get the temperature changes needed by running a Carnot heat pump or a Carnot engine between the surroundings and the water/ice.

AM

18. Oct 5, 2012

### DrDu

Ok, so all this confusing stuff about states A, B, C, D is really about freezing of water at -10 deg C.
You could have stated this at the very beginning!
I think there is no problem in supercooling water in that range.
The change in enthalpy is $\Delta H(263 K)=\Delta H(273 K)+\int_{273 K}^{263 K} dT (C_{P, solid}(T)-C_{P, liquid}(T))$.
The entropy change of the surrounding is ΔH/263K.
The change in free enthalpy of the system can also be calculated using the Gibbs-Helmholtz equation. From the free enthalpy change and the enthalpy change, also the entropy change of the system can be calculated.

19. Oct 5, 2012

### Andrew Mason

The path directly from A-D is irreversible but there is a reversible path from A-B-C-D.

$\Delta S_{AD} = \Delta S_{AB} + \Delta S_{BC} + \Delta S_{CD} = \int_{263}^{273} mC_{water}dT/T - mL/273 + \int_{273}^{263}mC_{ice} dT/T$

$\Delta S_{AD} = mC_{water}\ln{\frac{273}{263}} - mL/273 - mC_{ice}\ln{\frac{273}{263}}$

The change in entropy of the surroundings is just:

$\Delta S_{surr} = (-10mC_{water} + ml + 10mC_{ice})/273$

AM

20. Oct 5, 2012

### Bipolarity

Thank you all so much!!! I just want to make sure I understood this correctly:
- The entropy change of the system in going from A to D is simply the sum of the entropy changes of the system in each reversible step A to B, B to C, and C to D.
- The entropy change of the surroundings in going from A to D is NOT equal to the sum of the entropy changes of the surroundings in each reversible step A to B, B to C, and C to D.
- Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D.
- Since heat is a path function, we need some way to calculate it in this strange path of water freezing at 263K.
- We know that at constant pressure, heat = enthalpy which does happen to be a state function.
- We can easily calculate enthalpy for this reaction A to D. We do so, and equate with heat.
- Finally, we divide the enthalpy of A to D by the temperature of this reaction to get the entropy change of the surroundings.
- We note that whatever value we get must be such that the entropy of the universe is positive.

If anywhere I am mistaken pleease correct me! Thanks all for the help!

BiP