Conceptualizing entropy change of surroundings

[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

 

[Andrew Mason]

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).

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

 

[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.

 

[Bipolarity]

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

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

 

[DrDu]

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

 

[Bipolarity]

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

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

 

[Chestermiller]

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.

BiP

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

 

[Andrew Mason]

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

Did you mean 263K for D?

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?

Correct. What would a reversible path from A to B be? hint: try a combination of adiabatic and isothermal changes (compression or expansions).

AM

 

[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

 

[Studiot]

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

 

[Bipolarity]

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

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

 

[Studiot]

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

 

[Bipolarity]

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

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

 

[Studiot]

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

 

[Chestermiller]

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

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.

 

[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.

 

[Andrew Mason]

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.

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

 

[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.

 

[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

 

[Bipolarity]

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.

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

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

 

[Andrew Mason]

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.

Correct.

- 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.

How do you conclude that? Assume that you effect each of the changes by Carnot cycles between the water/ice and the constant temperature surroundings (surroundings have infinite heat capacity).

- Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D.

No. The entropy of the surroundings is the sum of all the heat flows to and from the water/ice in the reversible path A-B-C-D, divided by the temperature of the surroundings (273K).

- Since heat is a path function, we need some way to calculate it in this strange path of water freezing at 263K.

To find the entropy change you have to use the reversible path A-B-C-D.

- We note that whatever value we get must be such that the entropy of the universe is positive.

And it does - see my calculation: C_water>C_ice and ln(273/263)>10/273

AM

 

[DrDu]

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

You mean divided by 263, not 273, in the last step.

 

[Andrew Mason]

You mean divided by 263, not 273, in the last step.

I thought the surroundings were at 273K.

If the surroundings are at 263K, one has to postulate an arbitrarily large reservoir at 273K that is in thermal contact with the surroundings and the system (via Carnot devices) in order to have the reversible path A-B-C-D.

But you are right if the surroundings are at 263K. The change in entropy of the surroundings would be:

\Delta S_{surr} = (-10mC_{water} + mL + 10mC_{ice})/263

AM

 

[DrDu]

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

Sounds all correct to me.

 

[Andrew Mason]

- 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.

DrDu and I seem to disagree on this part. In my view, this cannot be correct. The entropy change of the surroundings in going from A to D cannot depend on the path.

AM

 

[Bipolarity]

I would like to quote the words of my textbook, in case it helps anyone:
"So far, our discussion of entropy changes has focused on the system. For a proper understanding of entropy, we must also examine what happens to the surroundings. Because of its size and amount of material it contains, the surroundings can be thought of as an infinitely large heat reservoir. Therefore, the exchange of hat and work between a system and its surroundings alters the properties of the surroundings by only an infinitesimal amount. Because infinitesimal changes are characteristic of reversible processes, it follows that any process has the same effect on the surroundings as a reversible process. Thus, regardless of whether a process is reversible or irreversible with respect to the system, we can write the heat change in the surroundings as

(dq_{sur})_{rev} = (dq_{sur})_{irev} = dq_{sur}



BiP

 

[Andrew Mason]

I would like to quote the words of my textbook, in case it helps anyone:
"So far, our discussion of entropy changes has focused on the system. For a proper understanding of entropy, we must also examine what happens to the surroundings. Because of its size and amount of material it contains, the surroundings can be thought of as an infinitely large heat reservoir. Therefore, the exchange of hat and work between a system and its surroundings alters the properties of the surroundings by only an infinitesimal amount. Because infinitesimal changes are characteristic of reversible processes, it follows that any process has the same effect on the surroundings as a reversible process. Thus, regardless of whether a process is reversible or irreversible with respect to the system, we can write the heat change in the surroundings as

(dq_{sur})_{rev} = (dq_{sur})_{irev} = dq_{sur}

Correct. The change in entropy of the surroundings is the negative of the actual heat flow into the system that occurs during the actual process divided by the temperature of the surroundings.

If that is what you mean by " Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D" then you are correct. But if the actual path is A-B-C-D for the system, which is what I understood from your example, then the change in entropy of the surroundings will be the sum of the heat flows out of the system/T_surr: \Delta S_{surr} = -(Q_{AB} + Q_{BC} + Q_{CD})/T_{surr}

AM

 

[Bipolarity]

Correct. The change in entropy of the surroundings is the negative of the actual heat flow into the system that occurs during the actual process divided by the temperature of the surroundings.

If that is what you mean by " Rather, the entropy change of the surroundings must be calculated by finding the heat of the reaction A to D" then you are correct. But if the actual path is A-B-C-D for the system, which is what I understood from your example, then the change in entropy of the surroundings will be the sum of the heat flows out of the system/T_surr: \Delta S_{surr} = -(Q_{AB} + Q_{BC} + Q_{CD})/T_{surr}

AM

I am sorry for the confusion Andrew, the "actual" reaction that the problem asked about was the irreversible direct path A-->D. I hope that resolves everything?

BiP

 

[Chestermiller]

I would like to quote the words of my textbook, in case it helps anyone:
"So far, our discussion of entropy changes has focused on the system. For a proper understanding of entropy, we must also examine what happens to the surroundings. Because of its size and amount of material it contains, the surroundings can be thought of as an infinitely large heat reservoir. Therefore, the exchange of hat and work between a system and its surroundings alters the properties of the surroundings by only an infinitesimal amount. Because infinitesimal changes are characteristic of reversible processes, it follows that any process has the same effect on the surroundings as a reversible process. Thus, regardless of whether a process is reversible or irreversible with respect to the system, we can write the heat change in the surroundings as

(dq_{sur})_{rev} = (dq_{sur})_{irev} = dq_{sur}



BiP

In my judgement, this is basically incorrect.

Of course, we must always have that

(dQ_sys) + (dQ_surr) =0

Thus, (dQ_sys)_rev + (dQ_surr)_rev =0

and (dQ_sys)_irrev + (dQ_surr)_irrev =0

But since, in general,
(dQ_sys)_rev≠ (dQ_sys)_irrev

we must also have that

(dQ_surr)_rev≠ (dQ_surr)_irrev

Also, none of these relationship say anything about the change in entropy of the surroundings, since we don't know the details of how the heat from the surroundings was supplied, which, of course, determines the entropy change of the surroundings. We can always think of processes in which

1. the path for the system and the path for the surroundings is reversible
2. the path for the system is irreversible, but the path for the surroundings is reversible
3. the path for the system is reversible, but the path for the surroundings is irreversible
4. the paths for the system and the the path for surroundings is irreversible

 

[Bipolarity]

In my judgement, this is basically incorrect.

Of course, we must always have that

(dQ_sys) + (dQ_surr) =0

Thus, (dQ_sys)_rev + (dQ_surr)_rev =0

and (dQ_sys)_irrev + (dQ_surr)_irrev =0

But since, in general,
(dQ_sys)_rev≠ (dQ_sys)_irrev

we must also have that

(dQ_surr)_rev≠ (dQ_surr)_irrev

Also, none of these relationship say anything about the change in entropy of the surroundings, since we don't know the details of how the heat from the surroundings was supplied, which, of course, determines the entropy change of the surroundings. We can always think of processes in which

1. the path for the system and the path for the surroundings is reversible
2. the path for the system is irreversible, but the path for the surroundings is reversible
3. the path for the system is reversible, but the path for the surroundings is irreversible
4. the paths for the system and the the path for surroundings is irreversible

Everything you say makes perfect sense Chet, which continues to boggle me because then it implies that the the entropy change of the universe must be 0 for all processes reversible and irreversible, which is a direct contradiction to the second law.

I thank everyone for their insights provided so far, but it isn't over yet. I encourage all to continue posting their thoughts so we can deliberate on this matter further and reach a ground where everyone can agree.

The question then, is it reconcile Chet's argument with the second law of thermodynamics, to come up with a precise definition of the surrounding's entropy change. Only then can I (or others) solve textbook problems involving the surrounding's entropy change without having a fuzzy notion of what we are talking about.

BiP

 

[Andrew Mason]

Everything you say makes perfect sense Chet, which continues to boggle me because then it implies that the the entropy change of the universe must be 0 for all processes reversible and irreversible, which is a direct contradiction to the second law.

That is not how you calculate entropy change. When the system process is irreversible, a reversible path between the beginning and end points results in a different heat flow than the actual path. You use a reversible path BETWEEN THE BEGINNING AND END STATES for the system to calculate entropy change for the system.

Similarly, you use a reversible path BETWEEN THE BEGINNING AND END STATES for the surroundings to calculate entropy change for the surroundings.

The reversible paths for the surroundings and the system are the SAME for both ONLY if the actual path is reversible. If the actual path is irreversible, the reversible path for the system between its beginning and end states IS DIFFERENT than the reversible path between the surrounding's beginning and end states.

Example: an adiabatic free expansion of an ideal gas from (P,V,T) to (P/2,2V,T). There is no Q and no work is done on or by the surroundings by or on the gas. Since ΔU = Q-W = 0, there is no change in T for the gas.

The change in entropy of the gas, however, is determined by the reversible path between the beginning and end states. The reversible path from (P,V,T) to (P/2,2V,T) for the ideal gas is a quasi-static isothermal expansion in which work is done against an external pressure. Since ΔU=Q-W = 0 and W>0 then Qrev>0 so ΔS = ∫dQrev/T > 0.

HOWEVER, for the surroundings the beginning and end states of the surroundings, UNLIKE the gas, ARE IDENTICAL. There has been no heatflow to/from the surroundings. So there is no change in entropy of the surroundings.

AM

 

[Bipolarity]

HOWEVER, for the surroundings the beginning and end states of the surroundings, UNLIKE the gas, ARE IDENTICAL. There has been no heatflow to/from the surroundings. So there is no change in entropy of the surroundings.

AM

Can't this argument be extended to any finite process which occurs in the laboratory, whether reversible or irreversible? In other words, wouldn't the entropy change of the surroundings always be 0 according to your logic, even for a reversible process?

BiP

 

[Studiot]

an adiabatic free expansion of an ideal gas from (P,V,T) to (P/2,2V,T).

Adiabatic or isothermal?

What do you mean by free expansion?

Please explain.

 

[Andrew Mason]

Can't this argument be extended to any finite process which occurs in the laboratory, whether reversible or irreversible? In other words, wouldn't the entropy change of the surroundings always be 0 according to your logic, even for a reversible process?

??No! If there is heatflow to/from the surroundings the entropy of the surroundings changes.

In the quasi static isothermal expansion of a gas, for example, there is heatflow from the surroundings to the system so the entropy of the surroundings decreases by the same amount that the entropy of the gas increases so the total entropy change is 0. In the free expansion of the gas, the entropy of the gas increases while the entropy of the surroundings is 0 so the total entropy change is > 0.

AM

 

[Chestermiller]

Everything you say makes perfect sense Chet, which continues to boggle me because then it implies that the the entropy change of the universe must be 0 for all processes reversible and irreversible, which is a direct contradiction to the second law.

BiP

It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say T_B. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/T_B for the system is always equal to -dQ/T_B for the surroundings. But this definitely does not mean that the entropy change for the system has to be minus the temperature change for the surroundings. The entropy change for the system is only equal to dQ/T_B if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/T_B if the surroundings are undergoing a reversible path. We can regard the system and the surroundings as really two separate systems that are mutually constrained at the boundary between them.

Now, for the system, we have that

(dS)_syst = dQ/T_B for a reversible path

(dS)_syst > dQ/T_B for an irreversible path

The second relationship represents the simplified version of the so-called Cauchy inequality.

And, for the surroundings,

(dS)_surr = -dQ/T_B for a reversible path

(dS)_surr > -dQ/T_B for an irreversible path

So if both system and surroundings undergo reversible paths, then

(dS)_syst + (dS)_surr = 0

If either the system or the surroundings or both undergo irreversible paths in the actual process, then

(dS)_syst + (dS)_surr > 0

BiP, yesterday I sent you an analysis I did of the irreversible heating of a metal bar by means of heat supplied by an isothermal reservoir (comprising the surroundings). This is a case where the system is undergoing a irreversible path, while the surroundings are experiencing a reversible path. I showed how to precisely calculate the change in entropy for both the system and the surroundings from the solution to the transient heat conduction equation, and how the positive definiteness of the sum total of the entropy changes for the system and the surroundings comes about (for this problem). I have attached that analysis to this reply for others involved in this thread to examine. This provides a specific example of how the Cauchy inequality evolved, and why the sum of the entropy changes for the system and surroundings is always greater than zero for an irreversible process.

Chet

 

[Andrew Mason]

It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say T_B. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/T_B for the system is always equal to -dQ/T_B for the surroundings.

Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?

The entropy change for the system is only equal to dQ/T_B if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/T_B if the surroundings are undergoing a reversible path.

This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ T_B

AM

 

[Bipolarity]

It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say T_B. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/T_B for the system is always equal to -dQ/T_B for the surroundings. But this definitely does not mean that the entropy change for the system has to be minus the temperature change for the surroundings. The entropy change for the system is only equal to dQ/T_B if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/T_B if the surroundings are undergoing a reversible path. We can regard the system and the surroundings as really two separate systems that are mutually constrained at the boundary between them.

Now, for the system, we have that

(dS)_syst = dQ/T_B for a reversible path

(dS)_syst > dQ/T_B for an irreversible path

The second relationship represents the simplified version of the so-called Cauchy inequality.

And, for the surroundings,

(dS)_surr = -dQ/T_B for a reversible path

(dS)_surr > -dQ/T_B for an irreversible path

So if both system and surroundings undergo reversible paths, then

(dS)_syst + (dS)_surr = 0

If either the system or the surroundings or both undergo irreversible paths in the actual process, then

(dS)_syst + (dS)_surr > 0

BiP, yesterday I sent you an analysis I did of the irreversible heating of a metal bar by means of heat supplied by an isothermal reservoir (comprising the surroundings). This is a case where the system is undergoing a irreversible path, while the surroundings are experiencing a reversible path. I showed how to precisely calculate the change in entropy for both the system and the surroundings from the solution to the transient heat conduction equation, and how the positive definiteness of the sum total of the entropy changes for the system and the surroundings comes about (for this problem). I have attached that analysis to this reply for others involved in this thread to examine. This provides a specific example of how the Cauchy inequality evolved, and why the sum of the entropy changes for the system and surroundings is always greater than zero for an irreversible process.

Chet

Thanks Chet! By the way I think you mean Clausius inequality and not Cauchy inequality

BiP

 

[Chestermiller]

Thanks Chet! By the way I think you mean Clausius inequality and not Cauchy inequality

BiP

Yes. You are right. I meant Clausius inequality. I realized that yesterday.

Chet

 

[Chestermiller]

Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?

This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ T_B

AM

Hi Andrew.

I don't have time to do it today, but in the next couple of days, I'm going to address all your questions to your satisfaction. Please bear with me. The reason that people struggle so much with these concepts is that the mathematics is done in such an ambiguous and imprecise way. I am also going to write up a document which explains things in a much more precise way mathematically. (That document may take another few days).

Chet

 

[DrDu]

I thought we calculated both the entropy change of the system and of the surrounding for the process in question? It can also be seen that the entropy increase of the surrounding is larger than the entropy decrease of the system, so that the total entropy clearly increases as should be for an irreversible process.
The change of entropy of the system is not problematic as both the start point and the end point are equilibrium states and entropy is a state function.
The entropy change of the surrounding we calculated as Q/T is only a lower bound. E.g. we could do some work on the system (stirring or electric heating) and this would produce additional heat transferred to the surrounding.

 

[Andrew Mason]

I thought we calculated both the entropy change of the system and of the surrounding for the process in question? It can also be seen that the entropy increase of the surrounding is larger than the entropy decrease of the system, so that the total entropy clearly increases as should be for an irreversible process.
The change of entropy of the system is not problematic as both the start point and the end point are equilibrium states and entropy is a state function.
The entropy change of the surrounding we calculated as Q/T is only a lower bound. E.g. we could do some work on the system (stirring or electric heating) and this would produce additional heat transferred to the surrounding.

The change in entropy of the system is ∫dQ/T over the reversible path A-B-C-D. That works out to (see post #19):

\Delta S_{sys} = m(C_{water}\ln{\frac{273}{263}} - L/273 - C_{ice}\ln{\frac{273}{263}})

The change in entropy of the surroundings is the actual heat flow to the surroundings divided by its temperature. If the actual process is freezing at 263K ie. from the supercooled water to ice at the same temperature, Q = mL (L being the latent heat of fusion) so:

\Delta S_{surr} = mL/263

There is no work done in this process. But if a process produces work (eg via a heat engine), the heat flow into the working substance will be greater than the heat flow out of the working substance but the temperatures at which the heat flows will always be such that |Q|/Th - |Q|/Tc ≤ 0

AM

 

[DrDu]

Why do you ignore the C_p dependent terms in Delta S_surr but not in Delta S_sys?

 

[Andrew Mason]

Why do you ignore the C_p dependent terms in Delta S_surr but not in Delta S_sys?

There is no change in the state of the surroundings except the heatflow to the surroundings from the system. We assume that its pressure, volume and temperature do not change. So change in entropy of the surroundings from A to D is determined by its temperature and the heatflow from the system to the surroundings. A reversible path from A to D is an isothermal process in which total heat flow Q flows to the surroundings where Q is the mass of the water x the heat of fusion for water. The change in entropy = ∫dQ/T over that path is just Q/T = mL/T.

To determine the change in entropy of the system, however, the reversible path from A-D is more complicated. It requires going through steps A-B-C-D. So you have to calculate ∫dQ/T for that path.

AM

 

[Darwin123]

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.

What they are probably saying is that the entropy being transferred into the surroundings has to be calculated by reversible processes. However, the surroundings can also gain entropy if entropy is created. The creation of entropy is an irreversible process.
I think the problem is with the phrase "heat transferred". This can mean one of two things, which you haven't specified.
"Heat transferred into the surroundings" can mean the changes in internal energy of the surroundings, This can include energy transferred by heat conduction or it can mean work.
I believe that this is what you are thinking. You are considering energy transferred into the environment by any and all means.
If the energy is transferred into the surroundings by work, then there are frictional forces involved. The energy transferred into the surroundings by friction create entropy. The creation of entropy is irreversible.
"Heat transferred into the surroundings" can mean energy transferred by "heat conduction". The energy transferred is then associated with entropy transfer, not the creation of entropy. Then that by nature is reversible.

 

[DrDu]

Andrew, what I meant is the following:
The heat of melting, L, is a function of temperature and in formulas for both entropies we should refer to the measured value at T=273K. You calculated the temperature dependence of L correctly already in #19, i.e.
<br /> L(263 K)=L(273 K)+(\Delta T)(C_{water}-C_{ice})<br />
where \Delta T=-10K.
Developing the logarithms, the C_p dependent terms cancel to lowest order in \Delta T and the total entropy change becomes \Delta S\approx -mL\Delta T/T.

 

[Andrew Mason]

Andrew, what I meant is the following:
The heat of melting, L, is a function of temperature and in formulas for both entropies we should refer to the measured value at T=273K. You calculated the temperature dependence of L correctly already in #19, i.e.
<br /> L(263 K)=L(273 K)+(\Delta T)(C_{water}-C_{ice})<br />
where \Delta T=-10K.

This is only true if the actual process by which the supercoolled water freezes is the reversible process A-B-C-D. If the water freezes at 263K the heat released to the surroundings would not be the same as in the reversible process.

Developing the logarithms, the C_p dependent terms cancel to lowest order in \Delta T and the total entropy change becomes \Delta S\approx -mL\Delta T/T.

To find the change in entropy of the surroundings one would have to measure the specific ΔH of supercooled water at 263K turning to ice at 263K and call that L₂₆₃ so that the change in entropy is mL₂₆₃/263. I don't know what the temperature dependence of L is but it has to be something that results in an increase in entropy overall.

You seem to be saying that the specific heat of fusion of supercooled water at temperature T is simply:

L(T) = L_{273} + \Delta T(C_{water} - C_{ice})

If that were true, you would be right but I don't know that it is.

AM

 

[Chestermiller]

Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?

This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ T_B

AM

In the present discussion of second law concepts, we have subdivided the world into two separate and distinct entities:

Entity 1 has been loosely referred to as "the system"

Entity 2 (which includes everything not included in Entity 1) has been loosely referred to as "the surroundings."

There is a "boundary" between the system and surroundings which does not permit mass to pass through, so that both the system and surroundings can be regarded as closed . However, the boundary does permit heat to transfer between the system and surroundings (across its interface), and the boundary can move and deform, so that work is done by the system on the surroundings, and vice versa.

The introduction to this picture of a third entity, say a Carnot device, muddies the waters a little, but only slightly. The third entity can be assimilated into either the system or the surroundings, and then we are back down to 2 entities again. In practice, it might be more convenient to assimilate the third entity into the surroundings, if the focus is more on the system.

In general, both the system and the surroundings can be undergoing irreversible processes, and the two regions will be coupled with one another at the boundary interface. What is the nature of this coupling? From our knowledge of transport phenomena and continuum mechanics, we know that certain parameters must be continuous across the interface: velocity vector, heat flux vector, temperature, and traction stress. The traction stress is equal to the stress tensor dotted with a unit normal vector to the interface. For an inviscid fluid, the traction stress is equal to the pressure times a unit normal, and the work being done by the system on the surroundings per unit area of the interface is equal to the pressure times the normal component of velocity.

Now back to our Entities. When most of us studied thermo, we learned it from a historical perspective. Early 2nd law pioneers worked with multiple entities, including heat reservoirs, Carnot engines, etc. As time progressed, and our understanding evolved, the concept of entropy developed, and the mathematics became more precise and refined. It was recognized that the number of entities could be reduced from many to just two, namely, the system and the surroundings. Then Clausius realized that he did not even need two entities to capture the essence of the 2nd law mathematically; he could do it all with just a single entity, the system. (He regarded the surroundings as just another system). Clausius' work formalized the 2nd law in a concise mathematical fashion.

Here is my attempt to paraphrase Clausius' thinking: I have a closed (constant mass) thermodynamic system that is enclosed within an interface boundary β. The system is initially in thermodynamic equilibrium state A, and I wish to bring it to thermodynamic equilibrium state B. I will consider all processes, both reversible and irreversible, that can accomplish this by applying heat transfer at various locations on β, and by moving and deforming β (i.e., doing work at the interface β). As I test these various processes, I will keep track of the local heat transfer rates I apply at the boundary of β (which can vary with position on β and time), and the corresponding temperatures at these locations along the boundary. Then I will calculate the ratios of the local heat transfer rates to the local temperatures, and sum these quantities over the entire boundary. I will then integrate this sum with respect to time, from time zero (corresponding to equilibrium state A), until the time that system has attained equilibrium state B. I will then compare the values I obtain for this integral for all the possible irreversible and reversible paths that I have imposed on the system. According to my understanding of the 2nd law, this integral will vary from path to path, but will never be greater than a certain limiting value. That limiting value is called the entropy change ΔS for the system. If I represents the value of the integral, then:

ΔS = I_max

The integral I attains this maximum value only along a reversible path. Otherwise I is less than I_max.

ΔS ≥ I

Mathematically, Clausius' representation of the second law constitutes a physical principle expressed using variational calculus.

Unfortunately, the mathematical from of the equation used by Clausius to capture the above powerful ideas is very imprecise and ambiguous, and has confused thermodynamics students over the years:

dS ≥ dQ/T

This equation does not specifically indicate where dQ and T are to be evaluated. For a system experiencing an irreversible path, the temperature (as well as pressure) will vary with spatial location and time within the system. So, what value of T should be used?
Also the heat flux within the system will be non-zero, and will vary with location and time. Our discussion above indicates that dQ and T should be evaluated exclusively at the boundary of the system. Our previous discussion also indicates that there can be multiple locations of heat flow in and corresponding temperatures at the boundary.

The continuum mechanics guys have formalized the mathematics of the Clausius inequality even more precisely by integrating the dot product of heat flux and unit normal divided by temperature over the entire boundary surface, and have used the divergence theorem to show mathematically why the change in entropy exceeds the integral by a positive definite amount. Google the key words continuum mechanics and Clausius inequality to see this.

 

[Studiot]

In the present discussion of second law concepts, we have subdivided the world into two separate and distinct entities:

Entity 1 has been loosely referred to as "the system"

Entity 2 (which includes everything not included in Entity 1) has been loosely referred to as "the surroundings."

There is a "boundary" between the system and surroundings which does not permit mass to pass through, so that both the system and surroundings can be regarded as closed . However, the boundary does permit heat to transfer between the system and surroundings (across its interface), and the boundary can move and deform, so that work is done by the system on the surroundings, and vice versa.

You obviously went to the same school as I did.

 

[DrDu]

You seem to be saying that the specific heat of fusion of supercooled water at temperature T is simply:

L(T) = L_{273} + \Delta T(C_{water} - C_{ice})

If that were true, you would be right but I don't know that it is.

AM

The heat of melting at 263K is equal to the enthalpy difference between states A and D.
As H is a state function it can be calculated along the path ABCD, in close analogy to the calculation of the entropy change of the system. Going from A to B and from C to D gives the terms \pm mC_{water/ice}ΔT, while the enthalpy change from B to C is -mL.

 

[Andrew Mason]

The heat of melting at 263K is equal to the enthalpy difference between states A and D.
As H is a state function it can be calculated along the path ABCD, in close analogy to the calculation of the entropy change of the system. Going from A to B and from C to D gives the terms \pm mC_{water/ice}ΔT, while the enthalpy change from B to C is -mL.

Ok. That makes sense. If there is constant pressure then L(T) = L(273) - (C_water - C_ice)(273-T).

AM