# Entropy change due to heat transfer between sources

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• cianfa72
Sorry, not sure to get your point. Do you mean that the thermodynamic state of each of the two sources does not change ? Hence the entropy of each source actually does not change.No. I'm saying that the entropy of the hot reservoir decreases by an amount ##Q/T_H ## because ##Q/T_H## amount of entropy is transferred out of the hot reservoir; and the entropy of the cold reservoir increases by an amount ##Q/T_C## because ##Q/T_C## amount of entropy is transferred into the cold reservoir. So there is more entropy entering the cold reservoir than leaving the hot reservoir; where was the extra entropy

#### cianfa72

Hi,

starting for this thread Question about entropy change in a reservoir consider the spontaneous irreversible process of heat transfer from a source ##A## at temperature ##T_h## to another source ##B## at temperature ##T_c## (##T_h > T_c##). The thermodynamic 'system' is defined from sources ##A## + ##B##. Since the process is not reversible, we can pick a reversible transformation between initial and final states to work out the entropy change between them (note that the entropy change is between the two states and it is not a property of any transformation/process).

For instance we can introduce two other sources ##C##, ##D## at ##T_h## and ##T_c## respectively. Now let's take two isothermals: the first transfers heat ##Q## from source ##A## to ##C## both at temperature ##T_h## , the second transfers the same heat ##Q## from ##D## to ##B## both at temperature ##T_c##.

At the end of this process the 'system' as defined above is actually in the same state we get from the spontaneous irreversible process.

If the above is correct, I was thinking about the possibility of employ a Carnot cycle to 'implement' a reversible transformation to calculate the entropy change between initial and final 'system' states.

Does it actually make sense ? Thank you.

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cianfa72 said:
the first transfers heat ##Q## from source ##A## to ##C## both at temperature ##T_h##
that's not going to happen !

BvU said:
that's not going to happen !
Why not ? If you have 2 sources at the same temperature, in principle you can transfer heat from one source to the other (at least from a theoretical point of view in order to use a reversibile process to calculate the entropy change).

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And how would you do that (in theory) ?

BvU said:
And how would you do that (in theory) ?
Ah ok, nevertheless the point is that the two sources (##A## and ##C## or ##B## and ##D##) have the same temperature (##T_h## or ##T_c## respectively). By definition a source is able to transfer heat (to or from) without changing its temperature.

On the other hand, which could be a reversible transformation useful for calculating the change in entropy between the initial and final state ? Thank you.

Remember: the 'system' is composed of source A and source B.

I agree with your approach. However, if you do it with a Carnot cycle, the amounts of heat transferred are not equal.

Are you assuming that you have two ideal constant temperature reservoirs transferring heat Q from the hot reservoir to the cold reservoir? If so, in your irreversible process, where there is a net entropy increase, where do you think the entropy generation takes place?

Chestermiller said:
I agree with your approach. However, if you do it with a Carnot cycle, the amounts of heat transferred are not equal.
Right, I think it is actually not feasible with a Carnot cycle.

Chestermiller said:
Are you assuming that you have two ideal constant temperature reservoirs transferring heat Q from the hot reservoir to the cold reservoir? If so, in your irreversible process, where there is a net entropy increase, where do you think the entropy generation takes place?
Yes, that is my assumption. Since the 'external ambient' is not involved in the irreversibile spontaneus process, the entropy is generated because for the same heat Q it results ##Q/T_c > Q/T_h##.

cianfa72 said:
Right, I think it is actually not feasible with a Carnot cycle.

Yes, that is my assumption. Since the 'external ambient' is not involved in the irreversibile spontaneus process, the entropy is generated because for the same heat Q it results ##Q/T_c > Q/T_h##.
If they are ideal constant temperature reservoirs, there is no entropy generated within either reservoir during the process. So, spatially, where do you think the irreversible entropy generation takes place?

Chestermiller said:
If they are ideal constant temperature reservoirs, there is no entropy generated within either reservoir during the process.
Sorry, not sure to get your point. Do you mean that the thermodynamic state of each of the two sources does not change ? Hence the entropy of each source actually does not change.

cianfa72 said:
Sorry, not sure to get your point. Do you mean that the thermodynamic state of each of the two sources does not change ? Hence the entropy of each source actually does not change.
No. I'm saying that the entropy of the hot reservoir decreases by an amount ##Q/T_H ## because ##Q/T_H## amount of entropy is transferred out of the hot reservoir; and the entropy of the cold reservoir increases by an amount ##Q/T_C## because ##Q/T_C## amount of entropy is transferred into the cold reservoir. So there is more entropy entering the cold reservoir than leaving the hot reservoir; where was the extra entropy (due to irreversibility) generated?

Chestermiller said:
So there is more entropy entering the cold reservoir than leaving the hot reservoir; where was the extra entropy (due to irreversibility) generated?
Ah ok, you're talking about generated entropy not the transferred one (from the hot reservoir to the cold reservoir).

I believe this 'extra' entropy is generated at the interface between the hot and cold reservoirs since there is there a gradient of temperature. I don't know if the aim of your question is to point out where the entropy is generated from a microscopic point of view.

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Chestermiller
cianfa72 said:
Ah ok, you're talking about generated entropy not the transferred one (from the hot reservoir to the cold reservoir).

I believe this 'extra' entropy is generated at the interface between the hot and cold reservoirs since there is there a gradient of temperature. I don't know if the aim of your question is to point out where the entropy is generated from a microscopic point of view.
Excellent. This is exactly what my aim was. Very perceptive (and correct).

I have the following doubt: the extra generated entropy can be associated to the new state of the cold source (source B) alone or it is actually a property of the new state of the overall system (i.e. source A + source B) ?

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cianfa72 said:
I have the following doubt: the extra generated entropy can be associated to the new state of the cold source (source B) alone or it is actually a property of the new state of the overall system (i.e. source A + source B) ?
All the generated entropy ends up in the cold sink B.
These two constant temperature reservoirs are just an idealization. In the real world, there would be a thin thermal boundary layer adjacent to the interface within both A and B. The extra entropy would be generated within each of these two thermal boundary layers. In A, the temperature within the thin thermal boundary layer would vary from the bulk temperature ##T_A## at the outer edge of the BL to ##T_I## at the interface. In B, the temperature within the thin thermal boundary layer would vary from the bulk temperature ##T_B## at the outer edge of the BL to ##T_I## at the interface. So the temperature distribution would be continuous.

We can simulate the BL effects by placing a thin low-thermal-conductivity layer of insulating material (having negligible heat capacity and mass) between the two reservoirs. The heat transfer rate would be very slow and essentially take place reversibly for each of the two reservoirs. All the entropy generation would occur within the thin layer, which would feature a high temperature gradient but slow rate of heat flow. In this idealization, the entropy transferred reversibly out of A would add to the entropy generated within the layer, and all this entropy would be transferred reversibly to B.

cianfa72
Chestermiller said:
In this idealization, the entropy transferred reversibly out of A would add to the entropy generated within the layer, and all this entropy would be transferred reversibly to B.
So basically by adding the thermal boundary layer (BL), we can break down the irreversibile process of heat Q tranfer from reservoir A to reservoir B in 3 processes: the reversible processes are the tranfer of heat Q from A to BL (1) and from BL to B (2) respectively. The irreversibile process, on the other hand, takes place inside the BL (3) and entropy is generated which is then transferred reversibly by (2) to reservoir B.

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cianfa72 said:
So basically by adding the thermal boundary layer (BL), we can break down the irreversibile process of heat Q tranfer from reservoir A to reservoir B in 3 processes: the reversible processes are the tranfer of heat Q from A to BL (1) and from BL to B (2) respectively. The irreversibile process, on the other hand, takes place inside the BL (3) and entropy is generated which is then transferred reversibly by (2) to reservoir B.
Yes. Exactly.