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When I want to calculate the entropy change of a system with two bodies with finite difference in temperature in an isolated system in thermal contact, do I calculate the whole difference in entropy as change in entropy generation or as change of entropy transfer or both?

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  • $\begingroup$ If the system is in thermal contact, it's not isolated anymore. $\endgroup$ – kevin012 Jul 23 at 7:19
  • $\begingroup$ I meant to say that there are two bodies in thermal contact this two bodies are placed in an isolated system. $\endgroup$ – Anna Dapont Jul 23 at 8:22
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You are determining the combination of both. If the two bodies are identical, then, at the contact surface between them, the temperature will be the arithmetic average of the original temperatures throughout the equilibration. In this case, it is easy to resolve the separate contributions of generation and transfer, because the entropy transfer is just equal to the heat transferred divided by the (constant) contact surface temperature.

This is for @Chemomechanics

What our discussion has shown is that, although the changes in entropy of our two blocks between the initial and final thermodynamics states is totally process path-independent, the spatial distribution of the entropy generation and the amounts of entropy transferred to and from our two blocks is highly process-dependent.

Another example to consider is a situation where we have an insulating medium of very low thermal conductivity present between our two blocks. In this case, the heat transfer rate through the medium will be extremely slow, and, as a result, the two blocks themselves will each experience a quasi-static (reversible) change, featuring essentially no entropy generation. All the entropy generation for this process will take place within the insulating medium.

The entire entropy decrease from the hot block will be transferred to the insulating medium at its interface with the hot block. And the entire entropy increase for the cold block will be transferred from the insulating medium at its interface with the cold block. And, since all the entropy generation for the process takes place within the insulating medium, the entropy transferred to the cold block will be greater than the entropy received from the hot block by an amount equal to the entropy generated within the medium.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Jul 24 at 14:29
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    $\begingroup$ Nice additional example! $\endgroup$ – Chemomechanics Jul 25 at 1:16
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This is for @Alecksy Druggist

Suppose that the objects are perfect cubes of side S. We will focus on the left cube which is the hot one, and which is insulated on its left side, x = 0, and is in contact with the cold cube on its right side x = S at constant temperature $T_S$. The transient temperature variation within the cube is described by the transient heat conduction equation:

$$\rho C\frac{\partial T}{\partial t}=k\frac{\partial^2 T}{\partial x^2}$$ If we divide this equation by the absolute temperature T, we obtain: $$\rho C\frac{\partial \ln{T}}{\partial t}=\frac{k}{T}\frac{\partial^2 T}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{k}{T}\frac{\partial T}{\partial x}\right)+\frac{k}{T^2}\left(\frac{\partial T}{\partial x}\right)^2$$The first term on the right hand side represents the rate of entropy transport per unit volume and the 2nd term, which is positive definite, represents the rate of entropy generation per unit volume at location x. If we integrate this equation over the volume of the cube, we obtain: $$\frac{d[\int_0^S{\rho CS^2 \ln(T/T_{init})dx}]}{ dt}=\left(S^2\frac{k}{T}\frac{\partial T}{\partial x}\right)_{x=S}+\int_0^S{\frac{kS^2}{T^2}\left(\frac{\partial T}{\partial x}\right)^2dx}$$ The left hand side of the equation represents the rate of change of entropy of the cube; note that, by the end of the process, term in brackets is mathematically equal to the entropy change calculated from step 4 of my procedure. The first term on the right hand side represents the rate of entropy transfer from the cold cube to the hot cube at their boundary, and the 2nd term on the right hand side represents the overall rate of entropy generation within the cube. So, at final equilibrium, the change in entropy as calculated from step 4 in my procedure is equal to the total entropy transferred from the cold cube to the hot cube at their boundary (a negative number in this specific instance) plus the total amount of entropy generated within the cube during the equilibration.

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  • $\begingroup$ thank you very much, everything in the explanation is consistent with the first principles of nonequilibrium thermodynamics; local equilibrium, fluxes, entropy production... $\endgroup$ – Aleksey Druggist Aug 1 at 13:09
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This is for @Chemomechanics

What our discussion has shown is that, although the changes in entropy of our two blocks between the initial and final thermodynamics states is totally process path-independent, the spatial distribution of the entropy generation and the amounts of entropy transferred to and from our two blocks is highly process-dependent.

Another example to consider is a situation where we have an insulating medium of very low thermal conductivity present between our two blocks. In this case, the heat transfer rate through the medium will be extremely slow, and, as a result, the two blocks themselves will each experience a quasi-static (reversible) change, featuring essentially no entropy generation. All the entropy generation for this process will take place within the insulating medium.

The entire entropy decrease from the hot block will be transferred to the insulating medium at its interface with the hot block. And the entire entropy increase for the cold block will be transferred from the insulating medium at its interface with the cold block. And, since all the entropy generation for the process takes place within the insulating medium, the entropy transferred to the cold block will be greater than the entropy received from the hot block by an amount equal to the entropy generated within the medium.

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Thank you for your answer. Does this mean that in an reversible heat transfer the entropy lost from the slightly hotter object is the same as the increase of entropy of the slightly colder object and when there is a finite temperature there is an additional entropy increase due to the irreversibility of the process.

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  • $\begingroup$ I don't quite understand why you are phrasing the correct (in my opinion) answer as a question?) $\endgroup$ – Aleksey Druggist Jul 24 at 13:56
  • $\begingroup$ If I understand you correctly, the assessment of the two specific cases you allude to would be correct. If you want to get a better understanding of how to determine the change in entropy for an irreversible process, see the following link which provides a cookbook recipe and worked examples: physicsforums.com/insights/grandpa-chets-entropy-recipe $\endgroup$ – Chet Miller Jul 24 at 15:38
  • $\begingroup$ @ChetMiller , Step 4: For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state. This will be the change of entropy This is not entirely clear to me. There is an expression in nonequilibrium thermodynamics for the local production of entropy due to heat conduction, which must be integrated over time and space in order to obtain a complete change in entropy. Physics is not a set of instructions or recipes, there must be a clear physical rationale... $\endgroup$ – Aleksey Druggist Jul 24 at 17:27
  • $\begingroup$ @AlekseyDruggist I assert that, if you determine the amount of entropy generated within the system by the procedure you indicated and add to it the entropy transferred to the system from the surroundings through its interface with the surroundings, you will exactly match the value calculated from step 4. Not only do I assert this, but I can also prove it. $\endgroup$ – Chet Miller Jul 24 at 20:11
  • $\begingroup$ @Chet Miller, I would be very interested to follow up on such proof $\endgroup$ – Aleksey Druggist Jul 27 at 15:32

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