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Suppose, we have a source at high temperature $T_\mathrm h$ and sink at temperature lower temperature $T_\mathrm l$.

If $Q_\mathrm h$ amount of heat flow from source to sink, then change in entropy of source is $\frac{-Q_\mathrm h}{T_\mathrm h}$ and change in entropy of sink is $\frac{+Q_\mathrm h}{T_\mathrm l}$. I saw a similar question here: https://physics.stackexchange.com/questions/358142/entropy-generation-during-heat-transfer-processes.

And while I understand that entropy is being generated in the partition, my question is how does heat conduction across a finite temperature gradient generates entropy

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  • $\begingroup$ Not sure what you mean by part b) $\endgroup$
    – Poutnik
    Jan 8, 2023 at 8:48
  • $\begingroup$ I meant that source and sink separated by some distance $\endgroup$
    – Natasha J
    Jan 8, 2023 at 9:21
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    $\begingroup$ Therefore thermally insulated(not considering radiation)? In such a case, entropy of reservoirs and whole system is constant, as there is no heat flow. $\endgroup$
    – Poutnik
    Jan 8, 2023 at 9:29
  • $\begingroup$ How are they thermally insulated? Won't heat still flow from source to sink? Thanks $\endgroup$
    – Natasha J
    Jan 8, 2023 at 9:37
  • $\begingroup$ Describe it better then. Confusing description leads to confused responses. How b) differs from having partition allowing heat flow between reservoirs? $\endgroup$
    – Poutnik
    Jan 8, 2023 at 9:45

1 Answer 1

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For steady state heat conduction in the slab between the source and sink, we have $$\frac{d}{d x}\left(k\frac{d T}{d x}\right)=0$$where k is the thermal conductivity. If we divide this equation by absolute temperature T, we obtain$$\frac{1}{T}\frac{d}{d x}\left(k\frac{d T}{d x}\right)=0$$Next, making use of the product rule for differentiation, we have:$$\frac{1}{T}\frac{d}{d x}\left(k\frac{d T}{d x}\right)=\frac{d}{d x}\left(\frac{k}{T}\frac{dT}{dx}\right)+\frac{k}{T^2}\left(\frac{dT}{dx}\right)^2=0$$Assuming that the source is at x = 0 and the sink is at x = L, if we integrate this equation between x = 0 and x + L, we obtain: $$-\frac{q}{T_C}+\frac{q}{T_H}+\int_0^L{\frac{q^2}{k}dx}=0\tag{1}$$ where the heat flux q is given by $$q=-k\frac{dT}{dx}$$If we rearrange Eqn. 1 slightly, we obtain:$$\frac{q}{T_C}=\frac{q}{T_H}+\int_0^L{\frac{q^2}{k}dx}\tag{2}$$The quantity $\frac{q}{T_H}$ is the rate at which entropy enters the slab per unit area, and the quantity $\frac{q}{T_C}$ is the rate at which entropy exits the slab per unit area of slab. Eqn. 2 tells us that the rate at which entropy exits the slab is equal to the rate at which entropy exits the slab plus the rate at which entropy is generated within the slab. According to Eqn.2, the rate of entropy generation within the slab per unit volume is $q^/k$. This entropy generation rate is, of course, positive definite.

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  • $\begingroup$ Thank you so much. I have one more question though, you showed above how entropy generation rate is positive definite, may I ask how is entropy being generated though? I mean, it is being generated this much is clear, but what physical processes are happening in the slab which generates entropy? I hope I'm being clear $\endgroup$
    – Natasha J
    Jan 8, 2023 at 13:08
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    $\begingroup$ The physical process is the irreversible transport of heat by conduction at finite rate. $\endgroup$ Jan 8, 2023 at 13:58
  • $\begingroup$ Thanks. I do not know much about transport of heat by conduction at finite rate, so I'll look it up now. Also, is there a thing such as reversible transport of heat? Or perhaps heat transferred through infinitesimal difference of temperature is the same thing as reversible transport of heat? $\endgroup$
    – Natasha J
    Jan 8, 2023 at 14:07
  • $\begingroup$ Also, is there a book which will help me understand how heat transfer generate entropy intuitively. Thank you $\endgroup$
    – Natasha J
    Jan 8, 2023 at 14:09
  • $\begingroup$ See Transport Phenomena, by Bird, Stewart, and Lightfoot, Chapter 11, Homework problem 11.D.1 $\endgroup$ Jan 8, 2023 at 14:46

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