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When two obects $A$ and $B$ exchange heat irreversibly ($T_A\neq T_B$), the entropy of $A+B$ increases. My question is where: in $A$, in $B$, on the boundary? Is there some entropy exchange between $A$ and $B$? For the moment, my question is not really clear. I'm going to try to give a better example.

Consider a one-dimensional metal rod with temperature $T(x)$ along the rod. The total entropy is:

$$S = \int s(x)dx$$

with $s(x) = \rho C\log(T(x))$ and $\rho$ (mass per unit of length) and $C$ (specifc heat capacity) are constants.

The heat diffuses according to the heat equation. The entropy $s(x)$ will change at each point. Is it possible to separate, at each point $x$, the entropy being created there from the entropy being "exchanged" with the neighborhood? Can you write something like:

$$ds(x)=\delta s_e(x) + \delta s_c(x)$$

where $\delta s_e(x)$ would be the entropy being "exchanged with the neighbourhood" and $\delta s_c(x)$ would be the entropy being "created" at $x$? If so, what would they be?

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You start out with the transient heat conduction equation, and divide by the absolute temperature to obtain: $$\frac{\rho C}{T}\frac{\partial T}{\partial t}=\frac{\partial s}{\partial t}=\frac{k}{T}\frac{\partial^2 T}{\partial x^2}$$where $\frac{\partial s}{\partial t}$ is the rate of change of entropy per unit volume. You then use the product rule for differentiation to express the right hand side as $$\frac{k}{T}\frac{\partial^2 T}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{k}{T}\frac{\partial T}{\partial x}\right)+\frac{k}{T^2}\left(\frac{\partial T}{\partial x}\right)^2$$ So you obtain: $$\frac{\partial s}{\partial t}=\frac{\partial}{\partial x}\left(\frac{k}{T}\frac{\partial T}{\partial x}\right)+\frac{k}{T^2}\left(\frac{\partial T}{\partial x}\right)^2$$The first term on the right hand side is the rate of entropy transport per unit volume and the 2nd term, which is positive definite, is the rate of entropy generation per unit volume.

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  • $\begingroup$ I found a way to derrive the second term on my side with my own method to understand how it works. I guess you were not worried about its correctness, but I was. Many thanks. $\endgroup$ – Benoit Sep 11 at 17:21
  • $\begingroup$ Sounds interesting. $\endgroup$ – Chet Miller Sep 11 at 19:06
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I'm addding a few details to Chet's answer to make sure the second term can be interpreted as the rate of entropy generation at point $x$.

The entropy generation caused by $\delta Q$ flowing from $T_A$ to $T_B$ is:

$$\delta S_c = \delta Q \left(\frac{1}{T_B} - \frac{1}{T_A}\right)$$

Applied to the metal rod at point $x$:

$$\delta s_c = \delta Q \frac{\partial}{\partial x}\left(\frac{1}{T}\right)$$

The heat flow is proprtionnal to the heat gradient : $ \delta Q = -k\frac{\partial T}{\partial x}dt $. Finally :

$$\frac{\delta s_c}{dt} = -k\frac{\partial T}{\partial x}\frac{\partial}{\partial x}\left(\frac{1}{T}\right)dt = \frac{k}{T^2}\left(\frac{\partial T}{\partial x}\right)^2$$

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