# What processes generate entropy as heat flows across temperature gradient

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

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

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.