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$$ k \frac{dT^2}{d^2 x} +heat generation =\rho c \frac{dT} {dt} $$

In steady state and in the absence of heat source, the equation becomes: $$ k \frac{dT^2}{d^2 x} =\ 0 $$

If the laplacian of temperature is zero, it means that the average of temperature at a certain point is equal to the average temperature at its neighboring points. So, this means that there's no heat transfer, and temperature is constant everywhere (no time change and no heat source), so it only means that gradient of temperature is zero as well, right?

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In a 1D problem, zero Laplacian in a uniform, linear region means that temperature is a linear function of position in this region. Even if no heat is generated within the region, there can still be a temperature gradient, and consequently, heat transfer through the region. This can happen when one end of this region is at a higher temperature than the other.

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  • $\begingroup$ what is heat generation? What do you consider the $TS$ term in the formula $U=TS-pV+....$ ? In a temperature gradient you do generate entropy at some temperature, so what was $\Delta S$ being transported and dropped to a lower temperature is now something else, In fact bigger. $\endgroup$
    – hyportnex
    Mar 10, 2022 at 13:08
  • $\begingroup$ @hyportnex I'm not sure what you mean. Are you equating entropy generation with heat generation? There is no heat generation because the heat source term is zero in the heat equation, meaning whatever heat enters the region exits out the other end. $\endgroup$
    – Puk
    Mar 10, 2022 at 23:46

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