# Will the heat flow of Joule heat be different, if the Joule heat is dissipated in a material that has a temperature gradient beforehand?

Let us assume one dimensional heat transfer, for example a finite length wire starting at point $0$ and ending at point $\ell$. If the current passes the wire, the Joule heat $I^{2}R$ will be generated and dissipated into the wire and its thermal surroundings. Had the wire had a constant temperature $T$, the half of the power $I^{2}R / 2$ will be passing the left end, the other half will be passing the right end.

Will the situation change if the non-zero temperature gradient $\nabla T$ is present before the Joule heating starts? I cannot grasp, which principle has "higher priority" in this case - be it either principle of dissipation of heat which should be considered "a random walk" or the second thermodynamic principle which states that on average more heat will flow from colder to hotter parts.

Motivation for this question are heat transfer equations in thermoelectricity. Thank you in advance for any answer of insightful comments!

## 2 Answers

As mentioned by Programmer, saying that if the wire temperature is constant then half of the heat will flow in either direction is incorrect. It really depends on the boundary conditions on either end of the wire (since it is of finite length). Assuming that the wire is at a spatially uniform temperature (not constant in time) and has same boundary conditions at both ends, the Joule heating will simply raise the temperature of the wire - uniformly. Heat cannot flow unless there is a axial temperature gradient in the wire. The temperature of the wire can then simply be modeled as:

\begin{equation} mc\frac{dT}{dt} = I^2 R \end{equation}

This assumes that the wire is not losing heat to its surrounding, otherwise a convection term needs to be added to the above equation. Next, say that the Joule heating starts with the wire already having a lengthwise temperature gradient. The flow of heat and development of temperature profile will be governed by:

\begin{equation} \frac{m}{l}c\frac{dT}{dt} = -kA\frac{dT}{dx} + I^2\frac{R}{l} \end{equation}

Again, the direction in which heat will flow really depends on your two boundary conditions.

Had the wire had a constant temperature T, the half of the power $I^2R/2$ will be passing the left end, the other half will be passing the right end.

I think this is a wrong statement. This is a common assumption used in a thermo-electric circuit theory to derive the equations. I would argue that this is valid in the case where the properties of the material are uniform, or assumed to be, and had nothing to do with the temperature gradient. After all, in a thermoelectric material, you would not have current flow to begin with if you did not have a temperature gradient, as that is how the seebeck effect works. So how can you now turn around and assume that the wire has a constant temperature.

Also, The joule heat is not something that is generated locally, it is generated over the entire volume. If you model the equations accurately, you can see it is far more complicated than you mention. I would encourage you to read, Callen, Thermodynamics and an Introduction to Thermostatistics ISBN-10: 0471862568

• To clarify, in the first paragraph I only assumed a case without thermal gradient, as sort of an introductory example. The reasoning that this is not the case in thermoelectrics by default is indeed the motivation to this question. – Jan Hirschner Nov 26 '13 at 17:48