I'm trying to apply a formula to calculate the thermal conductivity of a certain material.

$\lambda=\frac{\dot q_1}{4\pi\cdot \frac{\Delta T}{\Delta log(t)}}$

where $\lambda$ is thermal conductivity, and the fraction in the denominator is the linear slope of a plot that I have.

My setup is simply a hot wire embedded in said material, heating up the material around it via resistive/joule heating.

The one variable I'm not sure about, $\dot q_1$, from varying sources, is described as "the specific heat output of the linear heat source (emitted heat energy per unit time and unit length)" or also described as the "heat flux per unit length of the linear source".

Are these 2 terms the same? If so, how do I go about calculating this heat flux per unit length? I know the innate resistance of the wire, the current flowing through it and the voltage, as well as the temperature of the wire.

  • $\begingroup$ What is the exact experiment? Where are the temperatures measured? Are you sure that isn't a $\Delta ln(r) $ in the denominator? $\endgroup$ Jul 23, 2022 at 9:19
  • $\begingroup$ How accurate do you need your answer to be? These kinds of measurements can be tricky. If you look at the 3omega method or Angstrom method you will see that people will periodically heat the wire. The dimensions compared to the thermal wavelength can also be important. If it is a long wire in a cylinder then that helps. But keep in mind the resistance of the wire changes with temp. And if temperature differences are large the material will too. $\endgroup$
    – UVphoton
    Jul 23, 2022 at 17:50
  • $\begingroup$ The above comment aside, yes I think the two are a the same. Joule heating is VI. And that would be The heat flux for the whole length of wire. If the problem is one dimensional like the wire a cylinder then you can think about the problem in terms of series resistances. $\endgroup$
    – UVphoton
    Jul 23, 2022 at 17:57
  • $\begingroup$ Too be clear, thermal resistances in series perpendicular to the wire. $\endgroup$
    – UVphoton
    Jul 23, 2022 at 18:08
  • $\begingroup$ @ChetMiller The experiment consists of surrounding a wire in a uniform, cylindrical bed of particles. I heat up the wire via current flowing through it, and I try to measure the temperature of the surrounding particle bed to calculate thermal conductivity of the material that surrounds the wire. The temperature measured is simply the outer edge of the bed, at a constant r. As for the denominator, here is one source I am using showing how that formula is derived. tec-science.com/thermodynamics/heat/… $\endgroup$ Jul 24, 2022 at 9:25

1 Answer 1


I would not determine $\lambda$ the way they say. I would make a graph on semi-logarithmic paper of temperature T on the linear axis as a function of time t on the logarithmic scale axis, and determine the slope of the straight line section of the variation: $$slope=\frac{\dot{q}}{4\pi \lambda}$$That way you could use temperature data taken at many points in time, and also be certain that you are considering only the linear portion of the behavior. Any decent plotting software can take T vs t data and convert the t axis to logarithmic with a single click. It can also easily fit the data to a semilog relationship so that the slope is immediately available.

  • $\begingroup$ I have been fiddling with this equation and my experimental apparatus, but one thing stuck out to me. Doesn't this equation imply that, because q and 4pi are constant, the slope and the thermal conductivity are inversely proportional? As in, the higher the slope, the lower the thermal conductivity? If so, how does this make sense? The rise part of the m=rise/run is temperature, so the faster temperature rise-> higher slope. In that case, if the material heats up faster, shouldn't the thermal conductivity be higher? In short, shouldn't slope be directly proportional to slope, not inversely? $\endgroup$ Aug 8, 2022 at 1:05
  • $\begingroup$ No. The higher lambda, the more of the heat is transported beyond r, leaving less to produce heating at r. $\endgroup$ Aug 8, 2022 at 14:32

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