I have a two part question:

Part 1: I understand thermal conductivity as

The measure of a material's ability to allow thermal energy transfer to it, from it and within it via conduction

My very first question is,

is this definition of thermal conductivity correct?

This is something that I haven't read word to word anywhere, but have expressed it in my words according to what I understand about thermal conductivity.

Part 2: Consider two completely identical rods, except for their material. Both the rods are initially at the same temperature and are brought in contact, at the same time, with an isothermal surface at a higher temperature. Let thermal conductivity $k$, of material A $>$ material B

enter image description here

According to what I understand about thermal conductivity, after any time interval $\Delta t$ more energy will be transfer to rod A than rod B since A has a higher k than B. I have assumed some arbitrary values 50units and 10units for A and B respectively.

By just a knowledge of the thermal conductivity of both the materials can I tell which bar would've gained more energy in time $\Delta t$, or conversely the amount of heat transfer from which bar would've been higher?

I think we can (but strongly think I'm wrong). Since A has a higher k, it will allow thermal energy to get transferred within it, or get conducted within it more efficiently, which means the layers in A are not storing energy much but are passing to subsequent layers, which makes me conclude that energy coming out of bar A will be also higher than coming out of B.

  • $\begingroup$ Simonson: Engineering Heat Transfer is a good book. $\endgroup$
    – Solar Mike
    Commented Nov 2, 2021 at 13:58
  • $\begingroup$ IMHO, you are leaving out heat capacitance which plays a factor in your question ("... which means the layers in A are not storing energy much but are passing to subsequent layers,". Also you are not defining the boundary conditions i.e. what is the temperature difference that drives the heat transfer via conductivity. This is related to your diffusion question. $\endgroup$
    – NMech
    Commented Nov 2, 2021 at 14:00
  • $\begingroup$ @SolarMike Thanks, I'll check it out. $\endgroup$ Commented Nov 2, 2021 at 14:13
  • $\begingroup$ @NMech I see. Any sources you can recommend where thermal conductivity is explained in detail? $\endgroup$ Commented Nov 2, 2021 at 14:14
  • $\begingroup$ Also, can I be sure about one thing that the amount of heat transfer $to$ rod A will be greater than B, i.e. 50>10 in the figure? $\endgroup$ Commented Nov 2, 2021 at 14:43

1 Answer 1


Part one, yes.

Part two depends.

Let's say your bars have the same thermal conductivity, $k$ but different volumetric heat capacity, $\rho C_p \ \text{where} \ C_p= \text{specific heat capacity} $ then the one with lower volumetric heat capacity will transfer the heat faster because of higher diffusibility.

$$α = k/ρcp$$

So it is not just the $k \ $that sets the rate of heat transfer, Its $\alpha.$

In your case, if the density* specific heat capacity of the second bar, B, is less than 5 times the first it will transfer heat faster. This is for a short amount of time before both ends of A and B reach the T final.

In my diagram, the slope is K conductivity and the differential volumes of the bars are differential sections of the rod that have to get hot before the heat can be transferred to the next cell. Height is assumed one. so at the transient stage of the heating of the rods, the one with higher diffusibility will get hot first, but once the two rods are hot the one with larger k will transfer more heat per dt.


  • $\begingroup$ So just because material A has a higher k than material B doesn't necessarily mean that rod A will take more energy from the heated body (the hatched one in the figure) than rod B. The 50, 10 conclusion that I've made based solely on thermal conductivity is wrong, right? $\endgroup$ Commented Nov 2, 2021 at 18:06
  • $\begingroup$ at the bigging the rod B will load up with heat faster but not necessarily take more energy, because we assumed it was of lower volumetric heat capcity. and also once both the bars are at equilibrium with the hot surface the bar A will conduct heat faster. $\endgroup$
    – kamran
    Commented Nov 2, 2021 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.