# What is the difference between thermal conductivity and thermal diffusivity?

Please excuse me, for any inability in my way to frame the question. I myself had a hard time making myself understand what was the doubt I really had :P. Also, I'm studying heat transfer at Undergraduate Level.

I first really need to make sure I understand the concept of thermal conductivity correctly or not. This is what I understand about thermal conductivity:

Thermal conductivity as defined in many books is a measure of a material's ability to conduct heat. A material with a high thermal conductivity will propagate heat faster within it. It tells about the speed at which thermal energy travels in a medium. So, suppose I have two identical rods- A and B made up of different materials such that thermal conductivity of A > thermal conductivity of B. Both the rods are initially at the same temperature and are brought in contact at the same time with identical heat source and heat sink. The heat transfer is one dimensional. With the knowledge that thermal conductivity of A is greater than thermal conductivity B, I can only conclude that heat travels faster in A than in B, from one end to another. No conclusions can be made about the heat taken from the source by rod A and rod B (like I cannot tell by just a knowledge of thermal conductivity that which one of rod A and rod B will take more energy from source in any given time). Earlier I used to think that the one with a higher thermal conductivity will take more energy from the source in any given time.

Are there any other corollaries that I can make, if I know a material has a higher thermal conductivity? Corollary like if it conducts heat well, it will store less (I don't think so but..).

This is an excerpt from the book I'm referring to on the topic thermal diffusivity: The sentence in the green - isn't that what thermal conductivity tells, about how well thermal energy propagates or diffuses into the medium?

The sentence in blue - just can't make sense.

Edit: Even though my question mentions 'rods', a more general explanation for difference between k and $$\alpha$$ will be appreciated, which is applicable to solids, liquids and gasses. As pointed out in one of the answers, solids have high thermal conductivities and high thermal diffusivities implying a solid conducts heat well and stores less. But this is not the case always a material can conduct less energy and store less also. Consider water and air for example, water has a higher $$k$$ than air, which means it conducts well but it does not store less (contrary to what solids do, they conduct well and store less), because water has a higher specific heat than air.

• Can you please tell me why you unaccepted my answer? Nov 7, 2021 at 19:58
• Actually I feel like there is still something missing. Comparing water and air for eg, even though water has a higher conductivity than air it has a lower diffusivity than air. I acknowledge your answer was for solids only, but I feel an explanation which is valid for every material would be worth knowing. So I started a bounty even. Nov 7, 2021 at 20:25
• Frankly, it seems to me I answered your question as it was, and still is, framed since it clearly implies an interest in solids ( re: your example of two rods). Now, as an afterthought, you're thinking about liquids and gases. At a minimum you should edit your question to clarify what it is now that you are asking. Nov 7, 2021 at 20:46
• Regarding your edit, it's not clear to me what explanation you are looking for regarding the difference between k and $\alpha$ in the case of liquids and gases. Nov 8, 2021 at 16:24
• It’s more important to understand the relevant applications of k and $\alpha$. If you are interested in the use of $\alpha$ in the Fourier heat equation I gave keep in mind that equation only applies to conduction. It doesn’t account for the heat transfer by convection that occurs in liquids and gases Nov 8, 2021 at 17:39

In essense,

• thermal conductivity, $$\kappa$$, is how well the material passes on heat (thermal energy), while

• thermal diffusivity, $$\alpha$$, is how well the material passes on a temperature change.

The diffusivity namely takes into account the heat capacity and density as well which are proporties that influence the temperature change caused by the transferred thermal energy.

• Can you please explain 'is how well the material passes on a temperature change', may be with an example? Thank You Nov 8, 2021 at 18:20
• @HarshitRajput A material with a high thermal conductivity, $\kappa$, will suck in and absorb a lot of heat each second. But it might not heat up that fast. The absorbed energy might be spread over large amounts of the material (when the mass concentration, i.e. the density, $\rho$, is high) and might be absorbed in a high heat capacity, $C$ (how much energy each "portion" of the material can "carry" before heating up"). These factors will dampen the temperature increase. If you divide $\rho$ and $C$ away, then you thus have a measure that takes these factors "out"... Nov 8, 2021 at 18:56
• @HarshitRajput ... Such a measure is what we call thermal diffusivity: $$\alpha=\frac{\kappa}{\rho\cdot C}.$$ With this measure we can compare different materials across densities and heat capacities. The higher the $\alpha$, the faster will the temperature rise throughout the material when a temperature difference is set up on either side. Nov 8, 2021 at 18:56

The rate of heat flow per unit area through a material depends on the temperature gradient in the material.The higher the temperature gradient, the higher the rate of heat flow per unit area. The constant of proportionality is the thermal conductivity k: $$q=-k\frac{dT}{dx}$$

The thermal diffusivity of a material determines how fast a temperature change at the boundary propagates into the material. If the material has a very low thermal diffusivity, the speed of the thermal "wave" propagating into the material is lower than if the diffusivity is high. This is because more energy can be stored near the boundary so the wave travels slower.

• Maybe you can help me here, but I have found that materials with low thermal diffusivity almost always have low thermal conductivity as well. So wouldn't the speed of the thermal "wave" propagating into the material also be lower if the thermal conductivity is lower and since they both move in the same direction, what is the difference? It strikes me that the main value of the thermal diffusivity is mainly in solving the heat equation to determine temperature vs time and depth under transient conditions. Nov 3, 2021 at 16:26
• @BobD Well, even though, in many cases, low thermal diffusivity goes along with low thermal conductivity, I doubt that this is always the case. The product of heat capacity and density has to play a role too, especially quantitatively. With regard to the importance of diffusivity to transient heat conduction, I would say that it is also extremely important in laminar flow heat transfer situations, particularly for steady state flows. Nov 3, 2021 at 19:55
• Actually the data I have shows they almost always do go along. Regarding the volumetric heat capacity $\rho c$, the denominator of thermal diffusivity, it doesn't strongly correlate with either. Though materials like metals have lower specific heat they have higher density whereas materials such as plastics have higher specific heat but lower densities resulting in a range of volumetric heat capacities much narrower than the range in thermal conductivities for the same material. Nov 3, 2021 at 22:08
• For example, the volumetric heat capacity of aluminum is about 1.5 times that of general purpose polystyrene, whereas the thermal conductivity is about 1700 times greater. This suggests to me that thermal conductivity is the main driver of thermal diffusivity. If I can figure out how, I will include my table with my answer. Nov 3, 2021 at 22:08
• @BobD But to solve a problem quantitatively (for temperature profiles), you still need to use the specific value of the thermal diffusivity for your material. Nov 4, 2021 at 12:56

Are there any other corollaries that I can make, if I know a material has a higher thermal conductivity?

For one thing, if you know a solid material has higher thermal conductivity chances are it will also have higher thermal diffusivity. I base this on having made a table of various solids (metals, plastics, glass, etc.) sorted by thermal conductivity (highest to lowest) and found that they were almost all sorted by thermal diffusivity as well (highest to lowest).

Corollary like if it conducts heat well, it will store less (I don't think so but..).

Actually, a solid material that conducts heat well (high $$k$$) does store less per unit mass because the specific heats of high conductivity materials, like metal, are low. So metals will store much less than non metals (e.g. plastics) per unit mass because the specific heat of non metals is higher then metals.

But parts have volume. So if you are comparing two identical parts of different material, you need to look at their volumetric heat capacities, $$\rho c$$ in the denominator of the thermal diffusivity, not their specific heats. Since the density $$\rho$$ of metals is much greater than plastics, the volumetric heat capacity is often not very much different between metals and non metals. For example, the volumetric heat capacity of steel and Teflon is about the same.

The sentence in the green - isn't that what thermal conductivity tells, about how well thermal energy propagates or diffuses into the medium?

Since, as I stated above, solid materials with high thermal conductivity generally also have higher thermal diffusivity, it stands to reason that the diffusion of heat in solid materials should be higher with higher thermal conductivity. But you can't determine what it actually is based only on thermal conductivity because $$k\ne k/\rho c$$. In order to determine how temperatures in a solid material varies as a function of time $$t$$ and depth $$x$$ under transient conditions, you need to obtain the solution to the following heat equation

$$\frac{\partial T(x,t)}{\partial t}=\alpha\frac{\partial^2 T(x,t)}{\partial x^2}$$

For the given initial and boundary conditions, where $$\alpha=k/\rho c$$ = thermal diffusivity.

hope this helps.