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Why does specific heat capacity affect thermal diffusivity?

Physics Asked by the4naves on September 21, 2020

Sorry if this is a simple question, I did my research and couldn’t find much.

The equation for thermal diffusivity is: $a = frac{k}{pc_p}$
Where $k$ is thermal conductivity, $p$ is density, and $c_p$ is specific heat capacity.

The title pretty much says it all; Why does SHC have any effect on heat transfer? Say you have two objects, a and b, which for the purposes of this question are identical, only b has a higher SHC. According to the equation above, b would therefore have a lower thermal diffusivity.

Say you now applied some amount of heat (same for both) to one side of each object. Because b has a lower diffusivity, heat would transfer through it slower right? But to me, that doesn’t make any sense. As b has a higher SHC, it would have a lower average temperature. But why does it matter the temperature of the objects, wouldn’t the actual amount of energy contained matter more? It not like moving heat into an object with a high SHC just "destroys" that energy.

Have I just misunderstood a term? Missed something obvious?

I realize that (Assuming I’m wrong) there is without a doubt math that proves me wrong, though if you’re going to point me to known laws/equations, it’d help if you could also provide some sort of physical explanation if you have the time.

2 Answers

When you heat an object the way you say you are saying, the object does not heat uniformly, and the temperature gradient within the object is not uniform. The heat first begins getting stored in the side of the object you are applying the heat, penetrating only a small distance in. As time progresses, owing to the high temperature gradient near the side you are heating, and the lower temperature gradient further in, the heat penetrates further into the object. The shc determines how much can get stored in this thermal boundary layer, and, overall, the thermal diffusivity determines how far the penetration is at a given time.

Answered by Chet Miller on September 21, 2020

Let's discretize time and space so we can apply somewhat simpler equations than the spatiotemporal partial differential equation $alphadot{T}=T^{primeprime}$. Your example of a high-$c_P$ object and a low-$c_P$ object being heated is a great starting point; $c_P$ is of course the specific heat capacity, and let's assume the heating flux $q$ is applied in the $x$ direction.

Time step 1: In a small volume at the surface—call it element 1—the temperature (relative to the ambient temperature) increases from zero to $$Delta T=qADelta t/mc_P,$$ where $A$ is the element cross-sectional area moving inward, $Delta t$ is the time step duration, and m is the element mass.

(At this first time step, there's not yet any conduction inward from element 1 because this discretization scheme has us looking at the element temperatures at the beginning of the time step; the material starts out at a uniform relative temperature of zero.)

Clearly, the high-$c_P$ material heats up less.

Time step 2: Let's now analyze the inward heating by applying Fourier's law of conduction to element 1 and identical element 2 just inward: $$q_mathrm{cond}=kDelta T/L=qkADelta t/mc_PL,$$ where $k$ is the thermal conductivity and $L$ is the element depth moving inward.

We find that the inward heat flux is lower in the high-$c_P$ material, as all other variables are equal.

Additional time steps: Because of the lower incoming flux, the next block inward heats up less in the high-$c_P$ material, and the corresponding conduction through its inner side is subsequently lower. Analysis of $n$ subsequent time steps produces terms that scale with $c_P^{-n}$, emphasizing the importance of the specific heat. Equivalently, the heat wave has been effectively delayed and the speed of a given temperature change reduced in the high-$c_P$ material, as predicted by the lower thermal diffusivity.

So when you say, "It['s] not like moving heat into an object with a high SHC just 'destroys' that energy", you have a point—energy is never destroyed—but it's nevertheless true that the temperature increase in the material is suppressed, as is the propagating energy flux $q_mathrm{cond}$ that's driven by temperature gradients. In this sense, a higher-$c_P$ material both absorbs more energy and delays heat transfer.

Answered by Chemomechanics on September 21, 2020

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