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Which heat capacity is used in the heat equation?

Physics Asked on January 15, 2021

The heat equation is often written as $frac{partial T}{partial t} = frac{kappa}{c} nabla^2T$ where $kappa$ is the thermal conductivity and $c$ is a heat capacity per volume.

I often see $c$ written as $c_P$ implying that it is the heat capacity (per unit volume) for a system held at constant pressure, but I was wondering if this was necessary of not?

I understand that in most ‘everyday’ experimental examples, pressure will be the variable that is held constant, and that for liquids and solids there isn’t much difference between $c_P$ and $c_V$ anyway. However, in theory, can the heat capacity in this equation be with whatever variable you want to hold constant (so could be either $c_P$ or $c_V$ depending on your situation?

I also found this similar question but I couldn’t find a definitnive answer to my question in their answers: In deriving the heat transfer equation, why do we use heat capacity at constant pressure?.

One Answer

Both options ($C_p$ or $C_v$) are approximations.

In the general situation, a non-rigid solid material will expand non-uniformly and therefore will have a time-dependent stress distribution, which stores internal strain energy.

The simplest assumption is that the object is not mechanically constrained and that the non-uniform internal strain energy is small. Therefore the internal pressure in the solid remains approximately constant and $C_p$ is the relevant thermal capacity.

For a fluid, the internal pressure is uniform (ignoring the weight of the fluid and any other inertia forces) and the correct choice is more obviously dependent on whether the volume is constrained or not.

If the object is constrained to have constant volume, $C_v$ would be a better approximation, since it takes account of the internal strain energy (assuming the internal stress field is uniform, of course).

Answered by alephzero on January 15, 2021

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