TLDR: In the context of pans/cooking, is high density and high heat capacity basically interchangeable?

If I have two pans such that it takes approximately the same energy to heat them to the same temperature, one with a high heat capacity but a low density and one with a low heat capacity but a high density, which are otherwise similar, how will this affect the results when using the pan?

1 Answer 1


Sort of.

Specific heat capacity has units J kg^-1 K^-1; that is joules per Kelvin per kilogram, and is defined per material; (a given kind of) steel has one specific heat capacity, water has another, and so on, regardless of how much of that material you have. Heat capacity (note, no 'specific') is what you get when you have accounted for that mass; water, the material, has a specific heat capacity of a little over 4 kJ kg^-1 K^-1; one kilogram of water has a heat capacity of a little over 4 kJ K^-1, two kilograms has a heat capacity of a bit over 8 kJ K^-1, and so on.

There's also an issue of volume versus mass; for example, aluminium has a specific heat capacity about twice as great as that of steel, but is about one third as dense; as such, an aluminium pan of the same size (i.e. volume/displacement) as a steel pan will have about two thirds the heat capacity; the better SHC is not enough to make up for the lower density.

Finally, none of this information says anything about conductivity, which is the other important property re: heat transfer in a pan.

In your example, where you have two pans with similar heat capacities, one made of a high-SHC, low-density material and one made of a low-SHC, high-density material, I don't know that the question of how they'll perform differently can be adequately answered without knowing the conductivities of the two materials; as it stands, there are only so many materials and constructions that we generally make cookware out of, as such, it's probably most efficient to look at the properties of specific techniques over hypotheticals like this.

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.