# Heuristics for specific heat capacities of solids

A didactic question publish in The Physics Teacher (http://tpt.aapt.org/resource/1/phteah/v41/i1/p8_s1) asks which will melt more ice: 100g of metal at 100C or 100g of wood at 100C. (The particular metal is not specified. I have paraphrased the question.)

The solution given is that the wood will melt more ice because it has a higher specific heat capacity, but no explanation is given for why wood has the higher heat capacity.

Is there a heuristic that would reliably give you this information? Clearly, the specific heat capacity depends on the number of degrees of freedom per gram of material. Does the wood have a higher heat capacity simply because it is composed mostly of lighter elements? Are there other important considerations for an order-of-magnitude estimate?

The wikipedia article is good (http://en.wikipedia.org/wiki/Heat_capacity) and has tables of specific heats, including both metals and wood. I didn't see something there that completely answers this question, though.

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It's not just order of magnitude--- Dulong Petit is quantitatively ok at room temperature for most solid materials. – Ron Maimon Jun 4 '12 at 6:22

The heuristic is that every atom gives you 3R per mole specific heat (3k per atom). The reason is the equipartition theorem, which is reasonably accurate for solids at room temperature. The amount of energy at temperature T is $.5kT$ for each quadratic term in the Hamiltonian. For an atom, there are three independent kinetic energy quadratic terms, and three independent quadratic potential energy terms (from oscillations relative to other atoms on the approximate springs that connect them). So you get 6 times $.5kT$, or 3kT per atom, or 3R per mole specific heat. This is the Dulong-Petit law.