4
$\begingroup$

In a metal, how do free electrons conduct heat? Does the specific heat of a metal should be then related to the electrons and not the atom of the metal?

$\endgroup$
2
  • $\begingroup$ Wriggle! These ones wriggle here, which wiggles the wriggle of those ones there! $\endgroup$ Commented Nov 30, 2016 at 1:14
  • 1
    $\begingroup$ More seriously: heat is the energy in all randomized degrees of freedom, including the electrons. They're not exactly free in metals either, although a pure electron gas can be heated too. $\endgroup$ Commented Nov 30, 2016 at 1:16

3 Answers 3

1
$\begingroup$

I don't quite understand what answer you exactly want, since there are too many perspectives to deal with the problem.

In classical physics, you might use the degree of freedom to deal with this question, and you will find out the result fail to explain the experiment. And this problem comes from the "incorrect" Maxwell-Boltzman distribution when dealing with identical quantum particle like electron.

So here comes the problem, what type of model should I use will decide what conclusion you get.

Inside the metal, the conduction of heat is by phonon(you can think this as a quantum particle,if it makes you clear) while you can use free electron model dealing with the outer electron of the metal. And there is another type of model dealing with the outermost electron,called band structure. The three model above all have very different assumptions and priciples.

So you should first concentrate about where the electrons are,and then what type of model you wish. If you are only referring "free" electron, the free electron model is the only one you might want.

And if you are caring about specific heat inside the metal, phonon is the answer for you, while I think.it's not atom or electron will matter, but is another type of conduction~

Well, I'm a learner,too. So if there is something further hope others can tell~

$\endgroup$
1
  • $\begingroup$ The metal has a much higher ability to conduct heat than a non-electrically-conducting solid, so it seems to me that the free electrons do contribute to heat conduction in a metal. $\endgroup$ Commented Oct 26, 2016 at 21:29
1
$\begingroup$

Does the specific heat of a metal should be then related to the electrons and not the atom of the metal?

Specific heat (just like conductivity) depends on the molecular structure of the solid, and that depends on the atomic number, i.e. the shell of electrons. Whether the electrons are the ones conducting the heat or exchanges of vibrational excitations depends on the molecular structure. In metals the solid is like "one big molecule" as the outer shell electrons are shared in a band and cannot be identified with a specific molecule.

The specific heat of a solid depends on the vibrational levels of the molecules it consists of. (this helps understand differences in metals)

The mobility of electrons and changes in the energy level within the conduction band can transfer energy to the whole metal solid, i.e raise the vibrational level of the molecules of the metal fast, in contrast to non-conductors which have to transfer heat energy by vibrations only.

$\endgroup$
0
$\begingroup$

As the metal heats up at one location, eg, a laser pulse, the motions of the atoms in the crystal lattice increase. If these motions are coherent you have sound, but if they are random it is heat. Most motion starts as coherent, resulting in quantized sound (phonons) but they soon exchange momentum with the free electrons, and become randomized in a few nanoseconds.

The free electrons have in the newly hot region have more momentum than those in nearby, cooler regions, so they tend to disperse. This is essentially the Drude model; it works better with Fermi-Dirac statistics. It is semi-classical, but intuitive.

Thus phonons, electrons, and heat conduction. Different metals have different electron-phonon coupling strengths.

$\endgroup$

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.