0
$\begingroup$

Temperature is related to the random motion of atoms and molecules in a substance. More, specifically, temperature is proportional to the average "translational" kinetic energy of molecular motion. Molecules may also rotate or vibrate, with associated rotational or vibrational kinetic energy-- but these motions are not translational and don't directly affect temperature.

How well a solid object conducts heat depends on the bonding within its atomic or molecular structure. Solids built of atoms that have one or more "loose" outer electrons, which are free to carry energy by collisions throughout the metal. They are excellent conductors of heat and electricity for this reason.

The above paragraphs are from the book "conceptual physics by Hewitt". It is said that when we measure temperature by classical mercury thermometer what we measure is actually the average translational kinetic energy of the molecules within the substance. Thinking about the above paragraphs actually makes sense to me.

My Questions: Are there any free electrons in a block of wood or some other electrical insulators? If not, how do we define their temperatures if we cannot talk about translational motions within the body? And by the same logic, can't we measure the temperature of wood by a thermometer?

$\endgroup$
  • $\begingroup$ The paragraph talks about the movement of atoms and molecules. Why are you asking about free electrons? Of course electrons (especially free ones) contribute a lot to heat transfer, which is why metals transfer heat quickly and blocks of wood slowly. Why would you not be able to measure the temperature of wood by a thermometer? You can just take a thermometer and stick it in the wood... $\endgroup$ – Adomas Baliuka Oct 12 '18 at 8:56
  • $\begingroup$ If the temperature is a measure of average translational kinetic energy, how does this relate to wood example? $\endgroup$ – physicsguy19 Oct 12 '18 at 8:59
  • $\begingroup$ There are a lot of atoms and molecules that the wood is made of. The average kinetic energy of these is proportional to temperature. In particular one is talking about the motion of atomic cores, since electrons have a very small mass. $\endgroup$ – Adomas Baliuka Oct 12 '18 at 9:04
  • $\begingroup$ @Adomas Baliuka, Then why the translational kinetic energy assertion? $\endgroup$ – physicsguy19 Oct 12 '18 at 10:13
  • 1
    $\begingroup$ If Hewitt writes that, it is only true in the classical approximation. I wrote a bit about that here: physics.stackexchange.com/questions/413376/… $\endgroup$ – Pieter Oct 12 '18 at 14:11
2
$\begingroup$

I believe that the first paragraph, which distinguishes the internal vibrations and rotations of molecules from the translational motion of their centres of mass, may have led to some confusion. For gases, it is convenient to separate these different kinds of motion, and actually they are all characterized by the temperature. Most practical methods of measuring the temperature of a gas, though, will effectively be measuring the translational temperature.

In an atomic solid, or liquid, the atoms are still undergoing translational motion. It is hindered: they don't move in a straight line over any significant distance, but it still counts as translation. We can think of the motions in a solid as being vibrational in character (composed of phonons), but the atoms are still translating back and forth. If the solid is composed of molecules, they will also be rotating and will have internal vibrational degrees of freedom: however, these cannot be completely separated from the translational motion of the molecular centres of mass. Again, everyday methods of measuring the temperature of a liquid or a solid mostly amount to establishing thermal equilibrium between it and a thermometer, and this involves atomic translation.

If you measure the temperature of wood, one way or another you are probably sampling the kinetic energy of translation of the atoms in the wood, even though they are strongly bound to other atoms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.