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When we apply a voltage across a metallic conductor, the current starts to flow almost instantaneously. But when a temperature difference is established across the same conductor, the flow of heat is much slower. It takes larger time for the heat to reach from one end to the other than the current. Why is this so?

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  • $\begingroup$ Current is a result of drift velocity of electron that has a net direction. Thermal motion on the other hand is completely random. $\endgroup$ Commented Aug 24, 2020 at 16:04
  • $\begingroup$ During current flow, all the electrons in the wire are simultaneously accelerated but during heat flow heat has to be gradually transferred from the hotter end to the colder end. $\endgroup$
    – SRS
    Commented Aug 24, 2020 at 16:09
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    $\begingroup$ @JonCuster I think you are confusing dT/dt (or C_p) with the speed of heat. The latter is roughly equal to the speed of the electrons responsible for transport in metals, i.e. the Fermi velocity, i.e. about 1% of light speed. People (the OP included) seem to confuse the speed of heat propagation with the time taken for a local temperature to rise. $\endgroup$ Commented Aug 24, 2020 at 19:30
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    $\begingroup$ @JonCuster Diamond is an odd case. Substances with few free electrons are usually thermal insulators. Metals are usually good thermal conductors thanks to the electronic contribution of the electrons to kappa. Ans so are heavily doped semiconductors. $\endgroup$ Commented Aug 24, 2020 at 20:12
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    $\begingroup$ @Jon Custer that's not true and is discussed in classic textbooks like Kittel or Ziman. In good metals the electronic thermal conductivity is significantly larger than the phonon contribution even at room temperature. You can even take a look at calculations comparing the two in A. Jain Phys. Rev. B 93, 081206(R) (2016). $\endgroup$
    – KF Gauss
    Commented Aug 25, 2020 at 9:09

3 Answers 3

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The current flows almost instantaneously because it is driven by an electric field which appears across the conductor almost instantaneously (near the speed of light). All electrons in the conductor are set into motion by a chain reaction. Collectively they all move through the conductor at what is called the drift velocity at the same time.

By contrast, heat transfer by conduction requires the transfer of energy by collisions between particles in the material that starts at the high temperature end of the conductor and progresses gradually to the low temperature end of the conductor. In the case of metals, the particles are primarily electrons.

Hope this helps.

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    $\begingroup$ In metals heat is by far due to the electrons (i.e. the value of $\kappa$ is almost exclusively due to the electrons). Phonons play a minor role. $\endgroup$ Commented Aug 24, 2020 at 19:24
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    $\begingroup$ OK, but my main point is it is still collisions that need to transfer the energy from the hot to cold side and that takes time, be it due to electrons or atoms and molecules. But I will revise. $\endgroup$
    – Bob D
    Commented Aug 24, 2020 at 20:17
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    $\begingroup$ And thanks for the feedback. $\endgroup$
    – Bob D
    Commented Aug 24, 2020 at 20:20
  • $\begingroup$ I agree with your answer now (upvoted too!). I have deleted mine, as mycents pointed out, yours is correct and mine was not. $\endgroup$ Commented Aug 25, 2020 at 7:39
  • $\begingroup$ @thermomagneticcondensedboson Thanks, I appreciate it. I have found myself doing the same thing. $\endgroup$
    – Bob D
    Commented Aug 25, 2020 at 21:46
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Let me first point out that heat is not carried by electrons. In fact, the temperature of the electron gas is much higher than that of the metal itself (thousands of Kelvins), but it makes a small fraction of the total thermal energy.

Electric current is a response to the electric field, which propagates with the speed of light through the metal. It however takes time before this field is screened by the mobile electrons, since it involves physical movement of matter.

Heat transfer occurs via the interactions between the lattice ions, i.e. purely due physical movement. The equivalent of the speed of light here is the velocity of the lattice phonons.

As a useful fact, it is worth mentioning that the drift velocity of the electrons is much smaller than the velocity of their thermal motion. While Drude model and the Newton's equation with viscous friction seem to result in the same formula for the conductance, they describe rather different situations.

Correction and update
My answer above has incorrectly stated that the heat conductance in metals is due to movement of the lattice ions. This is true for semiconductors/insulators, but not for metals, where the heat also is carried by the electrons. Credit to @thermomagneticcondensedboson for pointing it.

The main point however remains essentially the same: heat conductance is a diffusive process, with its diffusion constant determined by the thermal speed of the particles and the characteristic collision time. On the other hand, electric current is a response to the application of an electric field, which, when turned on, propagates through the metal with the speed of light. On the other hand, the distribution of electrons which results in screening of this field and establishing a steady current-carrying state is a diffusive process, similar to the heat conduction, with a similar time scale.

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    $\begingroup$ In metals heat is by far due to the electrons (i.e. the value of κ is almost exclusively due to the electrons). Phonons play a minor role. The electrons responsible for heat conduction have energies near the Fermi energy, just like those of electrical conduction. $\endgroup$ Commented Aug 24, 2020 at 19:25
  • $\begingroup$ @thermomagneticcondensedboson not sure how this fits in with the Debye model... $\endgroup$
    – Roger V.
    Commented Aug 24, 2020 at 20:03
  • $\begingroup$ Debye's model deals with the specific heat and how phonons contribute to it (people seem to confuse it with the speed of heat, including the OP), not with the thermal conductivity. See www2.physics.ox.ac.uk/sites/default/files/BandMT_10.pdf for instance. $\endgroup$ Commented Aug 24, 2020 at 20:10
  • $\begingroup$ This answer is wrong and contradicted by the Wiedemann-Franz law. See en.m.wikipedia.org/wiki/Wiedemann%E2%80%93Franz_law. $\endgroup$
    – my2cts
    Commented Aug 24, 2020 at 21:18
  • $\begingroup$ I've removed several unfriendly comments. Please remember the difference between factual criticism and personal attacks and stick to the former, everyone. $\endgroup$
    – ACuriousMind
    Commented Aug 25, 2020 at 19:32
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In metals the thermal conductivity is determined by the "free" electrons, that is the electrons with energy with a range around $E_F$ of order $kT$. In the diffusive regime the length over which the heat conduction takes place exceeds the mean free path, which is of order of 50-100 nanometers. The heat diffusion constant is determined by this number and the Fermi velocity. The result is that heat diffuses at a "speed" much smaller than light $c/n$. At ranges of the order of the mean free path the heat transport is ballistic and occurs much faster.

There is also an effect which does propagate at $c/n$ and that is the appearance of a thermoelectric voltage across the metal. See https://en.wikipedia.org/wiki/Seebeck_coefficient

https://www.semanticscholar.org/paper/Electron-mean-free-path-in-elemental-metals-Gall/5fde93380cf4f2b5564c042d34b5610be16d18c3/figure/1

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  • $\begingroup$ A comment: Only very few of the free electrons are within kT of the Fermi energy. It is only those few free electrons that are responsible for heat transport (it is not very clear from your answer because of the sentence before the first comma). $\endgroup$ Commented Aug 25, 2020 at 12:04
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    $\begingroup$ Suggestion to improve your answer: could you give an estimate on the speed of heat? You claim it is much smaller than c/n, but how much, roughly? How does it compare for example to the speed of sound in the metal? $\endgroup$ Commented Aug 25, 2020 at 12:05
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    $\begingroup$ @ thermomagnetic condensed boson That would require some more work. However you should be able to find the diffusion constant from the $v_F$ and the mean free path $\lambda$ in the table. You can also put it out as a new question. $\endgroup$
    – my2cts
    Commented Aug 25, 2020 at 13:53

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