I have learnt that temperature is literally measure of KE. But I don't see a fan which increase gas's temperature. I was wondering, why?

  • $\begingroup$ Try to learn more details. What part of KE is relevant for temperature? $\endgroup$ – nasu May 1 '17 at 14:33
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    $\begingroup$ Why do you assume, a blowing fan does not increase a gas's temperature? $\endgroup$ – dudakl May 1 '17 at 14:40
  • $\begingroup$ Because I don't see increase in temperature in daily life, practically. $\endgroup$ – Brett Leigh May 1 '17 at 14:43
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    $\begingroup$ Possible duplicate of Does wind raise air temperature? $\endgroup$ – John Rennie May 1 '17 at 15:34
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    $\begingroup$ The canonical duplicate is physics.stackexchange.com/q/96327 $\endgroup$ – dmckee May 1 '17 at 16:39

A moving fan does increase the temperature of air in the room. This is because, the moving fan increases the average kinetic energy of gas molecules in the room. Assuming 3 degrees of freedom of gases in the room, KE = $3kT/2$, where k is the Boltzmann constant [$1.38 * 10^{-23} J/K$]. The fact is that the fan does not impart enough KE for the change in temperature to be noticeable.


Temperature is not a measure of kinetic energy, or literally its average. Temperature is a measure of how much more "ordered" a system becomes when it loses energy, how much entropy it loses.1 Or put another way, how easy it to spontaneously transfer energy between two systems while increasing total disorder (entropy).

A system with high temperature can give off much energy and become only a little more ordered. This concept comes from (i) energy can flow spontaneously between systems, i.e., heat can be transferred (ii) the second law of thermodynamics: the entropy must always increase.

So one introduces the quantity $\beta = \frac{dS}{dU}$, the change in entropy per change in energy. Since energy is conserved, $dU_A = -dU_B$ for two systems in thermal contact. The second law allows the energy transfer from A to B if and only iff $\beta_A \le \beta_B$. We then take $T = 1/\beta$ so we get the usual scale that heat flows from higher to lower temperatures.

Anyway, with that concept of temperature we see that it is not the average kinetic energy that matters but the distribution of it. If we accelerate a container of some gas, the relative distribution of velocities does not change, so the temperature is not different. [In fancy speech: temperature is Galilei invariant.]

Another way of stating this is that uniformly accelerating a gas, i.e., all molecules in it equally, is not heating it, but doing work on it.

Of course, in practice, there will be collisions with, say, the container walls. These are not uniform, so there will be heat transfer and the temperature changes. In fact, if you have a fluid -- gas or liquid -- and different parts of the fluid have different average velocities, collisions between those parts have the same effect, driving the whole fluid to have the same average velocity. (This is called viscosity, sometimes described as internal friction.) If you stir a cup of coffee, it will eventually come to rest. .

So the answer to your question, to summarize, is that a fan creates some heat, but only because anything that adds energy to a system will eventually create some heat beacuse of the second law. It is not due to increasing the average kinetic energy, because temperature is not a measure of average kinetic energy.

1 It's not entirely accurate to equate disorder and entropy, the former isn't a technical term. For a given volume, the maximum entropy system is a black hole which arguably is extremely ordered. It is perhaps more informative to think of entropy as the negative of information: a black hole gives off the least possible information about its interior.


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