In my undergrad physics classes, temperature was described as "the average kinetic energy of molecules". By simple, and probably naive, logic, one could then conclude that wind, which is basically just moving air, would raise the temperature of the air.

However, this seems too simple to be true. Does wind actually slightly raise air temperature?

I can see a couple potential arguments against this idea:

  1. It could be the case that the macroscopic wind and the microscopic kinetic energy of the air molecules don't really "interact", for lack of a better word.
  2. It could be the case that the macroscopic and the microscopic motions do "interact", but since the macroscopic wind is on such a large scale compared to the average kinetic energy of individual molecules, any system in which we use the average kinetic energy of the molecules is so small that the wind is effectively constant, and so it could be ignored.
  3. Regardless, I'm guessing that the temperature only depends on the relative motion of the air molecules, so the wind just isn't even relevant

Can anybody provide some more educated input?


3 Answers 3


Does wind actually slightly raise air temperature?

No and yes.

First, the no part of the answer. Temperature would not be a well-defined concept if that were the case. The temperature of a gas is the random motion of the atoms / molecules that comprise the gas. What you aren't taught is that that relation between temperature and kinetic energy is made from the perspective in which the total linear momentum of the gas is zero. There is no wind in a frame that is moving with the air mass.

However, consider a solid object such as an airplane or spacecraft flying through the air. An alternative point of view is that the object is stationary and that a wind is blowing past it. (All frames of reference are equally valid.) There is no flow right at the object/air boundary; as an aside, this is why the blades of fans get so dusty. The wind stops at the boundary, and this stoppage means that the kinetic energy due to motion is converted to thermal energy. This leads to an important concept in aerodynamics, stagnation temperature. Stagnation temperature can be a serious problem for hypersonic aircraft and reentering space vehicles. Things melt, or worse.

  • 1
    $\begingroup$ Interesting! I've always wondered why fan blades get dusty! $\endgroup$ Aug 1, 2016 at 4:57

It could be the case that the macroscopic and the microscopic motions do "interact", but since the macroscopic wind is on such a large scale compared to the average kinetic energy of individual molecules, any system in which we use the average kinetic energy of the molecules is so small that the wind is effectively constant, and so it could be ignored.

Yes, this is -in a certain way- the right answer.

Temperature is proportional to the average kinetic energy, thus it is proportional to the average molecular speed squared. The exact formula for an ideal gas is

$$\langle v \rangle = \sqrt{\frac{2RT}{M}}$$

where $M$ is the molecular mass (the molecular mass of air is $\simeq 29$ g/mol).

If you have a volume of "still" air at temperature $T=25$°C ($298$ K), the molecules in such a volume will be moving randomly at an average speed of 467 m/s.

If you superimpose to such a motion a translation in some direction, the average speed registered by a stationary thermometer will be higher. But how fast can such a motion be?

The highest wind speed ever registered on Earth was registered on Barrow Island during the passage of the Cyclone Olivia (1996), and it was 113 m/s. Average wind speeds are far below this extreme values; you can see from this table that the order of magnitude is 10 MPH, i.e. 4.5 m/s.

So the presence of "ordinary" wind can increase the average molecular speed in one spatial direction of $\simeq 1\%$. I specified "in one spatial direction" because wind usually blows in a certain specific direction.

Let's take a reference frame such that the direction of the wind is the $x$ direction. If the other two components of the molecular speed remain unchanged and the speed in the $x$ direction increases of $1\%$, and if we assume that the three components where initially on average equal, we can roughly estimate the variation in $\langle v \rangle$ to be

$$ \frac{v'-v}{v} = \frac{\sqrt{v_x'^2+v_y^2+v_z^2}-\sqrt{v_x^2+v_y^2+v_z^2}}{\sqrt{v_x^2+v_y^2+v_z^2}} = \frac{\sqrt{(1.01 u)^2 + 2 u^2}-\sqrt{3u^2}}{\sqrt{3u^2}} \simeq 0.0033 = 0.33\% $$

Since, from the previous equation,

$$T = \frac{M \langle v \rangle^2}{2R}$$

we see that the measured temperature will increase of $\simeq 0.67\%$ (i.e. $T'/T\simeq1.0067$), so if it was initially $25$°C $=298$K it will become $300$K = $27$°C.

So even in thi highly idealized situation the increase will not be very noticeable. In real life, we have at least two additional things to take into account:

  1. Air is not an ideal gas. The most important factor is that air usually contains a non negligible amount of water vapor, the presence of which can greatly change its thermodynamic properties

  2. In the atmosphere, wind is actually a large mass of air moving from regions of higher pressure to regions of lower pressure. Such a mass of air can be colder than the surrounding air, so the presence of wind can actually be linked to a decrease in temperature. The discussion becomes much more complicated in this cases, and a full understanding of it would require some knowledge of atmospheric science.

  • $\begingroup$ If air is initially at 25C, it is at 298K, 298 degrees above absolute zero. A 2% rise in temperature is about 6K. K degrees are the same size as C degrees, they just start at absolute zero instead of freezing point of water. So a 4.5 m/s breeze will raise the temperature of room temperature air about 6 C when the wind hits a stationary barrier and the extra kinetic energy is "thermalized," distributed among all the air molecules through collisions. $\endgroup$
    – mwengler
    May 26, 2017 at 2:40
  • $\begingroup$ @mwengler You are right, but I noticed I made another mistake. See updated answer. $\endgroup$
    – valerio
    May 26, 2017 at 8:47
  • $\begingroup$ Besides all that, wind blowing causes forced convection. People are warmer than air generally, so by making the air flow it will feel colder by taking heat away at a higher rate. $\endgroup$
    – JMac
    May 26, 2017 at 9:47
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    $\begingroup$ @JMac True, but we want to talk about the temperature measured by a thermometer. $\endgroup$
    – valerio
    May 26, 2017 at 10:52
  • $\begingroup$ @valerio92 Why do we only want to talk about thermometer temperature and not the sensible temperature? At the end of the day we are people, and we feel the heat transfer. The question didn't explicitly mention it, so it seemed worth noting the way heat transfer could affect human perception of the process. $\endgroup$
    – JMac
    May 26, 2017 at 11:14

If you gave every air molecule a momentum kick in one direction, then yeah $v_{rms}$ would be larger. However I don't think the temperature would actually be well-defined since the velocity distribution would no longer be a Boltzmann distribution. So yeah if someone referred to the temperature of windy air I would probably assume they meant in the air's "rest" frame.

Definitions for things like temperature tend to be a bit tricky, because they are usually only strictly speaking well defined for very particular types of systems. Usually I think you should just be more specific with what you want to know about a system.

Also keep in mind that the average kinetic energy of normal air is much larger than the velocity of wind.

  • $\begingroup$ Just to clarify, when you say "it's the speed of sound", is that just a rough approximation, or is that exact? If it's exact, then that must mean that the speed of sound changes in proportion to the average kinetic energy, which is interesting $\endgroup$ Aug 1, 2016 at 1:42
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    $\begingroup$ @BenSandeen: See the formula here...hyperphysics.phy-astr.gsu.edu/hbase/sound/souspe3.html#c1 $\endgroup$
    – DJohnM
    Aug 1, 2016 at 4:43
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    $\begingroup$ The average molecular speed is actually different from the speed of sound. $\endgroup$
    – valerio
    Aug 1, 2016 at 10:20

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