When we put hands on A/C it gives cold winds. These winds have high kinetic energy but low temperature. How ? *don't confuse with A/C being heat pump , just an example, take antarctic blizzards. I can't understand the paradox of low temperature winds. Temperature is something defined by kinetic energy
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21$\begingroup$ I think you misunderstand several things. First, it's not the total kinetic energy that relates to a temperature, but the mean kinetic energy in the bulk flow rest frame (i.e., in the frame where your "breeze" is not moving). It relates to the randomized velocities about the mean. Second, "cold" is a relative term. If air at $60^{\circ}$ F blows past you, it will feel "cold," but if that air is comparable to the ambient air and/or your body temperature, it will feel warm. $\endgroup$– honeste_vivereCommented May 30, 2017 at 14:24
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1$\begingroup$ Yes I do know , boltzman maxwell curve. Also wether I say hot or cold , I can still rephrase the question in whichever way you might like. Of course I am talking relative to STP conditions. $\endgroup$– user37060Commented May 30, 2017 at 14:27
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$\begingroup$ @honeste_vivere there can be no denial that average AC wind kinetic energy is lower than STP air mean $\endgroup$– user37060Commented May 30, 2017 at 14:28
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11$\begingroup$ The core misunderstanding in the question is addressed in Why am I not burned by a strong wind?. $\endgroup$– dmckee --- ex-moderator kittenCommented May 30, 2017 at 18:30
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3$\begingroup$ Suppose you are in a snowball fight. Your theory is that a snowball delivered to your target at 10 m/s is much colder than an identical snowball delivered to your target at 20 m/s, because the one has much lower kinetic energy than the other? $\endgroup$– Eric LippertCommented May 30, 2017 at 22:15
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4 Answers
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The average speed of an air molecule can be approximated by the following equation, which is exact only in the case of an ideal gas:
$$\langle v \rangle = \sqrt{\frac{2RT}{M}}$$
This means at $25$°C ($298$ K) air molecules will be moving randomly at an average speed of $\simeq 467$ m/s.
Let's say that the AC cools the air at $15$°C ($288$ K) before blowing it out into the room. From the above formula, the average molecular speed will then be $\simeq 459$ m/s.
When the AC blows out the air, it does so at $0.1-0.3$ m/s (1). This means that, in the worst case, a motion of $0.3$ m/s is superimposed to an average motion which is at around $460$ m/s, more than a thousand times faster.
You can then see how the movement of the air mass as a whole is negligible: what matters is the average molecular speed in the rest frame of the air mass.
Also, even if you use a simple fan instead of the AC you will perceive the air flux hitting your skin as colder. This is known as convective cooling. See for example this post for a simple explanation.
(1) Source
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4$\begingroup$ @Jeff the world record for wind speed is 231mph ~103 m/s. Normally air has to be supersonic to start heating - over 300m/s $\endgroup$– TimCommented May 30, 2017 at 22:34
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3$\begingroup$ @Jeff Yes it does, meaning that if you measure your temperature in a fixed reference frame or in a frame moving with the air mass you get different results. Anyway, the highest wind speed ever registered on Earth is 113 m/s (Cyclone Olivia, 1996). You can try to work out what change this would bring in the temperature, but be careful that the $v$ in $\langle v \rangle$ is the absolute value of the velocity, while the wind will (presumably and approximately) blow in only one spatial direction. $\endgroup$– valerioCommented May 30, 2017 at 22:46
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1$\begingroup$ @SGR no... the NET flow of the air, the actual "macro-scale" flow of air that you feel is the wind speed, but the air molecules are moving in all directions. Its abit like polystyrene beads on a surface of flowing water.. the net movement of the beads will follow the flow of the water, but the beads will move relative to one another too depending on what you fix as your frame of reference. $\endgroup$– James TCommented May 31, 2017 at 10:46
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2$\begingroup$ @JeopardyTempest Actually all I did was google max wind speed earth and look at the "one true answer" box :P I didn't feel there was much need for an especially reliable source, as long as what I found was significantly below the 300 m/s he'd suggested. $\endgroup$– TimCommented May 31, 2017 at 13:42
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3$\begingroup$ @Setop They are the same to 2sf to be fair - which is all we have the temperature measured to in the example. 460 is probably a better answer. $\endgroup$– TimCommented May 31, 2017 at 13:42
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There are two things at play here.
The Nature of Heat
Temperature, is to do with random, undirected motion. So for example, in room temperature gas, you can have individual molecules which themselves are moving at $100~\textrm{ms}^{-1}$, however on average, the velocity is zero.
When you have a body of colder air moving, it means on average their kinetic energy is lower, but they have a net "drift" velocity. That is the particles have random velocities, which corresponds to the heat, but because they've been "blown", they have an average velocity of maybe $5~\textrm{ms}^{-1}$ in one particular direction.
Why a breeze cools us down
The fact that a breeze cools us down has only a small amount to do with the temperature of the air. Consider your hand, sat in still air. Your hand will give some thermal energy to a layer of air surrounding it. (In the process of warming up that air, your hand will cool down). Now if the air is still, then that's the end of the story. However when there's a breeze, this layer of air that you've warmed up, gets carried away. It gets replaced with some new cold air, which your hand can then give heat to again, which cools your hand off again. When there's a breeze, your hand is constantly supplied with new air that it can give heat to. And so it's less that the air is a lower temperature, it's that as soon as it gets warmed up, it's replaced by a new set of cold air.
Just for fun, let's consider another similar effect. Say there's a little water on your hand. When this water evaporates, it requires heat from your hand, and so the evaporating water will cool you down. In still air, eventually the air will start to contain more water vapour, and so the evaporation process will be slower. However if we introduce a breeze, then we are bringing in fresh new dry air, so the evaporation process can occur faster, and so more heat can be taken away. This process is why if you lick your hand, and then blow on it, it cools it down.
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$\begingroup$ If temperature has little to do then how can there be "Hot winds"? $\endgroup$ Commented May 30, 2017 at 14:56
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$\begingroup$ Emphasis "little" to do with it. You still need there to be a temperature gradient, for heat to flow from your hand to the air. $\endgroup$– CDCMCommented May 30, 2017 at 15:06
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6$\begingroup$ @user154547 Hot winds are hot air masses moving along a pressure gradient. The speed of the moving air mass has little to do with its temperature. $\endgroup$– valerioCommented May 30, 2017 at 15:12
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$\begingroup$ @valerio92 to add: unless air is passing really (supersonic) fast $\endgroup$– jeanCommented May 30, 2017 at 16:52
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1$\begingroup$ With strong enough wind it is possible to die from hypotermia on a hot day - that is one of the reasons why motorbike riders ride with body folly covered. $\endgroup$– SumaCommented May 31, 2017 at 9:41
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The crux is that temperature is the undirected movement of particles.
Consider a very easy example of two air molecules flying with $100m/s$ away from each other, one in +$x$-direction, the other in $-x$-direction. The mean directed velocity is then $0$, however, the mean undirected velocity which corresponds to temperature is +$100m/s$.
Now suppose you accelerate both particles by $+50m/s$, so that one has a velocity of $-50m/s $ and the other $+150m/s$. The mean directed velocity is then $+50m/s$, which is the velocity of the breeze. In contrast the undirected remains at $(150+50)/2 = 100m/s$ - the temperature of the system has not changed.
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A breeze is not "a bunch of air molecules moving in that direction". A breeze is a wave of pressure.
The speed of air molecules is on the same order as the speed of sound. They are flying around smashing into things and mostly bouncing off. You are surrounded by a barrage of minute "meteors" that are smashing into your skin and bouncing off in all directions.
Well, these meteors are mostly bouncing off each other, because at 1 atm things are so dense that the average molecule travels 68 nm before hitting another molecule.
The only thing stopping it from knocking you flying is that you are being smashed into on all sides uniformly all of the time.
A perfect vacuum is just the effect of the other side smashing into you and pushing you that way, without the support of the molecules on the vacuum side. A vacuum cleaner reduces air pressure by a tiny amount, and it causes a pretty huge force this way.
A "wind" is a relatively tiny wave of higher pressure on top of this violet stew of air molecules. We feel it because on the side it is coming from, the pressure is higher, and on the other side it is lower. Even a hurricane is a relatively small change in pressure, and is enough to pick up and destroy houses, trees, cars and people. The strongest hurricane comes from about a 10% difference in atmospheric pressure! That pressure difference from the edge to the center is enough to power the winds of that entire storm.
A 10 km/h wind is traveling at 1% of the speed of sound, which is roughly how fast air molecules move. That macroscopic motion is a tiny contribution to the average kinetic energy of the molecules of air; it has roughly 2% more KE than the same air stationary.
A 100km/h wind is now going at 10% of the speed of sound. As KE is square of velocity, at that point we are talking about 20% more KE than the same air stationary.
You'd have to approach 700 km/h for the wind to have as much KE from its macroscopic "wind" motion as it does from it's microscopic "vibration" motion.
Now, as another poster has mentioned, the velocity of air molecules is actually a touch higher than the speed of sound; sound is a wave, so it isn't going to propogate at the exact same speed as air molecules are (it will travel a bit slower). But it gives you the right order of magnitude.
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5$\begingroup$ A breeze is a relatively small deviation of the mean velocity from zero. It is not a pressure phenomena or a wave. $\endgroup$ Commented May 30, 2017 at 18:33
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$\begingroup$ @dmckee I think he may be miswording it or misinterpreting it. Pressure differences are highly related to the wind (though I'm not a meteorologist and don't know enough about it to say if it's a thermal action creating pressure difference that drives the wind, or thermal variations drive movement which creates pressure differentials, etc.). Either way I agree it isn't accurate as is. $\endgroup$– JMacCommented May 31, 2017 at 11:39
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1$\begingroup$ @dmckee A breeze is a relatively small deviation of the mean velocity from zero with a corresponding pressure gradient that allows that velocity to propagate (in a wave like fashion). Saying a breeze is not a pressure phenomena is like saying light is only a small deviation in electric field, it's not a magnetic phenomena . $\endgroup$– EphCommented May 31, 2017 at 17:02
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$\begingroup$ @Rick It is only the relative motion between myself and the bulk of the air that make a breeze. That relative motion can be generated by weather pattern (which are indeed driven by pressure differences) or by the motion of a conveyance that I ride through a placid indoor space. Nor is wave-like behavior necessary: a steady-state flow is still a breeze. There is a lot here that could be part of a nice discussion of wind in general, but it's poking around "where does a natural wind come from" instead of the thermal transfer from moving air question the OP asked. $\endgroup$ Commented May 31, 2017 at 17:24
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$\begingroup$ @dmckee yeah, I agree that the pressure discussion is rather tangential to the OP's question. $\endgroup$– EphCommented May 31, 2017 at 17:29