How does the density of the air affect our perception of temperature?

The classic experiment where you have someone put one hand in a bowl of hot water and the other in a bowl of cold for a while then move their hands to a warm bowl is a classic demonstration that our perception of hot & cold is relative. If we assume that this means our perception of temperature is completely determined by the rate heat flows into or out of our body, then we can estimate that perception based on calculating heat flux through the air when its in contact with a $$37\,^\circ\mathrm{C}$$ $$1$$ meter diameter sphere, making this a physics question.

Based on this simplified model how does the net heat flux depend on air temperature and pressure (equivalently: temperature and density)? Don't forget to include the effects of blackbody radiation. Basically, I'm curious if differences in temperature feel less extreme at moderately higher altitude where the boiling point is only a little lower, but the pressure is down by an $$e$$-fold or so.

• Do you consider sweat and evaporation in your perception model? It seem that at higher altitudes there could be easier evaporations, and that could influence the perception. Commented Oct 2, 2022 at 21:39

A 1 meter diameter sphere is not bad, but I think I will skip factors of $$4\pi$$ and just take the surface area to be $$2m^{2}$$, which corresponds well with what doctors tend to use.

Your body is optically thick, so the blackbody approximation is fine. At 300K, you expect to emit 900 watts. This is much higher than basal (~100 watts), so it's fortunate that on Earth you are usually surrounded by 290K optically thick material, which provides you 800 watts. This ~10 Kelvin difference corresponds to finding room temperature roughly comfortable. In general, the blackbody flux will be $$\sigma (T_2^{4} - T_1^{4}) = \sigma ((T_1 + dT)^{4} - (T_1)^{4}) \sim 4\sigma T_1^{3} dT = (6 \times dT) W m^{-2} K^-1$$

Let's consider Conduction. Surprisingly, this list reveals that the conductivity of air depends very slowly on density! Why might that be?

Air molecules bounce around randomly. If the mean free path (the length of a single random drunken step) is $$\lambda$$, and the total length is $$L$$, it's well known from random-walks that the typical number of steps needed to cross will be $$(L/\lambda)^2$$ and the total path traversed will be $$(L^{2}/\lambda)$$, which will take time $$L^{2}/\lambda v$$. The mean free path gets shorter with higher density. If our area of collision is $$\sigma$$, we will collide with every particle in a tube of area $$\sigma$$ and length $$\lambda$$, the volume is $$\sigma \lambda$$ and the number of particles we will collide with is $$n \sigma \lambda \sim 1$$

So a higher density increases the rate that air molecules hit you, but each air molecule takes longer to deliver its energy payload, so conductivity (for gasses! liquids and solids are much more complicated) ends up not depending on density.

All this to say, we'll use a conductivity of $$0.02 W m^{-1} K^{-1}$$ for air. Re-writing a bit more suggestively, this is $$0.02 W m m^{-2} K^{-1}$$. The flux is $$\left(0.02 \frac{1m}{dx} \times dT \right) W m^{-2} K^{-1}$$. For conductivity to be large compared to radiation, the width of the temperature gradient dx must be smaller than $$\sim 3$$ millimeter. Does this depend on density? To understand $$dx$$, a boundary layer forms between your skin and the air depending on the air velocity and kinematic viscosity, and the width of that layer gets smaller if the air is moving faster, which causes wind chill. Well, to an extent the kinematic viscosity depends on density, but not strongly.

Lastly, consider the optically thin limit. In this limit, the blackbody radiation you receive from the air scales linearly with the ambient density. But optically thin vs thick does not only depend on density, but length AND composition: again, our tube of $$n \sigma \lambda$$ where this time the $$\sigma$$ is the cross section for light to interact with air (and water vapor, and in fact, mostly water vapor). Rather than do more detailed calculation, let's consider the very thin limit of outer space. In the shade, you receive very little blackbody radiation from your surroundings, though you may receive some from the Earth. Let's be generous and let the Earth give you half the blackbody radiation you are accustomed to (assuming it covers half of your visual sphere). In this limit, you will be losing 400 Watts, which will be more uncomfortable than usual.

For a 100 kg human comprised of 4 J/g K specific heat water, it will take a deficit of 1,000,000 Joules for hypothermia to set in, but this doesn't directly link to our wattage since your body will be vigorously burning food/fat to compensate for the losses. You'll need to eat roughly 344 kCal per hour (7200 in a day) to keep up.

tl;dr: blackbody most important, conduction not that density dependent after all. this why mountain and space are cold

• I think you more or less "killed" it, except for the fact that the optical "thickness" of the atmosphere depends mostly on the loading with water vapor and cloud reflectivity for the relevant IR wavelengths. Dry air therefor feels "colder" than moist air. Commented Oct 2, 2022 at 23:05
• I completely agree. Commented Oct 3, 2022 at 2:12
• To clarify, for most human experience we are in the optically thick regime when surrounded by walls and such so the dry air vs moist air factor people are used to feeling is probably mostly evaporation and humidity, but I agree that there is also the blackbody radiation component if you are outside and benefitting from e.g. cloud cover. Commented Oct 3, 2022 at 2:27
• Absolutely. Even the driest atmosphere adds a considerable amount of IR radiation. Clouds can actually increase the effective solar constant quite substantially, which is now being observed by solar farm operators regularly, it seems. That is not necessarily a good thing for those facilities. Commented Oct 3, 2022 at 3:29