The hypothesis is this: Water molecules, due to their asymmetry and electric dipole are particularly susceptible to radiation at specific frequencies. That suggests that excited water molecules will also radiate at those specific frequencies. Thus thermal energy in humid air will be disproportionately radiated at those frequencies most favorable to transfer kinetic energy to other water molecules. I am proposing that this effect is perceptible on a hot humid day.

By "thermal radiation" in this context I mean electromagnetic radiation in the spectral range which most effectively transfers heat energy between physically separate material bodies. Especially bodies containing a significant amount of $H_{2}O.$

I am confident that the theoretical answer is that thermal (black-body) radiation of the outdoor environment will transfer energy to the indoor environment. So the question really is, will this be significant enough to be perceptible? I have long had the impression that when it is hot outside, then I feel warmer inside a building than I otherwise would with the same indoor air temperature thermometer reading.

I will go so far as to say, this is more pronounced when the humidity is exceptionally high. And I hypothesize that this is not simply a result of higher indoor humidity. As anybody with experience with a microwave oven knows, water is far more susceptible to such radiation than are dry substances.

That means the human body is more susceptible to thermal radiation than is air. It also suggests that humid air will be a stronger black-body radiator of exactly those frequencies which would communicate heat energy to other $H_{2}O$ molecules via electromagnetic radiation.

Will thermal radiation from significantly hotter outdoor air cause a person indoor to feel warmer at the same indoor air temperature?

As another consideration along these lines, would hot air in an attic have the same perceptible effect?

I came across the following while looking into this topic. I don't have access to the entire article, but the abstract gives a bit of support for my hypothesis.

A device for the convenient quantitative measurement of the thermal radiation from the atmosphere has been developed. In the instrument the emission spectrum of the earth’s atmosphere as observed at ground level is compared automatically with a spectrum approximating that of a blackbody at the boiling point of liquid nitrogen at atmospheric pressure. The most prominent features of the atmospheric spectrum between 4 μ and 15.5 μ, observed during daylight and darkness when the sky is clear, are due to emission by carbon dioxide, ozone, and water vapor; the intensity of the water vapor emission shows pronounced variations with atmospheric temperature and humidity. The spectrum of an overcast sky resembles that of a blackbody. By comparing the recorder traces of the atmospheric spectra with similar traces obtained with a blackbody source, it is possible to estimate the effective radiation temperature of various portions of the sky for various atmospheric conditions.

© 1955 Optical Society of America

This paper appears to address the relevant physics applicable to my question. Unfortunately, I don't have time right now to read it.

Radiation and Heat Transfer in the Atmosphere: A Comprehensive Approach on a Molecular Basis

  • $\begingroup$ "As anybody with experience with a microwave oven knows, water is far more susceptible to such radiation than are dry substances". Just want to point out that microwave ovens do not cook by thermal radiation. They do not cook food by heat transfer. The chief mechanism involves the coupling of the 2450 MHz alternating electric field with the dipole moment of water molecules. The field does work on the molecules causing them to rotate and acquire rotational kinetic energy. That, in turn,increases translational kinetic energy which raises the temperature of the water in food. $\endgroup$
    – Bob D
    Commented May 29, 2019 at 17:25
  • $\begingroup$ I guess we have different definitions of thermal radiation. What you are describing is energy transfer at a specific frequency. That frequency falls into the spectrum of "thermal radiation" by the loose definition I gave when posting my question. $\endgroup$ Commented May 29, 2019 at 17:29
  • $\begingroup$ BTW, have you ever notice that honey is even more receptive than water? $\endgroup$ Commented May 29, 2019 at 17:42
  • $\begingroup$ But you realize that microwave ovens don't cook by radiant heat transfer, correct? $\endgroup$
    – Bob D
    Commented May 29, 2019 at 17:47
  • $\begingroup$ If your point is that the source of the radiation is not thermal, then, yes. I used to be a radar technician. I was merely pointing out that water molecules are more active participants in the exchanges of energy via EM-radiation than are most substances. $\endgroup$ Commented May 29, 2019 at 17:52

3 Answers 3


As a general rule gases don't behave like black bodies. Unless the wavelength of the light happens to coincide with an excitation of the gas molecules the gas is transparent to the light. I suppose you could describe it as a black body with an absorptivity and emissivity of zero, but this seems a rather eccentric description.

Gases normally exchange heat by convection. On the Earth that means the Sun heats the ground by radation, the ground heats the air by convection and the air heats you by convention.

You specifically mention water vapour, and water vapour does have vibrational transitions in the infra-red. This graph from Wikipedia shows the absorption spectrum of water vapour in the infra-red:

Water absorption spectrum

However the peak blackbody wavelength for 25°C is around 10 microns and as it happens there is a dip in the absorption at this wavelength.

So while there will be some black body radiation from hot air it's likely to be negligibly small compared to convective heating.

  • $\begingroup$ I agree with you that gases in the atmosphere aren't very "black", but as you pointed out, I am particularly interested in the case of humid air. I will speculate that, since water molecules have specific excitation states, what may be most significant is the radiation corresponding to those states. Hot, humid air will excite the water molecules through collisions. When the molecules transition back to their ground state, they will radiate at their excitation frequency. That will be exactly the kind of radiation that would heat water. $\endgroup$ Commented May 30, 2019 at 8:44
  • $\begingroup$ I up-voted your answer, but I didn't accept it. That indicates that the answer was useful toward a the goal of validating the hypothesis. More to the point, it helped me refine the hypothesis, and exposed the sloppiness of my original speculation. But, my objections still stand. What I am suggesting is that the water molecules in hot humid air are induced by collisions to radiate at frequencies which will be readily absorbed by other water molecules. I believe this effect will be more pronounced than simple black-body radiation and absorption. $\endgroup$ Commented May 31, 2019 at 3:57

When it is hot outside, it is generally sunny outside. In that case, the sun will enter through your windows and heat up the ground, which will make your house warmer. Of course, you're suggesting that it is when the house is the same temperature, to which I ask, how do you know it is the same temperature? Air conditioner thermometers are not a very accurate source of temperature reading for your entire house. As my source shows, the temperature is measured inside the AC unit, which I do not suspect is in the middle of your living room. It is probably tucked away somewhere in your basement, where it feels colder than the rest of your house.

If your home was at a reasonably low temperature, the rate at which you reradiate heat to your surroundings would be fast enough that any low energy black body radiation (that is, any radiation that isn't radiation from the sun) should be negligible, and it is unlikely that you'd notice its heating effects on yourself over the rest of the air around you.

  • $\begingroup$ I used to attribute the perception entirely to solar radiation. The first time I noticed it was while working the the control van of a missile system while it was bathed in solar radiation. But I've also notice the same kind of thing at night when the air in the attic above me was hot. I'm in an apartment with a local thermostat, and sitting .2 meters away from it. The temperature is ~6 F lower than usual, and I still feel warm. And it is hot an humid outside. No direct sunlight hitting this unit which is two floors from the roof. $\endgroup$ Commented May 29, 2019 at 18:20
  • $\begingroup$ I believe the most significant transfer of heat to my surroundings would be via thermal conduction, and not radiation. My intuitive assessment is in agreement with yours. I naively expect the radiative heat transfer from outside air to be negligible. The reason I am asking is that my perception seems to contradict that prediction. $\endgroup$ Commented May 29, 2019 at 18:57
  • $\begingroup$ @StevenThomasHatton I'm no expert in this field, I just provided the best explanation that I could. Hopefully someone else has a more clear explanation. $\endgroup$
    – Kraigolas
    Commented May 29, 2019 at 20:29

Thermal radiation from outdoor air can't DIRECTLY make you feel warmer indoors, because ordinary walls and windows are OPAQUE to thermal radiation. Not "translucent", but opaque.

Thermal radiation from outside gets absorbed by windows and walls, and then re-radiated both outside and inside. If the outside temperature is higher, the building will absorb heat, and its windows and walls will grow warmer.

Inside, the walls are radiating heat at you, and you're radiating heat back at them. The warmer the walls are relative to you, the warmer you'll feel, assuming the inside air temperature and humidity are held constant.

So, why did you feel warmer when it was more humid outside? It was probably more humid inside as well, and that will make you feel warmer.

  • $\begingroup$ Thermal radiation from the Sun in the visible range passes through window glass rather well. Since the relevant frequencies are likely near 2.45GHz this, assuming it is correct, is likely applicable. blog.ibwave.com/… $\endgroup$ Commented May 9, 2021 at 20:05
  • $\begingroup$ Well, in the very most technical sense I suppose visible light from the Sun is also thermal radiation. 2.45GHz microwaves can be, too, but they aren't a significant contributor to total energy -- remember, a thermal source peaking at 2.4GHz is in the single digits Kelvin. $\endgroup$
    – jeffB
    Commented May 10, 2021 at 1:08
  • $\begingroup$ Adding due to an unfortunate editing timeout: Thermal radiation at ordinary temperatures (293K) peaks around 10 microns. Glass is quite opaque at wavelengths in that range. From a 5000K source like the Sun, the proportion of radiant energy falling into the microwave range is negligible. $\endgroup$
    – jeffB
    Commented May 10, 2021 at 1:28

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