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On a chilly (but above freezing) temperatures on a clear night, you can freeze water outside because of radiative cooling.

By what mechanism do warm bodies on earth actually lose heat to space?

From the Stefan-Boltzmann Law, I know that a hot body loses heat energy at a rate proportional to the 4th power of the body's temperature minus the 4th power of the surrounding temperature.

Obviously that isn't a complete description of what's going on though, since the temperature immediately surrounding a hot fire is quite hot as well, but you (standing far away) can still feel the radiation.

I'm also not clear on how the temperature difference between a body and a space that is hundreds of miles away can affect the real-time radiation rate of that body?

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    $\begingroup$ Things radiate away photons based on their temperature, not on where the photons may end up. They don't get many back from the atmosphere or space, but with no expectation of where they came from. Thus, a net loss of energy under the right conditions, or a net gain of energy under different circumstances (daylight). $\endgroup$
    – Jon Custer
    May 31, 2016 at 17:48

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Bodies radiate based on their own temperature and properties. The water in your example is always radiating energy. Regardless of whether the night is clear or cloudy, it’s radiating away energy.

But the temperature change is based on the total flow of energy: the water gets colder if the energy it radiates away is more than the energy radiated toward it.

  • On a cloudy night, those clouds are also radiating energy, some of which comes toward your water: those are in rough balance, and not much cooling goes on.

  • On a clear, dry night the only thing radiating back at the water is deep space, and it’s not providing g much energy; the energy radiated by the water isn’t balanced by any inflow, and the water cools.

  • On a sunny day, more radiant energy comes in than leaves, and the water can heat up.

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Temperature is a measure of how much energy photons have $E = h\nu$ $= kT$, and the temperature reflects the energy of the photon incident on the detector. The total irradiance or EM power incident on an area $A$ is $P = \epsilon\sigma AT^4$ from Rayleigh-Jeans law. Here $\sigma$ is the Stefan-Boltzmann constant and $\epsilon$ the emissivity of the body. For the night sky $\epsilon = 1$ and the temperature is $20K$, a bit higher than the $2.7K$ of the CMB due to stars and such. So the temperature depends on the energy of photons and their flux. Even though the night sky is far away what counts is the energy and flux of photons coming in from it. This is pretty weak and so the night sky is cold. The Earth then radiates out to the night sky a lot more than it receives from it.

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    $\begingroup$ The night sky isn't anywhere close to 20K, not even in Antarctica, where it's over 240K, even in the depths of winter. The surface is considerably colder than is the night sky in Antarctica (temperature inversion). The atmosphere is quite opaque in the thermal infrared. $\endgroup$
    – David Hammen
    May 31, 2016 at 20:26
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    $\begingroup$ I agree, it's not that cold. See an earlier question and answer at physics.stackexchange.com/questions/153839/… See also asterism.org/tutorials/tut37%20Radiative%20Cooling.pdf. You have to take more into account. At radio frequencies the atmosphere is more transparent and the night sky effective temperature does go dowwn $\endgroup$
    – Bob Bee
    Jun 1, 2016 at 0:20

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