# Nocturnal freezing of water: Calculation of radiation through atmosphere?

I recently came across an article in Wikipedia, which claimed that medieval civilizations used to leave water out overnight in an insulated pot during clear and calm nights, which results in the water radiating away its heat and freezing into ice. Note that this method does not require the ambient atmospheric temperature to be below the freezing level.

Which brings me to my question: How is this possible, and what is the lowest temperature one can hope to cool a well insulated object during night? Let us assume, for instance, that it is a calm night where the ambient temperature of the atmosphere is 15C. While the object radiates its heat out to space, the atmosphere also radiates heat into the object (since we hope to cool the object to below the ambient temperature). How can this be calculated, ie., what is the 'radiative' temperature of the night sky, instead of the ambient temperature?

• Note that this method does not require the ambient atmospheric temperature to be below the freezing level. It does: heat ALWAYS flows from hot to cold. ALWAYS. An object not only radiates, it also RECEIVES radiation.
– Gert
Commented Feb 1, 2021 at 2:18
• If the humidity is low, evaporation may contribute to the cooling. I’d imagine the lowest temperature possible might depend on altitude—more rarefied air presumably radiates less. Commented Feb 1, 2021 at 3:12
• more rarefied air presumably radiates less The only thing that influences final temperature is the air's temperature.
– Gert
Commented Feb 1, 2021 at 19:14
• @Gert oh, my mistake. You realize then that under the assumption that air is at least somewhat transparent to IR, the temperature of space (~3K) matters. So not just the temperature of the air. Commented Feb 1, 2021 at 20:47
• The question seems to be picked up again here: Nocturnal Ice Making: is it possible and by what mechanism? Commented Feb 23, 2021 at 20:04

Which brings me to my question: How is this possible, and what is the lowest temperature one can hope to cool a well insulated object during night?

It is possible due to what's called radiative sky cooling. https://www.popularmechanics.com/technology/infrastructure/a29036147/radiative-sky-cooling/

To quote from the link:

“This effect occurs naturally all the time, especially on clear nights,” says study author Aaswath Raman, an assistant professor of materials science and engineering at the UCLA Samueli School of Engineering, in a press statement. “The result is that the object ejecting the heat, whether it’s a car, the ground or a building, will be slightly cooler than the ambient temperature.”

The statement is consistent with the following statement from the Wikipedia article:

"Provided the air was calm and not too far above freezing, heat gain from the surrounding air by convection was low enough to allow the water to freeze"

You have to keep in mind that the water radiates to the vacuum of outer space (several degrees K) not only to atmospheric air at above freezing temperature.

Here are a couple more relevant links:

https://aip.scitation.org/doi/10.1063/1.5087281

https://www.osti.gov/pages/servlets/purl/1424949

Hope this helps.

Let's model this as a thermodynamic system with three components: the water, the air, and outer space (a fourth component would be the ground, but we'll assume the insulation between the water and the ground below it is very good, and leave it out). For simplicity, let's pretend that the air is perfectly transparent to the infrared frequencies at which the water emits (it isn't--if it were there would be no "greenhouse effect"--but if the humidity is low, the air is transparent enough that water can be regarded as being in radiative thermal contact with the atmosphere at some higher altitude, where it is colder). We will also ignore the latent heat flux associated with evaporation.

In this model, the water is getting heat from the air by conduction (called the "sensible heat flux") and giving heat to outer space by radiation. Equilibrium is reached when these two fluxes balance. The radiative flux can be estimated by treating the water as a blackbody: $$\sigma T_{water}^4$$. The sensible heat flux is proportional to the air-water temperature difference, but also depends on wind speed and is pretty empirical. But let's say for the conditions on a particular night we can say the sensible heat flux into the water is $$C(T_{air}-T_{water})$$, where $$C$$ is a constant. Then we can set the two fluxes equal to each other and solve for $$T_{water}$$.

• I also thought along the same line. But what stopped me from writing an answer is that blackbody radiation at 0°C peaks at ~10 µm wavelength. So both, the water and the air radiate at this wavelength, but the water has strong absorption bands there. So it seems to me that overall the water would gain more energy from radiation than it loses. Commented Feb 1, 2021 at 21:12

With the air being very still and clear sky, the ground can radiate more heat than any possible absorption from the atmosphere above. The atmosphere up higher even in an inversion where it's warmer, is thinner (way lower my molecules per volume) and thus has a lower effective emissivity. Solid bodies such as the ground or water have high emissivity more similar to a "black body".

• At our house on a cold, clear, still winter night, we measure 20 degrees lower than the surrounding weather stations which are suffering from the heat island effect. All the solid surfaces of the buildings are emitting heat to the surrounding air and it takes longer for the air to cool down when there is so much solar heat stored up in the solid buildings. In rural areas it's just the ground emitting, a flat surface, rather than 3d buildings emitting stored daytime heat from 5 sides Commented Jul 18 at 16:02
• This one of the reasons that I think putting solar panels in a hot urban area is going to make the heat island effect worse. Solar panels are black and absorb solar energy, but only 20% is converted to electricity. The rest, 80% is emitted as waste heat to the surrounding urban air, emitted from both top and bottom of the panel since they're usually elevated. It would be better to locate them farther from cities and just bring the electricity into the city, similar to gas and coal power stations being outside the city. They have waste heat too. Commented Jul 18 at 16:23