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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?

Harsha
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3 Answers3

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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.

Bob D
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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 $T_{water}$.

Ben51
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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".

user415081
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