In reading the chapter on radiation for heat transfer I came across a brief section on the emissivity of the sky. A few days ago, I put the pieces together and answered a question I hadn't thought about in a while.
Dew and frost form when an object in contact with the air becomes cooler than the dew point of the air causing water vapor to condense. But since a tree or the grass is only in contact with the air and the ground, how can it become colder than the air? This would be like a warm glass of water making an ice cube grow.
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Turns out I just wasn't thinking of the heat transferred via radiation. Every object emits an particular amount of heat based on its temperature. The power (Watts) is a function of absolute temperature to the fourth power, so something twice as hot emits 16 times as much heat.
(Also, the wavelength range of the emitted radiation is a function of the object's temperature. In fact, if you see the right wavelength of light, you would be glowing. Your cooler surroundings would also be glowing, but at slightly longer wavelengths. This is how heat vision works. Also you can be certain that for the most part, every object that is just starting to glow is a certain temperature, 525 C to be exact. Moreover, this is why light bulb packaging has a temperature listed, the sun emits at about 5800 K so if you want blue/whiter light than sunlight, get 6000K+ or if you want dimmer/redder go lower.)
You might have heard of a 'black body'. A black body is an ideal emitter or absorber; in other words it does not reflect. Real objects only emit a certain fraction of the power a black body would. (Polished metals are least black, ~0.1, whereas matte organic surfaces are closer to black, 0.6-0.8+.)
Back to the dew, I wasn't accounting for the radiation exchange of the tree with the sky. There was an equation in the textbook for the emissivity of the sky. It's roughly 0.7 from the oxygen and nitrogen with a bit more based on how much water vapor is in the air.
This means that the sky emits as little as ~75% as much radiation as the temperature it really is. In other words, the effective temperature of the sky is significantly lower. At room temperature, the difference is roughly 20 C lower, about 38 F. An object in a 77 F night might be facing a 39 F sky!
Putting it all together, the heat transfer from the ground through even through a tree is negligible. Air is actually a really good insulator--the fluffy stuff in your walls or coat is mostly air--especially on a quiet night with mostly still air there won't be much heat transfer from convection. Hence the stuff outdoors really only interacts with the sky, which is effectively cooler than the air temperature.
Here are two tables I made with the actual equation for clear night sky emissivity. Clouds, by the way, would act like a radiation shield, reflecting the radiation of the ground so the sky wouldn't be colder than the ground.
The first is the effective sky temperature in F on a clear night. The blue area is typical winter nights in WI and the red is summer nights. Louisiana would be in the red area and to the left of it. It's in weird intervals because I did the calculations in C. [Click for enhanced legibility!]
swamp coolers already work find in those areas. On the other hand, if you want to get your pool as warm as possible, you should cover it at night. Doing that will also help you to not lose water to evaporation which additionally contributes to cold water.