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On a sunny day an outdoor swimming pool will heat up fairly quickly. My question is, what is the exact mechanism for this and can we put numerical figures on it?

Given that water is clear and colorless, it seems that it will absorb very little direct radiation from the sun. The sides and the bottom of the pool can however heat it by conduction.

Is it possible to estimate the influence of radiation versus conduction in this process or in general to give a full explanation?

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A large portion of the spectrum of sun light is around a wavelength of 1 $\mu$m (infrared). For that wavelength, more than half of the energy will be absorbed after 10 cm of water. This could very well be the main reason for heating the pool. You simply need to know the spectral intensity of the sun radiation, the absorption spectrum of the water and the heat capacity of the water. –  fffred Jun 8 '13 at 6:01
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There's a related question Does brown but transparent swimming pool water heat significantly faster than western style highly chlorinated pools?. The question isn't a duplicate, though the answers there are relevant.

At the equator the intensity of sunlight at the ground is about 1kW/m$^{2}$, of which about half is visible and half is IR (plus a few per cent in the UV). The As fffred says in the comment, water absorbs IR radiation so about half the energy is absorbed directly by the water. The visible light will pass through water unabsorbed and will heat the walls of the swimming pool. Because water absorbs IR you get a greenhouse effect that keeps the walls warm, and the walls then warm the water by conduction and convection.

Estimating exactly how fast the water heats will be hard because the swimming pools walls tend to painted a light colour that will reflect a lot of the light. We'd have to know the reflectance to calculate how much light the walls absorb and hence how fast they heat. Presumably the walls will lose some heat to the ground, though i'd guess this will be slow. Finally need to take into effect the fact that the sunlight intensity falls with latitude, and of course depends on atmospheric conditions. Assuming maximum absorption and a pool on the equator sunlight will heat a 2m deep pool at slightly under half a degree per hour.

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Thank you. I can't see directly why half the energy of daylight is IR from the spectrographs online. In fact it looks like very little is above 700nm. How do you know that is true and what wavelengths does water absorb? –  marshall Jun 8 '13 at 7:23
    
To be honest I just took that figure from the Wikipedia page. A quick Google found commons.wikimedia.org/wiki/File:Solar_Spectrum.png and this does show a lot of the energy lies in wavelengths longer than 700nm. –  John Rennie Jun 8 '13 at 7:35
    
A further quick Google found a spectrum of liquid water at backreaction.blogspot.co.uk/2009/01/… –  John Rennie Jun 8 '13 at 7:37
    
Thanks. Maybe I should ask a different question about how you get this figure of a half. I thought that the longer wavelengths carried a lot less energy. –  marshall Jun 9 '13 at 9:48
    
The absorption spectrum given by John shows that the opacity of water is about 0.1 cm$^{-1}$ for an infrared wavelength of 1 micron. You multiply by the distance, say 5 cm, to obtain the absorption coefficient $5\times 0.1=0.5$. The transmitted fraction is given by $\exp(-0.5)\sim 0.6$. This means about 60% is transmitted. –  fffred Jun 14 '13 at 22:43
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