# What exactly heats a pool of water on a sunny day

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?

-
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

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.
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