I created a rough spreadsheet (here) to calculate the amount of heating of pool water from sunlight. The absorption spectrum of water is shown in this post and the spectrum of sunlight at the surface of the Earth is described here (with averages computed in a spreadsheet here). The amount of absorption after traveling various depths is as follows:
1 inch -- 24%
1 foot -- 39%
3 feet -- 48%
6 feet -- 54%
12 feet -- 61%
16 feet -- 63.7%
17 feet -- 64.4%
You can see that about 1/4 of sunlight is absorbed within the first inch of water and about 40% in the first foot, but this is mostly infrared radiation so you don't see it (about half of solar energy at the Earth's surface is infrared and half is visible light plus some UV). Nevertheless, it explains why the surface of the water can be so much warmer than deeper water unless the circulation is very good. The sunlight that is absorbed after the first foot is more in the red (and some green) which is part of the reason why pool water has a cyan-blue color.
If I use 85% reflection (15% absorption) for white plaster (see this link which gives 91% but that's bright new plaster), then this implies an absorption in the shallow end with 3 feet of water (with reflection and double path length) of 1-(1-0.54)*0.85 = 61% while the deep end of 6 feet is 1-(1-0.61)*0.85 = 67% absorption. So around 64% of sunlight energy gets absorbed by a white plaster pool to heat it. If a pool had a darker bottom, then even more would be absorbed. [EDIT] Note that actual absorption will be higher due to diffuse reflection which will generally increase path lengths (so closer to 64% absorption in water which with 85% reflection would be 69%) and add additional absorptive reflections (two 85% reflections with my original calculation would get to 69% and combining the longer path length effect would get to 74%). Some assume overall absorption to be around 80%.[END-EDIT]
During the peak of summer with the sun nearly directly overhead at solar noon, the amount of power is around 1000 Watts per square meter which is 860,421 calories per hour per square meter. If the average depth is 4.5 feet, that's 1.37 meters so the rate of heating of 1.37 cubic meters or 1370 liters is 66% * 860,421 / (1370 * 1000 ml/l) = 0.41ŗC per hour or a heating of pool water by the sun at the rate of 0.75ŗF per hour. Of course, the pool can get cooled by evaporation (more in dry climates) and if the air is cool and there is wind then one doesn't usually see this amount of temperature rise. Use of a clear (or light clear blue) solar cover will cut down evaporation, shield from wind, and maximize heating even during the day.
Black mat unglazed plastic solar panels are around 80% efficient (i.e. they transfer 80% of the sun's energy into heating pool water) with no temperature difference between the air and water or with a small difference and minimal wind. If one had solar panels with the same amount of area as the pool surface, then this would be produce a temperature increase of around 0.9ŗF per hour at peak solar noon. A cloudy/overcast (but not really dark) day has roughly 1/3rd the solar power of a sunny day -- on the order of 300 Watts per square meter though this varies a lot on the thickness of the clouds. Since the clouds are water vapor, they absorb more of the infrared which means the pool surface heating effect on a cloudy/overcast day will be negligible though overall heating will still occur at a around 0.2ŗF per hour.
There are some websites that claim that evacuated tube solar panels are more efficient, but on clear sunny days they are not. I'll write more about various solar panel types (unglazed black plastic, glazed flat panel, evacuated tube) in another thread (here).