Vantablack Pool

Vantablack pool would be cool. Or rather the opposite? I came across this meme

The user "facts-i-just-made-up" makes an initial argument which seems quite reasonable. Black objects absorb the suns energy and radiate off heat. The things going beyond are a bit excessive.

So... let's overthink this.

How hot would a pool with Vantablack paint get?

To answer this, let's build a playground example. Ignoring the sides of the pool allows us to simplify the problem to a 1m x 1m square of Vantablack paint. Above that, we have a water column of 2 meters.

As water is very handily defined: 1m³ water is exactly 1000kg at 3.98°C. We want to know ballpark numbers and not go too crazy with it. So for further calculations we assume that our 2-meter column of water standing on the Vantablack square of 1m² at the bottom is 2000kg in mass at every temperature.

So... how much energy do we have hitting the square? To be exact it's fairly complicated. Because I'm lazy let's take the approximate number given in the article "[...] the Sun's rays are attenuated as they pass through the atmosphere, leaving maximum normal surface irradiance at approximately 1000 W /m² at sea level on a clear day. [...]"

So 1000 W/m² hit's our 1m² square of Vantablack. 1000W/m² / 1m² = 1000 W of energy we have here. Not bad. Vantablack absorbs 99.965% of that 1000W. Because I'm lazy, let's assume all of it.

Let's ignore the greater physics at play here and ask: how long would we need to sunlight our water with the 1000Watt of thermal energy to boil it? For this, we need to know water's specific heat capacity, or in terms I understand: How much energy we need to put into the water to make it 1°C warmer.

Luckily, water is an all-time favourite of Wikipedia article writers for examples and given in the summary of the article;  4179.6 J·kg⁻¹·K⁻¹  So again in terms I understand it, 4179.6 Joule are required to make 1kg of water 1°Kelvin (which is the same as 1°C) warmer.

So let's take the 1000Watt of sunlight the Vantablack absorbs. Handily, Watt is just Joule/s. So throwing numbers together:

4179.6 Joule/(kg*K) / 1000 Joule/s  = 4.1796 s/kg/K

So it takes 4.1796 seconds, or about 4 seconds to increase the temperature of 1kg of water by 1°C on our Vantablack pool surface. We have 2000 of these kilos to work with. So our column of water is heated by 1°C every 4.1796 s/kg/K * 2000kg = 8359s/kg = 139.3min/kg

So our theoretic Vantablack-pool heats up by about 6° in a day (of 12 hours).

So far so nice. Now let's go back to reality and see how messy it all is. First of all, we have to worry about the mixing of water and how the heat spreads in it. Next, we need even to consider the world outside the pool. How does the pool interact with its environment? Entropy dictates, that if the pool is warmer than the surrounding air, it will heat the air and the ground around it. Turned around, the air cools down the Vantablack pool! 

Relevant to this problem are concepts like Black Body Radiation and Kirchhof's Law of thermal radiation as well as the thermal diffusion of water, air and whatever the ground is made of. But that would be a real effort. I'm here to babble about things I have barely any idea about with Wikipedia. Not simulate complex systems.