What factors contribute to the expected temperature of a pool?

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The expected temperature of a pool is influenced by factors such as sunlight exposure, heat loss, and water volume. In a 60 ft by 30 ft by 10 ft pool, the initial temperature is 50 degrees Fahrenheit, and while sunlight can significantly warm the surface, it primarily affects only the top layer of water. The pool's large volume requires substantial energy—approximately 2.1 billion joules—to raise the temperature by just one degree Celsius, making rapid heating challenging. Despite the heat loss being minimal at 20 joules per hour, the pool will not warm significantly by 3 PM due to the depth and evaporation. Overall, achieving a comfortable swimming temperature will take considerable time and consistent sunlight exposure.
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Assume I have a pool that is 60 ft by 30 ft by 10 ft. The weatherman says the high tomorrow will be 84 degrees Fahrenheit. The pool temperature will be 50 degrees at 6 AM.
1. Ignoring heat loss, what will the temperature be at around 3PM?
2. The heat loss is around 20J/hour. What will the temperature be when I go for my afternoon swim at 4 PM?
 
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Such a deep pool won't warm up much from conduction with the atmosphere in a single day. The problem is this will only warm the top layer, also increasing cooling by evaporation. (the humidity as well as the temperature is important)
The only thing that will help is sunlight, which will shine into the water and heat the pool to a greater depth. It still won't heat quickly.

Your pool has 498 m^3 water, which will take 2.1 * 10^9 J to heat it 1 degree celsius. The surface is 165 m^2, which will catch about 165 Kw of sunlight with the sun overhead. It will still take 2.1*10^9 / 1650000 = 1.27 * 10^4s = 3.5 hours
to heat it by one degree celsius, and you want to heat it from 10 to 28.8 degrees celsius, and the sun won't be straight overhead all the time.

If the heat loss is really around 20J/hour, the pool would remain heated for the rest of your life.
 
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