What factors contribute to the expected temperature of a pool?

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SUMMARY

The expected temperature of a pool measuring 60 ft by 30 ft by 10 ft is influenced by several factors, including ambient temperature, sunlight exposure, and heat loss. At 6 AM, the pool temperature is 50 degrees Fahrenheit, and without heat loss, it would reach approximately 84 degrees Fahrenheit by 3 PM. However, accounting for a heat loss of 20J/hour, the temperature at 4 PM will be significantly lower due to the limited warming effect of sunlight and evaporation. The pool's large volume of 498 m³ requires 2.1 * 10^9 J to increase the temperature by 1 degree Celsius, making rapid heating impractical.

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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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