An atomic battle results in your swimming pool (10m long, 5m wide, and 1.5m deep) rising in temperature from 20 to 100°c. how many joules does the water absorb

Here's the Physics we need in order to answer the question:

The specific heat of water is . . . 4185.5 Joule per kilogram per °C.

That's IT !  Everything else is arithmetic.

Volume of the pool = (10m) x (5m) x (1.5m) = 75 cubic meters

Density of water = 1,000 kg/m²

Mass of water in the pool = (75 m³) x (1,000 kg/m³) = 75,000 kg

Change in temperature = (100°C) - (20°C) = 80°C .

Energy absorbed by the water in the pool to raise the temp 80°C:

Energy = (4185.5 J/kg-°C) x (75,000 kg) x (80°C)

Energy = (4185.5 · 75,000 · 80) (J · kg · °C / kg · °C)

Energy = 2.511 x 10¹⁰ J

Energy = 25.11 GigaJoules

But honestly ... the event that caused your swimming pool to get uncomfortably hot was the beginning of a nuclear war, so you probably have more important things to worry about right now.

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12000j is work done