Would it really require 44 car batteries to heat my pool? After doing some research and math, I 'discovered' that it would take 44 (give or take 20%) car batteries to heat 1,000 gallon pool by 10 degrees.  Is this right or am I missing something?  It seems a little crazy!


*

*It takes 8.3 BTUs to heat one gallon of water 1 degree F.

*Thats 83,000 BTUs to heat the entire pool 10 degrees.

*A car battery holds 2,000,000 joules.

*2,000,000 joules is 1900 BTUs.

*44 car batteries hold 38,600 BTUs.

 A: From wikipedia
Average private pool length = 7.3m
 Average private pool breadth = 3.7m
 Average private pool depth = 2m

This implies a top area of $27.01m^2$ and a volume of $54.02m^3$ which then means $54020 kg$ water.
emissivity of water is generally 0.98 Let your pool be $\text{t K}$ above ambient temperature. This would result in 
$$P_{rad} = 5.6704\times 0.98 \times 27.01 \times t^4 \times 10^{-8}\text{ Watt}$$
$$P_{rad} = 1.501 \times t^4 \times 10^{-6} \text{ Watt}$$
At least this much power will be required to maintain the pool at desired temperature. This seems manageable.
Now to raise the temperature of pool by $\text{t K}$
Specific heat of water is 4.187 kJ/kgK
$$E_{req} = 54020 \times 4.187 \times t \text{ kJ}$$
$$E_{req} = 2.26181.74 \times t \times 10^5\text{ kJ}$$
number of car batteries having $2000 \text{ kJ}$ energy required would be 
$$ \text{no. of batteries} = 113.09087 \times t$$
What you get with $44$ batteries, total energy = $88,000 \text{ kJ}$, how much water that heats to t K
$$m_{water} = \frac{8.8 \times 10^{4} \text{ kJ}}{4.187 \text{ kJ/kgK} \times t\text{ K}}$$
$$m_{water} = \frac{44901}{t} \text{kg}$$
$$vol_{water} = \frac{44.901}{t} m^3$$
Check for various temperatures yourself !
Note the amount of water $1000$Gallon if it were to be contained in a cube would be something like this

Clearly this does not look like a pool at all, maybe a hot tub or simply a bath tub.
