A power estimate for heating a pool.

It is spring break, so we are at my parents house for a couple of days. The kids like it because there is a pool, a heated pool even. It really isn't that cold outside, but yesterday the water measured at 62 oF. So, with some help from the kids, we cleaned out the pool and turned on the heater. We also put a cover on it, hopefully to help it heat up some more.

This is perfect for a quick calculation. Is it reasonable that the pool could get up to a swimable temperature by tomorrow? Let me first make some assumptions and data:

  • 15,000 gallons of water in the pool. This is about 57 m3.
  • Initial temperature of water = 62 F or 290 K.
  • Final temperature of water = 75 F or 297 K.
  • Time to heat = 24 hours = 86,400 seconds.
  • Specific heat of water = 4.186 J cm-3K-1 = 4.186 x 106 m-3K-1.
  • Assume the pool is thermally isolated (clearly, not true)

How much energy would this take? Note that I am using the volumetric specific heat.


Probably not the correct convention, but I wrote cvol to represent the volumetric specific heat. I think writing it that way typically means specific heat at constant volume. Now for the power:


That is a lot, but at least it is not 1.21 Jigawatts! Clearly my 2 kWatt portable generator could not do the job.

I started writing this post when the water temp was 62 F, sure enough the next day the water was about 73 F. Clearly this is possible. Also, the assumption that the pool is thermally isolated. I don't think this is a bad assumption. Probably the water looses some energy to the ground, but gains some from sunlight. Even stephen.

One more thing. How much does this cost and am I glad that I am not paying the bill? I will use a price of $0.12 per kiloWatt hour. Note that a kW hr is a unit of energy, not power. I could go through how to convert Joules to kilowatt hours, or you could just read my "unit conversions for mere mortals". I will use google to convert this. Just type in google "1.67e9 J in kilowatt hours" and BOOM. Google says 464 kilowatt hours. This would give a heating cost of:


Obviously, I can not remember how to put a $ in LaTeX.

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I don't say no such thing exists, but I've never seen an electric pool heater. The ones I've seen burn natural gas. Clearly that wouldn't change how much many joules it would require to raise the water temperature, but you'd be purchasing cubic feet of natural gas. The calculation would require determining how much heat energy is released by oxidation of an amount of methane.

But maybe I'm wrong, maybe it is an electric heater.

Paint the pool bottom black and let the ground level solar constant, about 500 watts/m^2, do the work for free. One suspects there is a law against it, so... reversibly lay down a sheet of black Geo-Textile.

To follow up on Uncle Al, choose a pool covering with a color that has a high emissivity in the visible (looks black to us), but a low emissivity in the infrared (looks white at longer wavelengths). Minimizing re-radiated power is sometimes as important as maximizing absorbed power.

And as for $, generating it is the same as all the other reserved symbols in LaTeX: put a slash in front (/$).

By Anonymous Coward (not verified) on 14 Apr 2009 #permalink