## Need a maths geek to answer a question about boiling water.

Hi all. I'm finally starting the prototype of my solar condenser this week, though this question doesn't relate to it directly, but to using it to power a steam distiller to purify water.
How much water will ten kilowatt hours boil?
ie, if I have one kilowatt of solar energy coming in over ten hours and 100% of it goes into heating the water.
Also, for extra credit, assume the water is in a black plastic container also in the sun, so the water is preheated to whatever temperature that would be likely to heat it to.
cheers!

active| newest | oldestkelseymh'svalues, I calculate that you could vapourise (at boiling-point) ~16Kg of water. Or you could heat more than 100Kg to 100^{o}.Combining the two ~14Kg of water. (~30 pounds, 3.7 US Gal)

L

kelseymhsaid you ought to use this as a learning opportunity and work it out. Which is why I didn't post the equation.But having the water pre-heated isn't going to make that much difference, as it's only 16Kg at 100

^{o}CL

Energy required to boil water

E_{V}= H_{v}* m(specific heat of vaporisation * mass)Energy required to heat water

E_{H}= SHC * dt * m(specific heat capacity of water * change in temperature * mass)Total energy

E_{T}= E_{V}+ E_{H}= (H_{v}+ (SHC * dt)) * mIf you want energy to be a known and mass to be an unknown, rearrange for "m="

m = E_{T}/ (H_{v}+ (SHC * dt))The ever-useful Wolfram Alpha tells me H

_{v}is 2.23*10^{6}J/kg, and SHC is 4180 J/kg K. This meansm = E_{T}/ (2230000 + (4180 * (100 - starting temperature)))For ten kilowatt hours (3600 * 1000 * 10 joules) this gives

m = 36000000 / (2230000 + (4180 * (100 - starting temperature)))Calculating the value of m for starting temperatures between 0

^{o}C and 100^{o}C shows the starting temperature doesn't make much difference, most of the energy goes into boiling the water. At 20^{o}C you could boil 13.8L, at 80^{o}C you could boil 15.6L.Yay algebra :)

Here in the west we use maybe 6-10 litres of water flushing the toilet half a dozen times a day and 50 litres having a shower. Don't get me started on washing the car with a hose and lawn sprinklers...

I suspect that while you are thinking of concentrated solar for boiling water, you don't need to go that high tech- distilling wated with plain solar energy is a well established concept, which could always do with better engineering to make sturdier, cheaper and higher-yield designs but doesn't need "overengineering".

- heat capacity of water (74.539 J/mol/K)
- heat of vaporization of water (40.65 kJ/mol)

The total energy to boil water is just the sum of the energy to raise water from an initial temperature to 100 C, plus the heat of vaporization.Conversion factors:

- liters/mol (18 mL/mol)
- watts = joules/s, so 1 kWh = 3.6 MJ

Initial conditions:- temperature of water (25 C ?)
- quantity of water

Now you can do the arithmetic yourself. Go buy a calculator.