Theoretical Limits of an Alcohol Stove

I’ve been playing with different alcohol stoves (lightweight backpacking ones, that is). Building them, researching them, etc. My main goal is minimal fuel to boil one pint of water (using some lightweight cooking set). Boil time isn’t much of a concern for me.

The real competitor to alcohol stoves in terms of fuel weight are Esbit stoves; they’ll boil a pint using a 14 gram fuel tablet, and can probably be marginally lighter than an alcohol cookset. (The negatives are filthier burning and the cost of the fuel.) So it’s my goal to boil a pint with 14g of alcohol (by which I mean ethyl alcohol). Ethyl alcohol (aka ethanol) has a density of 0.789 g/cm3, so that’s about 0.63 fluid ounces by volume. Quite doable, I’m sure, but I haven’t done it yet.

That got me thinking about the theoretical limit for a 100% efficient cookset. (And let me say right off the bat that I am not a chemist of any sort.) Alcohol burns per the reaction

C2H5OH(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

releasing −1409 kJ/mol. Once calorie is 4.184 J, and the molar mass of ethanol is 46.06844 g/mol, so that’s 7310 cal/g of ethanol burned.

Meanwhile, a pint of water is 472g. A calorie is (by definition) the energy need to raise one gram of H20 one degree Celsius. Let’s assume we’re at NIST standard temperature, or 20°C (68°F). At sea level, water boils at 100°C; higher up, boiling point goes down, but so does the starting temperature around dinner or breakfast time, broadly speaking, so we shouldn’t be too terribly far off from what might happen up in the mountains.

In any case, to raise 472g of water 80°C will take 37760 cal. A 100% efficient alcohol stove then would need 37760 cal / 7310 cal/g = 5.17g of alcohol fuel. The stove I’m looking for (which will burn 14g), then, would be 37% efficient. Stoves that consume 1 fluid oz of alcohol, or 23.3g (a fairly common case), are 22% efficient.

There you have it – if you can boil a pint of water with 1/4 fl oz of ethanol, you’ve got a hell of a stove on your hands that’s getting close to the theoretical maximum efficiency.

(One minor wrinkle in all this is that the water produced by the combustion is liquid. I suspect that that water is inevitably vaporized during combustion, so that might need to be subtracted out of the energy produced by the combustion. If so, the heat of vaporization of water is 0.65 kJ/mol. Each mole of alcohol burned produces three moles of liquid water, so the net energy produced my burning ethanol would be 1409 kJ/mol – 3 * 0.65 kJ/mol, or 1407 kJ/mol. It’s pretty much a non-issue.)

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