A simple way of looking at it is that you need ~5 grams of fuel to raise 2 cups of water from 70F to 212F so that is about 5 grams to raise the temperature by ~150 F or 30 F per gram of fuel. So for every 30F lower than 70 F (water at 40 F) you need to use 1 gram more fuel.
Stove Efficiency is simply the Energy required to boil water/Energy available. Once you know the Efficiency, you can calculate fuel consumption based upon water volume, water temperature etc.
Energy required to boil water Mw*cp*ΔT
Mw=Mass (g)
Cp= Specific Heat Capacity (4.186 J/gC)
ΔT =difference in temperature (C)
Energy Available (Isobutane) Specific Energy*Mass
Energy Density (isobutane) – (45208.8 J/g)
Mg = Mass of gas (g)
Efficiency = (Mw*cp*ΔT)/(Specific Energy*Mg)
A stove boiling 2 cups (473 ml) of 70 F water to 209 F using 4.9 grams of fuel has an efficiency of 70%.
Now what happens if you are boiling 500ml of 45 F water at 10,000 feet, how much fuel is required?
The boiling point of water at 10,000 feet is 194 F (or 90 C) with a starting point of 45 F (7.2 C)
0.74=(500g*(4.186 J/gC)*(90C-7.2C))/(( 45208.8 J/g)*(Mg))
0.74=(473*4.186*82.8)/(45.2088*Mg)
Solving for Mg = 5.5 g of fuel
My 2 cents
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