Heat of Combustion Calculator

Convert a fuel lower heating value into its heat of combustion.

Ten tabulated fuels, five energy-per-mass units, and mol or mmol mole counts.

Updated August 20, 2026

CreatorFrank Zhao

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What the heat of combustion means

Burn anything made of hydrogen and carbon and you get carbon dioxide, water vapour and heat. That water is the awkward part: turning liquid water into vapour takes energy, and in an open flame that energy leaves with the exhaust instead of warming anything useful. Engineers therefore quote a fuel on two different scales, and quoting the wrong one is the single most common way to miscompare fuels.

The lower heating value assumes the water leaves as vapour. The heat of combustion — also called the higher heating value or gross calorific value — assumes that vapour condenses back to liquid and hands its latent heat back.

This calculator does the arithmetic between the two. Give it a fuel's lower heating value and it adds the latent heat of the water the burn produced, working in moles so a fuel that makes more water per mole gets a bigger correction. It also runs the other way: type a measured heat of combustion and it recovers the lower heating value, the latent heat, or either mole count.

Fuel comparisons are usually made in British thermal units, which the U.S. Energy Information Administration describes as its standard unit for comparing energy sources. The calculator supports BTU per pound alongside the metric scales, and hydrogen is the extreme case: the EIA notes it has the highest energy content of any common fuel by weight.

Work out a heat of combustion

Choose the fuel

Pick a fuel from the menu and its tabulated lower heating value fills in, together with the burn that fuel actually performs — one mole of fuel against the moles of water its formula produces. Choose Custom to type a lower heating value of your own.

Check the heat of vaporization

Water is set to 2.257 MJ/kg, the value at its normal boiling point. Leave it alone unless you are modelling a different working fluid or a different temperature.

Adjust the mole counts if your burn differs

The two boxes hold moles of water vaporized and moles of fuel combusted. They may be given in millimoles; both scales are converted before the ratio is taken. Scaling both together changes nothing, because only their ratio matters.

Read the heat of combustion

It appears in the Combustion section in the same units you are working in. Typing a value there instead makes the calculator solve back for the lower heating value, the latent heat, or either mole count.

Check the number against the fuel

Select Methane, which loads a lower heating value of 50 MJ/kg and a burn of one mole of fuel against two moles of water. The heat of combustion comes out as:

H_{\mathrm{c}} = 50 + 2.257 \times \tfrac{2}{1}

=

54.514\ \mathrm{MJ/kg}

Condensing that water is worth 4.514 MJ per kilogram of methane, about nine percent on top of the lower heating value — the fraction a condensing appliance can chase and an open-flame appliance cannot.

Pro Tip: if your figures are in grams rather than moles, convert them first — the ratio in the equation counts particles, not mass. Our grams to moles calculatordoes that in one step, and the molar mass calculator supplies the molar mass when you need one.

The equation and its variables

The calculator applies one statement: the heat of combustion is the lower heating value plus the latent heat of the water that the burn produced, scaled by how many moles of water came out of each mole of fuel.

H_{\mathrm{c}} = \mathrm{LHV} + H_{\mathrm{v}} \times \frac{n_{\mathrm{H_2O}}}{n_{\mathrm{fuel}}}

The lower heating value, plus what condensing the exhaust water is worth.

\mathrm{LHV}

Lower heating value

Energy per unit mass released while the water stays vapour. Ten tabulated fuels are built in.

H_{\mathrm{v}}

Heat of vaporization

2257 J/g, which the menu reads as 2.257 MJ/kg — the energy one gram of water needs to evaporate.

n_{\mathrm{H_2O}}

Moles of water

How much water vapour the burn produced, in moles or millimoles.

n_{\mathrm{fuel}}

Moles of fuel

How much fuel went in, on the same mole scale as the water count.

\mathrm{n}/\mathrm{n}

Water per mole of fuel

the ratio

The two mole counts enter only as a ratio, so doubling both leaves the answer unchanged.

H_{\mathrm{c}}

Heat of combustion

The answer: the higher heating value, in the same energy-per-mass scale you entered.

Where the ratio comes from

A hydrocarbon burnt in excess oxygen follows a fixed pattern. For every mole of $\mathrm{C}_{a}\mathrm{H}_{b}$ fuel you get $a$ moles of carbon dioxide and $\tfrac{b}{2}$ moles of water, so the water yield is simply half the hydrogen count:

\mathrm{C}_{a}\mathrm{H}_{b} + \left(a + \tfrac{b}{4}\right)\mathrm{O_2} \rightarrow a\,\mathrm{CO_2} + \tfrac{b}{2}\,\mathrm{H_2O}

Methane $\mathrm{CH_4}$ gives 2, propane $\mathrm{C_3H_8}$ gives 4, and hydrogen $\mathrm{H_2}$ gives 1 — the numbers that fill the two mole boxes when a fuel is chosen.

Units, and how they are reconciled

Everything is normalised before the equation runs, so the menu's scales can be mixed freely:

kJ/kg and J/g are the same number

Both the numerator and the denominator scale by a thousand, so $1\ \mathrm{kJ/kg} = 1\ \mathrm{J/g}$ .

Megajoules and kilojoules

$1\ \mathrm{MJ/kg} = 1\ \mathrm{kJ/g} = 1000\ \mathrm{J/g}$

The imperial scale is exact

One BTU per pound is exactly $2.326\ \mathrm{J/g}$ — the BTU defined at 1055.05585262 J and the pound at 0.45359237 kg.

Millimoles are converted for you

$4000\ \mathrm{mmol} = 4\ \mathrm{mol}$ , so both counts can be quoted on whatever scale the exhaust analysis used.

Every direction of the same equation

Any one of the five boxes can be the unknown, as long as the other four are known — type the answer you measured and leave one box empty:

Recover the lower heating value

\mathrm{LHV} = H_{\mathrm{c}} - H_{\mathrm{v}} \times \tfrac{n_{\mathrm{H_2O}}}{n_{\mathrm{fuel}}}

Recover the latent heat

H_{\mathrm{v}} = \frac{H_{\mathrm{c}} - \mathrm{LHV}}{n_{\mathrm{H_2O}} / n_{\mathrm{fuel}}}

Recover either mole count

n_{\mathrm{H_2O}} = \frac{n_{\mathrm{fuel}} \left(H_{\mathrm{c}} - \mathrm{LHV}\right)}{H_{\mathrm{v}}}

Boundaries the calculator enforces

Every quantity must be greater than zero.
A specific energy may not exceed 150 kJ/g — hydrogen tops out near 120, and a hydrocarbon near 52.
A heat of vaporization may not exceed 5 kJ/g; water needs 2.257.

Numbers past those limits stay on screen but turn red with an explanation, so you can see what went wrong rather than watch a result vanish.

Getting an answer you can trust

The values the menu loads are the ones below. The water yield column is not a guess — it follows from each fuel's formula, and the last three entries are averages over the mixture that fuel actually is.

Fuel	Lower heating value (MJ/kg)	Water per mole of fuel	Formula
Hydrogen	119.96	1	H₂
Methane	50	2	CH₄
Natural gas	47.1	2	mainly CH₄
Ethane	47.62	3	C₂H₆
Propane	46.35	4	C₃H₈
Butane	45.75	5	C₄H₁₀
Pentane	45.35	6	C₅H₁₂
Kerosene	43	13	≈ C₁₂H₂₆
Diesel	43.4	17	≈ C₁₆H₃₄
Paraffin wax	41.5	26	≈ C₂₅H₅₂

Densest in energy: hydrogen at 119.96 MJ/kg, more than twice the next entry on the list.

Most water per mole: paraffin wax at 26 moles, the largest latent-heat correction on the list.

Four mistakes worth avoiding

Do not use a mass ratio

The equation wants moles of water per mole of fuel, not grams of water per gram of fuel. Methane burning 8 grams of water per 16 grams of fuel is the same as two moles of water per mole of fuel — but only the second reading is the one the formula expects.

Do not feed in a higher heating value

If the figure you already have is a gross calorific value, the water has been paid for once. Adding the latent term again double counts it, and the answer will read high by exactly that correction.

A named fuel locks its own value

While a fuel from the menu is selected, its tabulated lower heating value cannot be overwritten by the solver — the box is read-only. Switch to Custom to take ownership of it, which also opens the reverse solves for the lower heating value.

Blends move by a few percent

Natural gas, kerosene, diesel and wax are mixtures. Their tabulated heating values shift with composition and grade, so treat them as representative and replace them under Custom when you have a measured figure for your batch.

Related: once you know the burn, the air-fuel ratio calculator → gives the stoichiometric mass of air the same reaction demands.

Where this estimate stops

The vapour term is scaled by the mole ratio

This treats the latent heat as an energy per unit mass of fuel multiplied by the water yield, which is the convention this calculator follows. A full thermochemical balance also needs the fuel's molar mass, to turn the water yield into grams of water per gram of fuel. For a quick comparison of two fuels the convention is fine; for a certified calorific value, use a measured one.

Water is quoted at its normal boiling point

The 2.257 MJ/kg default is the value at 100 °C and one atmosphere. At other temperatures the latent heat is slightly different, and the calculator does not correct for that.

Complete combustion is assumed

Sooting, carbon monoxide or unburnt hydrocarbons all change the water yield and the energy released, and none of that is modelled here.

Tabulated values are typical figures

The ten fuels in the menu carry representative lower heating values. Real batches differ with grade, blend and moisture, so use Custom and your own measurement where the number matters to you.

An engineering estimate, not a certification

Do not use this as the sole basis for boiler sizing, emissions accounting, fuel purchasing or any safety decision. Where the result drives money or safety, have it confirmed by a laboratory calorimetry measurement.

Frequently Asked Questions

Why is the answer only a little above the lower heating value?

Because the correction is the latent heat of water — 2.257 MJ/kg — for each mole of water the burn produced per mole of fuel. Methane lifts 50 MJ/kg to 54.514 MJ/kg, about nine percent, while hydrogen lifts 119.96 MJ/kg to 122.217 MJ/kg, about 1.9 percent, even though hydrogen makes water twice as fast per mole. The ratio is the same, but hydrogen carries so much energy per kilogram that the same latent correction is a smaller share of it.

What is the difference between the heat of combustion and the lower heating value?

They differ by the latent heat of the water the fuel makes. The lower heating value assumes that water escapes as vapour, taking its evaporation energy with it; the heat of combustion assumes it condenses and that energy is recovered. The first is sometimes called the net calorific value, the second the gross calorific value or higher heating value.

Why does picking a fuel fill in two mole counts?

Because a named fuel states its own burn. Methane is $\mathrm{CH_4}$ , so one mole of fuel makes two moles of water; the boxes take those numbers so the ratio is the fuel's real stoichiometry without you having to look it up. You are free to rescale both — a batch of 3 moles of methane makes 6 moles of water — or to replace them entirely with a measured exhaust analysis.

Can it work backwards from a measured heat of combustion?

Yes. Type a heat of combustion and leave one other box empty, and the calculator solves for it: the lower heating value, the heat of vaporization, the moles of water or the moles of fuel. Choose Custom first if you want the lower heating value itself to be the unknown, since a named fuel keeps its tabulated value locked.

Why did my result turn red instead of disappearing?

The number is finite, so it is kept on screen — hiding it would tell you nothing. The red state means it broke one of the boundaries: a negative or zero input, a specific energy above 150 kJ/g, or a heat of vaporization above 5 kJ/g. Each case carries its own explanation next to the box, and the usual cause is a unit slip, such as a kilojoule figure entered where joules per gram belong.

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