How to calculate heat of combustion

Heat of combustion questions in the thermochemistry chapter catch students out for one reason: the mass you put into Q = mcθ is not the fuel. It is the mass of water being heated. Once you fix that, the calculation is short and the marks are easy. This guide gives you the definition, the method, a fully worked ethanol example, and the theory about why the number from a school experiment always comes out low.

The definition

Heat of combustion is the heat released when one mole of a substance is completely burned in excess oxygen. Combustion is exothermic, so the temperature of the water above the flame rises, and ΔH is negative. The fuels you meet at SPM are usually alcohols; the complete combustion of ethanol is:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

You can revise the reaction on our combustion of alcohols page and the energy idea on our heat of combustion page.

How the experiment works

A known volume of water sits in a copper can. A spirit burner of fuel is weighed, lit, and used to heat the water. You record the temperature rise of the water and the mass of fuel burned (from the fall in the burner’s mass). Two quantities matter:

• m in Q = mcθ is the mass of the water heated, not the fuel. Use density 1 g cm⁻³, so a volume of water in cm³ is its mass in g.
• n is the moles of fuel burned = mass of fuel burned ÷ molar mass of fuel.

The two formulas and the method

Heat absorbed by the water: Q = mcθ, with c = 4.2 J g⁻¹ °C⁻¹.

Per mole of fuel: ΔH = −Q ÷ n, in kJ mol⁻¹, negative because combustion is exothermic.

The steps are:

1. m = mass of water heated (from its volume).
2. θ = temperature rise of the water.
3. Q = mcθ, then divide by 1000 for kilojoules.
4. n = mass of fuel burned ÷ its molar mass.
5. ΔH = −Q ÷ n, with the negative sign.

Worked example: ethanol

A spirit burner of ethanol heats 250 cm³ of water. The temperature rises from 30.0 °C to 58.0 °C. The mass of the burner falls by 1.15 g. Find the heat of combustion of ethanol. (Molar mass of ethanol, C₂H₅OH = 46 g mol⁻¹.)

Step 1, mass of water. 250 cm³ of water has mass m = 250 g.

Step 2, temperature change. θ = 58.0 − 30.0 = 28.0 °C.

Step 3, heat released to the water.

Q = mcθ = 250 × 4.2 × 28.0 = 29 400 J = 29.4 kJ

Step 4, moles of ethanol burned.

n = 1.15 ÷ 46 = 0.025 mol

Step 5, heat of combustion.

ΔH = −29.4 ÷ 0.025 = −1176 kJ mol⁻¹

So the experiment gives a heat of combustion of about −1176 kJ mol⁻¹.

Why the experimental value is lower than the data-book value

This is a favourite explanation question. The value from a simple school experiment is always less than the accepted (data-book) value. The reasons are all forms of energy loss:

• Heat is lost to the surroundings, to the air, the copper can and the tripod, instead of all going into the water.
• Incomplete combustion produces some soot (carbon), so not all the fuel burns completely and less heat is released.
• Some fuel evaporates without burning.

Because heat escapes, the water heats up less than it should, Q is smaller, and the calculated ΔH is smaller in magnitude than the true value. Naming even two of these losses usually earns the mark.

The trend down the alcohol series

A common Paper 2 point: as you go up the alcohol homologous series (methanol, ethanol, propanol, butanol), the heat of combustion increases, it becomes more negative. Each extra –CH₂– group adds more carbon and hydrogen atoms, so more bonds form with oxygen during combustion and more heat is released per mole. You can be asked to plot this and describe the trend, so remember: bigger molecule, more energy released per mole.

Common mistakes to avoid

• Using the mass of fuel as m in Q = mcθ. m is the mass of water. The fuel mass is only used to find moles.
• Forgetting to convert Q to kilojoules before dividing by moles.
• Dropping the negative sign, combustion is exothermic.
• Explaining a low value with “the thermometer is wrong”. Examiners want heat loss, incomplete combustion or evaporation.

Practise the full method

Heat of combustion is quick once the roles are clear: water gives you m, the fuel gives you n. Work one alcohol question all the way through, including the “why is it lower” explanation, and the pattern sticks. If your child keeps putting the fuel mass into Q = mcθ, that is a two-minute fix with a teacher. Our online one-to-one lessons run in English from RM50 an hour, with a paid one-hour trial, see how it works to try this against real thermochemistry questions.