Heat of combustion

This page covers one Form 5 Thermochemistry content standard: the heat of combustion. It applies the same Q = mcθ method you used for neutralisation, displacement and precipitation, but here the fuel burns in air and heats water rather than reacting inside a solution. The calculation is a firm favourite in the exam, and the experiment also carries a classic “why is the value low?” discussion that examiners like to test.

What the heat of combustion is

The heat of combustion is the heat released when one mole of a substance is completely burned in excess oxygen. Combustion is the reaction of a substance with oxygen that releases heat, and complete combustion means there is enough oxygen for the fuel to burn fully, for a hydrocarbon or an alcohol, the products are carbon dioxide and water.

For example, the complete combustion of ethanol is:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

Combustion releases heat to the surroundings, so it is strongly exothermic, and ΔH is negative. The definition fixes the amount to one mole of the substance burned, so different fuels can be compared fairly.

Note the two conditions that must appear in a full definition: one mole of the substance, and complete combustion in excess oxygen. If oxygen is limited, incomplete combustion occurs instead, giving carbon monoxide or soot (carbon) and releasing less heat, a common comparison question.

The experiment and the calculation

In the school experiment, a measured mass of fuel in a spirit lamp is burned to heat a known volume of water in a copper can. You record the mass of fuel burned and the temperature rise of the water, then work through three steps.

1. Heat gained by the water, Q = mcθ, m is the mass of water in grams (density 1 g cm⁻³, so cm³ of water equals grams), c is the specific heat capacity of water (4.2 J g⁻¹ °C⁻¹), and θ is the temperature rise in °C.
2. Moles of fuel burned = mass of fuel burned ÷ molar mass of the fuel.
3. Heat of combustion, ΔH = −Q ÷ moles of fuel, in kJ mol⁻¹, written with a negative sign.

Worked example

Question. In an experiment, 0.92 g of ethanol (C₂H₅OH) is burned completely and the heat is used to raise the temperature of 200 cm³ of water from 28.0 °C to 55.0 °C. Calculate the heat of combustion of ethanol. (c = 4.2 J g⁻¹ °C⁻¹; density of water = 1 g cm⁻³; relative atomic masses: H = 1, C = 12, O = 16.)

Step 1, Mass of water. m = 200 × 1 = 200 g

Step 2, Temperature rise. θ = 55.0 − 28.0 = 27.0 °C

Step 3, Heat gained by the water. Q = mcθ = 200 × 4.2 × 27.0 = 22 680 J = 22.68 kJ

Step 4, Moles of ethanol burned. Molar mass of C₂H₅OH = (2 × 12) + (6 × 1) + 16 = 46 g mol⁻¹ Moles = 0.92 g ÷ 46 g mol⁻¹ = 0.020 mol

Step 5, Heat of combustion. ΔH = −Q ÷ moles = −22.68 kJ ÷ 0.020 mol = −1134 kJ mol⁻¹

Answer. The heat of combustion of ethanol is −1134 kJ mol⁻¹; the negative sign shows the reaction is exothermic.

Practice question

Question. 0.64 g of methanol (CH₃OH) is burned completely and heats 250 cm³ of water from 27.0 °C to 45.0 °C. Calculate the heat of combustion of methanol. (c = 4.2 J g⁻¹ °C⁻¹; density = 1 g cm⁻³; H = 1, C = 12, O = 16.)

Answer. Mass of water = 250 × 1 = 250 g. Temperature rise θ = 45.0 − 27.0 = 18.0 °C. Heat gained Q = mcθ = 250 × 4.2 × 18.0 = 18 900 J = 18.9 kJ. Molar mass of CH₃OH = 12 + (4 × 1) + 16 = 32 g mol⁻¹, so moles = 0.64 ÷ 32 = 0.020 mol. ΔH = −18.9 kJ ÷ 0.020 mol = −945 kJ mol⁻¹. The negative sign confirms combustion is exothermic.

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

A very common exam question asks why the heat of combustion measured in the school experiment is lower (less exothermic) than the accepted value. The standard reasons are all forms of heat loss:

• heat is lost to the surroundings, to the air and the copper can, rather than all going into the water,
• incomplete combustion of the fuel produces soot on the can and releases less heat,
• some fuel evaporates without burning, and heat is also lost by the water as it warms.

To reduce these errors you can use a windshield to reduce draughts, place the flame close to the can, and use a lid, but some loss always remains, which is why the experimental value falls short of the true value. Being able to state these reasons clearly is worth several marks.

Exam tip

Two mass values appear in a combustion question, and mixing them up is the classic error: use the mass of water in Q = mcθ, and the mass of fuel to find the moles of fuel. Keep the steps in order, mass of water, temperature rise, Q = mcθ, moles of fuel, then ΔH, and always finish with a negative sign and the unit kJ mol⁻¹. Learn the molar-mass calculation for common alcohols (methanol 32, ethanol 46, propanol 60) so you are quick under pressure. And be ready for the follow-up: if asked why your value is lower than the true value, give heat loss to the surroundings, incomplete combustion and evaporation of fuel as the reasons.

Where this fits

This is content standard 11.5 of the Thermochemistry chapter, and it is often the final calculation type in the chapter. It uses the exothermic idea and the Q = mcθ method from the earlier standards, and it links to fuels and alcohols from Carbon Compounds. Reinforce the definition and the heat-loss discussion with the chapter revision notes, and rehearse the two-mass calculation with the worked examples. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers make sure students never confuse the mass of water with the mass of fuel, because that single slip is the most common way marks are lost here, and this quantity is tested throughout SPM Chemistry.

Quick recap

• Definition: heat released when one mole of a substance is completely burned in excess oxygen.
• Combustion is exothermic, so ΔH is negative.
• Calculation: Q = mcθ (using the mass of water), then ΔH = −Q ÷ moles of fuel, in kJ mol⁻¹.
• The experimental value is lower than the data-book value because of heat loss, incomplete combustion and evaporation.