Spirit-burner combustion: how far can a school calorimetry result be trusted?
In a school laboratory, a candidate burns ethanol in a spirit burner beneath a copper can holding 50.0 g of water. The water temperature rises from 21.0 °C to 55.5 °C while the burner loses 0.92 g of ethanol. The candidate takes c(water) = 4.18 J g⁻¹ K⁻¹ and Mr(ethanol) = 46.0, calculates an enthalpy of combustion, and compares it with the data-book value of −1367 kJ mol⁻¹. The flame is visibly yellow, the base of the can is coated with black deposit, and there is no draught shield. A second candidate repeats the procedure with propan-1-ol using the same can, burner and water volume, keeping the burner-to-can distance fixed, then argues that the two experimental values can be compared with each other even though both are far below the data-book values. The teacher points out that the can itself is heated and that some ethanol vapour escapes from the wick without burning.
1. Calculate the experimental enthalpy of combustion of ethanol from the candidate's data and explain the sign of the value.
Read the worked answer
Heat gained by the water q = m c ΔT = 50.0 g × 4.18 J g⁻¹ K⁻¹ × (55.5 − 21.0) K = 50.0 × 4.18 × 34.5 = 7210 J = 7.21 kJ
Amount of ethanol burned n = 0.92 g ÷ 46.0 g mol⁻¹ = 0.0200 mol
Enthalpy of combustion ΔHc = −q ÷ n = −7.21 kJ ÷ 0.0200 mol = −361 kJ mol⁻¹ (about 3 significant figures)
The sign is negative because the water temperature rises: the combustion is exothermic and heat flows from the system (ethanol + oxygen) to the surroundings. The measured q is the heat gained by the water, so it must be given a negative sign when attributed to the fuel. Note that the calculation assumes all the heat released is absorbed by the water only - it is not, which is why the value is only about 26% of −1367 kJ mol⁻¹.
2. Identify the two most significant sources of error and argue, using the observations, which one you judge to be dominant. State one measurement that would settle the question.
Read the worked answer
Candidate errors
- Heat losses to the surroundings and to the apparatus. The copper can, the air around it and the bench all absorb heat; the can's heat capacity is ignored in q = mcΔT, and an unshielded flame loses much of its heat by convection past the can.
- Incomplete combustion and unburned fuel. The yellow, luminous flame and the black soot on the can show that some carbon is not converted fully to CO₂ (soot and CO release far less heat per mole than complete combustion to CO₂). Ethanol vapour escaping from the wick is also unburned.
Judgement I would judge heat loss dominant, because the apparatus is open on all sides and the flame is much wider than the base of the can; only a fraction of the hot gas stream actually touches the can. The soot deposit shows incomplete combustion is occurring, but a luminous flame still burns most of the carbon to CO₂, so this alone is unlikely to remove ~74% of the energy. Heat loss, by contrast, can easily account for the bulk of the shortfall.
Discriminating measurement Burn the same mass of ethanol in a bomb calorimeter (sealed, excess O₂, known total heat capacity), or analyse the exhaust gas for CO and soot. If the bomb value is close to −1367 kJ mol⁻¹, the discrepancy lies in heat loss from the school apparatus, not in the fuel.
3. Evaluate the second candidate's claim that the two experimental values can still be compared with each other.
Read the worked answer
What the claim gets right Both fuels are burned in the same apparatus with the same water volume and the same burner-to-can distance, so the systematic losses (convection past the can, heating the can) are of a similar type. If those losses were exactly proportional for the two fuels, the ratio of the two experimental values would approximate the ratio of the true values.
Why it is nonetheless weak The loss is not proportional: propan-1-ol is less volatile and burns with a sootier flame than ethanol, so the fraction of heat lost and the degree of incomplete combustion differ between the two fuels. Each experimental value is therefore biased by a different factor, and a 25%-of-true figure for one fuel does not imply a 25% figure for the other. Any error that depends on the fuel (volatility, sootiness, water produced condensing on the can) breaks the comparison.
Verdict and condition for fairness The claim is only defensible as a rough ranking of the two fuels, not as a quantitative comparison. To make it fair: use a shielded burner at a fixed distance, burn similar masses for similar times, plot ΔT against moles burned for each fuel, and compare the gradients obtained under identical conditions.