Spirit-Burner Calorimetry: When the Data Book Is Not Wrong
A student burns ethanol from a spirit burner beneath a copper calorimeter containing 200.0 g of water. The burner weighs 185.42 g before the experiment and 183.96 g afterwards. The water temperature rises from 18.5 °C to 47.0 °C. Taking the specific heat capacity of water as 4.18 J g⁻¹ K⁻¹, she calculates the heat gained by the water and divides by the number of moles of ethanol burned, obtaining a value of about −750 kJ mol⁻¹. The data book gives −1367 kJ mol⁻¹. In her report she writes: 'The experiment is invalid and the data-book value must be wrong.' Her partner points out that the calorimeter itself becomes too hot to hold, that a black deposit has formed on the base of the calorimeter, and that the flame flickered in the draught of the fume cupboard.
1. Reproduce the student's calculation and state clearly what the sign of the answer means.
Read the worked answer
Temperature change: ΔT = 47.0 − 18.5 = 28.5 K
Heat gained by the water: q = m c ΔT = 200.0 g × 4.18 J g⁻¹ K⁻¹ × 28.5 K = 23 826 J = 23.83 kJ
Moles of ethanol burned: mass burned = 185.42 − 183.96 = 1.46 g n = 1.46 g / 46 g mol⁻¹ = 0.03174 mol
Enthalpy of combustion: ΔH = −q/n = −23.83 kJ / 0.03174 mol = −751 kJ mol⁻¹
The sign is negative because the reaction releases heat to the surroundings - the water's temperature rose, so the system (ethanol) lost energy. The calculation itself is arithmetically correct; the question is whether it measures what it claims to measure.
2. Identify the main sources of error and explain why the experimental value is numerically much smaller in magnitude than the data-book value.
Read the worked answer
Every significant error in this apparatus makes the measured heat smaller than the heat actually released, so the experimental figure is a lower limit, not a wrong data-book figure:
- Heat loss to the surroundings: much of the flame's heat warms the air, the bench and the burner, not the water. Only a fraction of q reaches the water that is being measured.
- Heat absorbed by the calorimeter: the copper calorimeter also heats up, but the calculation only counts the water. Correcting would need q = (m_water c_water + C_calorimeter) ΔT.
- Incomplete combustion: the black deposit is soot, i.e. carbon that was not fully oxidised to CO₂. Less energy is released per mole of ethanol vaporised, and some ethanol may escape unburned or evaporate without burning.
- Draught: the flickering flame loses heat sideways and burns less efficiently.
All of these lower the measured heat release while the moles burned are still counted in full, so the magnitude falls from 1367 to about 750 kJ mol⁻¹. The data book is not invalidated by a calorimeter that loses heat.
3. Suggest two specific improvements, and explain for each one whether it would move the result towards −1367 kJ mol⁻¹ and why.
Read the worked answer
Lag/insulate the calorimeter and shield the flame from draughts (e.g. a draught shield and a lid). This cuts conduction, convection and radiation losses, so a larger fraction of the heat released is captured by the water. The measured temperature rise increases and the calculated magnitude moves towards 1367 kJ mol⁻¹.
Ensure complete combustion and better heat transfer - a larger air supply and positioning the burner so the flame touches the base of the calorimeter. Complete oxidation to CO₂ releases the full energy per mole, and less soot forms; both raise the measured heat and therefore the magnitude of ΔH.
Include the calorimeter's heat capacity in the calculation. This increases q for the same ΔT, raising the magnitude.
A bomb calorimeter under excess oxygen would give a value close to the data-book figure because it removes both the heat-loss and the incomplete-combustion problems; a school spirit-burner set-up can realistically only give a value well below the true one.