Spirit Burners and Cost per Megajoule: Choosing a Camping Fuel
A hillwalking club wants to move away from methylated-spirit stoves and asks a chemistry student to compare ethanol and butan-1-ol as fuels.
The student burns each fuel in a spirit burner under a copper can holding 150 g of water, using a draught shield and continuous stirring. The calorimeter was calibrated electrically beforehand and has a heat capacity of 85 J K⁻¹ in addition to the water.
| Fuel | Mass burned | Temperature rise |
|---|---|---|
| Ethanol (Mr 46.1) | 1.20 g | 42.0 °C |
| Butan-1-ol (Mr 74.1) | 0.95 g | 38.5 °C |
Data book standard enthalpies of combustion: −1367 kJ mol⁻¹ (ethanol) and −2676 kJ mol⁻¹ (butan-1-ol).
The club needs about 15 MJ of useful heat for a three-day trip and can buy ethanol at £4.50 per litre (density 0.789 g cm⁻³) or butan-1-ol at £12.00 per litre (density 0.810 g cm⁻³). The student's draft report claims the results "prove" that butan-1-ol is the better fuel.
1. Calculate the experimental enthalpy of combustion of each fuel, including the calorimeter heat capacity, and express the discrepancy with the data book as a percentage. Explain the direction of the error.
Read the worked answer
Use q = (m_water × c_water + C_calorimeter) × ΔT.
Combined heat capacity = (150 × 4.18) + 85 = 627 + 85 = 712 J K⁻¹.
Ethanol
- q = 712 × 42.0 = 29 904 J = 29.90 kJ
- n = 1.20 / 46.1 = 0.02603 mol
- ΔHc = −29.90 / 0.02603 = −1149 kJ mol⁻¹
- Error = (1367 − 1149)/1367 × 100 = 15.9 %
Butan-1-ol
- q = 712 × 38.5 = 27 412 J = 27.41 kJ
- n = 0.95 / 74.1 = 0.01282 mol
- ΔHc = −27.41 / 0.01282 = −2138 kJ mol⁻¹
- Error = (2676 − 2138)/2676 × 100 = 20.1 %
Both magnitudes are too small - the measured values are always less exothermic than the true value, never more. Reasons:
- Heat loss to the surroundings by convection and radiation, plus heat used to warm the can, thermometer and air. This heat is produced by the fuel but never reaches the water.
- Incomplete combustion - a spirit burner gives a luminous, slightly smoky flame and deposits soot on the can, so some carbon is oxidised only to CO or to soot.
- Evaporation of fuel from the wick contributes to the measured mass loss without contributing heat, so n is overestimated and |ΔH| underestimated.
- Non-standard conditions: products are not returned to 298 K and some water leaves as vapour.
Note the value of the calibration: ignoring the 85 J K⁻¹ would have given −1012 and −1883 kJ mol⁻¹ (errors of 26 % and 30 %). Correcting for the calorimeter's own heat capacity is not optional.
2. Using the theoretical enthalpies of combustion and the quoted prices, calculate the volume and cost of each fuel needed to deliver 15 MJ, and recommend one.
Read the worked answer
Convert molar enthalpy to energy per cm³.
Ethanol: 1367 / 46.1 = 29.7 kJ g⁻¹; × 0.789 = 23.4 kJ cm⁻³ Butan-1-ol: 2676 / 74.1 = 36.1 kJ g⁻¹; × 0.810 = 29.3 kJ cm⁻³
Volume and cost for 15 000 kJ
| Ethanol | Butan-1-ol | |
|---|---|---|
| Volume needed | 15 000 / 23.4 = 641 cm³ | 15 000 / 29.3 = 512 cm³ |
| Cost | 0.641 × £4.50 = £2.88 | 0.512 × £12.00 = £6.14 |
| Cost per MJ | £0.19 | £0.41 |
Recommendation. On cost, ethanol wins by more than a factor of two. Butan-1-ol's advantage is energy density: about 25 % more energy per gram and per cm³, so the club carries roughly 130 cm³ (≈105 g) less fuel. That saving is irrelevant for a three-day trip, but it matters if the trip is long or cold, because fuel mass and canister volume then dominate.
Practical points that should also be weighed:
- Butan-1-ol has the higher flash point (≈35 °C vs ≈13 °C for ethanol), so it is safer to carry and to refill in warm weather; ethanol's flame is also pale blue and easy to miss.
- The two percentage errors (16 % vs 20 %) are similar and both in the same direction, so the experimental data would rank the fuels the same way as the theoretical values - the comparison is still valid even though the absolute numbers are not.
3. Critique the student's claim that the results "prove" butan-1-ol is the better fuel.
Read the worked answer
The word prove is unjustified on three separate grounds.
1. The absolute values are wrong by 16–20 %. A method that loses a fifth of the heat cannot establish anything numerically; it can only make a relative comparison, and only because both fuels were burned in the same apparatus with the same systematic losses. Even that comparison needs the same ΔT range and the same flame position, which were not controlled.
2. The comparison is being made per mole, which is the wrong basis for a fuel. Butan-1-ol has the larger |ΔHc| per mole simply because a mole of it contains four carbons against ethanol's two. The user buys litres and carries grams. Per gram the gap narrows (29.7 vs 36.1 kJ g⁻¹), and once price is included the ranking reverses: ethanol is less than half the cost per megajoule.
3. "Better fuel" is not a single property. It depends on the criterion chosen - cost per MJ (ethanol), energy per unit mass or volume (butan-1-ol), flash point and handling safety (butan-1-ol), soot and can cleanliness (ethanol burns cleaner in this setup), or availability. The student's report should state the criterion explicitly before making a recommendation.
What a defensible report would say: "Under the conditions used, both fuels gave combustion enthalpies 16–20 % below the data book values; on a cost-per-megajoule basis ethanol is preferable (£0.19 vs £0.41 per MJ), while butan-1-ol offers 25 % more energy per unit volume and a higher flash point."