Spirit-burner calorimetry: 5% low - bad data book or bad apparatus?
In a school laboratory a spirit burner is filled with methanol (M = 32 g/mol; the accepted standard enthalpy of combustion of liquid methanol is −726 kJ/mol). The burner with its fuel weighs 148.50 g. A copper calorimeter can of mass 180 g (specific heat capacity 0.385 J g⁻¹ K⁻¹) is filled with 500 g of water at 24.0 °C and set on a tripod above the burner, with a draught shield around the flame. The burner is lit and the stirred water is heated until the thermometer reads 39.0 °C; the flame is then put out and the burner is reweighed at 147.00 g. The flame was yellow and left a black deposit on the base of the can. The student computes the enthalpy of combustion, obtains a value about 5% smaller in magnitude than −726 kJ/mol, and concludes that "the textbook value must be wrong". His partner argues that the apparatus, not the textbook, is at fault. Take the specific heat capacity of water as 4.18 J g⁻¹ K⁻¹.
1. Calculate the experimental enthalpy of combustion, using both the water and the can, and state the percentage difference from the accepted value.
Read the worked answer
Heat gained by the water: 500 × 4.18 × 15.0 = 31,350 J Heat gained by the copper can: 180 × 0.385 × 15.0 = 1,040 J Total heat released ≈ 32,390 J = 32.39 kJ
Methanol burned = 148.50 − 147.00 = 1.50 g = 1.50/32.0 = 0.046875 mol
ΔcH = −32.39/0.046875 = −691 kJ/mol
Comparison: (726 − 691)/726 = 4.8% smaller in magnitude - the "5%" the student noticed.
Note what this calculation establishes: the arithmetic is internally sound. The discrepancy is real, but its cause has not yet been identified, and nothing here says the accepted value is wrong.
2. Judge which explanation fits the gap. Quantify the effect of ignoring the can, and decide whether the partner's conclusion is justified.
Read the worked answer
Ignoring the can: −31.35/0.046875 = −669 kJ/mol, i.e. 7.9% low. So including the can recovers 2.9 percentage points; the remaining gap of about 5% must come from physical effects, not arithmetic.
Every observation points the same way - each lowers the magnitude:
- Heat loss through the can walls and by convection/draught, despite the shield.
- Water evaporating from the open can, carrying away latent heat that the thermometer never registers.
- Unburnt methanol vapour escaping, so the 1.50 g mass loss overstates the fuel actually oxidised.
- Yellow flame with soot on the base: incomplete combustion to CO and carbon releases less energy per mole than complete combustion to CO2.
Since all four are loss mechanisms, the partner is right: a value ~5% low is exactly what this apparatus should give. The textbook value is not in question at this precision - the correct conclusion is "my apparatus has systematic losses", supported by the specific evidence of the sooty flame and the open can.
3. The partner proposes repeating the run with a lid and lagging. Should the result be expected near −700, −720, or −750 kJ/mol? Explain the reasoning, and say what a definitive measurement would require.
Read the worked answer
Expect something close to but not exceeding the accepted value - roughly −710 to −720 kJ/mol.
Reasoning: lagging and a lid remove the heat-loss term, and a lid also cuts evaporation, so the magnitude rises. But incomplete combustion and escaping methanol vapour remain, so the value should still fall slightly short of −726 kJ/mol.
A result of −750 would be larger in magnitude than the accepted value, and no flaw in this apparatus can produce that - every error here destroys heat or destroys fuel, so it can only shrink the measured magnitude. A number beyond −726 is therefore a warning that something has been miscalculated (for example heat absorbed from the surroundings, or an over-estimate of fuel burned).
A definitive value requires a bomb calorimeter: combustion in excess oxygen (so complete), a sealed constant-volume vessel, and a heat capacity calibrated electrically by the same procedure, so that heat loss is measured rather than guessed.