Jet fuel releases roughly 43,000 kJ per kilogram. Wood releases about 15,000 kJ. That 3× difference isn't guesswork — it's measured in a calorimeter. The principle is simple: burn the fuel, heat water, and watch the thermometer.
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Why does a small block of chocolate contain more kilojoules than a whole bowl of lettuce? The answer isn't just about quantity — it's about what the molecules are made of and how much energy is stored in their bonds.
If you burned both in a laboratory, you could measure exactly how much heat each released. How would you design that experiment? What would you actually be measuring, and what would you need to keep constant to make a fair comparison?
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📚 Core Content
The spirit burner calorimeter is a simple, imperfect device — understanding its setup is inseparable from understanding why experimental ΔHc values are always lower than the true value.
Components: spirit burner containing the fuel (e.g. ethanol, methanol), a copper calorimeter holding a known mass of water, a thermometer with its bulb fully submerged, and a draught shield around the apparatus.
Procedure (in order):
Copper is chosen for the calorimeter because it has high thermal conductivity — heat transfers quickly from the flame to the water, minimising lag between the flame and the thermometer reading.
Every spirit burner experiment gives a ΔHc that is less negative than the true value — the key skill is identifying exactly why, and stating the direction of each error's effect.
The experimental ΔHc is always less negative (lower in magnitude) than the accepted value because heat is lost before it can be transferred to the water. In HSC questions, you must name the specific physical source of error and state its directional effect on the result.
🔧 Worked Examples
A student burns ethanol (C₂H₅OH, M = 46.07 g mol⁻¹) in a spirit burner. The mass of the burner decreases from 152.34 g to 151.62 g. The calorimeter contains 200.0 g of water. The water temperature rises from 19.5°C to 31.2°C. Calculate the molar enthalpy of combustion of ethanol.
The accepted molar enthalpy of combustion of methanol (CH₃OH) is −726 kJ mol⁻¹. A student's experiment gave −412 kJ mol⁻¹. (a) Calculate the percentage error. (b) Suggest two specific sources of error that would explain why the experimental value is less negative.
🧪 Activities
a Calculate the mass of propan-1-ol burned and convert to moles.
b Calculate the heat absorbed by the water (q in kJ).
c Calculate the experimental ΔHc and the percentage error compared to the accepted value.
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1 The student used m = 0.80 g (mass of fuel) instead of m = 200.0 g in the q = mcΔT step. What ΔHc do they calculate, and is it more or less negative than the correct value?
2 Due to forgetting to cap the burner, an extra 0.10 g of butanol evaporated after the flame was extinguished. The student records mass of fuel burned as 0.90 g (instead of 0.80 g). How does this affect ΔHc?
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Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: Incomplete combustion is safer than complete combustion because it produces less CO₂.
Right: Incomplete combustion produces toxic carbon monoxide (CO) and particulate carbon (soot), which are deadly. Complete combustion produces CO₂ and H₂O, which are non-toxic. The goal is always complete combustion for safety and efficiency.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Explain why, in combustion calorimetry using a spirit burner, the value of 'm' in the formula q = mcΔT must be the mass of water and not the mass of fuel. In your answer, state what q represents and why the specific heat capacity of water is used. 3 MARKS
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7. A student burns 0.85 g of methanol (CH₃OH, M = 32.04 g mol⁻¹) and heats 180.0 g of water from 21.0°C to 34.5°C. (a) Calculate q in kJ. (b) Calculate ΔHc. (c) The accepted value is −726 kJ mol⁻¹. Calculate the percentage error and identify one specific source of error that explains why the experimental value is less negative. 5 MARKS
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8. Real-World Application: Fuel engineers compare fuels by their molar enthalpy of combustion (ΔHc) per gram (energy density). The table below shows data for three fuels.
| Fuel | Formula | M (g mol⁻¹) | ΔHc (kJ mol⁻¹) | Energy per gram (kJ g⁻¹) |
|---|---|---|---|---|
| Methanol | CH₃OH | 32.04 | −726 | Calculate |
| Ethanol | C₂H₅OH | 46.07 | −1367 | Calculate |
| Octane (petrol proxy) | C₈H₁₈ | 114.23 | −5471 | Calculate |
(a) Complete the energy per gram column (kJ g⁻¹ = |ΔHc| ÷ M). (3 marks)
(b) Using your calculated values, explain which fuel is most efficient per gram and suggest why octane is preferred in internal combustion engines over methanol. (2 marks)
5 MARKS
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Go back to your Think First response. Now that you've studied combustion calorimetry:
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(a) m(fuel) = 188.45 − 187.81 = 0.64 g; n = 0.64 ÷ 60.10 = 0.01065 mol
(b) ΔT = 32.6 − 18.2 = 14.4°C = 14.4 K; q = 250.0 × 4.18 × 14.4 = 15,048 J = 15.048 kJ
(c) ΔHc = −15.048 ÷ 0.01065 = −1413 kJ mol⁻¹; % error = |−1413 − (−2021)| ÷ 2021 × 100 = 608/2021 × 100 = 30.1%
Correct values: n = 0.80/74.12 = 0.01079 mol; q = 200.0 × 4.18 × 15.0 = 12,540 J = 12.54 kJ; ΔHc(correct) = −12.54/0.01079 = −1162 kJ mol⁻¹
Scenario 1 (wrong m): q(wrong) = 0.80 × 4.18 × 15.0 = 50.16 J = 0.05016 kJ; ΔHc = −0.05016/0.01079 = −4.65 kJ mol⁻¹. Massively underestimated — 250× too small.
Scenario 2 (evaporation): n(wrong) = 0.90/74.12 = 0.01214 mol; q unchanged = 12.54 kJ; ΔHc(wrong) = −12.54/0.01214 = −1033 kJ mol⁻¹. Less negative than correct −1162 — overestimated n means smaller |ΔHc|.
1. B — q = mcΔT calculates heat absorbed by the water. 'm' is the mass of water, not fuel.
2. C — Spirit burner calorimeters are open systems; significant heat escapes to the surrounding air before reaching the water, making q (and therefore |ΔHc|) smaller than the true value.
3. B — n = 0.500/100 = 0.00500 mol; q = 150.0 × 4.18 × 8.4 = 5267 J = 5.267 kJ; ΔHc = −5.267/0.00500 = −1053 kJ mol⁻¹.
4. A — Water absorbs heat (q > 0, positive), but the reaction releasing that heat is exothermic (ΔH < 0, negative). The negative sign converts the positive q into the correct negative ΔH.
5. D — Balance imprecision would produce random scatter (some too high, some too low), not a systematic 50% shortfall always in the same direction. The consistent direction of the error — always less negative — points to a systematic source: heat loss. This is the dominant error in spirit burner experiments, far exceeding any contribution from balance precision.
Q6 (3 marks): q represents the heat energy absorbed by the water during combustion [1]. 'm' must be the mass of water because q = mcΔT calculates the thermal energy transferred to the water — not to the fuel or the calorimeter [1]. The specific heat capacity of water (c = 4.18 J g⁻¹ K⁻¹) is used because water is the substance being heated, and c tells us how many joules are needed to raise 1 g by 1 K [1].
Q7 (5 marks): (a) ΔT = 34.5 − 21.0 = 13.5°C; q = 180.0 × 4.18 × 13.5 = 10,157 J = 10.157 kJ [1]. (b) n = 0.85/32.04 = 0.02653 mol; ΔHc = −10.157/0.02653 = −383 kJ mol⁻¹ [2 — 1 for n, 1 for ΔHc]. (c) % error = |−383 − (−726)| / 726 × 100 = 343/726 × 100 = 47.2% [1]. Source: heat loss to surroundings — a large proportion of the combustion energy heats the surrounding air rather than the water, making q smaller than the true heat released and ΔHc less negative [1].
Q8 (5 marks): (a) Methanol: 726/32.04 = 22.7 kJ g⁻¹ [1]; Ethanol: 1367/46.07 = 29.7 kJ g⁻¹ [1]; Octane: 5471/114.23 = 47.9 kJ g⁻¹ [1]. (b) Octane is most efficient per gram at 47.9 kJ g⁻¹ — more than twice methanol's 22.7 kJ g⁻¹ [1]. Octane (petrol) is preferred in internal combustion engines because it releases more energy per gram of fuel, allowing vehicles to travel further on the same tank mass — this is the same reason jet fuel (kerosene) is denser in energy than wood or alcohol [1].
Answer questions on Calorimetry — Combustion before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–2.
Tick when you've finished all activities and checked your answers.