Calorimetry, Combustion
In 1780, Antoine Lavoisier and Pierre-Simon Laplace used an ice calorimeter at the French Academy of Sciences to measure the heat of combustion of charcoal at approximately 34 kJ g⁻¹, one of the first quantitative measurements of chemical energy ever made. Their device surrounded the burning sample with ice and measured the mass of meltwater, turning the spirit of q = mcΔT into a physical scale. Every school calorimetry experiment today follows the same logic they established 245 years ago.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
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?
Key Facts
- The formulas q = mcΔT and ΔHc = −q/n
- The components of a spirit burner calorimeter
- Five sources of error in combustion calorimetry
Concepts
- Why 'm' in q = mcΔT is the mass of water, not fuel
- Why the negative sign appears in ΔHc = −q/n
- Why experimental ΔHc is usually less negative than the accepted value
Skills
- Calculate q and then ΔHc from experimental data in 5 steps
- Calculate percentage error between experimental and accepted values
- Identify sources of error and state their directional effect on ΔHc
The spirit burner calorimeter is a simple, imperfect device, understanding its setup is inseparable from understanding why experimental ΔHc values are usually lower than the accepted value.
Components: spirit burner containing a school-approved fuel (ethanol is the usual choice), 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):
- Record initial mass of burner + fuel (balance reading before)
- Record initial water temperature
- Light the burner; burn fuel until a target temperature rise is reached (e.g. ΔT ≈ 10°C)
- Extinguish the flame; record final water temperature
- Record final mass of burner + fuel (balance reading after)
- Calculate: mass of fuel burned = initial mass − final mass
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.
In combustion calorimetry, q = mcΔT (m = mass of water, c = 4.18 J g⁻¹ K⁻¹); then ΔHc = −q/n (n = moles of fuel). This simple energy balance assumes all the heat reaches the water and neglects the calorimeter's heat capacity, heat loss and incomplete combustion, so it is an experimental approximation: the dominant systematic errors usually make the measured ΔHc less negative than the accepted value.
Pause, copy the highlighted definition into your book before moving on.
Mini-task: A student sets up a spirit burner experiment but forgets to cap the burner after extinguishing the flame. Will this make their calculated ΔHc too high, too low, or unchanged? Explain the chain of reasoning in two sentences.
We just saw the spirit burner setup and how ΔHc = −q/n is calculated. That raises a question: why does the experimental ΔHc usually come out less negative than the textbook value? This card answers it → by identifying the specific systematic errors that reduce the measured heat.
In a school spirit burner the measured ΔHc is usually 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 usually less negative (lower in magnitude) than the accepted value because the dominant systematic errors all reduce the heat reaching the water; random measurement errors, by contrast, can shift a result in either direction. In HSC questions, you must name the specific physical source of error and state its directional effect on the result.
These systematic errors usually make a spirit burner ΔHc less negative than the accepted value. Five named sources: (1) heat loss to surroundings, (2) incomplete combustion, (3) fuel evaporation (uncapped burner), (4) heat absorbed by copper walls, (5) thermometer not submerged. Each reduces |ΔHc|; remember that random measurement errors can scatter either way.
Add the highlighted point to your notes before the check below.
Explain it: In your own words, explain why a school spirit burner usually gives a ΔHc that is less negative than the accepted value. Use the phrase "systematic error" and name at least two specific physical sources.
Worked examples · reveal as you go
Calculating q and ΔHc. 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.
m(fuel) = 152.34 − 151.62 = 0.72 g
$n = \dfrac{m}{M} = \dfrac{0.72}{46.07} = 0.01562 \text{ mol}$
ΔT = 31.2 − 19.5 = 11.7°C = 11.7 K
$q = mc\Delta T = 200.0 \times 4.18 \times 11.7 = 9781 \text{ J} = \mathbf{9.781 \text{ kJ}}$
$\Delta H_c = \dfrac{-q}{n} = \dfrac{-9.781}{0.01562} = \mathbf{-626 \text{ kJ mol}^{-1}}$
The accepted value is −1367 kJ mol⁻¹. The experimental value is less than half, showing very significant heat loss, typical of spirit burner experiments.
Percentage error & sources of discrepancy. 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.
$\% \text{ error} = \dfrac{|\text{experimental} - \text{accepted}|}{|\text{accepted}|} \times 100$
$= \dfrac{|-412 - (-726)|}{|-726|} \times 100 = \dfrac{314}{726} \times 100 = \mathbf{43.3\%}$
(1) Heat loss to surroundings: A large proportion of combustion energy heats the surrounding air and copper walls rather than the water, making q smaller than the true heat released, giving a ΔHc less negative than the accepted value.
(2) Incomplete combustion: Insufficient oxygen at the flame produces CO instead of CO₂ for some of the combustion. CO releases less energy per mole than complete combustion to CO₂, reducing the total heat released and making ΔHc less negative.
(a) 43.3% error. (b) Heat loss to surroundings; incomplete combustion. Both make ΔHc less negative than the accepted value.
Put the calorimetry combustion procedure in the correct order. Click two steps to swap them, then click Check order.
- Calculate q = mcΔT for the water in the calorimeter.
- Measure the initial mass of the fuel + spirit burner.
- Divide q by the moles of fuel burned to get ΔHcomb.
- Record the initial and final temperature of the water.
- Find the mass of fuel burned (initial mass − final mass).
Formula Reference, This Lesson
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Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Common errors · the 3 traps that cost marks
"m in q = mcΔT is the mass of fuel"
Students substitute the mass of fuel burned into q = mcΔT instead of the mass of water in the calorimeter.
Fix: The 'm' in q = mcΔT is always the mass of the substance being heated, in a combustion calorimeter, that is the water. The fuel is the energy source, not the substance whose temperature you're measuring. q tells you how much energy the water absorbed, and c is the specific heat capacity of water.
"The negative sign in ΔHc = −q/n is a mistake"
Students omit the negative sign or write ΔHc = +q/n, giving a positive value for combustion.
Fix: Combustion is exothermic, ΔHc must be negative. The formula ΔHc = −q/n converts the positive q (heat absorbed by water) into the correct negative enthalpy change for the reaction. If you get a positive ΔHc for combustion, you've made a sign error.
"Human error explains the discrepancy"
Students write "human error" or "inaccurate measurements" as sources of error in an HSC answer.
Fix: On their own, these vague phrases usually earn no marks. Always name the specific physical process (e.g. "heat loss to the surrounding air through the open sides of the apparatus") and state its directional effect on ΔHc (e.g. "this makes q smaller than the true heat released, so ΔHc is less negative than the accepted value").
Quick-fire practice · 5 reps +2 XP per reveal
Calorimetry Data Processing, Propan-1-ol: A student burns propan-1-ol (C₃H₇OH, M = 60.10 g mol⁻¹). Burner mass: before 188.45 g, after 187.81 g. Water mass: 250.0 g. Temperature: initial 18.2°C, final 32.6°C.
(a) Calculate mass and moles of propan-1-ol burned.
(b) Calculate q in kJ.
(c) Calculate ΔHc and the percentage error (accepted value = −2021 kJ mol⁻¹).
(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%
Error analysis, butan-1-ol: A student burns 0.80 g of butan-1-ol (C₄H₉OH, M = 74.12 g mol⁻¹) and heats 200.0 g of water from 18.0°C to 33.0°C (ΔT = 15.0°C).
Scenario 1, wrong m: The student uses m = 0.80 g (mass of fuel) instead of 200.0 g in q = mcΔT. Calculate the incorrect q and ΔHc.
Scenario 2, evaporation: The burner was not capped; recorded mass loss is 0.90 g instead of 0.80 g. Calculate the resulting ΔHc.
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|.
Short calculation: A student burns 0.500 g of an unknown fuel (M = 100 g mol⁻¹) and records a temperature rise of 8.4°C in 150.0 g of water. Calculate q in kJ and then ΔHc in kJ mol⁻¹.
Sign and direction reasoning: Water absorbs heat during combustion calorimetry, so q (for the water) is positive. The combustion reaction releases heat, so ΔH (for the reaction) is negative. A student writes ΔHc = +q/n. Identify the error and explain the correct sign convention using the formula ΔHc = −q/n.
Extension, systematic error direction: A student's spirit burner experiment consistently gives ΔHc values about 50% below the accepted NESA data sheet values. A classmate suggests the balance is imprecise and introduces 0.01 g random error in the fuel mass reading. Could balance imprecision explain the 50% shortfall? Explain your reasoning.
Go back to your Think First response. Now that you've studied combustion calorimetry, return to Lavoisier and Laplace's 1780 ice calorimeter at the French Academy of Sciences:
- Their ice calorimeter measured mass of meltwater rather than ΔT. How does this differ from the spirit burner method? Write the formula they would have used, relating heat to mass of ice melted and the latent heat of fusion (334 J g⁻¹).
- Lavoisier and Laplace got ~34 kJ g⁻¹ for charcoal. If you burned 1.50 g of charcoal in a spirit burner and heated 200 g of water, what ΔT would you expect if no heat was lost? Show your working.
- Their result was a lower bound on the true value, just like your spirit burner result. What was their main source of heat loss, and what was yours?
Pick your answer, then rate your confidencethat tells the system what to drill next.
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 immediately dangerous to health. Complete combustion produces CO₂ and H₂O, which are far less acutely toxic, though CO₂ is an asphyxiant at high concentration and a greenhouse gas. We still aim for efficient, complete combustion for safety and efficiency.
Wrong: "m in q = mcΔT is the mass of fuel burned."
Right: m is the mass of water in the calorimeter, the substance being heated. The specific heat capacity c = 4.18 J g⁻¹ °C⁻¹ is the value for water, not the fuel.
Q1. 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
Q2. 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
Q3. 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
Show comprehensive answers ▼
Activity 1, Propan-1-ol Calorimetry
(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%
Activity 2, Error Analysis (Butanol)
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|.
Multiple Choice
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, less negative each time, points to a systematic source: heat loss. This is the dominant error in spirit burner experiments, far exceeding any contribution from balance precision.
Short Answer Model Answers
Q1 (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].
Q2 (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].
Q3 (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].
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