Chemistry • Year 11 • Module 4 • Lesson 2
Calorimetry — Combustion
Apply the q = mcΔT and ΔHc = −q/n relationships to real data, interpret a temperature–time graph, and evaluate sources of error in combustion calorimetry.
1. Temperature–time graph — ethanol combustion
A student burns ethanol in a spirit burner under 200.0 g of water. The temperature of the water is recorded every 30 seconds from ignition (t = 0) to extinction (t = 150 s) and then for one minute of cooling. The data are plotted in the SVG graph below. 8 marks
(a) Read off Tinitial and Tfinal from the graph and calculate ΔT. 1 mark
(b) Calculate q in kJ using the graph’s Tinitial and Tfinal values. Show full working. 2 marks
(c) The burner mass decreased from 148.92 g to 148.13 g. Calculate n(ethanol) and then ΔHc. M(ethanol, C2H5OH) = 46.07 g mol−1. 2 marks
(d) The temperature continues to drop slightly after extinction (from 31.8°C to 30.9°C). Explain why this occurs and describe how you could use the post-extinction data to correct for heat loss during burning. 3 marks
2. Data-table interpretation — alcohol fuel comparison
The table below shows experimental results for four alcohol fuels, each burned under the same conditions (200.0 g water, spirit burner). Accepted ΔHc values are from a bomb calorimeter. Use the data to answer the questions. 7 marks
| Fuel | Formula | M (g mol−1) | Mass burned (g) | ΔT (°C) | Experimental ΔHc (kJ mol−1) | Accepted ΔHc (kJ mol−1) | % error |
|---|---|---|---|---|---|---|---|
| Methanol | CH3OH | 32.04 | 0.80 | 6.8 | — | −726 | — |
| Ethanol | C2H5OH | 46.07 | 0.79 | 9.9 | — | −1367 | — |
| Propan-1-ol | C3H7OH | 60.10 | 0.85 | 12.4 | — | −2021 | — |
| Butan-1-ol | C4H9OH | 74.12 | 0.92 | 16.1 | — | −2676 | — |
(a) Calculate the experimental ΔHc for each fuel. Show full working for methanol; you may show working more briefly for the others. 4 marks (1 per fuel)
(b) Describe the trend in accepted ΔHc as the number of carbon atoms increases from methanol to butan-1-ol and provide a molecular explanation. 2 marks
(c) Predict whether pentanol (C5H11OH, M = 88.15 g mol−1) would have a ΔHc more or less negative than −2676 kJ mol−1. Justify your prediction. 1 mark
3. Cause-and-effect chain — uncapped burner error
A student forgets to cap the spirit burner immediately after extinguishing the flame. Complete the cause-and-effect chain to show how this error propagates through the calculation and affects the final ΔHc value. 4 marks
Overall outcome (so…): The experimental ΔHc is an underestimate of even the already low spirit-burner value because …
Stuck? Think about what evaporation of fuel from the wick does to the “recorded mass of fuel burned” and therefore n.4. Case study — Karratha LNG export and fuel energy density
Australia is the world’s largest exporter of liquefied natural gas (LNG), with major processing facilities at Karratha, Western Australia (Woodside’s North West Shelf and Pluto operations). LNG is primarily methane (CH4), which is compressed and cooled to −162°C for export by tanker. The table below gives the standard molar enthalpy of combustion and molar mass of methane and two alternative fuels. 5 marks
| Fuel | Formula | M (g mol−1) | Accepted ΔHc (kJ mol−1) | Energy density (kJ g−1) |
|---|---|---|---|---|
| Methane (LNG) | CH4 | 16.04 | −890 | calculate |
| Ethanol (biofuel) | C2H5OH | 46.07 | −1367 | calculate |
| Octane (petrol proxy) | C8H18 | 114.23 | −5471 | calculate |
(a) Calculate the energy density (kJ g−1) for each fuel using energy density = |ΔHc| ÷ M. 3 marks
(b) Using your calculated values, explain why LNG (methane) is highly competitive as an export fuel despite having a lower energy density than octane. In your answer, refer to at least one other property of methane relevant to the Karratha export context. 2 marks
5. Predict and justify — bomb calorimeter vs spirit burner
A researcher measures the ΔHc of eucalyptus oil (a mixture dominated by 1,8-cineole, approximated here as C10H18O, M = 154.25 g mol−1, accepted ΔHc ≈ −5960 kJ mol−1) using (i) a bomb calorimeter and (ii) a spirit burner. Eucalyptus oil is highly volatile and burns with an orange soot-producing flame in open air. 4 marks
(a) Predict whether the spirit burner result for eucalyptus oil will show a larger or smaller percentage error than a typical alcohol fuel such as ethanol. Justify your prediction with reference to at least two specific sources of error. 3 marks
(b) Eucalyptus oil is present in leaf litter across much of south-east Australia. Predict how its high ΔHc per gram would affect the calorific energy released per tonne of dry leaf litter during a bushfire compared with leaf litter dominated by non-volatile species. 1 mark
Q1 — Temperature–time graph
(a) Tinitial = 19.5°C; Tfinal = 31.8°C; ΔT = 31.8 − 19.5 = 12.3°C.
(b) q = mcΔT = 200.0 × 4.18 × 12.3 = 10,283 J = 10.28 kJ.
(c) m(ethanol) = 148.92 − 148.13 = 0.79 g; n = 0.79/46.07 = 0.01714 mol; ΔHc = −10.28/0.01714 = −600 kJ mol−1. (Accepted: −1367 kJ mol−1; % error ≈ 56%.)
(d) After extinction, no new heat is generated, but the water continues to lose heat to the cooler surroundings (copper walls, air), causing a slow temperature drop. To correct for heat loss during burning, the student could extrapolate the post-extinction cooling curve back to the moment of extinction to estimate the “true” Tmax that would have been reached with no heat loss. This corrected Tmax gives a larger ΔT and a more negative ΔHc closer to the accepted value. [1 for explaining post-extinction drop as continued heat loss; 1 for describing back-extrapolation method; 1 for linking corrected Tmax to reduced error.]
Q2 — Alcohol fuel comparison
Methanol: n = 0.80/32.04 = 0.02497 mol; q = 200.0 × 4.18 × 6.8 = 5685 J = 5.685 kJ; ΔHc = −5.685/0.02497 = −228 kJ mol−1.
Ethanol: n = 0.79/46.07 = 0.01714; q = 200.0 × 4.18 × 9.9 = 8,276 J = 8.276 kJ; ΔHc = −8.276/0.01714 = −483 kJ mol−1.
Propan-1-ol: n = 0.85/60.10 = 0.01414; q = 200.0 × 4.18 × 12.4 = 10,367 J = 10.37 kJ; ΔHc = −10.37/0.01414 = −733 kJ mol−1.
Butan-1-ol: n = 0.92/74.12 = 0.01241; q = 200.0 × 4.18 × 16.1 = 13,460 J = 13.46 kJ; ΔHc = −13.46/0.01241 = −1085 kJ mol−1.
(b) Accepted ΔHc becomes more negative as chain length increases (methanol −726 → butan-1-ol −2676 kJ mol−1). Each additional –CH2– unit adds more C–H and C–C bonds that are oxidised to CO2 and H2O during complete combustion, releasing additional bond-energy per mole.
(c) Pentanol will have a ΔHc more negative than −2676 kJ mol−1 because it contains one additional –CH2– unit, adding approximately an extra −600 kJ mol−1 (the approximate increment per CH2 group).
Q3 — Cause-and-effect chain
Box 2: Fuel evaporates from the uncapped wick between weighings → Box 3: Recorded mass of fuel “burned” is larger than the mass actually combusted; n is overestimated → Outcome: Because ΔHc = −q/n, a larger n gives a smaller (less negative) ΔHc, so the result is an underestimate of even the true spirit-burner value.
Q4 — Karratha LNG case study
(a) Methane: 890/16.04 = 55.5 kJ g−1; Ethanol: 1367/46.07 = 29.7 kJ g−1; Octane: 5471/114.23 = 47.9 kJ g−1.
(b) Methane actually has the highest energy density per gram (55.5 kJ g−1) of the three. In the Karratha export context, methane is liquefied at −162°C, dramatically increasing its volumetric density, allowing enormous quantities to be shipped by tanker efficiently. Its very low molar mass means each kilogram of LNG contains more moles of combustible material than an equivalent mass of octane or ethanol, making it commercially and energetically competitive per unit mass shipped.
Q5 — Eucalyptus oil prediction
(a) The spirit burner result for eucalyptus oil will show a larger percentage error than ethanol. Two reasons: (1) Evaporation before ignition: eucalyptus oil is highly volatile; significant vapour escapes from the wick before and after burning, further overestimating the mass of fuel “combusted” and reducing n, giving a less negative ΔHc. (2) Incomplete combustion: the visible orange soot-producing flame indicates incomplete combustion — some carbon is oxidised only to CO or remains as soot (C), releasing far less energy per mole than full oxidation to CO2, which decreases q and makes ΔHc less negative still. [1 mark per specific, directional error explanation; 1 mark for correctly concluding larger error.]
(b) Leaf litter rich in eucalyptus oil would release significantly more calorific energy per tonne during a bushfire than non-volatile species, because each gram of eucalyptus material contains more high-energy C–H/C–C bonds per unit mass and has a higher ΔHc per gram. This contributes to the intensity and speed of spread of eucalyptus-fuelled bushfires in south-eastern Australia.