Bomb calorimeter: glucose pellet to calorific value
SCH4U Grade 12 Chemistry · Thermochemistry
A bomb calorimeter is built to contain a combustion completely: a thick-walled steel vessel — the bomb — submerged in a stirred pail of water, with one thermometer reading the pail. A 1.200 g pellet of glucose (C₆H₁₂O₆, M = 180.16 g/mol) is pressed in a die, wired for ignition with a fine fuse wire, and sealed into the bomb's steel vessel, which is then charged with oxygen so the burn goes to completion. A calibration burn of benzoic acid, done beforehand, established that the bomb, pail and its 2.000 L of water together absorb heat like 2.48 kg of water — the calorimeter constant, quoted as a water equivalent. The stirrer runs until the thermometer holds steady; then the firing circuit closes, the wire glows, and the pellet burns in a fraction of a second inside the sealed steel. Over the next minutes the stirred water carries the heat outward, and the thermometer climbs from 24.10 °C to 25.90 °C and levels off. Water's specific heat is 4,186 J/(kg·K). The pellet is whatever your lab burns: swap the mass and molar mass for sucrose (342.30 g/mol), ethanol (46.07) or a chip of paraffin, and the whole chain re-runs.
Every number in this problem is editable, the material included — change any value below and the whole chain recalculates.
- m_p = 1.2 g — Mass of the glucose pellet
- m_eq = 2.48 kg — Calorimeter water equivalent (calibrated)
- T₁ = 24.1 °C — Temperature before firing
- T₂ = 25.9 °C — Temperature after firing
- M = 180.16 g/mol — Molar mass of glucose
- c_w = 4,186 J/(kg·K) — Specific heat of water
- (a)the heat the burn released into the calorimeter
- (b)the calorific value of glucose per unit mass
- (c)the amount of glucose burned, in moles
- (d)the molar heat of combustion of glucose
Everything the burn released, the calorimeter caught: Q = m_eq·c·ΔT with the CALIBRATED 2.48 kg, not the 2.000 kg of pail water alone. Forgetting the bomb and pail's share — the calorimeter constant — is the classic error, and it reads every fuel about 19% too lean.
Carried onward at full precision, not this rounded figure.
A calorific value is an energy per unit mass — the same Q/m shape as a latent heat, so the same solver does the division. The 18.7 kJ spread over just 1.200 g of pellet is what makes the number large: about 15.6 kJ from every gram, the figure printed per 100 g on food labels after the kilo-conversion.
Carried onward at full precision, not this rounded figure.
Chemistry counts in moles, so the pellet's 1.200 g becomes n = m/M — about 6.66 mmol. Keep this number at full precision: it is about to sit under a division, where a rounded 0.0067 would shift the final answer by half a percent all by itself.
Carried onward at full precision, not this rounded figure.
ΔH = q/n converts the burn to the chemist's currency: energy per mole of reaction as written, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. Strictly the tabulated ΔH_c carries a negative sign — combustion is exothermic — and the calorimeter measures its magnitude; dropping that convention in a Hess's-law problem later flips answers wholesale.
Carried onward at full precision, not this rounded figure.
Therefore the pellet released 18.69 kJ into the calorimeter, glucose delivers about 15.6 kJ per gram, and the 6.661 mmol burned put its molar heat of combustion at 2,805 kJ/mol — right beside the handbook's 2,803 kJ/mol for ΔH_c of glucose.
Why this order
A bomb calorimeter reduces all of thermochemistry to one thermometer: the order of the chain is the order in which that single ΔT is converted into ever more portable currencies. First joules (the calorimeter's own units, via the calibrated water equivalent), then joules per gram (the engineer's and the nutritionist's number), then moles, then joules per mole (the chemist's, ready for Hess's law). The step that separates a right answer from a plausible one is (a): the calorimeter constant. The steel bomb, the pail and the stirrer all warm by the same 1.80 K as the water, and calibration with a standard — benzoic acid, whose heat of combustion is certified — is how their share gets folded into one effective 2.48 kg. Run the arithmetic with the 2.000 kg of visible water and every answer downstream lands 19% low, consistently and convincingly wrong.
The cross-check built into the chain is that parts (b) and (d) must agree through the molar mass: 15.57 kJ/g × 180.16 g/mol = 2,805 kJ/mol, the same number part (d) reaches by its own route — one burn, two currencies, one exchange rate. And the magnitudes deserve their moment: 15.6 kJ/g is the 3.7 kcal/g behind the "4 calories per gram of carbohydrate" on every nutrition panel (bomb values run a shade high because a body does not fully oxidize what a bomb does), and 2,805 kJ/mol is precisely the energy budget respiration spends 38 ATP at a time. The derived numbers here are this page's own, from its own readings — close to the handbook because the scenario was built honestly, not copied from it.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.