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General Chemistry I · Thermochemistry

Bomb Calorimetry (Constant-Volume Calorimetry)

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Study tools
  7. Sources & references

In 30 seconds

A bomb calorimeter is a sealed, heavy-walled steel vessel ("the bomb") submerged in a known mass of water inside an insulated jacket. The sample is ignited electrically and burns completely in an atmosphere of excess oxygen. Because the rigid steel bomb cannot expand or contract, the reaction occurs at constant volume, so no pressure–volume work is done (w = 0 for the system) and the measured heat equals the change in internal energy: q_v = ΔE. Bomb calorimetry is the standard method for measuring heats of combustion — the energy content of fuels and foods.

Why this matters

Every Calorie printed on food packaging is ultimately traceable to bomb calorimetry: food is burned in a bomb, and the released heat (converted from joules to Calories) is its energy content. The same technique rates fuels — gasoline, coal, natural gas, biofuels — and lets engineers compute how much energy a fuel delivers per gram. Understanding ΔE vs. ΔH also prepares you for the difference between constant-volume and constant-pressure conditions throughout thermodynamics.

The college version

Key Ideas

  • Constant volume → no P–V work → q_v = ΔE. The device measures internal-energy change directly.
  • The whole apparatus is the "surroundings." Heat released by the reaction warms both the water and the steel bomb, thermometer, and stirrer. We account for everything at once with the calorimeter's heat capacity, C_cal.
  • C_cal is the heat required to raise the entire calorimeter assembly by 1 °C (units J/°C or kJ/°C). It is determined by calibration with a substance of known heat of combustion (usually benzoic acid).
  • q_reaction = −q_calorimeter (energy conservation).
  • Combustion reactions release large amounts of heat, so q is typically reported in kJ, not J.
  • ΔH vs. ΔE: for a combustion reaction producing fewer moles of gas than it consumes, ΔH and ΔE differ slightly; they are related by ΔH = ΔE + Δ(n_gas)RT.

Equations and Variables

SymbolMeaningUnits
C_calHeat capacity of the calorimeterkJ/°C or J/°C
ΔTTemperature rise of the calorimeter°C
q_calHeat absorbed by the calorimeterkJ
q_reaction (q_rxn)Heat released by the reactionkJ
ΔEChange in internal energykJ
ΔHChange in enthalpykJ
Δ(n_gas)Change in moles of gasmol
RGas constant8.314 J/(mol·K)
TAbsolute temperatureK
  • q_cal = C_cal · ΔT
  • q_reaction = −q_cal
  • ΔE = q_v (constant volume)
  • ΔH = ΔE + Δ(n_gas)RT

How It Works

  1. Calibrate: Burn a known mass of benzoic acid (its heat of combustion is 26.42 kJ/g). Measure ΔT and solve C_cal = q_known / ΔT.
  2. Measure: Load the sample of interest into the bomb, charge it with ~25 atm of O₂, seal it, and submerge it in a measured mass of water. Ignite electrically.
  3. Record the temperature rise of the water (and thus the whole assembly).
  4. Compute: q_cal = C_cal · ΔT, then q_reaction = −q_cal.
  5. Convert to ΔE per mole by dividing by moles of sample.

Because the bomb's volume is fixed, the gas produced by combustion simply raises the pressure inside rather than expanding against the atmosphere. With no expansion, the system does no P–V work, so all the energy change shows up as heat.

Worked Example

A bomb calorimeter is calibrated by burning 1.000 g of benzoic acid, which raises the temperature by 4.26 °C. Benzoic acid releases 26.42 kJ/g. Then 1.000 g of an unknown compound is burned, raising the temperature by 2.98 °C. If the compound's molar mass is 58.0 g/mol, find ΔE of combustion per mole.

Step 1 — Calibrate. Heat from benzoic acid: q_known = (1.000 g)(26.42 kJ/g) = 26.42 kJ.

C_cal = q_known / ΔT = 26.42 kJ / 4.26 °C = 6.20 kJ/°C

Step 2 — Heat from the unknown. q_cal = C_cal · ΔT = (6.20 kJ/°C)(2.98 °C) = 18.48 kJ.

q_reaction = −18.48 kJ (combustion releases heat).

Step 3 — Moles of unknown: 1.000 g / 58.0 g/mol = 0.01724 mol.

Step 4 — ΔE per mole: ΔE = −18.48 kJ / 0.01724 mol = −1072 kJ/mol

The heat of combustion is about −1070 kJ/mol (negative, exothermic). If we also wanted ΔH, we would use ΔH = ΔE + Δ(n_gas)RT — usually a small correction of a few kJ.

Common Confusions

  • ΔE vs. ΔH. Constant volume → ΔE; constant pressure → ΔH. This is the #1 exam distinction.
  • Using c_water instead of C_cal. In a bomb calorimeter you must use the whole calorimeter's heat capacity, not just the water's specific heat. The steel bomb itself absorbs significant heat.
  • Units. Combustion heats are large; report kJ, not J. Keep C_cal in kJ/°C to match.
  • Sign. q_reaction is negative (heat released); the calorimeter's q is positive (it absorbed heat).
  • Δ(n_gas) matters. The small ΔH − ΔE difference comes from the change in moles of gas; don't ignore it when the question asks for both quantities.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of a bomb calorimeter as a really tough metal lunchbox sealed shut with a tiny fire inside. You can't open it, and it can't stretch, so when something burns inside, the gas can't push out — it just gets squished (the pressure goes up, but the volume stays the same). Because the box can't expand, the fire can't do any "pushing" work; all its energy comes out as heat, which warms the water around the box. That's why the bomb measures "internal energy" (ΔE) while the open coffee-cup measures "enthalpy" (ΔH) — the cup lets gas push against the air, the bomb doesn't. The same trick is how scientists figure out how many Calories are in a candy bar: they literally burn it in a metal box and see how much it warms the water.

Key takeaways

  • Bomb = constant volume = measures ΔE (not ΔH).
  • No expansion work because the rigid vessel fixes volume.
  • The calorimeter's heat capacity C_cal includes the water + bomb + thermometer + stirrer.
  • Calibrated with benzoic acid (26.42 kJ/g).
  • q_reaction = −C_cal · ΔT.
  • ΔH = ΔE + Δ(n_gas)RT relates the two for gas-producing reactions.
  • Combustion is always exothermic (ΔE < 0).
  • What thermodynamic quantity does a bomb calorimeter measure directly, and why?
  • A bomb calorimeter has C_cal = 5.10 kJ/°C. Burning a sample raises the temperature 3.20 °C. What is q_reaction?
  • Why is benzoic acid used for calibration?
  • Write the equation relating ΔH and ΔE for a reaction that changes the number of gas moles.
  • Answers: (1) ΔE, because constant volume means no P–V work; (2) q_cal = 5.10 × 3.20 = 16.32 kJ, so q_reaction = −16.3 kJ; (3) it burns completely and has a precisely known heat of combustion (26.42 kJ/g); (4) ΔH = ΔE + Δ(n_gas)RT.

Keep learning

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Practice General Chemistry I

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

  • Describe the design of a bomb calorimeter and why it operates at constant volume.
  • Explain why constant-volume calorimetry measures ΔE (internal energy) rather than ΔH.
  • Use a calorimeter's heat capacity (C_cal) to measure the heat of combustion.
  • Relate ΔE and ΔH for combustion reactions.

Sources & references

  1. OpenStax, *Chemistry 2e*, §5.2 Calorimetry (bomb calorimetry).
  2. NIST — benzoic acid calorimetric standard and heat-capacity reference data.
  3. NIST Chemistry WebBook — heats of combustion.

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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