Calorimetry and the Bomb Calorimeter
On this page
Direct answer
A bomb calorimeter burns a weighed sample in excess oxygen inside a sealed steel vessel immersed in a measured water jacket, and because the volume cannot change, the heat it records is the internal energy change: q_v = delta-U = C(cal) × delta-T, where C(cal) is the heat capacity of the entire calorimeter assembly in joules per kelvin. The open coffee-cup calorimeter, by contrast, works at constant atmospheric pressure and delivers delta-H directly. The two quantities connect through delta-H = delta-U + dn(gas)RT, counting only gaseous species; for combustions with no change in gas moles, like graphite burning to carbon dioxide, the two are identical.
What you must remember
- Constant volume versus constant pressure: bomb calorimetry measures delta-U; coffee-cup measures delta-H — the first fact of every question in this chapter.
- The conversion equation: delta-H = delta-U + dn RT, dn counting only gaseous moles (products minus reactants); use R = 8.314 J K^-1 mol^-1 when delta-U is in joules.
- Calorimeter constant: q = C(cal) × delta-T applies to the whole apparatus (vessel plus water); instruments are calibrated with a standard such as benzoic acid.
- Combustion conventions: standard enthalpy of combustion is defined for one mole of substance with water as liquid product in tabulated values.
- Neutralisation benchmark: strong acid plus strong base in the coffee-cup releases about −57 kJ per equivalent of water formed, a number worth carrying into the exam hall.
- Specific heat route: q = m × s × delta-T for a known mass of one substance, usually water at 4.18 J g^-1 K^-1.
One combustion, done honestly
Burn 1.00 g of graphite in a bomb calorimeter whose calorimeter constant is 20.7 kJ K^-1, and suppose the temperature rises 1.58 K. Heat released: q_v = 20.7 × 1.58 = 32.7 kJ. Moles burned: 1.00/12 = 0.0833 mol. So delta-U per mole = −32.7/0.0833 = −393 kJ/mol. Now the gas-mole check: C(graphite) + O2(g) → CO2(g) has two moles of gas on each side, dn = 0, so delta-H = delta-U = −393 kJ/mol — matching the tabulated −393.5 kJ/mol and closing the loop beautifully.
Contrast that with methane: CH4 + 2O2 → CO2 + 2H2O(l) has 3 moles of gas reactant and 1 mole of gas product, dn = −2. At 298 K, delta-H = delta-U − 2RT = delta-U − 4.96 kJ/mol, so the enthalpy change is about 5 kJ more negative than the bomb's reading — precisely the correction a JEE numerical will ask you to apply.
How the exam frames it
JEE Main favours the integer-type arithmetic: heat released, temperature rise, or molar mass of a fuel from calorimeter data, with the trap being unit bookkeeping — heat capacity in kJ K^-1 versus specific heat in J g^-1 K^-1, and answers demanded in joules after data given in kilojoules. The second trap is sign discipline: the sample loses the heat the calorimeter gains, so q(sample) = −C × delta-T whenever delta-T is positive. JEE Advanced adds the conceptual split: which device measures delta-U and which measures delta-H, and why the difference vanishes exactly when dn(gas) = 0 — water forming as liquid rather than vapour changes dn by +1 per mole of water, a subtlety that has decided more than one multi-correct question. Count gas moles on both sides before touching R.
Frequently asked questions
Why does a bomb calorimeter measure delta-U and not delta-H?
The sealed vessel fixes the volume, so no pressure-volume work is done (w = 0) and the heat exchanged equals the internal energy change.
When are delta-H and delta-U equal for a reaction?
When the gaseous mole count is unchanged, dn = 0, as in graphite + oxygen → carbon dioxide or benzene hydrogenation with matching gas moles.
What is a calorimeter constant?
The heat capacity of the entire calorimeter assembly, joules per kelvin, found by burning a standard sample — it converts a measured temperature rise into heat.
Why is neutralisation enthalpy about −57 kJ per equivalent for any strong acid-strong base pair?
Because the only net reaction is H+ + OH- → H2O, and the spectator ions are identical whatever the acid-base pair.
What happens to delta-H if the product water is vapour instead of liquid?
The reaction has one extra mole of gas per mole of water, so delta-H exceeds delta-U by RT per mole — combustion values with liquid water are correspondingly more negative.