What it does
A bomb calorimeter measures the energy released when something burns. The sample sits in a sealed steel container (the “bomb”) full of oxygen, surrounded by water. A spark sets it alight, and the heat warms the bomb and the water by a small, carefully measured amount ΔT.
Because the bomb is sealed, the burning happens at constant volume, so the heat measured is the internal energy change ΔU. To get the enthalpy change ΔH (constant pressure), add the work term for the change in moles of gas: ΔH = ΔU + Δn(gas)RT.
This virtual practical walks you through it: first calibrate the calorimeter with benzoic acid, then burn a sample and work out ΔU and ΔH per mole. The temperature trace is drawn live as if from a fine thermometer.
It follows thermodynamics in Class 11 chemistry (measuring ΔU and ΔH). Go on to the gibbs explorer to see whether a reaction is spontaneous, the bond enthalpy estimator estimator to predict ΔH from bonds, or the thermochemistry calculator for any reaction.
How to use it
- Step 1, calibrate. Type the mass of benzoic acid (about 1 g) and press Fire. It releases 26.43 kJ per gram, so C = q ÷ ΔT gives the heat capacity of this calorimeter.
- Step 2, burn your sample. Choose glucose, sucrose, naphthalene or urea, type its mass and press Fire. The page uses your C to turn ΔT into heat, then into ΔU per mole and ΔH.
- Use a different calorimeter gives one with another heat capacity. Calibrate again; your final ΔU should not change.
Worked example
Calibration: 1.000 g of benzoic acid gives q = 26.43 kJ and a rise of ΔT = 2.604 K, so C = 26.43 ÷ 2.604 = 10.150 kJ/K.
Glucose: 1.200 g gives ΔT = 1.839 K. Heat q = 10.150 × 1.839 = 18.665 kJ. Moles = 1.200 ÷ 180.16 = 0.0066607 mol, so ΔU = −18.665 ÷ 0.0066607 = −2802.3 kJ/mol. For glucose, C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), Δn(gas) = 6 − 6 = 0, so ΔH = ΔU.
Naphthalene (1.000 g, ΔT = 3.960 K) gives ΔU = −5151.5 kJ/mol; there are 12 mol of O₂ before and 10 mol of CO₂ after, so Δn = −2 and ΔH = −5151.5 − 2 × 8.314 × 298 ÷ 1000 = −5156.5 kJ/mol.
Check yourself
Three questions with fresh numbers: the heat capacity from a calibration, ΔU per mole from a temperature rise, and ΔH from ΔU. Nothing is saved.
FAQ
Why calibrate? The bomb, water, stirrer and thermometer all soak up heat. Rather than adding them up, burn something with a known energy and measure.
Why is ΔU negative? Heat leaves the reacting chemicals and warms the water, so the chemicals lose energy.
Is this from my textbook? It covers the standard school topic with its own examples and pictures. It is an independent study aid, not part of NCERT.
Where the numbers come from
- Standard thermodynamics: constant-volume calorimetry q = CΔT, ΔU = q_V, ΔH = ΔU + Δn(gas)RT. R from CODATA 2018. The benzoic acid calibration value (26.43 kJ/g) and the samples' combustion energies are rounded standard values, labelled as rounded; the readings are simulated. Written for fibub; no textbook text, figures or exercises are used.
- Checked by independent hand calculation in Python: calibration C = 26.43 / 2.604 = 10.150 kJ/K; 1.2 g of glucose gives ΔT = 1.839 K and ΔU = −2802 kJ/mol; ΔH for naphthalene (Δn = −2) and urea (Δn = +½).
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.