On this page
A bomb calorimeter question gives you a temperature rise and asks for the enthalpy of combustion. Two steps are hiding in it. The calorimeter measures internal energy change, not enthalpy, and the conversion depends on counting gas moles correctly. This post explains why the bomb gives ΔU, how its heat capacity is found, and how to get from a thermometer reading to ΔH without sign or unit slips.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists the first law of thermodynamics with the "concept of work, heat, internal energy and enthalpy, heat capacity", and the NEET (UG) 2026 syllabus has the same wording. The CBSE 2026-27 Chemistry curriculum names "Measurement of ∆U and ∆H: Calorimetry" in its thermodynamics unit. The NCERT Class 11 Chemistry Thermodynamics unit (2026-27 reprint) covers it in Section 5.3, with the bomb calorimeter in Fig. 5.7 and the graphite combustion in Problem 5.6. Check the current year's syllabus.
Why does a bomb calorimeter measure ΔU?
The "bomb" is a sealed steel vessel, immersed in water. A sample burns inside it in excess oxygen, and the heat released warms the bomb and water. NCERT's point is that the bomb's volume does not change, so no pressure-volume work is done, "even for reactions involving gases". With w = 0, the first law ΔU = q + w becomes ΔU = q at constant volume.
Most reactions in the lab, and most tabulated values, are at constant pressure instead, where the heat equals ΔH. NCERT links the two:
ΔH = ΔU + Δn(g)RT
where Δn(g) is moles of gaseous products minus moles of gaseous reactants.
How is the calorimeter calibrated?
The calorimeter's heat capacity C (bomb, water, stirrer and all) must be known. It is found by burning a substance whose energy of combustion is known accurately. Benzoic acid is the usual standard: NIST's SRM 39j certificate certifies its heat of combustion as 26,434 J per gram, within 3 J. NIST's predecessor laboratory described benzoic acid's role in standardising bomb calorimeters as early as Jessup (1942).
Worked example: from thermometer to ΔH
The temperature rises in this example are sample numbers chosen for the calculation.
Step 1: calibrate. Burning 1.000 g of benzoic acid raises the temperature by 2.500 K.
C = 26.434 kJ ÷ 2.500 K ≈ 10.57 kJ/K
Step 2: burn the sample. In the same calorimeter, 0.920 g of liquid ethanol (molar mass 46.07 g/mol, so 0.920/46.07 ≈ 0.01997 mol) raises the temperature by 2.577 K.
Heat absorbed by the calorimeter: q = 10.57 × 2.577 ≈ 27.25 kJ.
Step 3: ΔU per mole, with the right sign. The reaction released that heat, so for the reaction it is negative.
ΔU = −27.25 kJ ÷ 0.01997 mol ≈ −1364.5 kJ/mol
Step 4: count gas moles.
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
Δn(g) = 2 − 3 = −1.
Step 5: convert. RT = 8.314 × 298 J/mol ≈ 2.48 kJ/mol.
ΔH = −1364.5 + (−1)(2.48) ≈ −1367 kJ/mol
That matches the enthalpy of combustion of ethanol NCERT uses in its Hess's law example, −1367 kJ/mol. The correction is only about 0.2% here, but questions are built around it.
When are ΔU and ΔH the same?
When Δn(g) = 0. NCERT's Problem 5.6 burns graphite, C(s) + O₂(g) → CO₂(g), with one gas mole on each side, so ΔH = ΔU = −2.48 × 10² kJ/mol. Glucose behaves the same way: six O₂ in, six CO₂ out, with liquid water. For benzoic acid, C₇H₆O₂(s) + 7½O₂ → 7CO₂ + 3H₂O(l), Δn(g) = −½, so ΔH is about 1.2 kJ/mol more negative than ΔU.
Which mistakes should you avoid?
Counting liquid water as gas. At 298 K it is liquid unless the question says otherwise.
Mixing J and kJ. R = 8.314 J mol⁻¹ K⁻¹, so RT is about 2.48 kJ/mol, not 2480 kJ/mol.
Losing the sign. The calorimeter gains heat; the reaction's ΔU is negative.
Reporting per gram when the question asks per mole, or the reverse.
Calling the bomb reading ΔH. It is ΔU until you convert it.
If you mix up the physics and chemistry forms of the first law, read our guide to thermodynamics sign conventions. For using combustion enthalpies to find others, see Hess's law in thermochemistry.
A practice routine
Rework NCERT Problem 5.6 and explain why ΔH equals ΔU.
Redo the worked example for a 0.640 g sample of methanol (CH₃OH, liquid), with a temperature rise you choose.
Write Δn(g) for five combustions, including one with a solid fuel.
Run the calculation backwards: from a known ΔH, predict the temperature rise.
When the link between ΔU and ΔH still feels like a formula rather than an idea, the Fermi Concept Library covers every concept in chemistry, so you can revise internal energy and enthalpy before returning to calorimetry.
Start with step 1, and write Δn(g) before any other number.
