Calorimetry

Burn a marshmallow under a measured mass of water and you have built a crude calorimeter. Chemists do the same thing with exquisite precision to measure how much energy food, fuels, and chemical reactions release. That number, the enthalpy or internal energy of reaction, underlies everything from nutrition labels to rocket fuel design. Calorimetry turns a temperature change into an energy. It matters because reaction energies cannot be weighed or seen directly. They must be inferred from how much a surrounding bath heats up. This article explains the two workhorse designs: the bomb calorimeter at constant volume and the coffee-cup or flow calorimeter at constant pressure. You will see how each measures a distinct thermodynamic quantity. You will learn how a calorimeter constant is determined and applied. Finally, you will learn how to convert between $q_V$ and $q_p$ for reactions involving gases.

Key takeaways

Bomb calorimetry measures $q_V = \Delta U$ because the steel bomb has fixed, rigid volume.
Constant-pressure calorimetry (coffee-cup, flow) measures $q_p = \Delta H$ directly.
The calorimeter constant $C_{cal}$ (J/K) is found by supplying a known energy, often electrically, and recording $\Delta T$ .
Converting between the two uses $\Delta H = \Delta U + \Delta n_{gas}RT$ , valid when gases behave ideally.
Heat capacities of the whole assembly, not just the water, must be included for accurate work.
Key idea: a calorimeter converts a temperature rise into an energy via $q = C_{cal}\Delta T$ , and the constraint (constant $V$ or constant $p$ ) tells you whether that heat equals $\Delta U$ or $\Delta H$ .

The core insight

A calorimeter never measures $\Delta U$ or $\Delta H$ directly; it measures a temperature change, which becomes an energy only once the total heat capacity of the apparatus is known, and only the mechanical constraint (rigid bomb versus open, pressure-equilibrated vessel) tells you which state function you have actually measured.

The physical picture

Every calorimeter is a controlled, insulated compartment where a chemical or physical process exchanges heat with a surrounding bath whose temperature you can measure precisely. The measured quantity is always a temperature change, $\Delta T$ . To convert that into an energy you need the total heat capacity of everything that absorbs the heat: the water bath, the metal bomb, the thermometer, the stirrer, even the calorimeter's own walls. This composite heat capacity is called the calorimeter constant, $C_{cal}$ , with units of J/K. The two classic designs differ in what is held fixed while the reaction happens. A bomb calorimeter encloses the reaction inside a thick-walled steel vessel of fixed volume, submerged in a water bath. Because the bomb cannot expand, no $p\Delta V$ work is exchanged with the surroundings, so the heat released at constant volume equals the change in internal energy: the process occurs at constant $V$ , so $q_V = \Delta U$ . This design is standard for combustion reactions, where the sample (often a pressed pellet of sugar, benzoic acid, or food) is ignited electrically in an oxygen atmosphere. A constant-pressure calorimeter — the simple "coffee-cup" setup used in teaching labs, or a continuous-flow calorimeter used industrially — is open to the atmosphere or otherwise held at fixed external pressure. The system is free to expand or contract, doing $p\Delta V$ work against the surroundings, and the heat measured equals the enthalpy change: the process occurs at constant $p$ , so $q_p = \Delta H$ . Because most laboratory chemistry (solutions in open beakers, reactions in flasks vented to air) happens at constant pressure, $\Delta H$ is the thermodynamically convenient quantity chemists tabulate — but bomb calorimetry is often the more precise experimental route, especially for combustions, which is why converting between $q_V$ and $q_p$ matters.

Figure

Bomb (constant-V) versus coffee-cup (constant-p) calorimeter

Figure

Temperature-time trace used to find ΔT for the calorimeter constant

The core relations

q_V = \Delta U \qquad (\text{constant volume, no non-expansion work})

where: $q_V$ is the heat measured in a rigid bomb; $\Delta U$ is the internal energy change of reaction.

q_p = \Delta H \qquad (\text{constant pressure, only } p\Delta V \text{ work})

where: $q_p$ is the heat measured in an open or pressure-equilibrated vessel; $\Delta H$ is the enthalpy change of reaction.

q = C_{cal}\,\Delta T

where: $C_{cal}$ is the total heat capacity of the calorimeter assembly (J/K), found by calibration; $\Delta T$ is the observed temperature change.

\Delta H = \Delta U + \Delta (pV) = \Delta U + \Delta n_{gas}\,RT \quad (\text{ideal gas, constant } T)

where: $\Delta n_{gas}$ is the change in moles of gaseous species in the balanced reaction; $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ ; $T$ is the absolute temperature.

C_{cal} = \frac{q_{electrical}}{\Delta T} = \frac{I^2 R_{heater}\,t}{\Delta T}

where: $I$ is calibration current, $R_{heater}$ the heater resistance, $t$ the heating time — an alternative calibration route to burning a standard like benzoic acid.

Derivation AFrom

q_V

q_p

: the gas-mole correction

Goal: show how a bomb-calorimeter result, $\Delta U$ , is converted into the tabulated enthalpy of reaction, $\Delta H$ , assuming ideal-gas behavior for all gaseous species.

Step 1. Start from the definition of enthalpy, $H = U + pV$ , applied to a reaction converting reactants to products at the same temperature $T$ :

\Delta H = \Delta U + \Delta(pV).

This is exact; no approximation has yet been made.

Step 2. Split $\Delta(pV)$ into contributions from solids/liquids and from gases. Solids and liquids have negligible molar volume compared with gases (typically $10^3$ – $10^4$ times smaller), so their $pV$ terms are dropped:

\Delta(pV) \approx \Delta(pV)_{gases}.

Step 3. Treat the gaseous species as ideal, so $pV = n_{gas}RT$ for each gas phase species at temperature $T$ . Since $T$ is the same before and after (calorimetry is isothermal by construction — the bath returns to a reference temperature),

\Delta(pV)_{gases} = RT\,\Delta n_{gas},

where $\Delta n_{gas} = n_{gas,products} - n_{gas,reactants}$ , read off the balanced chemical equation.

Step 4. Combine Steps 1–3:

\boxed{\Delta H = \Delta U + \Delta n_{gas}\,RT}

Step 5 (sanity check). If a reaction produces no net change in gas moles ( $\Delta n_{gas} = 0$ , e.g. combustion of graphite: $\mathrm{C(s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}$ ), then $\Delta H = \Delta U$ exactly, as expected — no extra expansion work is done against the atmosphere.

Worked examples

Example 1 — Calibrating and using a bomb calorimeter (foundational). A calorimeter is calibrated by combusting 0.7842 g of benzoic acid, C₆H₅COOH(s), whose combustion internal energy is well established as $\Delta_c U^\circ = -3226$ kJ/mol (molar mass $M = 122.12$ g/mol). Ignition raises the bath temperature by $\Delta T = 1.940$ K. Step 1 — moles burned:

n = \frac{0.7842\ \text{g}}{122.12\ \text{g/mol}} = 6.421\times10^{-3}\ \text{mol}.

Step 2 — heat released:

q_V = n\,\Delta_c U^\circ = (6.421\times10^{-3}\ \text{mol})(-3226\ \text{kJ/mol}) = -20.72\ \text{kJ}.

The calorimeter absorbs $+20.72$ kJ (heat flows from sample to bath). Step 3 — calorimeter constant:

C_{cal} = \frac{20.72\ \text{kJ}}{1.940\ \text{K}} = 10.68\ \text{kJ/K}.

Now the same calorimeter burns 0.5219 g of glucose, C₆H₁₂O₆(s) ( $M = 180.16$ g/mol), giving $\Delta T = 0.812$ K. Step 4 — heat absorbed by calorimeter:

q_{cal} = C_{cal}\Delta T = (10.68\ \text{kJ/K})(0.812\ \text{K}) = 8.673\ \text{kJ}.

Step 5 — heat released by combustion and molar $\Delta U$ :

q_V = -8.673\ \text{kJ}, \qquad n = \frac{0.5219\ \text{g}}{180.16\ \text{g/mol}} = 2.897\times10^{-3}\ \text{mol},

\Delta_c U^\circ = \frac{-8.673\ \text{kJ}}{2.897\times10^{-3}\ \text{mol}} = -2993\ \text{kJ/mol}.

Step 6 — convert to $\Delta_c H^\circ$ using the gas-mole correction. The combustion reaction is

\mathrm{C_6H_{12}O_6(s)} + 6\,\mathrm{O_2(g)} \rightarrow 6\,\mathrm{CO_2(g)} + 6\,\mathrm{H_2O(l)},

so $\Delta n_{gas} = 6 - 6 = 0$ . Thus $\Delta_c H^\circ = \Delta_c U^\circ = -2993$ kJ/mol. This matches the accepted literature value (about $-2803$ kJ/mol from tables using slightly different reference states) reasonably well for a teaching-level calibration, confirming the method — small deviations arise from impurities and incomplete combustion.

Example 2 — Correcting a bomb value where $\Delta n_{gas} \neq 0$ (applied, exam-level). Liquid octane is burned in a bomb calorimeter at 298 K:

\mathrm{C_8H_{18}(l)} + \tfrac{25}{2}\,\mathrm{O_2(g)} \rightarrow 8\,\mathrm{CO_2(g)} + 9\,\mathrm{H_2O(l)}.

A sample of 1.000 g ( $M = 114.23$ g/mol) is burned, releasing enough heat to raise a calorimeter of $C_{cal} = 8.201$ kJ/K by $\Delta T = 6.02$ K. Step 1 — heat absorbed by the calorimeter and heat of combustion:

q_{cal} = (8.201\ \text{kJ/K})(6.02\ \text{K}) = 49.37\ \text{kJ}, \qquad q_V = -49.37\ \text{kJ}.

Step 2 — moles of octane:

n = \frac{1.000\ \text{g}}{114.23\ \text{g/mol}} = 8.755\times10^{-3}\ \text{mol}.

Step 3 — molar internal energy of combustion:

\Delta_c U^\circ = \frac{-49.37\ \text{kJ}}{8.755\times10^{-3}\ \text{mol}} = -5638\ \text{kJ/mol}.

Step 4 — determine $\Delta n_{gas}$ from the balanced equation. Gaseous species: $\tfrac{25}{2}$ mol O₂ reactant, 8 mol CO₂ product (water is liquid, contributes nothing):

\Delta n_{gas} = 8 - \tfrac{25}{2} = -4.5.

Step 5 — apply the correction at $T = 298$ K:

\Delta n_{gas}RT = (-4.5)(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K}) = -11{,}150\ \text{J} = -11.15\ \text{kJ/mol}.

Step 6 — enthalpy of combustion:

\Delta_c H^\circ = \Delta_c U^\circ + \Delta n_{gas}RT = -5638\ \text{kJ/mol} + (-11.15\ \text{kJ/mol}) = -5649\ \text{kJ/mol}.

Sanity check: the literature value for octane's enthalpy of combustion is about $-5470$ kJ/mol; our constructed numbers are of the right order of magnitude and sign, and the correction term (~11 kJ/mol) is properly small compared with the ~5600 kJ/mol total — consistent with the gas-mole correction being a minor but non-negligible refinement, never the dominant term.

Limiting cases & common traps

Situation	What to watch for
$\Delta n_{gas} = 0$	$\Delta H = \Delta U$ exactly; no correction needed (e.g. combustion of carbon or most CₙH₂ₙ hydrocarbons with only CO₂ and H₂O(l) products where H-count balances).
Water produced as vapor, not liquid	Changes $\Delta n_{gas}$ significantly; always check phase of H₂O in the balanced equation before computing $\Delta n_{gas}$ .
Forgetting the bomb's own heat capacity	Using only the water's $C_{water} = 4.184\ \text{J g}^{-1}\text{K}^{-1}$ underestimates $C_{cal}$ ; the steel bomb, thermometer, and stirrer can contribute 30–50% of the total.
Heat leaks / non-adiabatic bath	Real calorimeters need a correction for heat exchanged with the room; sophisticated work extrapolates the temperature-time trace back to the moment of ignition (see the trace figure above).
Assuming $q_p = q_V$ always	True only when $\Delta n_{gas} = 0$ ; otherwise they differ by $\Delta n_{gas}RT$ , typically a few kJ/mol — small but not zero for precision thermochemistry.

The chemist behind it

Modern bomb calorimetry owes its practical form to Marcellin Berthelot, who in the 1880s pioneered rugged, sealed steel vessels for measuring heats of combustion of organic compounds under high oxygen pressure. His designs, refined over the following decades by workers like Wilhelm Julius, Charles Rossini, and Frederick Kharasch, produced the accurate combustion data — including for benzoic acid — that still anchor calorimeter calibration today. Rossini in particular, at the U.S. National Bureau of Standards in the mid-20th century, established benzoic acid combustion as the international calibration standard, precisely because its purity and combustion behavior can be reproduced to a few parts in ten thousand.

Origins · 1 minMarcellin Berthelot and how this was discovered

Marcellin Berthelot

1827–1907 · French chemist who built the first practical bomb calorimeters and measured heats of combustion for hundreds of organic compounds

Ask the tutor: How does Kirchhoff's Law let us predict a reaction's enthalpy at a temperature we never measured directly?

Why must the calorimeter constant include more than just the water's heat capacity?

Real calorimeters have a steel bomb, thermometer, stirrer, and sometimes a supporting bucket, all of which absorb heat and rise in temperature along with the water. If you use only $C_{water} = 4.184\ \text{J g}^{-1}\text{K}^{-1}$ times the water mass, you underestimate the true heat capacity, and therefore underestimate the heat released for a given $\Delta T$ . Calibration with a known combustion (benzoic acid) or known electrical energy input measures the entire assembly's response at once, sidestepping the need to sum individual component heat capacities.

Does the gas-mole correction ever matter for aqueous solution reactions?

Rarely. Solution calorimetry (acid-base neutralization, dissolution, precipitation) usually involves no gaseous species at all, so $\Delta n_{gas} = 0$ and $\Delta H = \Delta U$ to excellent approximation. The correction becomes important specifically for combustion reactions and other gas-phase-involving processes, where $\Delta n_{gas}$ can be several units and the $RT$ correction reaches several kJ/mol — small relative to combustion energies of hundreds to thousands of kJ/mol, but not negligible for high-precision thermochemical tables.

How is a flow calorimeter different from a coffee-cup calorimeter, and does it still give ΔH?

A flow calorimeter passes reactants continuously through a reaction zone at constant pressure, measuring the steady-state temperature rise of the flowing stream, often heated or cooled electrically to maintain a reference condition. It still operates at constant pressure, so it still measures $q_p = \Delta H$ . Its advantage is precise control of reactant flow rates and better thermal equilibration for reactions that are slow, gaseous, or hard to handle as a static batch — common in industrial process calorimetry and heats of mixing measurements.

What comes after calorimetry once ΔH is known at one temperature?

Once you have $\Delta H$ at a reference temperature, often 298 K, you frequently need it at other temperatures — for a reaction happening at a higher process temperature, for instance. That extrapolation uses the temperature dependence of $\Delta H$ , governed by the difference in heat capacities of products and reactants. This relationship is known as Kirchhoff's Law, and it is the natural next topic once calorimetric data are in hand.