Combustion releases gases and can't be done in an open cup. Instead the sample burns in pure oxygen inside a sealed steel "bomb" submerged in water. The method is Calorimetry for Reactions with one change: the whole apparatus is calibrated as a unit.
Key idea
The bomb, its water bath and fittings together have a heat capacity $C_\text{cal}$ (in kJ/°C), found by burning a standard. Then $$q_\text{cal} = C_\text{cal}\,\Delta T, \qquad q_\text{rxn} = -C_\text{cal}\,\Delta T.$$ Because the volume is fixed, no PV work is done, so this heat is $\Delta U$; for most combustions it is close to $\Delta H$.
Why it works
Water, steel, thermometer and stirrer all warm together, and adding up each one's $mc$ separately would be tedious and error-prone. Instead, burn a known amount of a standard with a known energy release (benzoic acid gives 26.38 kJ per gram) and see how much the temperature rises: that gives the calorimeter's heat capacity once and for all. In a sealed bomb $w = 0$, so the first law gives $\Delta U = q_V$. The difference from $\Delta H$ is $\Delta n_\text{gas}RT$, only a few kJ per mole.
Worked example: Calibrating
Burning $1.000\ \text{g}$ of benzoic acid ($26.38\ \text{kJ/g}$ released) raises a calorimeter's temperature by $2.637^\circ\text{C}$. Find $C_\text{cal}$.
- Heat released: $26.38\ \text{kJ}$, all absorbed by the calorimeter.
- $C_\text{cal} = \dfrac{26.38\ \text{kJ}}{2.637^\circ\text{C}} = 10.00\ \text{kJ/°C}$.
Answer: $10.00\ \text{kJ/°C}$
Worked example: Energy in sugar
In the same calorimeter, $0.500\ \text{g}$ of sucrose ($342.30\ \text{g/mol}$) raises the temperature by $0.826^\circ\text{C}$. Find the energy released per gram and per mole, and in food Calories per gram.
- $q_\text{cal} = (10.00)(0.826) = 8.26\ \text{kJ}$; $q_\text{rxn} = -8.26\ \text{kJ}$.
- Per gram: $8.26/0.500 = 16.5\ \text{kJ/g}$. Per mole: $16.52 \times 342.30 = 5650\ \text{kJ/mol}$.
- A food Calorie is a kilocalorie, $4.184\ \text{kJ}$: $16.5/4.184 = 3.95$, about 4 Calories per gram, as on food labels for sugars.
Answer: $16.5\ \text{kJ/g}$; $5.65 \times 10^{3}\ \text{kJ/mol}$; about 4 Cal/g.
How different are $\Delta U$ and $\Delta H$? $\Delta H = \Delta U + \Delta n_\text{gas}RT$. For sucrose, $$\mathrm{C_{12}H_{22}O_{11}} + 12\,\mathrm{O_2} \to 12\,\mathrm{CO_2} + 11\,\mathrm{H_2O(l)}$$ the moles of gas don't change ($\Delta n_\text{gas} = 0$), so the two are equal. Even when they differ, $RT$ is only 2.5 kJ per mole of gas at room temperature.
Common mistake
Using water's specific heat and a mass when the calorimeter's heat capacity is given in kJ/°C. $C_\text{cal}$ already includes the water and the steel; just multiply it by $\Delta T$. Check the units: kJ/°C times °C is kJ, with no grams anywhere.