Measurement of $\Delta U$ and $\Delta H$

Calorimetry rests on one physical fact: energy is conserved. Heat that a reaction releases does not vanish — it flows into the calorimeter and its contents, usually water, and shows up as a measurable rise in temperature. Run the arithmetic backwards, from the temperature rise to the heat, and from the heat (with the right constant-volume or constant-pressure condition) to $\Delta U$ or $\Delta H$.

The bomb calorimeter measures $\Delta U$. A combustion reaction is sealed, with excess oxygen, inside a thick-walled steel vessel called a "bomb," which is then submerged in a known mass of water inside an insulated jacket. The rigid steel walls cannot expand or contract, so the volume of the system is fixed throughout the reaction. Because $w = -P\Delta V$ and $\Delta V = 0$, no work is done ($w = 0$), and the first law $\Delta U = q + w$ collapses to $\Delta U = q_v$ — the heat measured at constant volume is the internal-energy change, with no correction needed. The heat itself is found from how much the surrounding water and the calorimeter hardware warmed up:

where: $m_{water}$ is the mass of water surrounding the bomb, $c_{water} = 4.18\ \text{J g}^{-1}\text{K}^{-1}$ is its specific heat capacity, $C_{calorimeter}$ is the heat capacity of the bomb assembly itself (obtained separately by calibration, and often small enough to neglect), and $\Delta T$ is the observed temperature rise.

The open (coffee-cup) calorimeter measures $\Delta H$. Reactions in solution — neutralisations, dissolutions, dilutions — are usually carried out in a simple insulated cup open to the atmosphere, so the pressure stays fixed at atmospheric pressure throughout. At constant pressure, the heat exchanged with the surroundings equals the enthalpy change directly: $q_p = \Delta H$. The heat is again read from a temperature change, this time of the solution itself:

where: $m$ and $c$ are the mass and specific heat capacity of the solution (commonly taken as those of water, since the solutions are dilute and aqueous), and $\Delta T$ is the temperature change measured in the cup. The figure below lines the two setups up side by side: the sealed bomb pins volume constant and reads out $\Delta U$; the open cup pins pressure constant and reads out $\Delta H$.

The key distinction to hold onto is which variable is pinned down: a rigid bomb pins volume, so its heat reading is $\Delta U$; an open vessel pins pressure, so its heat reading is $\Delta H$. Whenever a question hands you a value from one type of calorimeter but asks for the other quantity, a conversion is needed, and that conversion is the derivation below.

Problem: When $1.0\ \text{g}$ of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$, molar mass $180\ \text{g mol}^{-1}$) is completely burnt in a bomb calorimeter, the temperature of $1000\ \text{g}$ of surrounding water (specific heat $4.18\ \text{J g}^{-1}\text{K}^{-1}$) rises by $3.14\ \text{K}$; the heat capacity of the calorimeter itself is negligible. Find the molar internal energy of combustion, and then use it to find the molar enthalpy of combustion.

Step 1. Heat absorbed by the water: $q = mc\Delta T = (1000\ \text{g})(4.18\ \text{J g}^{-1}\text{K}^{-1})(3.14\ \text{K}) = 13{,}125\ \text{J} \approx 13.1\ \text{kJ}$.

Step 2. This heat was released by burning $1.0\ \text{g}$ of glucose, so the heat released per gram is $-13.1\ \text{kJ g}^{-1}$ (negative because the reaction releases heat to the surroundings).

Step 3. Since the bomb operates at constant volume, $\Delta U = q_v$ for the reaction. Scaling to one mole ($180\ \text{g}$): $\Delta U = -13.1\ \text{kJ g}^{-1} \times 180\ \text{g mol}^{-1} = -2358\ \text{kJ mol}^{-1}$.

Step 4. Convert to $\Delta H$ using the derivation above. Glucose combustion has $\Delta n_g = 0$, so the correction term $\Delta n_g RT$ vanishes and $\Delta H = \Delta U$. The molar internal energy of combustion of glucose is $-2358\ \text{kJ mol}^{-1}$, and because $\Delta n_g = 0$ for this reaction, the molar enthalpy of combustion is also $-2358\ \text{kJ mol}^{-1}$.

Competitive stretch

If the same $-2358\ \text{kJ mol}^{-1}$ had instead come from a reaction with $\Delta n_g = +2$ at $T = 300\ \text{K}$, the correction would be $\Delta n_g RT = (2)(8.314)(300) = 4988\ \text{J mol}^{-1} \approx +5.0\ \text{kJ mol}^{-1}$, giving $\Delta H = -2358 + 5.0 = -2353\ \text{kJ mol}^{-1}$ — enthalpy is slightly less negative than internal energy here because expanding gas does work on the surroundings, an energy cost that $\Delta U$ (constant volume) never has to pay.

• Assuming a bomb calorimeter measures $\Delta H$ directly — it measures $\Delta U$, because volume is held constant, not pressure.
• Forgetting the sign convention: heat released by the reaction is heat absorbed by the calorimeter and water, so $q_{reaction} = -q_{water}$.
• Applying $q = mc\Delta T$ with the mass of the reacting sample instead of the mass of the water or solution — the temperature-change term belongs to the calorimeter contents, not the burning or reacting sample.
• Using $\Delta H = \Delta U + \Delta n_g RT$ with $\Delta n_g$ counting all species instead of only gaseous ones, or forgetting to convert temperature to kelvin before multiplying by $R$.

Ask the tutor: Want a mixed set of bomb-calorimeter and open-calorimeter problems, including a few where $\Delta n_g \neq 0$ so you get practice applying the correction term?