Calorimetry
Calorimetry is a technique used to determine the energy evolved from a chemical reaction or physical process by measuring changes in temperature. These techniques are carried out in a calorimeter.
Constant Pressure Calorimetry
Constant-pressure calorimetry measures enthalpy changes (ΔH) for chemical reactions or physical processes. The simplest version uses a covered Styrofoam™ cup (a “coffee cup calorimeter”) filled with an aqueous solution. Because the cup is open to the atmosphere, the pressure remains constant at atmospheric pressure throughout the experiment.
Styrofoam™ is a trademark of DuPont (formerly Dow Chemical) for extruded polystyrene foam, used mainly in building insulation. The white foam cups commonly called “Styrofoam cups” are actually expanded polystyrene (EPS), a different material. Both have excellent thermal insulating properties due to trapped air pockets.
How the Coffee Cup Calorimeter Works
The coffee cup calorimeter has five components:
- Nested cups - two cups with an air gap between them, mimicking a thermos flask for thermal insulation
- Lid - blocks convective heat loss; holds instruments in place
- Thermometer - measures ΔT with sufficient precision (typically ±0.1 °C resolution)
- Stirrer - keeps the solution homogeneous so temperature is uniform throughout
- Reaction mixture - the aqueous solution where the reaction occurs
The primary advantage of this design is its simplicity and low cost. Styrofoam™ cups are inexpensive and disposable, making coffee cup calorimetry accessible for educational laboratories and routine measurements. The open design also means the system remains at atmospheric pressure throughout the experiment, which is essential for directly measuring enthalpy changes.
Strictly speaking, a coffee cup calorimeter is an open system since the lid doesn’t form a perfect seal. In practice, we treat it as approximately closed (mass loss through evaporation is small) and approximately adiabatic (the lid blocks convective heat loss and the insulation minimizes conductive heat exchange). Under the adiabatic approximation, heat released by the reaction is absorbed entirely by the solution, causing a measurable temperature change.
Why Constant Pressure Gives Us ΔH
Recall from the First Law of Thermodynamics that under constant pressure conditions, the heat transferred equals the enthalpy change:
\[ \Delta H = q_{\mathrm{p}} \]
where qp is the heat transferred at constant pressure. Since the coffee cup is open to the atmosphere, maintaining constant external pressure, measuring the heat flow directly gives us ΔH for the process. This makes coffee cup calorimetry ideal for studying reactions in aqueous solution under typical laboratory conditions.
Heat Balance and Approximations
When a reaction occurs in the calorimeter, the heat evolved by the reaction (qreaction) is absorbed by the solution (qsolution). These are commonly abbreviated as qrxn and qsoln:
\[ -q_{\mathrm{rxn}} = q_{\mathrm{soln}} \]
The heat absorbed by the solution depends on its mass, specific heat capacity, and temperature change:
\[ q_{\mathrm{soln}} = m_{\mathrm{soln}} c_{\mathrm{p},\mathrm{soln}} \Delta T \]
For dilute aqueous solutions, we typically make two approximations:
- Specific heat capacity: cp,soln ≈ cp,water = 4.184 J g−1 °C−1 (the solution behaves thermally like pure water)
- Density: dsoln ≈ dwater ≈ 1.00 g mL−1 (allows conversion from volume to mass)
These approximations are reasonable for dilute solutions where the solute concentration is low enough that the thermal properties remain close to those of pure water.
Example: An acid-base neutralization reaction is carried out in a coffee cup calorimeter. A 100.0 mL sample of 1.00 M HCl at 21.0 °C is mixed with 75.0 mL of 1.33 M NaOH also at 21.0 °C. The reaction is:
\[\mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)}\]
Upon reaction, the temperature of the solution increases to 26.2 °C. Find the molar enthalpy of reaction (in kJ mol−1).
Solution
The heat evolved by the reaction is absorbed by the solution:
\[-q_{\mathrm{rxn}} = q_{\mathrm{soln}}\]
Applying the water approximations and recognizing that mass can be found from density and volume:
\[ \begin{align*} q_{\mathrm{rxn}} &= -q_{\mathrm{soln}} \\[1.5ex] &= -m_{\mathrm{soln}}~c_{\mathrm{p},\mathrm{soln}}~\Delta T_{\mathrm{soln}} \\[1.5ex] &= -\rho_{\mathrm{soln}}~V_{\mathrm{soln}}~c_{\mathrm{p},\mathrm{soln}}~\Delta T_{\mathrm{soln}} \\[1.5ex] &= - \left ( \dfrac{1.0\bar{0}~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{175.\bar{0}~\mathrm{mL}}{} \right ) \left ( \dfrac{4.18\bar{4}~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( \dfrac{26.\bar{2}~^{\circ}\mathrm{C} - 21.\bar{0}~^{\circ}\mathrm{C}}{} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\[1.5ex] &= - \left ( \dfrac{73\bar{2}.2~\mathrm{J}}{^{\circ}\mathrm{C}} \right ) \left ( \dfrac{5.\bar{2}~^{\circ}\mathrm{C}}{} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right )\\[1.5ex] &= -3.\bar{8}07~\mathrm{kJ} \\[1.5ex] &= -3.8~\mathrm{kJ} \end{align*} \]
This is the total heat released (ΔH = −3.8 kJ). To find the molar enthalpy of reaction, we need the number of moles that reacted.
The limiting reagent determines how many moles reacted:
- n(HCl) = (0.1000 L)(1.00 mol L−1) = 0.100 mol
- n(NaOH) = (0.0750 L)(1.33 mol L−1) = 0.100 mol
Both reagents are present in stoichiometric amounts, so 0.100 mol of reaction occurred. The molar enthalpy of reaction is:
\[ \begin{align*} \Delta_{\mathrm{r}} H &= \dfrac{\Delta H}{n} \\[1.5ex] &= \dfrac{-3.\bar{8}07~\mathrm{kJ}}{0.10\bar{0}~\mathrm{mol}} \\[1.5ex] &= -3\bar{8}.07~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -38~\mathrm{kJ~mol^{-1}} \end{align*} \]
The negative sign indicates the reaction is exothermic. The literature value for the molar enthalpy of neutralization of a strong acid with a strong base is approximately −57 kJ mol−1; the discrepancy is due to heat loss to the surroundings in this simple apparatus.
Coffee cup calorimetry can also determine the specific heat capacity of unknown materials by measuring heat transfer between a hot sample and cold water. For a detailed example of this technique, see Heat Transfer and Thermal Equilibrium in the Heat chapter.
Constant Volume Calorimetry
Constant volume calorimetry measures internal energy changes (ΔU) for chemical reactions or physical processes. These measurements are performed in a bomb calorimeter, an insulated container housing a rigid steel vessel (the “bomb”) submerged in water. A combustion reaction is carried out inside the bomb, which is pressurized with oxygen gas.
The exothermic combustion releases heat that is absorbed by the water and the bomb apparatus itself. A thermometer submerged in the water records the temperature increase, allowing determination of ΔU for the reaction.
Why Constant Volume Gives Us ΔU
The rigid steel vessel prevents volume change during the reaction (ΔV = 0). Recall from the First Law that when no PV work can occur, the heat transferred at constant volume equals the internal energy change:
\[ \Delta U = q_{\mathrm{V}} \]
where qV is the heat transferred at constant volume. Since the bomb cannot expand or contract, all the energy released by combustion appears as heat absorbed by the calorimeter system.
Relationship Between ΔH and ΔU
For combustion reactions, we often want the enthalpy change (ΔH), not the internal energy change (ΔU). These differ by the PV work term. For reactions involving gases at constant temperature and pressure:
\[ \Delta H = \Delta U + \Delta n_{\mathrm{gas}} RT \]
where Δngas is the change in moles of gas (products minus reactants). For reactions with no change in gas moles, ΔH ≈ ΔU. For reactions with significant Δngas, the correction term becomes important.
Heat Capacity of the Calorimeter
The bomb calorimeter absorbs heat through two components: the bomb itself (steel vessel, ignition wire, sample holder) and the surrounding water.
One way to handle this is to characterize the entire calorimeter system (bomb plus water) by a single heat capacity Ccalorimeter (often abbreviated as Ccal), determined through calibration with a substance of known combustion enthalpy. The heat absorbed is then:
\[ q_{\mathrm{cal}} = C_{\mathrm{cal}} \Delta T \]
Alternatively, the bomb and water heat capacities can be treated separately:
\[ \begin{align*} q_{\mathrm{total}} &= q_{\mathrm{bomb}} + q_{\mathrm{water}} \\[1.5ex] &= C_{\mathrm{bomb}} \Delta T + m_{\mathrm{water}} c_{\mathrm{p,water}} \Delta T \end{align*} \]
Why Ccal and Cbomb rather than CV,cal? Although the reaction occurs at constant volume, the calorimeter components (steel bomb, water bath) are solids and liquids for which Cp ≈ CV. The water bath is actually at constant pressure (open to the atmosphere). Since the subscript distinction is negligible for condensed phases, we simply use Ccal or Cbomb. These are empirically calibrated values for the apparatus, not fundamental thermodynamic quantities.
The combined approach is simpler if the calorimeter has been calibrated. The separate approach is more fundamental and allows tracking individual contributions.
Before a bomb calorimeter can be used for unknown samples, it must be calibrated to determine its heat capacity. This is done by combusting a standard reference material with precisely known thermodynamic properties.
Standard Calibration Procedure:
- Combust a known mass of benzoic acid (C6H5COOH), the certified standard
- ΔcU° = −3228 kJ mol−1 (NIST certified value)
- Measure the temperature change ΔT
- Calculate the calorimeter heat capacity:
\[ C_{\mathrm{calorimeter}} = \frac{|q_{\mathrm{rxn}}|}{\Delta T} \]
Once calibrated, this Ccalorimeter value is used for all subsequent measurements with that calorimeter. This connects to the broader principle of reference standards in thermochemistry: we determine unknown values by comparison to precisely known reference materials.
Example (Combined Heat Capacity): A 0.6475 g sample of naphthalene (C10H8) is combusted in a bomb calorimeter. The calorimeter has a combined heat capacity of 8.0931 kJ °C−1. The temperature rises from 22.00 °C to 25.21 °C. Find the molar internal energy of combustion, ΔcU.
\[\mathrm{C_{10}H_{8}(s)} + 12~\mathrm{O_2(g)} \longrightarrow 10~\mathrm{CO_2(g)} + 4~\mathrm{H_2O(l)}\]
Solution
The heat released by combustion is absorbed by the calorimeter:
\[ \begin{align*} q_{\mathrm{rxn}} &= -q_{\mathrm{cal}} \\[1.5ex] &= -C_{\mathrm{cal}} \Delta T \\[1.5ex] &= -\left ( \dfrac{8.093\bar{1}~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( 25.2\bar{1}~^{\circ}\mathrm{C} - 22.0\bar{0}~^{\circ}\mathrm{C} \right )\\[1.5ex] &= -\left ( \dfrac{8.093\bar{1}~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( 3.2\bar{1}~^{\circ}\mathrm{C} \right )\\[1.5ex] &= -25.\bar{9}78~\mathrm{kJ} \end{align*} \]
This is ΔU for the combustion of 0.6475 g naphthalene. To find the molar quantity, divide by the number of moles:
\[ \begin{align*} \Delta_{\mathrm{c}} U &= q_{\mathrm{rxn}}~n^{-1} \\[1.5ex] &= q_{\mathrm{rxn}}~\left ( m~M^{-1}\right )^{-1} \\[1.5ex] &= \left ( -25.\bar{9}78~\mathrm{kJ} \right ) \left ( \dfrac{0.647\bar{5}~\mathrm{g}}{128.1\bar{8}~\mathrm{g~mol^{-1}}}\right )^{-1}\\[1.5ex] &= -51\bar{4}2.6~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -5.14 \times 10^{3}~\mathrm{kJ~mol^{-1}} \end{align*} \]
The molar internal energy of combustion for naphthalene is −5.14 × 103 kJ mol−1.
Example (Separated Components): A bomb calorimeter contains 1.500 kg of water and has a bomb with heat capacity Cbomb = 0.850 kJ °C−1. A 1.250 g sample of glucose (C6H12O6) is combusted, causing the temperature to rise from 20.00 °C to 23.65 °C. Find the molar enthalpy of combustion, ΔcH.
\[\mathrm{C_6H_{12}O_6(s)} + 6~\mathrm{O_2(g)} \longrightarrow 6~\mathrm{CO_2(g)} + 6~\mathrm{H_2O(l)}\]
Solution
Calculate the heat absorbed by the water:
\[ \begin{align*} q_{\mathrm{water}} &= m_{\mathrm{water}} c_{\mathrm{p,water}} \Delta T \\[1.5ex] &= \left ( 1.50\bar{0} \times 10^3~\mathrm{g} \right ) \left ( \dfrac{4.18\bar{4}~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( \dfrac{23.6\bar{5}~^{\circ}\mathrm{C} - 20.0\bar{0}~^{\circ}\mathrm{C}}{} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= \left ( 1.50\bar{0} \times 10^3~\mathrm{g} \right ) \left ( \dfrac{4.18\bar{4}~\mathrm{J}}{\mathrm{g~^{\circ}C}} \right ) \left ( \dfrac{3.6\bar{5}~^{\circ}\mathrm{C}}{} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= 22.\bar{9}07~\mathrm{kJ} \end{align*} \]
Calculate the heat absorbed by the bomb:
\[ \begin{align*} q_{\mathrm{bomb}} &= C_{\mathrm{bomb}} \Delta T \\[1.5ex] &= \left ( \dfrac{0.85\bar{0}~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( \dfrac{23.6\bar{5}~^{\circ}\mathrm{C} - 20.0\bar{0}~^{\circ}\mathrm{C}}{} \right ) \\[1.5ex] &= \left ( \dfrac{0.85\bar{0}~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( \dfrac{3.6\bar{5}~^{\circ}\mathrm{C}}{} \right ) \\[1.5ex] &= 3.1\bar{0}25~\mathrm{kJ} \end{align*} \]
The total heat absorbed equals the heat released by combustion (with opposite sign):
\[ \begin{align*} q_{\mathrm{rxn}} &= -(q_{\mathrm{water}} + q_{\mathrm{bomb}}) \\[1.5ex] &= -(22.\bar{9}07~\mathrm{kJ} + 3.1\bar{0}25~\mathrm{kJ}) \\[1.5ex] &= -26.\bar{0}09~\mathrm{kJ} \end{align*} \]
This is ΔU for the sample. Converting to molar internal energy:
\[ \begin{align*} \Delta_{\mathrm{c}} U &= q_{\mathrm{rxn}}~n^{-1}\\[1.5ex] &= q_{\mathrm{rxn}}~\left ( m~M^{-1}\right )^{-1}\\[1.5ex] &= \left ( -26.\bar{0}09~\mathrm{kJ} \right ) \left ( \dfrac{1.25\bar{0}~\mathrm{g}}{180.1\bar{8}~\mathrm{g~mol^{-1}}}\right )^{-1} \\[1.5ex] &= -37\bar{4}9.0~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -3.75\times 10^{3}~\mathrm{kJ~mol^{-1}} \end{align*} \]
Now convert ΔcU to ΔcH. The change in stoichiometric coefficients for gases is:
\[ \Delta\nu_{\mathrm{gas}} = 6 - 6 = 0 \]
Since Δνgas = 0, the RT correction vanishes:
\[ \begin{align*} \Delta_{\mathrm{c}} H &= \Delta_{\mathrm{c}} U + \Delta\nu_{\mathrm{gas}}RT \\[1.5ex] &= \Delta_{\mathrm{c}} U + 0\\[1.5ex] &= -37\bar{4}9.0~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -3.75\times 10^{3}~\mathrm{kJ~mol^{-1}} \end{align*} \]
The molar enthalpy of combustion for glucose is −3.75 × 103 kJ mol−1. When the number of moles of gaseous products equals the number of moles of gaseous reactants, ΔH and ΔU are equal.
Practice
The combustion of 6.491 g methanol (CH3OH) is carried out in a bomb calorimeter.
\[\mathrm{CH_3OH(l)} + \tfrac{3}{2}~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(l)}\]
The calorimeter has a heat capacity of 14.402 kJ °C−1 and a temperature change of 3.74 °C is recorded. Find the standard molar enthalpy of combustion.
Solution
Find Internal Energy of Combustion
The heat released by combustion is absorbed by the calorimeter:
\[ \begin{align*} q_{\mathrm{rxn}} &= -C_{\mathrm{cal}}\Delta T \\[1.5ex] &= -\left ( \dfrac{14.40\bar{2}~\mathrm{kJ}}{^{\circ}\mathrm{C}} \right ) \left ( 3.7\bar{4}~\mathrm{^{\circ}C} \right ) \\[1.5ex] &= -53.\bar{8}63~\mathrm{kJ} \end{align*} \]
This is ΔU for the sample. Converting to molar internal energy:
\[ \begin{align*} \Delta_{\mathrm{c}} U &= q_{\mathrm{rxn}}~n^{-1}\\[1.5ex] &= q_{\mathrm{rxn}}~\left ( m~M^{-1}\right )^{-1}\\[1.5ex] &= \left ( -53.\bar{8}63~\mathrm{kJ}\right ) \left ( \dfrac{6.49\bar{1}~\mathrm{g}}{32.0\bar{5}~\mathrm{g~mol^{-1}}}\right )^{-1} \\[1.5ex] &= -26\bar{5}.954~\mathrm{kJ~mol^{-1}} \end{align*} \]
Convert to Enthalpy
The molar relationship between enthalpy and internal energy is:
\[\Delta_{\mathrm{c}} H = \Delta_{\mathrm{c}} U + \Delta\nu_{\mathrm{gas}}RT\]
where Δνgas is the change in stoichiometric coefficients for gaseous species:
\[ \begin{align*} \Delta\nu_{\mathrm{gas}} &= 1 - \frac{3}{2} \\[1.5ex] &= -\frac{1}{2} \end{align*} \]
Using T = 298.15 K for standard conditions:
\[ \begin{align*} \Delta_{\mathrm{c}} H^{\circ} &= \Delta_{\mathrm{c}} U + \Delta\nu_{\mathrm{gas}}RT \\[1.5ex] &= -26\bar{5}.954~\mathrm{kJ~mol^{-1}} + \left ( -\frac{1}{2} \right ) \left ( \dfrac{8.31\bar{4}~\mathrm{J}}{\mathrm{mol~K}} \right ) \left ( \dfrac{298.1\bar{5}~\mathrm{K}}{} \right ) \left ( \frac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= -26\bar{5}.954~\mathrm{kJ~mol^{-1}} - 1.23\bar{9}40~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -26\bar{7}.193~\mathrm{kJ~mol^{-1}} \\[1.5ex] &= -267~\mathrm{kJ~mol^{-1}} \end{align*} \]
The standard molar enthalpy of combustion for methanol is −267 kJ mol−1.
Limitations
No calorimeter is perfectly adiabatic. Heat leaks to the surroundings, and the measured ΔT is always smaller than it would be in an ideal system. This effect worsens with longer experiments, larger temperature differences, and poorer insulation. Coffee cup calorimeters are particularly susceptible since the Styrofoam™ provides limited insulation and the lid doesn’t seal.
In coffee cup calorimetry, we typically ignore the heat capacity of the cup and thermometer, assuming only the solution absorbs heat. This approximation is reasonable for dilute aqueous solutions where the solution mass dominates, but introduces error in precise work. Evaporation from open cups also carries away energy during exothermic reactions.
Bomb calorimetry has its own issues. Incomplete combustion leaves fuel unreacted, releasing less heat than expected. This is minimized by pressurizing with excess oxygen and using finely divided samples to maximize surface area.
The thermometer in a bomb calorimeter measures the temperature of the water bath surrounding the steel bomb. As combustion occurs, heat transfers from the bomb to the water, raising its temperature. However, heat also leaks from the warm water bath to the cooler outer jacket, causing the measured ΔT to be slightly lower than it would be in a perfectly insulated system.
Real bomb calorimeters handle this heat leak in one of two ways:
Isoperibol mode (most common in teaching laboratories): The outer jacket is passive insulation held at roughly constant (room) temperature. Since the water bath is warmer than the jacket, heat continuously leaks outward. The measured ΔT is corrected mathematically by extrapolating what the temperature change would have been without heat loss.
Adiabatic mode (research-grade instruments): The outer jacket has an active heating/cooling system that adjusts the jacket temperature to match the water bath temperature throughout the experiment. With no temperature gradient between water and jacket, no heat leaks, and the measured ΔT requires no correction.
The examples in this chapter assume ideal conditions where heat leak is negligible, which is a reasonable approximation for well-insulated calorimeters with fast reactions.
Summary
Calorimetry Overview
Calorimetry measures energy changes in chemical reactions or physical processes by recording temperature changes.
Constant Pressure Calorimetry
At constant pressure, heat equals enthalpy change: ΔH = qp
The heat balance for a reaction in solution:
\[ -q_{\mathrm{rxn}} = q_{\mathrm{soln}} = m_{\mathrm{soln}} c_{\mathrm{p},\mathrm{soln}} \Delta T \]
For dilute aqueous solutions: cp,soln ≈ 4.184 J g−1 °C−1 and ρsoln ≈ 1.00 g mL−1
Constant Volume Calorimetry
At constant volume, heat equals internal energy change: ΔU = qV
\[ -q_{\mathrm{rxn}} = q_{\mathrm{cal}} = C_{\mathrm{cal}} \Delta T \]
To convert ΔU to ΔH for reactions involving gases:
\[ \Delta H = \Delta U + \Delta n_{\mathrm{gas}} RT \]
where Δngas is the change in moles of gas (products minus reactants).
For molar quantities:
\[ \Delta_{\mathrm{r}} H = \Delta_{\mathrm{r}} U + \Delta\nu_{\mathrm{gas}} RT \]
where Δνgas is the change in stoichiometric coefficients for gaseous species.