How to Calculate the Heat of Dissolution in a Calorimeter

A calorimeter experiment measures how the temperature of a solution changes when a solute dissolves. The solution and the calorimeter exchange heat with the dissolving process, so their measured heat change has the opposite sign from the heat of dissolution.

You can calculate the heat for the amount of solute used, or divide by the solute amount to report a molar enthalpy of dissolution. At constant pressure, that molar heat is the enthalpy change, provided the experiment is insulated well enough that heat exchange with the room is negligible and pressure-volume work is the only work.

The heat-balance equation

Use the temperature change

$$\Delta T = T_{\text{final}} - T_{\text{initial}}.$$

For a solution whose specific heat capacity is approximately known,

$$q_{\text{solution}} = m_{\text{solution}} c_{\text{solution}}\Delta T.$$

Here, $m$ is the solution mass and $c$ is its specific heat capacity. If the calorimeter itself absorbs or releases heat, include it:

$$q_{\text{calorimeter}} = C_{\text{cal}}\Delta T,$$

where $C_{\text{cal}}$ is the calorimeter heat capacity in joules per degree Celsius or joules per kelvin. A temperature difference has the same numerical value in °C and K.

The heat of dissolution is the negative of the combined heat change of the solution and calorimeter:

$$q_{\text{dissolution}} = -\left(q_{\text{solution}}+q_{\text{calorimeter}}\right).$$

If the calorimeter heat capacity is negligible or has not been measured, the introductory approximation is $q_{\text{dissolution}}\approx -q_{\text{solution}}$. This approximation can make the result less accurate.

Step-by-step calculation

1. Record the initial temperature and the final temperature after the solution has mixed and reached its measured temperature.
2. Calculate $\Delta T$ as final minus initial. Keep the sign: cooling gives a negative value, and warming gives a positive value.
3. Calculate the solution heat using its mass and specific heat capacity. Add the calorimeter heat if its heat capacity is known.
4. Reverse the sign of the combined heat to obtain the heat of dissolution for the sample.
5. To report molar enthalpy, divide the sample’s heat of dissolution by the moles of solute dissolved.

For dilute aqueous solutions, labs often approximate the solution’s specific heat capacity as that of water, $4.184\ \text{J/(g·°C)}$. Use the value specified by your lab when available. The solution mass is usually the mass of water plus the dissolved solute, unless the lab instructions prescribe another model.

Worked example: ammonium nitrate dissolving

Suppose $5.00\ \text{g}$ of ammonium nitrate dissolves in $100.0\ \text{g}$ of water. The temperature falls from $22.0\ ^\circ\text{C}$ to $18.7\ ^\circ\text{C}$. The calorimeter heat capacity is $35\ \text{J/K}$. Assume the final solution mass is $105.0\ \text{g}$, the solution’s specific heat capacity is $4.184\ \text{J/(g·°C)}$, and the solute fully dissolves.

First, find the temperature change:

$$\Delta T = 18.7 - 22.0 = -3.3\ ^\circ\text{C}.$$

Then calculate the heat change of the solution and calorimeter:

$$\begin{aligned} q_{\text{solution}} &= (105.0)(4.184)(-3.3) = -1449.8\ \text{J},\\ q_{\text{calorimeter}} &= (35)(-3.3) = -115.5\ \text{J},\\ q_{\text{dissolution}} &= -(-1449.8-115.5) = +1565.3\ \text{J}. \end{aligned}$$

These are unrounded arithmetic values. The positive sign means the dissolving process absorbed heat. The solution and calorimeter cooled because they supplied that heat.

Ammonium nitrate has a molar mass of about $80.04\ \text{g/mol}$, so the amount dissolved is

$$n = \frac{5.00\ \text{g}}{80.04\ \text{g/mol}} = 0.06247\ \text{mol}.$$

Therefore, the molar enthalpy of dissolution for these experimental conditions is

$$\Delta H_{\text{diss}} \approx \frac{1.5653\ \text{kJ}}{0.06247\ \text{mol}} = 25.06\ \text{kJ/mol} \approx +25\ \text{kJ/mol}.$$

Because $\Delta T$ is reported to $0.1\ ^\circ\text{C}$, report the final results to two significant figures: the sample absorbed about $1.6\ \text{kJ}$, and the molar result is approximately $+25\ \text{kJ/mol}$. The sample heat and molar enthalpy describe different quantities, so include the units and make clear which one you are reporting.

Common mistakes

• Dropping the minus sign: A cooling solution has negative $q_{\text{solution}}$, but the dissolution heat is positive after you reverse the sign. A warming solution gives the opposite result.
• Forgetting the calorimeter: If its heat capacity is known, its heat change contributes to the total heat transferred. Do not add it twice if your lab’s stated calorimeter constant already accounts for the equipment in another way.
• Using the wrong mass: Check whether your procedure expects the solvent mass or total solution mass. Do not mix the mass convention with an incompatible specific heat assumption.
• Reporting heat instead of molar enthalpy: The heat for one experimental sample is in joules or kilojoules. Divide by moles to obtain a value per mole.
• Overstating precision: Uncertainty in temperature readings, heat capacity, heat loss, and the specific-heat approximation limits the number of meaningful digits. Report a result consistent with the precision of your measurements.

A calorimeter constant is commonly determined in a separate calibration experiment. Use that measured value rather than assuming the calorimeter absorbs no heat when the procedure provides one.