3.1.4.2 - Calorimetry

What q = mcΔT actually measures

The equation $q = mc\Delta T$ links the energy transferred to a substance with the temperature change you observe. In practice, that substance is often water or a reaction mixture in a cup calorimeter. If the temperature rises, that substance has gained energy. If the temperature falls, it has lost energy.

That definition explains the units in the equation. $m$ is the mass of the substance whose temperature changes, in grams. $c$ is the specific heat capacity, usually given in $\mathrm{J}\,\mathrm{g^{-1}}\,\mathrm{K^{-1}}$. $\Delta T$ is the temperature change, found from final temperature − initial temperature. For temperature changes, a rise of $1\,\mathrm{K}$ is the same size as a rise of 1 °C, so you can use either scale for $\Delta T$.

Here, the value of $c$ is usually supplied, so the important thing is using it correctly rather than memorising it. A very common value is $4.18\,\mathrm{J}\,\mathrm{g^{-1}}\,\mathrm{K^{-1}}$ for water or dilute aqueous solutions.

Suppose $100.0\,\mathrm{g}$ of water warms by $5.0\,\mathrm{K}$ and $c = 4.18\,\mathrm{J}\,\mathrm{g^{-1}}\,\mathrm{K^{-1}}$. Then

That is the energy transferred to the water. The equation is simple, but it only works if the mass, units, and temperature difference are all chosen carefully.

From heat change to molar enthalpy change

In calorimetry, $q$ is only the first step. To find a molar enthalpy change, you then need to work out how many moles of reaction that energy change corresponds to.

For reactions in solution, the mass is usually taken from the total volume mixed, because dilute solutions have a density close to $1.0\,\mathrm{g}\,\mathrm{cm^{-3}}$. So $50.0\,\mathrm{cm^{3}}$ of acid mixed with $50.0\,\mathrm{cm^{3}}$ of alkali is treated as $100.0\,\mathrm{g}$ of solution. That total mass is used in $q = mc\Delta T$, not the volume of just one reactant.

If the reaction is exothermic, the solution gets warmer. The measured $q$ for the solution is positive because the solution gained energy, but the enthalpy change of the reaction is negative because the reaction released that energy. A reliable method is:

1. calculate $q$
2. convert to kJ if needed
3. find the number of moles reacting from the balanced equation
4. apply the sign that matches the reaction

Worked example:

$40.0\,\mathrm{cm^{3}}$ of $1.50\,\mathrm{mol}\,\mathrm{dm^{-3}}$ hydrochloric acid is mixed with $40.0\,\mathrm{cm^{3}}$ of $1.50\,\mathrm{mol}\,\mathrm{dm^{-3}}$ sodium hydroxide. The temperature rises by $10.0\,\mathrm{K}$. Take $c = 4.18\,\mathrm{J}\,\mathrm{g^{-1}}\,\mathrm{K^{-1}}$.

Mass of solution $= 80.0\,\mathrm{g}$

Reaction:

Moles of $\ce{HCl}$ reacting:

The alkali is also $0.0600\,\mathrm{mol}$, so the reaction as written occurs for $0.0600\,\mathrm{mol}$. If the reactants are not in the equation ratio, compare each amount divided by its coefficient. The smaller value is the amount of reaction as written; use it when dividing the heat change to find molar enthalpy.

The diagram below shows a simple cup calorimeter and highlights why the total mass of the mixed solution is used in the calculation.

[DIAGRAM: asset_name: 1.4.2 - Calorimetry - Diagram 1; asset_slug: 1.4.2 - Calorimetry - Diagram 1; recommended_method: retained_png; description: An insulated solution calorimeter with lid and immersed thermometer; the measured solution mass is used in q = mcΔT.]

Notice the three key details here: the total mass was used, the moles came from the balanced equation, and the final answer was given a negative sign because the reaction was exothermic.

Related calculations and details that matter

The same equation can be rearranged for related calculations. For example,

That rearrangement lets you find the temperature rise produced by a known energy change, or the mass of water needed to absorb a certain amount of energy. The chemistry does not change; only the subject of the equation changes.

A few details are worth checking every time:

• Use the mass of the substance whose temperature changes.
• Treat $\mathrm{cm^{3}}$ as g only when the solution can reasonably be taken to have a density of about $1.0\,\mathrm{g}\,\mathrm{cm^{-3}}$.
• Keep energy units consistent. If $q$ is in joules and your answer needs $\mathrm{kJ}\,\mathrm{mol^{-1}}$, divide by 1000 before or after the mole step.
• Match the sign to the reaction: exothermic means negative $\Delta H$, endothermic means positive $\Delta H$.

In an exothermic experiment, heat loss and warming of the cup both make the measured temperature rise smaller, giving a less negative calculated enthalpy. An endothermic mixture instead gains heat from warmer surroundings, reducing the measured fall and making the calculated value less positive. Insulation and a lid reduce heat transfer; they do not remove it completely.

For a combustion experiment, use the mass of water in $q=mc\Delta T$, but the mass of fuel burned to calculate the fuel amount. For example, burning 0.460 g ethanol (molar mass 46.0 g mol⁻¹) warms 100.0 g water by 20.0 K. The water gains 8.36 kJ and 0.0100 mol fuel burns, so the measured combustion enthalpy is −836 kJ mol⁻¹. Incomplete combustion or fuel evaporation makes this value less negative: less energy reaches the water for each measured mole lost from the burner.

If you can identify the correct mass, calculate $q$, find the relevant number of moles, and then choose the correct sign, you have covered the core skill of this topic. That same routine works for a cup calorimeter, for burning a fuel, or for a rearranged version of the equation.