Measuring ΔH: q = mcΔT and the per-mole step

8 minAS Level5.1.7

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can calculate an enthalpy change from a temperature change, using $q = mc\Delta T$ and then $\Delta H = -\dfrac{mc\Delta T}{n}$ , and you can say what makes such an experiment inaccurate.

The thermometer is a spy in the water

You cannot put a thermometer inside a reaction to see its enthalpy. What you can do is surround the reaction with a known mass of water and watch the water's temperature.

The energy the water gains is the energy the reaction lost (minus whatever leaked away). So:

Calculate the energy that went into the water: $q = mc\Delta T$ .
Work out how many moles reacted.
Divide to get the energy per mole, and fix the sign.

Definition

The calorimetry equation

$q = mc\Delta T$

$q$ = energy transferred, in joules;
$m$ = mass of the substance being heated (usually water or a solution), in grams;
$c$ = specific heat capacity, $4.18\ \text{J g}^{-1}\text{K}^{-1}$ for water;
$\Delta T$ = temperature change, in kelvin (the same size as a Celsius degree).

Then:

$\Delta H = -\frac{mc\Delta T}{n}$

where $n$ is the number of moles of the limiting reagent (or of water formed, for neutralisation).

Why the minus sign is there

If the water gets hotter, $\Delta T$ is positive, so $q$ is positive — but the reaction itself released that energy, so its $\Delta H$ must be negative.

The minus sign makes that flip automatically: a temperature rise gives a negative $\Delta H$ .

For an endothermic process, the water cools, $q$ is negative, and the minus sign makes $\Delta H$ positive. One sign rule, both directions.

Worked example: Worked example: neutralisation

$25.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ hydrochloric acid is mixed with $25.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ sodium hydroxide. The temperature rises by $6.8\ \text{K}$ .

Calculate $\Delta H_\text{neut}$ , taking the density of the solution as $1.00\ \text{g cm}^{-3}$ and $c = 4.18\ \text{J g}^{-1}\text{K}^{-1}$ .

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