How to Calculate Enthalpy Change from Calorimetry (q = mcΔT)

The Short Answer

To calculate an enthalpy change from a calorimetry experiment:

Find the temperature change of the solution, $\Delta T$ — ideally the corrected value from a temperature–time graph.
Calculate the heat change of the solution: $q = mc\Delta T$ , using the mass of the solution (1 cm³ ≈ 1 g) and $c = 4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ .
Convert $q$ from J to kJ.
Find $n$ , the amount in moles of the limiting reagent.
Divide and reverse the sign:

\Delta H = -\frac{q}{n} \quad (\text{kJ mol}^{-1})

If the temperature rose, the reaction is exothermic and $\Delta H$ is negative. If the temperature fell, it is endothermic and $\Delta H$ is positive.

Learning Goals: By the end of this guide, you should be able to:

Use $q = mc\Delta T$ with the correct mass and units.

Convert $q$ into $\Delta H$ in kJ mol⁻¹ with the correct sign.

Extrapolate a cooling curve to find a corrected $\Delta T$ (AQA Required Practical 2).

Explain why measured values are usually less exothermic than data-book values.

What a Calorimeter Measures

A simple "coffee-cup" calorimeter is a polystyrene cup, usually with a lid, standing in a beaker for support. The reaction happens in the solution, and the solution itself is what changes temperature. So the thermometer measures the energy that the reaction transferred to (or took from) the water in the cup.

That is why every calorimetry calculation makes the same standard assumptions:

Assumption	Value used
Density of the solution	$1.00\ \text{g cm}^{-3}$ (so 50 cm³ has a mass of 50 g)
Specific heat capacity of the solution	$4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ (same as water)
Heat absorbed by the cup and thermometer	Ignored

A temperature change of 1 °C is the same as a change of 1 K, so $\Delta T$ can be used directly in $q = mc\Delta T$ without converting to kelvin.

Worked Example 1: Enthalpy of Neutralisation

50.0 cm³ of 1.00 mol dm⁻³ HCl is placed in a polystyrene cup. 50.0 cm³ of 1.00 mol dm⁻³ NaOH at the same temperature is added. The temperature rises by 6.6 K. Calculate the enthalpy change of neutralisation.

HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)

Step 1 — mass. The solution heated is both solutions together: $50.0 + 50.0 = 100.0\ \text{cm}^3$ , so $m = 100.0\ \text{g}$ .

Step 2 — heat change.

q = mc\Delta T = 100.0 \times 4.18 \times 6.6 = 2759\ \text{J} = 2.759\ \text{kJ}

Step 3 — moles. HCl and NaOH react 1 : 1 and are present in equal amounts, so either gives the moles of water formed:

n = cV = 1.00 \times \frac{50.0}{1000} = 0.0500\ \text{mol}

Step 4 — enthalpy change. The temperature rose, so the reaction is exothermic:

\Delta H = -\frac{q}{n} = -\frac{2.759}{0.0500} = -55.2\ \text{kJ mol}^{-1}

The data-book value for a strong acid with a strong base is about $-57.1\ \text{kJ mol}^{-1}$ (Chinese textbooks quote $-57.3$ ). The percentage error is

\frac{|-55.2 - (-57.1)|}{57.1} \times 100 = 3.3\%

and the measured value is less exothermic — the usual result, because some heat escaped.

Correcting for Heat Loss: The Temperature–Time Graph

If the reaction is fast, you can simply read the highest temperature. But for slower reactions — such as zinc powder reacting with copper(II) sulfate — the solution is already cooling while the reaction is still releasing heat. The highest reading you see is then lower than the true temperature rise.

The AQA Required Practical 2 method fixes this with a graph:

Measure the temperature of the solution in the cup every 30 seconds for 3 minutes to establish a steady starting temperature.
At 3 minutes, add the second reagent and stir. Do not take a reading at 3 minutes — you are busy mixing.
Continue taking readings every 30 seconds until about 10 minutes.
Plot temperature against time. Draw a horizontal line through the readings before mixing (the initial temperature).
Draw a straight line of best fit through the readings after the peak, where the solution is cooling steadily.
Extrapolate this cooling line back to t = 3 minutes, the moment of mixing.
The corrected $\Delta T$ is the vertical gap between the two lines at 3 minutes.

Why does this work? The cooling line shows how fast heat is being lost. Extending it back to the time of mixing estimates the temperature the solution would have reached if all the heat had been released instantly, with no time to escape.

Calorimetry Lab

Run neutralisation, zinc + copper(II) sulfate and ammonium nitrate dissolving in a virtual calorimeter. Record readings every 30 s, add the reagent at 3 minutes, drag the extrapolation line onto the cooling curve, and compare a lidded polystyrene cup with a glass beaker.

Open the Calorimetry Lab

Worked Example 2: Displacement with Extrapolation

25.0 cm³ of 0.500 mol dm⁻³ CuSO₄ is placed in a polystyrene cup and excess zinc powder is added at 3 minutes.

Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)

From the graph: initial temperature = 20.0 °C, highest reading = 43.9 °C, extrapolated temperature at 3 minutes = 45.2 °C.

Corrected $\Delta T = 45.2 - 20.0 = 25.2\ \text{K}$ (compared with only $23.9\ \text{K}$ from the highest reading).

Mass. Only the copper(II) sulfate solution is heated: $m = 25.0\ \text{g}$ . The zinc is not included.

q = 25.0 \times 4.18 \times 25.2 = 2633\ \text{J} = 2.633\ \text{kJ}

Moles. Zinc is in excess, so CuSO₄ is the limiting reagent:

n(CuSO_4) = 0.500 \times \frac{25.0}{1000} = 0.0125\ \text{mol}

\Delta H = -\frac{2.633}{0.0125} = -211\ \text{kJ mol}^{-1}

Using the uncorrected $23.9\ \text{K}$ would give $-200\ \text{kJ mol}^{-1}$ . The extrapolation brings the result much closer to the data-book value of about $-217\ \text{kJ mol}^{-1}$ .

Worked Example 3: An Endothermic Process

5.00 g of ammonium nitrate, $NH_4NO_3$ ( $M = 80.0\ \text{g mol}^{-1}$ ), is dissolved in 50.0 cm³ of water. The temperature falls by 7.2 K, so $\Delta T = -7.2\ \text{K}$ .

NH_4NO_3(s) \rightarrow NH_4^+(aq) + NO_3^-(aq)

Mass. Use the water, $m = 50.0\ \text{g}$ — not the 5.00 g of solid.

q = 50.0 \times 4.18 \times (-7.2) = -1505\ \text{J} = -1.505\ \text{kJ}

The negative $q$ means the solution lost heat.

n = \frac{5.00}{80.0} = 0.0625\ \text{mol}

\Delta H = -\frac{q}{n} = -\frac{(-1.505)}{0.0625} = +24.1\ \text{kJ mol}^{-1}

The positive sign shows the process is endothermic; the data-book value is about $+25.7\ \text{kJ mol}^{-1}$ . Many students prefer to use the size of $\Delta T$ and then decide the sign from "temperature rose → negative, temperature fell → positive". Both approaches give the same answer.

Common Mistakes

Getting the sign wrong. $q = mc\Delta T$ is the heat gained by the solution. The reaction's enthalpy change is the opposite: a temperature rise means $\Delta H$ is negative. Always write a sign (+ or −) in front of a final $\Delta H$ .
Using the mass of the solid. In $q = mc\Delta T$ , $m$ is the mass of the solution whose temperature changes. Do not add the mass of zinc or ammonium nitrate, and never use the mass of the limiting reagent itself.
Forgetting to combine volumes. When two solutions are mixed, the mass is the total volume: 50 cm³ + 50 cm³ gives 100 g, not 50 g.
Mixing J and kJ. $q = mc\Delta T$ gives joules because $c$ is in J g⁻¹ K⁻¹. Divide by 1000 before dividing by $n$ , or your answer will be 1000 times too big.
Using moles of the reagent in excess. Only the limiting reagent tells you how much reaction happened. In the zinc example, the moles of zinc added are irrelevant — use the moles of CuSO₄.
Using the highest reading for a slow reaction. If the question gives a temperature–time graph, extrapolate. The highest reading underestimates $\Delta T$ .

Why Measured Values Are Less Exothermic

Experimental enthalpy changes almost always have a smaller magnitude than data-book values. The main reasons are:

Heat exchange with the surroundings. This is the largest error. Heat escapes through the walls and the open surface. A lid and a polystyrene cup reduce it; a glass beaker makes it much worse.
Heat absorbed by the apparatus. The cup and thermometer warm up too, but the calculation ignores them.
Approximate solution properties. The density and specific heat capacity of the solution are not exactly those of water.
Incomplete reaction in slow reactions, or solid left undissolved.

The thermometer also limits precision. A thermometer read to ±0.1 K gives an uncertainty of about ±0.1 K in $\Delta T$ ; for $\Delta T = 6.6\ \text{K}$ that is $\frac{0.1}{6.6} \times 100 = 1.5\%$ . Using larger quantities, so that $\Delta T$ is bigger, reduces this percentage uncertainty.

Exam Tips (A-Level / AP / Chinese High School)

AQA / OCR / Edexcel: expect to describe the extrapolation method in words and draw it on a given graph. State that the reading at the moment of mixing is skipped.
AP Chemistry: the same calculation appears as "heat gained by the solution equals heat released by the reaction", $q_{\text{soln}} = -q_{\text{rxn}}$ . Watch for questions that use $\text{J g}^{-1}\ ^\circ\text{C}^{-1}$ — the numbers are identical to $\text{J g}^{-1}\ \text{K}^{-1}$ .
中和反应反应热的测定: use the same method, with the data value of 57.3 kJ mol⁻¹ and the emphasis on insulation, stirring and fast mixing.
Give $\Delta H$ to an appropriate number of significant figures (usually 3) with the unit kJ mol⁻¹.

Frequently Asked Questions

Why is a polystyrene cup used instead of a glass beaker?

Polystyrene is a poor thermal conductor and has a very small heat capacity, so less heat passes through the walls and very little is absorbed by the cup. Adding a lid also stops heat escaping from the surface by convection and evaporation.

Do I need to convert °C to K for ΔT?

No. A change of 1 °C equals a change of 1 K, so $\Delta T$ is the same number on both scales. Only absolute temperatures need converting.

What if the two solutions start at different temperatures?

Use the mean of the two starting temperatures as the initial temperature, provided the volumes are equal. This is why both solutions are usually left to reach room temperature first.

Hess's Law — Combine measured enthalpy changes to find ones that cannot be measured directly.
Born–Haber Cycles — Apply enthalpy cycles to ionic lattices.
Gibbs Free Energy — Combine $\Delta H$ with entropy to predict feasibility.