You cannot measure the energy of a reaction directly. What you can do is let the reaction warm up a known mass of water and measure that instead — because energy is conserved, whatever the water gained, the reaction must have lost.
📚 What you need to know
Calorimetry measures enthalpy changes by recording a temperature change in a known mass of water or solution.
q = mcΔT, where q is in J, m in g, c in J g–1 K–1 and ΔT in K or °C.
The specific heat capacity of water is 4.18 J g–1 K–1.
ΔH = –q / n, where n is the moles of the substance the value refers to. Convert J to kJ by dividing by 1000.
Assumptions: the solution behaves like water (density 1 g cm–3, c = 4.18), the container absorbs nothing, and no heat is lost.
A temperature correction graph extrapolates the cooling line back to the moment of mixing to allow for heat loss.
The equation
Heat transferred
q = m × c × ΔT
q — heat energy transferred, in J
m — mass of water or solution being heated, in g
c — specific heat capacity, 4.18 J g–1 K–1 for water
ΔT — temperature change, in K or °C
The specific heat capacity is the energy needed to raise the temperature of 1 g of a substance by 1 K. Water’s is unusually high, which is why it makes such a good calorimetry fluid.
You never need to convert °C to K here. A rise of 12 °C is a rise of 12 K — the divisions are the same size, and only the change appears in the equation.
The single most common error in the whole topic: m is the mass of the water or solution being heated, never the mass of the fuel or the solid. Burning 1.2 g of ethanol to heat 200 g of water? m = 200.
Reactions in solution
Cheap, and better than it looks. Polystyrene conducts heat poorly and the lid stops losses from the top.
A polystyrene cup makes a surprisingly good calorimeter: it is a poor conductor, so relatively little heat escapes through the walls, and a lid cuts losses from the top. The method is to record a steady starting temperature, add the second reactant, and follow the temperature until it peaks.
To make the numbers workable we make some deliberate assumptions:
The solution has the same density as water, so 1 cm3 weighs 1 g
The solution has the same specific heat capacity as water, 4.18 J g–1 K–1
The cup itself absorbs no heat
The reaction goes to completion
There are no heat losses to the surroundings
Every one of those is slightly untrue, which is why measured values are never quite right — and why “state an assumption” is such a common exam question.
🧩 The calculation, every time
Find m — the total mass of solution being heated, in grams.
Find ΔT — the temperature change.
Calculate q = mcΔT, in joules.
Find n — moles of the substance the enthalpy change refers to.
ΔH = –q / n, then divide by 1000 for kJ mol–1. Add the sign: negative if the temperature rose.
WORKED EXAMPLE
50.0 cm3 of 1.00 mol dm–3 HCl is mixed with 50.0 cm3 of 1.00 mol dm–3 NaOH in a polystyrene cup. The temperature rises by 6.8 °C. Calculate the enthalpy of neutralisation.
Step 1 — mass of solution being heated50.0 + 50.0 = 100.0 cm³ → m = 100.0 gStep 2 — calculate qq = 100.0 × 4.18 × 6.8 = 2842.4 JStep 3 — moles of water formedn(HCl) = 0.0500 × 1.00 = 0.0500 mol → 0.0500 mol H₂OStep 4 — divide and convert2842.4 ÷ 0.0500 = 56 848 J molΔH neut = −56.8 kJ molNegative because the temperature rose. Both solutions are heated, so m is 100 g, not 50 g.
Enthalpy of combustion
Far more heat escapes here than in a polystyrene cup, which is why combustion values measured this way always come out too small.
Here the fuel is burnt in a spirit burner underneath a copper can of water. Copper is used because it conducts heat well. You weigh the burner before and after, so the mass difference tells you exactly how much fuel burned.
This experiment is considerably less accurate than the solution method, for two reasons that you should be able to name:
Heat loss — a lot of the flame’s energy heats the air, the can and the tripod rather than the water
Incomplete combustion — soot on the bottom of the can is carbon that released less energy than it should have
Both errors point the same way: less energy reaches the water than the reaction really released, so the measured ΔH comes out less exothermic than the true value. Losses can be reduced with a lid, a draught shield, and keeping the flame close to the can.
WORKED EXAMPLE
0.720 g of ethanol (M = 46.08 g mol–1) is burnt and heats 150 g of water by 28.0 °C. Calculate the enthalpy of combustion, and comment on your answer given the accepted value of –1367 kJ mol–1.
Step 1 — q for the WATERq = 150 × 4.18 × 28.0 = 17 556 JStep 2 — moles of ethanol burntn = 0.720 ÷ 46.08 = 0.015625 molStep 3 — divide and convert17 556 ÷ 0.015625 = 1 123 584 J molΔH c = −1120 kJ mol (3 s.f.)CommentAbout 18% less exothermic than the accepted value — consistent with heat loss to the surroundings and incomplete combustion.
Temperature correction graphs
Some reactions are not instant. While you wait for the peak temperature, the mixture is already losing heat to the room — so the highest reading you actually see is lower than the true maximum. Extrapolation fixes this.
The peak you actually observe is already too low. Extrapolating the cooling line back recovers the temperature you would have seen with no heat loss.
🧩 How to do it
Record the temperature for a few minutes before adding the second reactant, to establish a steady baseline.
Add the second reactant, noting the time, and keep recording as the temperature rises and then falls.
Plot temperature against time and draw a line of best fit through the cooling points.
Extrapolate that cooling line back to the moment of mixing.
Read off the corrected maximum, and take ΔT from the baseline up to it.
The assumption here is that the rate of cooling is constant. The same technique works for endothermic reactions — you just extrapolate a warming line back instead, as the mixture returns towards room temperature.
WORKED EXAMPLE
Excess zinc powder is added to 100.0 cm3 of 0.250 mol dm–3 copper(II) sulfate. Extrapolation gives a corrected temperature rise of 12.6 °C. Calculate ΔH for the reaction.
Step 1 — q for the solutionq = 100.0 × 4.18 × 12.6 = 5266.8 JStep 2 — moles of the LIMITING reactantZinc is in excess, so the copper(II) sulfate limits the reaction.n = 0.1000 × 0.250 = 0.0250 molStep 3 — divide and convert5266.8 ÷ 0.0250 = 210 672 J molΔH = −211 kJ mol (3 s.f.)Always use the reactant that is NOT in excess — the excess one never fully reacts.
💡 Exam tip
m is the water or solution, not the fuel and not the solid. This is the most frequently lost mark in the topic.
When two solutions are mixed, add their volumes to get m.
Use the limiting reactant for n. If a question says “excess”, that substance is not the one to use.
q is in joules; ΔH is in kJ mol–1. Divide by 1000 and check your answer looks sensible.
Asked to explain a difference from the data book value? Say heat loss and, for combustion, incomplete combustion — and say which way they push the answer.
⚠️ Common mix-up
Using the mass of fuel as m. The equation describes the water being heated.
Forgetting the minus sign when the temperature rises. The surroundings gained energy, so the system lost it.
Forgetting to divide by moles. q on its own is not an enthalpy change.
Leaving the answer in joules. ΔH is quoted in kJ mol–1.
Reading the highest thermometer value instead of the extrapolated one when a correction graph is provided.
That completes Measuring Enthalpy Change. You can now say what an enthalpy change is, put a sign on it, read it off a profile, define it precisely enough to compare with anyone else’s, and measure it yourself — which is everything you need before moving on to calculating values you cannot measure at all.
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