Enthalpy lives in the chemicals, where you cannot reach it. So we do something sneaky: let the reaction warm up some water, measure the water, and work backwards. That is the whole of calorimetry — one equation, one minus sign, and a handful of assumptions you will be asked to criticise.
📘 What you need to know
q = mcΔT, where q is energy in J, m is the mass of the water or solution in g, c is specific heat capacity in J g−1 K−1, and ΔT is the temperature change in K.
c for water (and, by assumption, for dilute solutions) is 4.18 J g−1 K−1.
ΔH = −q / n. The minus sign converts “the surroundings gained” into “the system lost”.
n is the amount of the substance the answer is asked per mole of — usually the limiting reagent.
For solution reactions, m is the mass of the whole solution; assume its density is 1.00 g cm−3.
For combustion, m is the mass of water in the can, and n is the moles of fuel burnt.
Main errors: heat loss to the surroundings, heat absorbed by the apparatus, and for combustion, incomplete combustion and evaporation of fuel.
A temperature correction graph extrapolates the cooling line back to the moment of mixing to recover the temperature rise lost to cooling.
The one equation
Every calorimetry question, without exception, starts here:
Energy transferred to the surroundingsq = m × c × ΔT
Each symbol earns its place:
m — the mass, in grams, of the stuff whose temperature you measured. This is the water or the solution. It is never the mass of the fuel or the solid you added.
c — the specific heat capacity: the energy needed to raise 1 g of the substance by 1 K. For water it is 4.18 J g−1 K−1, and water’s unusually high value is exactly why we use it.
ΔT — the temperature change. Same number in °C and K, so no conversion is needed.
The single most common calorimetry error in the world is putting the mass of the magnesium ribbon, or the mass of the fuel, into m. Ask yourself: what did the thermometer actually have its bulb in? That is your mass.
From q to ΔH: mind the minus sign
The value of q you calculate is the energy that arrived in the surroundings. But ΔH describes the system. Since whatever one gains the other loses, the sign has to flip. Then divide by the amount so the answer is per mole.
Turning a measurement into an enthalpy change
ΔH = −q ÷ n
If the mixture got hotter, q is positive and ΔH comes out negative — exothermic, as it should be. If the mixture got colder, ΔT is negative, so q is negative and ΔH comes out positive. The sign looks after itself if you keep ΔT signed and never drop the minus.
Which n? Divide by the moles of the substance the question asks about — and if one reactant is in excess, that is a strong hint that the other one is limiting and is the one to use. The word “excess” in a calorimetry question is never decoration.
Enthalpy changes for reactions in solution
This is the polystyrene-cup experiment: neutralisation, displacement, dissolving. The reaction happens in the water, so the water is both solvent and thermometer.
Cheap, ugly and surprisingly good. Expanded polystyrene traps air, and trapped air is one of the worst conductors of heat available in a school laboratory.
🧩 Method: enthalpy change of a reaction in solution
Measure a known volume of one solution into a polystyrene cup and record its temperature for a couple of minutes until it is steady.
Add the second reactant in one go, with one of the two in excess so the other is fully used up.
Stir and record the temperature every 30 seconds, through the maximum and well beyond it.
Find ΔT — ideally by extrapolation (see below), otherwise as maximum minus initial.
Calculate q = mcΔT using the total mass of solution.
Calculate n for the limiting reagent, then ΔH = −q/n, and convert to kJ mol−1.
The assumptions you are expected to know
The specific heat capacity of the solution is the same as pure water, 4.18 J g−1 K−1.
The density of the solution is the same as water, 1.00 g cm−3, so 50.0 cm3 has a mass of 50.0 g.
The heat capacity of the cup and thermometer is ignored.
The reaction goes to completion.
There are no heat losses to the surroundings.
Every one of these is slightly false, and that is the point — exam questions love asking which assumption explains why your value is smaller than the data booklet value.
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 solution50.0 + 50.0 = 100 cm³ → m = 100 gboth liquids get warm, so the mass is the totalStep 2: Energy transferredq = 100 × 4.18 × 6.8 = 2842 JStep 3: Moles of water formedn(HCl) = 0.0500 × 1.00 = 0.0500 mol1 : 1 reaction, neither in excess, so 0.0500 mol of water formsStep 4: Enthalpy change per moleΔH = −2842 ÷ 0.0500 = −56 840 J mol⁻¹ΔH = −56.8 kJ mol⁻¹close to the accepted value of about −57 kJ mol⁻¹ − a good sign
WORKED EXAMPLE
Excess zinc powder is added to 25.0 cm3 of 0.200 mol dm−3 copper(II) sulfate solution. The temperature rises by 10.6 °C. Calculate ΔH for Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s).
Step 1: Which mass, which moles?mass = the solution only (25.0 g); zinc is in excess, so CuSO₄ is limitingStep 2: Energy transferredq = 25.0 × 4.18 × 10.6 = 1108 JStep 3: Moles of the limiting reagentn = 0.0250 × 0.200 = 0.00500 molStep 4: Enthalpy changeΔH = −1108 ÷ 0.00500 = −221 600 J mol⁻¹ΔH = −222 kJ mol⁻¹the zinc powder is never in the mass, however tempting it looks
Enthalpy of combustion experiments
Here the reaction happens outside the water. A known mass of fuel is burnt in a spirit burner underneath a metal can of water, and the water’s temperature rise tells you how much energy arrived. Metal is used for the can because we want the heat to get through.
Notice how much of this apparatus exists purely to stop energy escaping. Even so, values from this experiment are typically twenty per cent below the accepted ones.
WORKED EXAMPLE
0.615 g of methanol (M = 32.05 g mol−1) is burnt in a spirit burner and heats 150.0 g of water by 18.0 °C. Calculate the enthalpy of combustion, and comment on the value given that the data booklet quotes −726 kJ mol−1.
Step 1: Energy gained by the waterq = 150.0 × 4.18 × 18.0 = 11 286 JStep 2: Moles of fuel burntn = 0.615 ÷ 32.05 = 0.01919 molStep 3: Enthalpy of combustionΔH = −11 286 ÷ 0.01919 = −588 100 J mol⁻¹ΔHₐ = −588 kJ mol⁻¹Step 4: Commentabout 19% less energy than expected: heat lost to the air and to the can, plus some incomplete combustionthe experimental value is always less negative, never more
Why combustion values come out too small
Heat loss to the surroundings — the flame heats the air, the tripod and the room, not just the water.
Heat absorbed by the apparatus — the copper can itself warms up, and that energy is not counted.
Incomplete combustion — soot on the base of the can is visible proof that some carbon became CO or C instead of CO2, releasing less energy.
Evaporation of the fuel — volatile fuels lose mass without burning, so n is overstated.
Water evaporating from the can, and the reaction not being at standard conditions.
Notice they all push the same way: less energy measured, so a less negative answer. If your experimental value comes out more negative than the accepted one, suspect a calculation error rather than a lucky experiment.
Temperature correction graphs
Slow reactions create a problem. While you are waiting for the maximum temperature, the mixture is already cooling to the room. The peak you record is therefore lower than the peak that would have occurred if the reaction had been instant.
The fix is graphical. Record the temperature before mixing, mix, then keep recording well into the cooling. Draw a best-fit line through the cooling section and extend it backwards to the exact time you mixed. Where it crosses is the temperature you would have reached with no heat loss.
The recorded peak here was only about 19 K above the start; the extrapolated value is 20.6 K. Using the recorded peak would have made the enthalpy change roughly 7 per cent too small.
🧩 Method: building a temperature correction graph
Record the temperature every 30 s for two or three minutes before adding the second reactant, to establish a steady baseline.
Add the second reactant, noting the exact time, and keep stirring.
Keep recording well past the maximum, into a clear steady cooling pattern.
Plot temperature against time and draw a best-fit straight line through the cooling points only.
Extrapolate that line back to the time of addition. Read off the temperature.
ΔT = extrapolated temperature − steady starting temperature. Use this in q = mcΔT.
Endothermic reactions work the same way. The temperature drops, then warms back towards room temperature. Draw the best-fit line through the warming section and extrapolate it back to the time of mixing to find the lowest temperature that would have been reached.
WORKED EXAMPLE
Excess zinc is added to 50.0 cm3 of 0.400 mol dm−3 copper(II) sulfate. The extrapolated temperature rise is 20.6 K, but the highest reading actually recorded was 19.2 K above the start. Calculate ΔH using each value and comment.
Step 1: Moles of the limiting reagentn(CuSO₄) = 0.0500 × 0.400 = 0.0200 molStep 2: Using the corrected riseq = 50.0 × 4.18 × 20.6 = 4305 JΔH = −4305 ÷ 0.0200 = −215 kJ mol⁻¹Step 3: Using the recorded maximumq = 50.0 × 4.18 × 19.2 = 4013 JΔH = −4013 ÷ 0.0200 = −201 kJ mol⁻¹−215 kJ mol⁻¹ corrected, −201 kJ mol⁻¹ uncorrectedthe uncorrected value is 7% too small because the mixture was already cooling
💡 Exam tip
Write out q = mcΔT before substituting anything. Method marks are available even if the arithmetic goes wrong.
Keep q in joules right through the calculation and convert to kJ only at the final step.
State the units of your answer: kJ mol−1 for an enthalpy change, kJ for a quantity of energy.
If the word excess appears, immediately identify the other reactant as limiting and use it for n.
Give your answer to a sensible number of significant figures — match the least precise data, usually 2 or 3.
When asked to improve the experiment, give practical answers: insulate the container, add a lid, use a draught shield, keep the can close to the flame, stir continuously.
When asked why the value differs from the data booklet, name a specific loss and say which way it moves the answer.
⚠ Common mix-up
Using the mass of the solid or the fuel as m.m is always the water or solution whose temperature you measured.
Forgetting to add the two solution volumes together in a neutralisation. Both liquids warmed up.
Dropping the minus sign in ΔH = −q/n, and reporting an exothermic reaction as positive.
Dividing by the wrong n — using the reagent in excess instead of the limiting one.
Leaving the answer in J mol−1 when the question asked for kJ mol−1.
Drawing the best-fit line through every point on a correction graph. Only the cooling points count.
Saying “heat was lost” as the only evaluation. Say where it went and how it changed the result.
Claiming heat loss makes the value too negative. Heat loss always makes the measured energy, and so the value, too small.
Up next: Bond Enthalpy Calculations — some enthalpy changes cannot be measured in a cup at all, so we will learn to work them out from the bonds instead.
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