The big idea: You cannot read enthalpy off a meter — so chemists measure the temperature change of the surroundings instead. This is calorimetry.
A reaction releases (or absorbs) heat, and that heat warms (or cools) a known mass of water. Measure how much the water's temperature changes and you can work out the heat transferred.
Two key terms: - Specific heat capacity, c — the energy needed to raise 1 g of a substance by 1 K (1 °C). For water, c = 4.18 J g⁻¹ K⁻¹. - Temperature change, ΔT — the final temperature minus the initial temperature. A 1 °C change is the same size as a 1 K change, so ΔT is the same number in either unit.
Watch the sign: If the water gets warmer, the reaction released heat → exothermic → ΔH is negative.
If the water gets cooler, the reaction absorbed heat → endothermic → ΔH is positive.
Getting this sign right is worth a mark on its own.
The first step of every calorimetry calculation is the same: find the heat that flowed into (or out of) the water using the given equation. Almost always the substance heated is water, so use c = 4.18 J g⁻¹ K⁻¹, and take the mass of water in grams (1 cm³ of water ≈ 1 g).
- heat energy transferred (J)
- mass of the substance heated, usually water (g)
- specific heat capacity (J g⁻¹ K⁻¹); for water c = 4.18
- temperature change, T_{final} − T_{initial} (K or °C)
Worked example — heat gained by the water
Burning a fuel raises the temperature of 100.0 g of water from 21.5 °C to 46.5 °C. Calculate the heat energy gained by the water.
Solution
- Find ΔT — final minus initial:
- Formula first — write the given equation:
- Substitute (m = 100.0 g, c = 4.18, ΔT = 25.0):
- Work it out — convert to kJ by dividing by 1000:
Final answer
Q = 1.05 × 10⁴ J (10.45 kJ) of heat is gained by the water.
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Q is the heat for the actual amount reacted. Enthalpy change ΔH is quoted per mole, so divide Q by the amount in moles, n, and attach the sign: exothermic = negative.
- enthalpy change per mole (kJ mol⁻¹)
- heat transferred (kJ)
- amount of the substance that reacted (mol)
Worked example — ΔH per mole
In the experiment above, the heat released (10.45 kJ) came from burning 0.0250 mol of the fuel. Calculate the enthalpy of combustion, ΔHc, in kJ mol⁻¹.
Solution
- The water warmed up, so the reaction is exothermic → ΔH will be negative.
- Formula first — heat per mole:
- Substitute (Q = 10.45 kJ, n = 0.0250 mol):
- Work it out — keep the negative sign and the unit:
Final answer
ΔHc = −418 kJ mol⁻¹ (negative because the reaction is exothermic).
Order of operations: Always: ΔT → Q = mcΔT → ÷ 1000 for kJ → ÷ n for per mole → add the sign. Use the mass of water, not the mass of fuel, in Q = mcΔT.
How this is tested: Calorimetry is a guaranteed quantitative question.
- Paper 1A (MCQ): a one-step 'find ΔH per mole from the temperature change'. - Paper 2: a multi-part calculation — often 'find the temperature rise', then 'calculate ΔH', then 'state an assumption or source of error'.
Markers reward the right sign and the correct per-mole step, and they love to ask why the measured value is less exothermic than the data-booklet value (heat loss).
IB-style question — combustion of propan-1-ol (a)
A spirit burner of propan-1-ol, C3H7OH, is used to heat 150.0 g of water. The temperature of the water rises from 19.0 °C to 41.0 °C. (a) Calculate the heat energy, in kJ, transferred to the water. (c(water) = 4.18 J g⁻¹ K⁻¹.)
Solution
- Find ΔT:
- Formula first:
- Substitute:
- Convert to kJ:
Final answer
Q = 13.8 kJ transferred to the water.
IB-style question — combustion of propan-1-ol (b)
(b) The mass of propan-1-ol burned was 0.360 g (M = 60.10 g mol⁻¹). Calculate the enthalpy of combustion, ΔHc, in kJ mol⁻¹.
Solution
- Find the amount burned:
- The water warmed, so combustion is exothermic → ΔH negative. Formula first:
- Substitute (Q = 13.8 kJ):
- Work it out — keep the sign:
Final answer
ΔHc = −2.30 × 10³ kJ mol⁻¹.
IB-style question — combustion of propan-1-ol (c)
(c) The data-booklet value for the enthalpy of combustion of propan-1-ol is more exothermic than the value found above. Outline two reasons for this difference. [2]
How to score the marks
- Heat loss to the surroundings: not all the heat released reaches the water — some warms the air, the beaker and the apparatus, so the measured temperature rise (and ΔH) is too small.
- Incomplete combustion / evaporation of fuel: soot (incomplete combustion) and some unburned fuel evaporating mean less than the assumed amount releases heat to the water.
- (Either two distinct, valid reasons score the 2 marks; both make the measured ΔH less exothermic than the true value.)
Final answer
Heat is lost to the surroundings and the apparatus, and combustion is incomplete (soot) / some fuel evaporates — so the measured value is less exothermic.