All subjects · Chemistry · WCH12 (Unit 2)

Measuring enthalpy changes by calorimetry

You can measure an enthalpy change from the temperature change it causes in a known mass of water or solution.

The equation

Energy transferred: Q = mcΔT

m: mass of the water or solution heated (g); for a solution take 1 cm³ = 1 g
c: specific heat capacity, 4.18 J/g/K for water
ΔT: temperature change (K, which has the same size as °C)

Then ΔH = -Q / n, in kJ/mol, where n is the moles of the limiting reactant. The minus sign is because a temperature rise means an exothermic reaction.

Reactions in solution

Mix reactants in an insulated polystyrene cup with a lid, stirring, and record the temperature before and at intervals after mixing. Extrapolate the cooling curve back to the time of mixing to find the maximum temperature change, correcting for heat lost.

Combustion

Burn a fuel in a spirit burner under a copper can of water. Weigh the burner before and after to get the mass of fuel burned. Results are usually much less exothermic than data-book values because of heat loss to the surroundings, incomplete combustion and evaporation of the fuel.

Uncertainty

Percentage uncertainty = (uncertainty / measured value) x 100. A small ΔT gives a large percentage uncertainty, so use amounts that give a reasonable temperature change.

Worked example

50.0 cm³ of 1.00 mol/dm³ HCl is mixed with 50.0 cm³ of 1.00 mol/dm³ NaOH. The temperature rises by 6.8 °C. Find ΔneutH.

Q = 100 x 4.18 x 6.8 = 2842 J
n(H2O) = 0.0500 mol
ΔH = -2.842 / 0.0500 kJ/mol

Answer: ΔneutH = -57 kJ/mol (2 s.f., because ΔT has 2 s.f.)

Key idea: Q = mcΔT (m is the mass of what is heated); ΔH = -Q/n in kJ/mol; extrapolate to correct for heat loss.

Check yourself

How much energy, in kJ, raises the temperature of 150 g of water by 12.0 °C? (c = 4.18 J/g/K)
Show answer
7.52 kJ. Q = 150 x 4.18 x 12.0 = 7524 J = 7.52 kJ.
Burning 0.690 g of ethanol (M = 46.0 g/mol) heats 250 g of water by 13.0 °C. Find ΔcH in kJ/mol. (c = 4.18 J/g/K)
Show answer
-906 kJ/mol. Q = 250 x 4.18 x 13.0 = 13 585 J = 13.585 kJ; n = 0.690 / 46.0 = 0.0150 mol; ΔcH = -13.585 / 0.0150 = -906 kJ/mol (the data-book value is about -1367, because heat is lost).
Excess zinc is added to 50.0 cm³ of 0.200 mol/dm³ CuSO4. The temperature rises by 10.2 °C. Find ΔrH in kJ/mol. (c = 4.18 J/g/K)
Show answer
-213 kJ/mol. Q = 50.0 x 4.18 x 10.2 = 2132 J; n(CuSO4) = 0.0100 mol; ΔH = -2.132 / 0.0100 = -213 kJ/mol.
Why is a measured enthalpy of combustion usually less exothermic than the data-book value?
- Heat is lost to the surroundings
- The thermometer is too accurate
- Too much oxygen is present
- The water has a low specific heat capacity
Show answer
Heat is lost to the surroundings. Not all the energy released heats the water.
In Q = mcΔT for a reaction in solution, what is m?
- The mass of the solid reactant
- The molar mass of the product
- The mass of the solution being heated
- The mass of the polystyrene cup
Show answer
The mass of the solution being heated. m is the mass of whatever absorbs the heat: the solution.
Why is the temperature-time graph extrapolated back to the time of mixing?
- To estimate the maximum temperature change allowing for heat loss
- To find the starting temperature
- To make the graph neater
- To measure the rate of reaction
Show answer
To estimate the maximum temperature change allowing for heat loss. Heat is lost while the reaction is still happening, so the highest reading is too low.
A temperature change of 6.8 °C is measured with a thermometer with an uncertainty of ±0.5 °C per reading. What is the percentage uncertainty in ΔT? (Two readings are used.)
Show answer
14.7 %. Total uncertainty = 2 x 0.5 = 1.0 °C; 1.0 / 6.8 x 100 = 14.7%.

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Last updated 7 November 2026. Spotted a mistake? Tell us.