Calorimetry: Measuring Enthalpy Changes
A concise revision guide to measuring enthalpy changes by calorimetry: q = mcΔT, converting to kJ mol⁻¹, the cooling-curve correction, and evaluating errors and assumptions, for AQA A Level Chemistry.
The Key Equation: q = mcΔT
A calorimetry experiment measures the temperature change of a known mass of water or solution and uses it to calculate the energy transferred:
Key equation: q = m × c × ΔT
| Symbol | Meaning | Units |
|---|---|---|
| q | energy transferred to or from the surroundings | J |
| m | mass of the water or solution being heated or cooled (not the mass of the reactants) | g |
| c | specific heat capacity; 4.18 J g⁻¹ K⁻¹ for water and dilute solutions | J g⁻¹ K⁻¹ |
| ΔT | temperature change; a change of 1 K equals a change of 1 °C | K |
For dilute aqueous solutions it is assumed that the density is 1.00 g cm⁻³, so 50.0 cm³ of solution has a mass of 50.0 g, and that the specific heat capacity is the same as that of water.
Common mistake: Using the mass of the solid added as m. The energy heats the water or solution, so m is the mass of the liquid.
From q to the Enthalpy Change
q gives the energy for the amounts used in the experiment. To find the enthalpy change in kJ mol⁻¹:
- Calculate q = mcΔT in joules.
- Calculate the moles of the reactant that is not in excess.
- Divide q by those moles and convert to kJ by dividing by 1000.
- Add the sign: negative if the temperature rose (exothermic), positive if it fell.
Exam sentence: Both a sign and units are required in the final answer, for example ΔH = −56.8 kJ mol⁻¹. An answer without a sign is incomplete.
Worked Example 1: Reaction in Solution
50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid is mixed with 50.0 cm³ of 1.10 mol dm⁻³ sodium hydroxide in a polystyrene cup. The temperature rises by 6.8 K. Calculate the enthalpy change of neutralisation.
- Mass of solution = 100.0 g; q = 100.0 × 4.18 × 6.8 = 2842 J
- n(HCl) = 1.00 × 50.0 ÷ 1000 = 0.0500 mol; the alkali is in excess, so 0.0500 mol of water forms
- ΔH = −2842 ÷ 0.0500 = −56 840 J mol⁻¹ = −56.8 kJ mol⁻¹
Key idea: Add the volumes of both solutions to get the mass heated. Divide by the moles of the reagent that is used up, not the one in excess.
Worked Example 2: Combustion of a Fuel
Burning 0.460 g of ethanol, C₂H₅OH, from a spirit burner raises the temperature of 100 g of water in a copper can by 20.0 K. Calculate the enthalpy change of combustion.
- q = 100 × 4.18 × 20.0 = 8360 J
- n(C₂H₅OH) = 0.460 ÷ 46.0 = 0.0100 mol
- ΔH = −8360 ÷ 0.0100 = −836 000 J mol⁻¹ = −836 kJ mol⁻¹
The data book value is −1367 kJ mol⁻¹. The experimental value is far less exothermic because much of the energy heats the air and the can rather than the water, and some ethanol burns incompletely. Combustion calorimetry nearly always underestimates the enthalpy change.
Exam focus: When asked why an experimental value is less exothermic than the data value, give heat loss to the surroundings first, then incomplete combustion or evaporation of the fuel.
Correcting for Heat Loss: the Cooling Curve
In a slower reaction the solution starts losing heat before the maximum temperature is reached, so the highest reading is too low. The correction is to take readings every minute before and after mixing, plot temperature against time, draw a best-fit line through the cooling points and extrapolate it back to the time of mixing. The temperature change is read from the graph at that time.
Extrapolating the cooling line back to the moment of mixing gives the temperature rise that would have been reached with no heat loss.
Remember: Record the starting temperature for a few minutes before mixing; if two solutions are used, measure both and take the mean.
Errors, Assumptions and Technique
Questions often ask you to evaluate the method. The main sources of error and the assumptions made are:
| Source | Effect | Improvement |
|---|---|---|
| Heat lost to the surroundings | ΔT too small, so ΔH not exothermic enough | lid on the cup, insulate, use the cooling-curve correction |
| Heat absorbed by the cup or can | ignored in the calculation | use a polystyrene cup, which has a very low heat capacity |
| Solution assumed to have c = 4.18 J g⁻¹ K⁻¹ and density 1.00 g cm⁻³ | small systematic error | accept as a stated assumption |
| Incomplete reaction or incomplete combustion | less energy released | stir; use excess of one reagent; ensure a good oxygen supply |
| Thermometer resolution | large percentage uncertainty in small ΔT | use a 0.1 °C thermometer or a larger ΔT |
The calorimetric method in one view: the equation, the practical steps, the calculation and the main limitations.
This method is assessed in the enthalpy change required practical, which uses measured temperature changes to find an enthalpy change.
Exam sentence: The value is less exothermic than expected because heat is lost to the surroundings, so the measured temperature rise is smaller than the true value.
Common Exam Points
Calculate an enthalpy change from a temperature rise
q = mcΔT with the mass of solution, divide by moles of the reagent not in excess, convert to kJ, add a negative sign.
Explain the extrapolation on a temperature-time graph
It corrects for heat lost while the reaction is still taking place, giving the temperature at the moment of mixing.
Suggest why the experimental value differs from the data value
Heat loss to the surroundings, heat absorbed by the apparatus, incomplete reaction or combustion.
Do not say
“Human error”; “m is the mass of the solid”; an answer with no sign.
Check Your Understanding
Work through new calorimetry data rather than the worked examples above.
FAQs
Use these quick answers to check the calorimetry calculations and practical points.
What mass goes into q = mcΔT?
The mass of the water or solution being heated. For dilute solutions, 1 cm³ is taken to have a mass of 1 g.
Why is the answer negative when the temperature rises?
A temperature rise shows energy was released to the surroundings, so the reaction is exothermic and ΔH is negative.
Which moles do I divide by?
The moles of the reactant that is not in excess, because that decides how much reaction takes place.
Why are experimental values usually less exothermic than data book values?
Heat is lost to the surroundings and absorbed by the apparatus, so the measured temperature change is smaller than the true value.
Why use a polystyrene cup?
It is a good insulator and has a very low heat capacity, so little energy is lost through it or used to heat it.
Copyright and author footprint: This OLS revision page was written for Online Learning System by Dr. Mohammed Al-Fatah. It is designed for A Level Chemistry revision and should not be copied or redistributed without permission.
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