Lesson 2 of 5
Calorimetry
Convert the surroundings temperature change into energy per mole of reaction.
A-Level 9476 (2026-2027)
I can calculate enthalpy from experimental heat measurements.Syllabus
Syllabus 9476, 7(d). Calculate enthalpy from experimental heat measurements.
The thermometer measures the surroundings, not the reaction directly
Find q for what warms or cools, reverse its sign, then divide by the reacting amount.
q = mcΔT, where m is the mass being warmed, c is its specific heat capacity and ΔT is final minus initial temperature. With m in g and c in J g-1 K-1, q is in J. A temperature interval of 1 K equals an interval of 1 °C. For an insulated arrangement, qreaction = -qsurroundings.
Worked exampleNeutralisation measured in an insulated cup
50.0 cm3 of 1.00 mol dm-3 HCl is mixed with 50.0 cm3 of 1.00 mol dm-3 NaOH. The temperature rises by 6.80 K. Assume density 1.00 g cm-3, c = 4.18 J g-1 K-1 and negligible cup heat capacity.
Try it first. Then show the worked stepsHide the worked steps
- Total solution mass is 100.0 g, not 50.0 g.
- q(solution) = 100.0 × 4.18 × 6.80 = +2842.4 J, so q(reaction) = -2.8424 kJ.
- Both acid and base supply 0.0500 mol; 0.0500 mol water forms.
- Divide by the water amount: delta H = -2.8424/0.0500.
ΔHneut = -56.8 kJ mol-1 to three significant figures.
Try it step by step
Work it out in steps: Heat taken in or given out by the solution, q = mc delta T, then Moles of the limiting reactant, then Enthalpy change, with its sign.
Show the answer
Problem 1.
Heat taken in or given out by the solution, q = mc delta T: The solution mass is 80.0 g, both volumes together. q = 80.0 x 4.18 x 10.2 = 3411 J, gained by the solution.
Moles of the limiting reactant: n(HCl) = n(NaOH) = 1.50 x 40.0/1000 = 0.0600 mol, so 0.0600 mol of water forms.
Enthalpy change, with its sign: delta H = -3.411 / 0.0600 = -56.8 kJ mol-1. The sign is negative because the solution warmed: the reaction gave out heat.
Problem 2.
Heat taken in or given out by the solution, q = mc delta T: q = 50.0 x 4.18 x 10.0 = 2090 J, gained by the solution.
Moles of the limiting reactant: Zinc is in excess, so CuSO4 is limiting: n = 0.200 x 50.0/1000 = 0.0100 mol.
Enthalpy change, with its sign: delta H = -2.090 / 0.0100 = -209 kJ mol-1.
Problem 3.
Heat taken in or given out by the solution, q = mc delta T: q = 50.0 x 4.18 x 5.90 = 1233 J, lost by the water, which is 1230 J to 3 significant figures.
Moles of the limiting reactant: n(NH4NO3) = 4.00 / 80.0 = 0.0500 mol.
Enthalpy change, with its sign: delta H = +1.233 / 0.0500 = +24.7 kJ mol-1. The sign is positive because the water cooled: the dissolving took heat in.
If the calorimeter heat capacity C is supplied, include its heat gain: qsurroundings = (mc + C)ΔT. In a combustion experiment, use the mass of water warmed and the amount of fuel actually burnt. In a dissolution experiment, use the appropriate solution mass and heat capacity, and divide by the amount of solute dissolved.
| Effect | Consequence for an exothermic result |
|---|---|
| Heat lost to the room or absorbed by an ignored vessel | Measured heat release is too small in magnitude; the calculated enthalpy is less negative. |
| Incomplete combustion | Less energy is released than for complete combustion to the stated products. |
| Fuel evaporates without burning | Apparent fuel consumption is too large; energy per measured mole is too small in magnitude. |
| Reactant concentrations or final temperature are mismeasured | Determine the direction from the actual calculation; there is no universal error direction. |
A temperature-time record before and after mixing can support extrapolation to estimate the temperature at mixing, reducing a systematic heat-loss error. Repeated trials reveal scatter but do not by themselves remove the same heat loss from every trial. Always identify the limiting reagent before dividing q by an amount.