Calorimetry

Learning Objectives (OCR Spec 3.2.1)

By the end of this lesson you should be able to:

• Explain the principle of calorimetry as the measurement of heat transfer
• Apply the equation q = mcΔT to calculate the energy change in a reaction
• Calculate enthalpy changes per mole from experimental data for neutralisation, dissolution and combustion reactions
• Evaluate sources of error in simple calorimetry experiments
• Sketch temperature-time graphs and perform graphical extrapolation to correct for heat loss

The Principle of Calorimetry

Calorimetry uses the fact that when a reaction releases (or absorbs) heat, the heat is transferred to (or from) a surrounding mass of liquid - usually water - causing a measurable temperature change. By measuring the temperature change of a known mass of water, we can calculate the energy transferred:

q = mcΔT

where:

Symbol	Meaning	Units
q	Heat energy transferred	J
m	Mass of the solution (usually water)	g
c	Specific heat capacity of the liquid	J g⁻¹ K⁻¹
ΔT	Temperature change	K or °C

For water, c = 4.18 J g⁻¹ K⁻¹. This is provided in the OCR data booklet.

Remember: a temperature change of 1 K is identical in size to a temperature change of 1 °C, so the two units are interchangeable in ΔT.

Approximations Made

OCR accepts two simplifying assumptions in calorimetry calculations:

1. The density of aqueous solutions is assumed to be 1 g cm⁻³, so the volume in cm³ equals the mass in g.
2. The specific heat capacity of dilute aqueous solutions is assumed to be that of water (4.18 J g⁻¹ K⁻¹).

Always state these assumptions if a question asks you to evaluate the method.

Converting q to ΔH

q gives the heat exchanged in joules. To convert to ΔH in kJ mol⁻¹ for the reaction as written:

1. Convert q from J to kJ (divide by 1000)
2. Divide by the number of moles of the limiting reagent
3. Apply the correct sign (negative for exothermic, positive for endothermic)

ΔH = −q / (n × 1000) (exothermic, where n is moles of limiting reagent)

Worked Example 1 – Neutralisation

Q: 50.0 cm³ of 1.00 mol dm⁻³ HCl(aq) is added to 50.0 cm³ of 1.00 mol dm⁻³ NaOH(aq) in a polystyrene cup. The temperature rises from 20.5 °C to 27.3 °C. Calculate ΔneutH°.

A:

Step 1 - Find the total mass of solution: Total volume = 100 cm³ → m = 100 g (density approximation)

Step 2 - Find ΔT: ΔT = 27.3 − 20.5 = 6.8 K (note: positive because the reaction is exothermic, but this is just the magnitude)

Step 3 - Calculate q: q = mcΔT = 100 × 4.18 × 6.8 = 2842.4 J = 2.8424 kJ

Step 4 - Find moles of water formed: n(HCl) = c × V = 1.00 × 0.0500 = 0.0500 mol n(NaOH) = 1.00 × 0.0500 = 0.0500 mol HCl + NaOH → NaCl + H2O, so n(H2O) = 0.0500 mol

Step 5 - Calculate ΔneutH° and apply the sign: ΔH = −2.8424 / 0.0500 = −56.8 kJ mol⁻¹ (3 s.f.)

This is very close to the literature value of −57.1 kJ mol⁻¹.

Worked Example 2 – Combustion (Flame Calorimeter)

Q: A spirit burner containing ethanol (M = 46.0 g mol⁻¹) is used to heat 200 g of water from 18.0 °C to 42.5 °C. The mass of ethanol burned is 1.15 g. Calculate the experimental ΔcH°(ethanol).

A:

q = mcΔT = 200 × 4.18 × (42.5 − 18.0) q = 200 × 4.18 × 24.5 = 20482 J = 20.482 kJ

n(ethanol) = 1.15 / 46.0 = 0.0250 mol

ΔcH° = −20.482 / 0.0250 = −819 kJ mol⁻¹

The data-book value for ethanol is −1367 kJ mol⁻¹, so only about 60% of the heat from combustion reached the water.

Why the large discrepancy? Heat loss to the surroundings; incomplete combustion (soot formation); evaporation of ethanol without burning; heat absorbed by the glass beaker and copper calorimeter rather than just the water. Flame calorimetry is notoriously inaccurate.

Worked Example 3 – Dissolution (Endothermic)