Calorimetry

Spec Mapping — OCR H432 Module 3.2.1 — Enthalpy changes (calorimetry), covering the experimental determination of enthalpy changes from temperature changes using $q = mc\Delta T$q=mcΔT, evaluation of sources of error in simple polystyrene-cup and flame calorimetry, the use of cooling-curve extrapolation to correct for heat loss, and the principle of bomb calorimetry as the high-precision reference method. This is the practical core of Module 3.2.1 and is directly assessed by PAG 3 — Enthalpy determination (refer to the official OCR H432 specification document for exact wording).

Calorimetry is the experimental art of measuring enthalpy changes. Every $\Delta H$ΔH value tabulated in your OCR data booklet — every $\Delta_f H^\ominus$Δf​H⊖, every $\Delta_c H^\ominus$Δc​H⊖, every $\Delta_{neut} H^\ominus$Δneut​H⊖ — was ultimately measured by allowing the reaction to occur in contact with a known mass of liquid (almost always water) and recording the temperature change. The technique rests on a single equation, $q = mc\Delta T$q=mcΔT, and on careful accounting for the heat that escapes through every leaky boundary. This lesson takes you from the equation, through three worked examples covering neutralisation, combustion, and endothermic dissolution, to the experimental subtleties (heat loss, calorimeter heat capacity, cooling-curve extrapolation, bomb calorimetry) that examiners reward at the top of the mark scheme. You will encounter the system–surroundings sign convention in its full glory: $q$q measured on the surroundings (water warming up by $+x$+x J) corresponds to $\Delta H$ΔH on the system of opposite sign ($-x$−x J), and getting this single accounting step right is the single most common dividing line between a Grade C and a Grade A answer in this module.

Key Equation: the heat absorbed (or released) by an aqueous solution is given by: $q = m c \Delta T$q=mcΔT where $m$m is the mass of solution in g, $c$c is the specific heat capacity (4.18 J g$^{-1}$−1 K$^{-1}$−1 for water), and $\Delta T$ΔT is the temperature change in K (numerically identical to °C). To obtain a molar enthalpy change: $\Delta H = -\frac{q}{n \times 1000} \quad \text{kJ mol}^{-1}$ΔH=−n×1000q​kJ mol−1 with the sign reversed because $\Delta H$ΔH refers to the chemical system while $q$q was measured on the surroundings.

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The principle of calorimetry

In a calorimetry experiment, the reaction takes place in contact with a known mass of liquid (usually water) inside a thermally insulating container (typically a polystyrene cup for solution reactions, or a copper-bottomed glass flask with a spirit burner for combustion). When heat is released by the reaction, the temperature of the water rises; when heat is absorbed, the temperature falls. By measuring $\Delta T$ΔT on a calibrated thermometer (precision $\pm 0.1$±0.1 K is typical for a school experiment, $\pm 0.01$±0.01 K for a research-grade instrument), $q$q can be calculated directly.

Symbol	Meaning	Units
$q$q	Heat energy transferred between system and surroundings	J
$m$m	Mass of the solution (or water heated)	g
$c$c	Specific heat capacity of the liquid	J g$^{-1}$−1 K$^{-1}$−1
$\Delta T$ΔT	Temperature change of the liquid	K (= °C for a difference)

For pure water $c = 4.18$c=4.18 J g$^{-1}$−1 K$^{-1}$−1 (often rounded to 4.2). The OCR data booklet supplies this. For dilute aqueous solutions $c$c is taken to be the same as water — a deliberate simplification.

Common mark-scheme trap: $\Delta T$ΔT is a difference. The kelvin and the celsius are numerically identical for differences (1 K = 1 °C), so you do not add 273 to $\Delta T$ΔT. A 6.8 °C rise is exactly a 6.8 K rise; converting it to 279.8 K is a sign that you have misunderstood the question.

Standard approximations used in school calorimetry

OCR mark schemes accept two simplifying assumptions:

1. The density of dilute aqueous solutions is taken as 1 g cm$^{-3}$−3, so the volume in cm³ equals the mass in g.
2. The specific heat capacity of dilute aqueous solutions is taken as $c_{water} = 4.18$cwater​=4.18 J g$^{-1}$−1 K$^{-1}$−1.

Always state these assumptions explicitly if the question asks you to evaluate the method, and identify them as a source of small systematic error (typically ≲ 1%) compared with literature reference data.

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From $q$q to $\Delta H$ΔH — the sign-flip and the per-mole step

$q$q is the heat exchanged in joules, measured on the surroundings. To convert to $\Delta H$ΔH in kJ mol$^{-1}$−1 for the reaction as written:

1. Convert J to kJ by dividing by 1000.
2. Divide by the moles of the species the question asks about — usually the limiting reagent, or moles of water formed for neutralisation, or moles of fuel for combustion.
3. Flip the sign: if the temperature rose ($q_{surr} > 0$qsurr​>0), the reaction is exothermic ($\Delta H < 0$ΔH<0); if the temperature fell ($q_{surr} < 0$qsurr​<0), endothermic ($\Delta H > 0$ΔH>0).

A compact formula: $\Delta H = -\frac{q}{n \cdot 1000}$ΔH=−n⋅1000q​ kJ mol$^{-1}$−1.

m = ρV ≈ 100 g c = 4.18 J g⁻¹ K⁻¹ q = mcΔT

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Worked example 1 — neutralisation in a polystyrene cup

Q: 50.0 cm³ of 1.00 mol dm$^{-3}$−3 HCl(aq) is added to 50.0 cm³ of 1.00 mol dm$^{-3}$−3 NaOH(aq) in a polystyrene cup. The temperature rises from 20.5 °C to 27.3 °C. Calculate $\Delta_{neut} H^\ominus$Δneut​H⊖.

A:

Step 1 — total mass: total volume $= 100$=100 cm³ ⇒ $m = 100$m=100 g using the density-1 approximation.

Step 2 — $\Delta T$ΔT: $\Delta T = 27.3 - 20.5 = 6.8$ΔT=27.3−20.5=6.8 K. (Temperature rose, so the reaction is exothermic — $\Delta H$ΔH will be negative.)

Step 3 — calculate $q$q:

$q = m c \Delta T = 100 \times 4.18 \times 6.8 = +2842 \text{ J} = +2.842 \text{ kJ}$q=mcΔT=100×4.18×6.8=+2842 J=+2.842 kJ

Step 4 — moles of water formed:

$n(\text{HCl}) = c \times V = 1.00 \times 0.0500 = 0.0500 \text{ mol}, \quad n(\text{NaOH}) = 0.0500 \text{ mol}$n(HCl)=c×V=1.00×0.0500=0.0500 mol,n(NaOH)=0.0500 mol

Since 1 mol HCl forms 1 mol H$_2$2​O, $n(\text{H}_2\text{O}) = 0.0500$n(H2​O)=0.0500 mol.

Step 5 — molar enthalpy with sign flip:

$\Delta_{neut} H^\ominus = -\frac{q}{n} = -\frac{2.842}{0.0500} = \mathbf{-56.8 \text{ kJ mol}^{-1}} \text{ (3 s.f.)}$Δneut​H⊖=−nq​=−0.05002.842​=−56.8 kJ mol−1 (3 s.f.)

The literature value for HCl + NaOH is $-57.1$−57.1 kJ mol$^{-1}$−1 — an experimental deviation of ~0.5%, well within polystyrene-cup precision. The small underestimate is consistent with the dominant systematic error in this geometry: heat loss to the surroundings through the cup walls and lid.

Worked example 2 — flame calorimetry of ethanol

Q: A spirit burner containing ethanol ($M = 46.07$M=46.07 g mol$^{-1}$−1) is used to heat 200 g of water in a copper-bottomed flask from 18.0 °C to 42.5 °C. The mass of ethanol burned is 1.15 g. Calculate the experimental $\Delta_c H^\ominus$Δc​H⊖(ethanol).

A:

$q = m c \Delta T = 200 \times 4.18 \times (42.5 - 18.0) = 200 \times 4.18 \times 24.5 = +20\,482 \text{ J} = +20.48 \text{ kJ}$q=mcΔT=200×4.18×(42.5−18.0)=200×4.18×24.5=+20482 J=+20.48 kJ

$n(\text{C}_2\text{H}_5\text{OH}) = \frac{1.15}{46.07} = 0.02497 \text{ mol}$n(C2​H5​OH)=46.071.15​=0.02497 mol

$\Delta_c H^\ominus_{exp} = -\frac{q}{n} = -\frac{20.48}{0.02497} = \mathbf{-820 \text{ kJ mol}^{-1}} \text{ (3 s.f.)}$Δc​Hexp⊖​=−nq​=−0.0249720.48​=−820 kJ mol−1 (3 s.f.)

The data-book value is $-1367$−1367 kJ mol$^{-1}$−1, so only about 60% of the heat from complete combustion has reached the water. Why the dramatic shortfall?

• Heat loss to the surroundings through the flask walls and the gap between burner and flask base (the dominant error).
• Incomplete combustion (soot on the flask shows unburnt carbon → $\Delta H$ΔH from C → CO not C → CO$_2$2​, the former liberates less heat).
• Evaporation of ethanol from the wick before combustion (small mass loss not corresponding to heat in the water).
• Heat absorbed by the apparatus (the glass flask, the copper bottom, and the thermometer have heat capacities, not the water alone).
• Convection-driven heat loss above the flame and around the rim of the flask.

Flame (or "spirit-burner") calorimetry is notoriously inaccurate; bomb calorimetry (below) is the high-precision substitute.

Worked example 3 — endothermic dissolution of ammonium nitrate

Q: 5.00 g of NH$_4$4​NO$_3$3​ ($M = 80.04$M=80.04 g mol$^{-1}$−1) is dissolved in 50.0 cm³ of water in a polystyrene cup. The temperature falls from 22.0 °C to 18.1 °C. Calculate $\Delta_{sol} H$Δsol​H.

A:

$q = 50.0 \times 4.18 \times (18.1 - 22.0) = 50.0 \times 4.18 \times (-3.9) = -815 \text{ J}$q=50.0×4.18×(18.1−22.0)=50.0×4.18×(−3.9)=−815 J

(Negative because the water lost heat to the dissolving salt — its temperature fell.)

$n(\text{NH}_4\text{NO}_3) = \frac{5.00}{80.04} = 0.0625 \text{ mol}$n(NH4​NO3​)=80.045.00​=0.0625 mol

$\Delta_{sol} H = -\frac{q}{n \times 1000} = -\frac{-815}{0.0625 \times 1000} = +\frac{815}{62.5} = \mathbf{+13.0 \text{ kJ mol}^{-1}}$Δsol​H=−n×1000q​=−0.0625×1000−815​=+62.5815​=+13.0 kJ mol−1

Endothermic, as expected — the dissolution of NH$_4$4​NO$_3$3​ is the classic example used in cold-packs for sports injuries.

Subtle point: strictly the calorimeter mass is the water plus solute (= 55.0 g); using this slightly changes the answer to $+14.3$+14.3 kJ mol$^{-1}$−1. For very dilute solutions the difference is negligible; for concentrated solutions OCR mark schemes specify the convention to use.

Worked example 4 — bomb-calorimeter constant

Q: A bomb calorimeter has been calibrated with benzoic acid ($\Delta_c H = -3227$Δc​H=−3227 kJ mol$^{-1}$−1, $M = 122.12$M=122.12 g mol$^{-1}$−1) such that 1.000 g of benzoic acid produced a temperature rise of 1.972 K in the water bath. Calculate the calorimeter heat capacity $C_{cal}$Ccal​ (J K$^{-1}$−1).

A: Moles of benzoic acid $= 1.000 / 122.12 = 8.190 \times 10^{-3}$=1.000/122.12=8.190×10−3 mol; heat released $= 8.190 \times 10^{-3} \times 3227 = 26.43$=8.190×10−3×3227=26.43 kJ $= 26\,430$=26430 J. With $\Delta T = 1.972$ΔT=1.972 K, $C_{cal} = q / \Delta T = 26\,430 / 1.972 = 13\,400$Ccal​=q/ΔT=26430/1.972=13400 J K$^{-1}$−1.

Subsequent enthalpies of combustion are then calculated as $\Delta_c H = -C_{cal} \Delta T / n$Δc​H=−Ccal​ΔT/n — no need to track $m$m or $c$c of water separately.

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Sources of error in simple calorimetry

| Error | Effect on $|\Delta H|$∣ΔH∣ | Mitigation | |-------|------------------------|------------| | Heat loss to surroundings (walls, top) | Too small (under-reads exothermic) | Polystyrene cup, lid, draught shield, cooling-curve extrapolation | | Heat absorbed by apparatus (calorimeter heat capacity $C_{cal}$Ccal​) | Too small | Use $q = (mc + C_{cal})\Delta T$q=(mc+Ccal​)ΔT, or calibrate with a standard | | Incomplete combustion (in flame calorimetry) | Too small | Bomb calorimeter at excess O$_2$2​ pressure | | Evaporation of volatile fuel | Too small | Seal burner, weigh before/after | | Non-standard conditions ($T \neq 298$T=298 K, $p \neq 100$p=100 kPa) | Systematic offset | Correct to 298 K using $\Delta C_p$ΔCp​ (university-level) | | Density and $c$c approximations | Small (< 1%) | State explicitly; quote pure-water values | | Thermometer precision ($\pm 0.1$±0.1 K) | Random scatter | Take multiple readings; use digital thermometer |

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Cooling-curve extrapolation — correcting for heat loss

For a slow reaction (or a reaction with a long mixing time), heat is already escaping while the temperature is still rising. The result: the measured peak temperature is less than the temperature the solution would have reached in a perfectly insulated calorimeter. Cooling-curve extrapolation corrects for this:

flowchart LR
  A[Record T every 30 s for 3-5 min before mixing] --> B[Add reagent at marked time t1]
  B --> C[Continue recording T every 30 s for 10-15 min after]
  C --> D[Plot T vs t]
  D --> E[Identify cooling region on RHS]
  E --> F[Extrapolate cooling straight line backwards to t1]
  F --> G[Read off corrected T_max at t1]
  G --> H[Use corrected ΔT in q = mcΔT]

The corrected $\Delta T_{extrap}$ΔTextrap​ is the temperature change the system would have experienced if all the heat had remained in the water — the geometry assumes that the cooling rate after the reaction is the same as the cooling rate that was already happening during the reaction, so extending the cooling line backwards undoes the heat loss.