Step 1. Calculate Heat Transferred (q)
\( q = mc\Delta T \)
- m = mass of the solution being heated/cooled (g). not the mass of solid added
- c = specific heat capacity of water = 4.18 J g⁻¹ K⁻¹
- ΔT = temperature change (°C or K. Same magnitude)
Step 2. Scale to Molar Enthalpy Change
\( \Delta H = -\dfrac{q}{n} \)
n = moles of the limiting reactant. The negative sign ensures the correct sign convention (temperature ↑ = exothermic = ΔH negative).
Convert J → kJ by dividing by 1000.
Standard Assumptions (Solution Calorimetry)
| Assumption | Why |
|---|---|
| Density of solution = 1.00 g cm⁻³ | Treat volume (cm³) as mass (g) directly |
| Specific heat capacity = 4.18 J g⁻¹ K⁻¹ | Approximate dilute solution as pure water |
| No heat loss to surroundings | Polystyrene cup assumed to be a perfect insulator |
⚠️ Examiner Traps
- Mass confusion: m = mass of the water/solution, NOT the solid reagent
- Mole error: Divide q by the moles of the limiting reactant only, not total moles
- Units: q is in Joules. Divide by 1000 to get kJ mol⁻¹
- Sign: If temp rises, q is positive → ΔH must be negative (exothermic)
- °C to K: ΔT in °C ≡ ΔT in K. No conversion needed. Don't waste exam time!