Calorimetry Energy Change Calculations
This topic covers how to calculate the heat energy transferred in a reaction by measuring the temperature change of a known mass of a substance, typically water in a simple calorimeter. Mastering the core formula and being careful with units and signs is key for the exam.
Part of the ESAT Chemistry syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The fundamental calculation uses the formula q = m c ΔT, which links heat energy transferred to mass, specific heat capacity, and temperature change.
- In experiments with aqueous solutions, 'm' is the mass of the water or solution, not the mass of the chemical reactant.
- The specific heat capacity 'c' for water is a common value provided in questions, typically 4.18 or 4.2 J / g / °C.
- A crucial step is converting the calculated heat energy (q) for the experiment into the molar enthalpy change (ΔH) by scaling to one mole of the reactant and applying the correct sign.
- If the water temperature increases, the reaction is exothermic, and the calculated enthalpy change (ΔH) for the system must be negative.
- If the water temperature decreases, the reaction is endothermic, and the calculated enthalpy change (ΔH) for the system must be positive.
Formulae
q = m c ΔT Use this to find the heat energy (q) absorbed or released by the surroundings (usually water) in a calorimetry experiment. Here, 'm' is mass, 'c' is specific heat capacity, and 'ΔT' is the change in temperature.
Definitions
- Specific Heat Capacity (c)
- The amount of heat energy required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin).
- Calorimetry
- The experimental technique used to measure the quantity of heat energy absorbed or released during a chemical or physical process.
- Enthalpy Change (ΔH)
- The overall heat energy change for a reaction occurring at constant pressure. A negative value indicates an exothermic process, and a positive value indicates an endothermic process.
Worked example
0.02 moles of a salt is dissolved in 50 cm³ of water, causing the temperature to drop from 22.0 °C to 20.0 °C. Calculate the molar enthalpy change of solution for this salt in kJ/mol. (Use the specific heat capacity of water, c = 4.2 J / g / °C, and the density of water, ρ = 1.0 g / cm³).
- 1
1.
Find the temperature change:
ΔT = Tfinal - Tinitial = 20.0 - 22.0 = -2.0 °C - 2
2.
Find the mass of the water:
m = volume × density = 50 cm³ × 1.0 g/cm³ = 50 g - 3
3.
Calculate heat change of the water:
q = m c ΔT = 50 g × 4.2 J/g/°C × (-2.0 °C) = -420 J - 4
4.
Determine the heat change of the reaction:
The water lost 420 J of energy, so the reaction must have absorbed 420 J.
Therefore, the enthalpy change for the amount reacted is +420 J (endothermic).
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5.
Calculate the molar enthalpy change:
This energy was absorbed by 0.02 moles.
So, ΔH = (+420 J) / 0.02 mol = +21000 J/mol - 6
6.
Convert to kJ/mol:
ΔH = +21000 J/mol / 1000 J/kJ = +21 kJ/mol
Answer: +21 kJ/mol
Common mistakes
- ×Sign errors: If the surroundings get hotter (exothermic reaction), ΔH must be negative. If they get colder (endothermic reaction), ΔH must be positive. Don't forget this final sign flip.
- ×Unit conversion errors: Questions often ask for the answer in kJ/mol, but the q = mcΔT calculation gives a result in Joules. Always check if you need to divide by 1000 at the end.
- ×Forgetting to scale to one mole: The calculated 'q' is the energy for the specific amount of reactant used. You must divide 'q' by the number of moles to find the molar enthalpy change.
- ×Arithmetic mistakes with decimals: Errors often occur when multiplying or dividing by small decimal numbers (e.g., the number of moles). Take care with place values.
No-calculator tips
- ✓When multiplying by 4.2, you can multiply by 4 then add on 20% (or 0.2) of the original number. For example, 50 × 4.2 = (50*4) + (50*0.2) = 200 + 10 = 210.
- ✓Dividing by a decimal is the same as multiplying by its reciprocal. For example, dividing by 0.02 is the same as multiplying by 50. Dividing by 0.05 is the same as multiplying by 20.
- ✓Look for ways to simplify the multiplication in q = mcΔT. For instance, if m=50 and ΔT=2, calculate 50*2=100 first, which makes multiplying by 4.2 straightforward (420).