Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

3.1.3Physical Chemistry

Energetics Calculations — Edexcel International GCSE Science Double Award 4SD0 - Chemistry Revision Notes & Topical Questions

5 min read 12 parts Lesson 4 of 19

📚 Edexcel International GCSE Science Double Award 4SD0 - Chemistry Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read

Part 1 of 5The calculation toolkit

Heat energy change
  • Heat energy change Q = m × c × ΔT, with Q in joules, m in grams, c in J/g/°C and ΔT in °C.
  • The specific heat capacity of water is 4.18 J/g/°C.
Temperature units
  • Temperature units A temperature change in degrees Celsius is numerically the same as the change in kelvin, so no conversion is needed for ΔT.
Mass of solution
  • Mass of solution Assume the solution has the density of water, 1 g/cm3.
    • A volume of 100 cm3 therefore has a mass of 100 g.

Part 2 of 5A routine for every question

1
Write the equation Q = m × c × ΔT.
2
List the known values: m of the water or solution, c, and ΔT (final minus initial temperature).
3
Substitute the values and calculate Q in joules.
4
Apply the sign: a temperature rise means the reaction is exothermic, so the enthalpy change is negative; a fall means endothermic, so it is positive.
5
Convert from joules to kilojoules by dividing by 1000 if the question asks for kJ.

Part 3 of 5Worked example: reaction in solution

1The data
  • The data Excess iron was added to 100.0 cm3 of copper(II) sulfate solution and the temperature rose by 7.5 °C.
2Substitute
  • Substitute m = 100 g, c = 4.18 J/g/°C, ΔT = 7.5 °C, so Q = 100 × 4.18 × 7.5.
3Solve
  • Solve Q = 3135 J of energy released.
4Sign and units
  • Sign and units The temperature rose, so the reaction is exothermic and the value is negative: –3135 J ÷ 1000 = –3.14 kJ (3 significant figures).

Part 4 of 5Worked example: burning a fuel

The question
  • The question 0.01 mol of propan-1-ol (Mr = 60.11) heats 250 g of water from 25 °C to 37 °C.
    • Find the energy per gram and the molar enthalpy change.
Energy transferred
  • Energy transferred ΔT = 37 – 25 = 12 °C, so Q = 250 × 4.18 × 12 = 12 540 J = 12.54 kJ released, and the sign is negative because the temperature rose.
Energy per gram
  • Energy per gram Mass burned = 0.01 × 60.11 = 0.601 g.
    • Energy per gram = 12.54 ÷ 0.601 = 20.9 kJ/g released, so –20.9 kJ/g to 3 significant figures.
Molar enthalpy change
  • Molar enthalpy change ΔH = Q ÷ n = –12.54 kJ ÷ 0.01 mol = –1254 kJ/mol.

Part 5 of 5Formulae and units at a glance

QuantityFormulaUnits
Heat energy changeQ = m × c × ΔTJ (divide by 1000 for kJ)
Moles of fueln = mass ÷ Mrmol
Energy per gramQ ÷ mass of fuel burnedkJ/g
Molar enthalpy changeΔH = Q ÷ nkJ/mol

Key terms

TermMeaning
Heat energy change (Q)The energy transferred to or from the surroundings, calculated from the temperature change.
Molar enthalpy change (ΔH)The energy change per mole of a substance reacting or burning, in kJ/mol.
Specific heat capacityThe energy needed to raise 1 g of a substance by 1 °C.
ExothermicA reaction releasing energy; its enthalpy change is negative.
Significant figuresThe number of meaningful digits in an answer, normally matching the least precise data given.

Avoid these mistakes

1
Leaving the answer in joules when the question wants kJ/mol. Always divide by 1000.
2
Giving a positive enthalpy change for a combustion or any experiment where the temperature rises. Exothermic changes are negative.
3
Using the mass of the fuel as m in Q = m × c × ΔT, rather than the mass of water heated.
4
Forgetting to convert grams of fuel into moles before finding the molar enthalpy change.
5
Subtracting temperatures the wrong way round, giving a negative ΔT and confusing the sign.

Last updated:

\n