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6.4Thermochemistry

Heat Capacity and Calorimetry — AP CHEMISTRY Revision Notes & Topical Questions

5 min read 9 parts Lesson 9 of 14 1 quiz

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Lesson summary

6.4 Heat Capacity and Calorimetry — AP CHEMISTRY notes: Calculate heat transfer using q = mcΔT, and use calorimetry to determine enthalpy changes.Heat capacity C — the heat needed to raise the temperature of an object by 1 K, in J K-1Specific heat capacity c — the heat needed to …

  • Molar heat capacity
  • What the symbols mean
  • Calorimeter
  • Heat of the reaction
  • Molar enthalpy change
  • Sources of error

Section 1 of 9Heat capacity

Heat capacity C
  • Heat capacity C the heat needed to raise the temperature of an object by 1 K, in J K-1
Specific heat capacity c
  • Specific heat capacity c the heat needed to raise 1 g of a substance by 1 K, in J g-1 K-1
  • Water has c = 4.18 J g-1 K-1.
Molar heat capacity
  • Molar heat capacity heat needed to raise 1 mol of a substance by 1 K

Section 2 of 9Heat transfer equation

q
q mcΔT
  • q is heat in J, m is mass in g and ΔT = Tfinal − Tinitial
    What the symbols mean: E, Q or q = energy transferred (J) · m = mass · c = specific heat capacity (J kg−1 °C−1; water 4.18 J g−1 °C−1) · ΔT or Δθ = temperature change (°C or K)
    • (kg; g in chemistry calorimetry)
  • ΔT has the same value in K and °C.
Substances with high c change temperature less for the same energy input

Section 3 of 9Constant-pressure calorimetry

Calorimeter
  • Calorimeter an insulated container such as a polystyrene cup, vacuum flask or metal can
Heat of the reaction
  • Heat of the reaction qreaction = −qsolution = −msolutioncΔT
  • The solution is assumed to have the density and specific heat of water.
  • The mass is the total mass of the solution.
Molar enthalpy change
  • ΔH = qreaction/n, where n is the moles of the reacting substance

Section 4 of 9Method for a reaction in solution

1
Measure the mass of solution and the temperature change ΔT.
2
Calculate q = mcΔT for the solution.
3
Reverse the sign to obtain q for the reaction.
4
Divide by moles to obtain ΔH in kJ mol-1.

Section 5 of 9Combustion calorimetry

Principle
  • Principle the heat released by burning a fuel raises the temperature of a known mass of water
  • Propan-1-ol: 250 g water rising 12 K gives q = 12 540 J for 0.01 mol, so ΔH = −1254 kJ mol-1.
Sources of error
  • Sources of error heat loss to the surroundings and the apparatus, and incomplete combustion
  • A lid, shielding from draughts and a calorimeter close to the flame reduce heat loss.
  • Heat loss makes the calculated ΔH less exothermic than the true value.

Section 6 of 9Dissolution experiments

Temperature rise
  • Temperature rise the dissolving process is exothermic
Temperature fall
  • Temperature fall the dissolving process is endothermic

Section 7 of 9More from the specification

More from the specification
Condensation
  • Gas to liquid on cooling, releasing energy.
Kinetic energy
  • The energy of motion of a particle.
Internal energy
  • Sum of kinetic and potential energies of the particles in a system.
Heating curve
  • A graph of temperature against time as a substance is heated.
Intermolecular forces
  • Weak attractions between separate molecules, overcome on melting or boiling.
Liquid
  • A state with a fixed volume that flows, particles sliding past each other.
Particle
  • A tiny portion of matter, too small to see.
Melting
  • Solid to liquid at the melting point, absorbing energy.

Section 8 of 9Key terms

TermMeaning
Heat capacityHeat needed to raise the temperature of an object by 1 K.
Specific heat capacityHeat needed to raise 1 g of a substance by 1 K.
CalorimeterInsulated device used to measure heat changes.
q = mcΔTEquation for heat transferred when a substance changes temperature.
Molar enthalpy changeHeat change per mole of reacting substance.

Section 9 of 9Exam technique

1
Using the mass of the solute only — the mass in q = mcΔT is the total mass of solution.
2
Forgetting to reverse the sign — the reaction gains the heat that the solution loses.
3
Ignoring heat loss to the surroundings — it makes the magnitude of ΔH too small.
4
Dividing by mass instead of moles when calculating ΔH.

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