3.6.6

Calorimetry

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Calorimetry

Calorimetry is used to physically measure changes in enthalpy.

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Bomb calorimetry

  • Bomb calorimetry uses a machine called a bomb calorimeter to measure enthalpy changes of combustion.
  • This process involves burning a sample of a compound in a sealed vessel and measuring the temperature change.
    • Often the calorimeter will just determine the temperature change in the vessel and you will have to calculate the enthalpy change of combustion.
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Inaccuracy

  • Bomb calorimetry can be inaccurate due to:
    • Heat lost to the surroundings.
    • Any incomplete combustion that may take place.
    • Loss of some reactant that evaporates before it combusts.
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Calculating enthalpy changes

  • The equation to calculate enthalpy changes from temperature changes is:
    • q = m × c × ΔT
      • q is the heat energy.
      • m is the mass of the sample.
      • c is the specific heat capacity.
      • ΔT is the temperature change.
  • If the pressure is constant, q = ΔcH
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Calculating enthalpy changes

  • We have calculated q, the energy given off to the surroundings (the enthalpy change).
    • The units of q are Joules.
  • To determine the enthalpy change of combustion, we must convert Joules into the unit of enthalpy change, Joules per mole.
    • Calculate the number of moles:
      • Moles = mass ÷ Mr
      • ΔH = qmoles\frac{q}{moles}

Example Calculations - Calorimetry Experiment

Below are some example calculations based on the calorimetry experiment.

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Combustion of cyclohexane

  • Cyclohexane fuel is burned completely in a calorimeter.
    • There are 200 g of water in the calorimeter.
    • There are 0.5 moles of cyclohexane burnt.
    • The temperature of the water was raised from 298 K to 368 K.
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The calculation

  • The calculation:
    • q = mcΔT
    • q = 200 g x 4.18 Jg-1K-1 × 70 K
    • q = 58520 Joules
  • Enthalpy change of combustion = q ÷ moles
    • ΔH = −58520 J ÷ 0.5 moles
      • Note the minus sign added. This is because we know the reaction is exothermic since the water's temperature was increased.
    • ΔH = −117040 Jmol-1
    • ΔH = −117.04 kJmol-1
      • Note the final units of kJmol-1 as this is more standard.
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Neutralisation reaction

  • Calculate the heat lost/gained during the reaction between H2SO4(aq) and NaOH(aq):
    • 20 cm3 of the acid is added to an insulate container.
    • 30 cm3 of the base is then added.
    • The temperature change is recorded to be 40 K.
      • Assume the density of the solutions to be the same as water, 1 gcm-3.
      • Assume the specific heat capacity is the same as water's, 4.18 Jg-1K-1.
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The calculation

  • Because we have assumed that the density is the same as water, we can calculate the mass of the solution as:
    • 1 cm3 = 1 g
    • (20 + 30) cm3 = 50 g
      • The heat change:
    • q = mcΔT
    • q = (50) g x 4.18 Jg-1K-1 x 40 K
    • q = 8360 Joules
    • q = 8.36 kJ

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1Principles of Science I

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2Practical Scientific Procedures and Techniques

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4Principles of Science II

4.1Extracting Elements

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4.3Organic Chemistry

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4.5The Circulatory System

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4.7Urinary System

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5Contemporary Issues in Science

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