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Heat Capacity Calculations

Key Concepts

  • The amount of heat energy (q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature - initial temperature)

    q = m x Cg x (Tf - Ti)

  • Specific Heat Capacity (Cg) of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1oC (or by 1 K).

  • The units of specific heat capacity are J oC-1 g-1 or J K-1 g-1
Specific Heat Capacities of Some Substances
[Cg (J K-1 g-1 or J oC-1 g-1)]
aluminiumCg = 0.90 waterCg = 4.18
carbonCg = 0.72 ethanol (ethyl alcohol)Cg = 2.44
copperCg = 0.39 sulfuric acid (liquid)Cg = 1.42
leadCg = 0.13 sodium chloride solidCg = 0.85
mercuryCg = 0.14 potassium hydroxide solidCg = 1.18

  • The amount of heat energy gained or lost by a substance can also be calculated for moles of substance:
    q = n x Cn x (Tf - Ti)
    q = amount of heat energy gained or lost by substance
    n = moles of substance
    Cn = molar heat capacity (J oC-1 mol-1 or J K-1 mol-1)
    Tf = final temperature
    Ti = initial temperature

  • Molar Heat Capacity (Cn) of a substance is the amount of heat required to raise the temperature of 1 mole of the substance by 1oC (or by 1 K).
    The units of molar heat capacity are J oC-1 mol-1 or J K-1 mol-1

  • Heats of Reaction can be measured in the laboratory using a simple calorimeter made up of a polystyrene cup (these are appropriate because of their good insulating properties) that is fitted with a lid and a thermometer.

Examples

  1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC.

    q = m x Cg x (Tf - Ti)
    m = 250g
    Cg = 4.18 J oC-1 g-1 (from table above)
    Tf = 56oC
    Ti = 20oC

    q = 250 x 4.18 x (56 - 20)
    q = 250 x 4.18 x 36
    q = 37 620 J = 38 kJ

  2. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25o to 60o.

    q = m x Cg x (Tf - Ti)
    q = 204.75 J
    m = 15g
    Ti = 25oC
    Tf = 60oC

    204.75 = 15 x Cg x (60 - 25)
    204.75 = 15 x Cg x 35
    204.75 = 525 x Cg
    Cg = 204.75 ÷ 525 = 0.39 JoC-1 g-1

  3. 216 J of energy is required to raise the temperature of aluminium from 15o to 35oC. Calculate the mass of aluminium.
    (Specific Heat Capacity of aluminium is 0.90 JoC-1g-1).

    q = m x Cg x (Tf - Ti)
    q = 216 J
    Cg = 0.90 JoC-1g-1
    Ti = 15oC
    Tf = 35oC

    216 = m x 0.90 x (35 - 15)
    216 = m x 0.90 x 20
    216 = m x 18
    m = 216 ÷ 18 = 12g

  4. The initial temperature of 150g of ethanol was 22oC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol?
    (Specific heat capacity of ethanol is 2.44 JoC-1g-1).

    q = m x Cg x (Tf - Ti)
    q = 3240 J
    m = 150g
    Cg = 2.44 JoC-1g-1
    Ti = 22oC

    3240 = 150 x 2.44 x (Tf - 22)
    3240 = 366 (Tf - 22)
    8.85 = Tf - 22
    Tf = 30.9oC


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