Saturday, November 20, 2010

Worked Examples on Heat Capacity and Specific Heat Capacity

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 ÷ 204.75 = 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

1 comment:

  1. thank you so much for this it really helped

    ReplyDelete