IB Physics SL

Revision Notes

3.1.5 Specific Latent Heat

Specific Latent Heat

  • During a phase change (i.e. a change of state) thermal energy is transferred to a substance or removed from it, while the temperature of the substance does not change
  • In this case, the thermal energy is calculated as follows:

E = mL

  • Where:
    • m = mass of the substance in kilograms (kg)
    • L = specific latent heat of the substance in J kg–1
  • The specific latent heat of a substance is defined as:

The amount of energy required to change the state of 1 kg of a substance without changing its temperature

  • This definition can be explained when the above equation is rearranged for L:

  • This means that the higher the specific latent heat of a substance, the greater the energy needed to change its state
    • Note that the specific latent heat is measured in J kg–1
  • The amount of energy required to melt (or solidify) a substance is not the same as the amount of energy required to evaporate (or condense) the same substance
  • Hence, there are two types of specific heat:
    • Specific latent heat of fusion, Lf
    • Specific latent heat of vaporisation, Lv
  • Specific latent heat of fusion is defined as:

The amount of energy needed to melt or solidify a material at its melting point

  • This applies to the following phase changes:
    • Solid to liquid
    • Liquid to solid
  • Specific latent heat of vaporisation is defined as:

The amount of energy needed to evaporate or condense a material at its boiling point

  • This applies to the following phase changes:
    • Liquid to gas
    • Gas to liquid
  • For the same substance, the value of the specific latent heat of vaporisation is always much higher than the value of the specific latent heat of fusion
    • In other words, Lv > Lf
  • This is because much more energy is needed to evaporate (or condense) a substance than it is needed to melt it (or solidify it)

Specific Latent Heat Table, downloadable IB Physics revision notes

Worked Example

Determine the energy needed to melt 200 g of ice at 0°C.

  • The specific latent heat of fusion of water is 3.3 × 105 J kg–1
  • The specific latent heat of vaporisation of water is 2.3 × 106 J kg–1

Step 1: Determine whether to use latent heat of fusion or vaporisation

    • We need to use the specific latent heat of fusion because the phase change occurring is from solid to liquid

Step 2: List the known quantities

    • Mass of the ice, m = 200 g = 0.2 kg
    • Specific latent heat of fusion of water, Lf = 3.3 × 105 J kg–1

Step 3: Write down the equation for the thermal energy 

E = mLf

Step 4: Substitute numbers into the equation 

E = 0.2 kg × (3.3 × 105) J kg–1

 = 6.6 × 104 J = 66 kJ

Worked Example

Energy is supplied to a heater at a rate of 2500 W.

Determine the time taken to boil 0.50 kg of water at 100°C. Ignore energy losses.

  • The specific latent heat of fusion of water is 3.3 × 105 J kg–1
  • The specific latent heat of vaporisation of water is 2.3 × 106 J kg–1

Step 1: Determine whether to use latent heat of fusion or vaporisation

    • We need to use the specific latent heat of vaporisation because the phase change occurring is from liquid to gas

Step 2: Write down the known quantities

    • Power, P = 2500 W
    • Mass, m = 0.50 kg
    • Specific latent heat of vaporisation of water, Lv = 2.3 × 106 J kg–1

Step 3: Recall the equation linking power P, energy E and time t

E = Pt

Step 4: Write down the equation for the thermal energy E 

    • The energy E in the previous equation is the thermal energy transferred by the heater to the water

E = mLf

Step 5: Equate the two expressions for energy 

Pt = mLf

Step 6: Solve for the time t 

t = 460 s

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