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:
Q?=?mL
Where:
Q?= heat energy transferred (J)
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 energy released when 1 kg of liquid freezes to become solid at constant temperature
This applies to the following phase changes:
Solid to liquid
Liquid to solid
Therefore, the definition for specific latent heat of fusion could also be:
The energy absorbed when 1 kg of solid melts to become liquid at constant temperature
Specific latent heat of vaporisation is defined as:
The energy released when 1 kg of gas condenses to become liquid at constant temperature
This applies to the following phase changes:
Liquid to gas
Gas to liquid
Therefore, the definition for specific latent heat of vaporisation could also be:
The energy absorbed when 1 kg of liquid evaporates to become gas at constant temperature
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)
In melting, the intermolecular bonds only need to be weakened to turn from a solid to a liquid
When evaporating, the intermolecular bonds need to be completely broken to turn from liquid to gas. This requires a lot more energy.
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?
Q?=?mLf
Step 4: Substitute numbers into the equation?
Q?= 0.2 kg?× (3.3 × 105) J kg–1
Q = 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?Q?transferred by the heater to the water
Q=?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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