Latent Heat
Thermal Physics Lesson 2
Learning Objectives
Define specific latent heat.
Perform calculations using ∆Q=ml.
Describe how specific latent heat of
fusion/vaporisation can be measured in the lab.
Explain why energy is needed to evaporate a
liquid/melt a solid
Safety
Students must not sit down to watch this
experiment - serious scalding has occurred when
the beaker breaks or falls and the pupil has been
unable to move away instantly.
Heating/Cooling Curves
Remember from GCSE?
Demonstration
What tells you the water is boiling?
So energy is being supplied but the temperature
is not rising?
What is going on?
Work is being done to separate the particles
against intermolecular attractive forces.
shc vs. latent heat (simple terms)
Specific Heat Capacity:Energy needed to heat something
Latent Heat:Energy needed to change phase
Definition
The specific latent heat (l) of fusion or
vaporisation is the quantity of thermal energy
required to change the phase of 1kg of a
substance.
Fusion (liquid  solid)
Vaporisation (liquid  gas)
…or the other way (Melting/Condensing)
Equation
Q  ml
where:∆Q is the energy change in J
m is the mass of substance changing phase in kg
lv is the latent heat of vaporisation in J kg-1
lf is the latent heat of fusion in J kg-1
Worked Example 1
The specific latent heat of fusion (melting) of
ice is 330,000 J kg-1. What is the energy needed
to melt 0.65 kg of ice?
Worked Example 1
The specific latent heat of fusion (melting) of ice
is 330,000 J kg-1. What is the energy needed to
melt 0.65 kg of ice?
Q  ml
∆Q = ml = 0.65 kg × 330,000 J kg-1 = 210,000 J (2 s.f.)
Worked Example 2
The power of the immersion heater in the
diagram is 60 W. In 5 minutes, the top pan
balance reading falls from 282g to 274g.
What is the specific latent heat of
vaporisation of water?
(resourcefulphysics.org)
Worked Example 2
The power of the immersion heater in the diagram is 60 W. In 5
minutes, the top pan balance reading falls from 282g to 274g.
What is the specific latent heat of vaporisation of water?
P = 60 W
∆ t = 5 minutes = (5 × 60)s = 300 s
m = m2 - m1 = 282g – 274g = 8g = 0.008 kg
lv = ?
Energy
P
t
Q  ml
∆Q = P∆ t = 60 W × 300s = 18,000 J
lv = ∆Q/m = 18,000 J / 0.008 kg = 2.3 × 106 J kg-1 (2 s.f.)
(resourcefulphysics.org)