Topic 3 – Lessons 3 and 4
Thermal physics
Today’s lesson
• Define specific heat capacity and thermal
capacity.
• Solve problems involving specific heat
capacities and thermal capacities.
Imagine if……..
Two beakers containing different amounts of
water were heated by identical heaters for an
equal amount of time.
Imagine if……..
The beaker with less water gets hotter. WHY?
Imagine if……..
Two beakers, one containing water and one
containing kerosene (equal masses) were
heated by identical heaters for an equal amount
of time.
The beaker containing the kerosene would be
hotter! WHY?
Since the amount of heat energy supplied
is the same to both substances, it seems
that different substances require different
amounts of heat energy to cause the same
temperature rise.
Heat Capacity
The relationship between the amount of
heat energy a substance requires to raise
its temperature by a given amount is
called its thermal capacity. It is measured
in J.°C-1 or J.K-1.
Definition to learn
• Thermal capacity is the amount of energy
needed to raise the temperature of a
substance/object by 1K.
Why are thermal capacities different?
• When a substance is heated, its internal
energy increases (potential and kinetic).
The stronger the force between the
particles in the substance, the more heat
energy goes into potential energy (and less
into kinetic), so the temperature rise is less
than in substances with little force between
particles. Obviously the more particles there
are too, the more heat energy can be
absorbed.
Calculations using Thermal
capacity
Energy absorbed = Thermal capacity x Temp rise
J
J.°C-1
NOT in the data booklet!
Q = CΔT
°C
Specific heat capacity
Specific heat capacity is the amount of
energy needed to raise the temperature of
unit mass of a substance by 1K
Specific heat capacity of water = 4186 J.kg-1.°C-1
Specific heat capacity of kerosene = 2010 J.kg-1.°C-1
Specific heat capacity of mercury = 140 J.kg-1.°C-1
Calculations using S.H.C.
Energy absorbed = Mass x Specific Heat capacity x Temp rise
J
kg
J.kg-1.K-1
Q = mcΔT
K
For example
500 g of olive oil is heated until its temperature rises by
120 K. If the specific heat capacity of olive oil is 1970
J.kg-1.K-1, how much heat energy was used?
Energy absorbed = Mass x Specific Heat capacity x Temp rise
Energy absorbed = 0.5 x 1970 x 120
Energy absorbed = 118200 J
An analogy: Water and wetness
“This analogy is
one of my ideas!”
Richard Feynmann – Nobel
prize winning Physicist,
lock-picker and bongo
player
Two towels – same size/mass
• You can add the same amount of water (heat), but the cheaper
towel will be “wetter” (temperature). They have different capacities
for absorbing water
Let’s try some
questions!
Investigation time!
Let’s do an experiment to measure
specific heat capacities
Specific heat capacities
• Cwater = 4181.3 +/- 0.1 J.kg.°C-1
• Caluminium = 897 +/- 1 J.kg.°C-1
Things to note:
• If specific heat capacity is constant, the
temperature will rise at a uniform rate so long
as the power input is constant and no energy
is lost to the outside.
• There are large potential heat losses if the
substance is not well insulated. These can be
accounted for in some experiments (how?).
• You should be able to think of a number of
reasons why your value does not match that
in the data book.