Topic 3 – Lessons 3 and 4
Thermal physics
Lesson Outcomes
• State the equation for specific heat
capacity.
GRADE D
• Use the equation for specific heat
capacity to calculate energy
transferred.
GRADE C
• Explain what it means for a material
to have a high specific heat capacity.
GRADE B
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.
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
Specific Heat Capacity
The specific heat capacity of a material is
the amount of energy needed to raise the
temperature of 1kg of the material by 1°C.
Units J/kg°C.
SPECIFIC HEAT
CAPACITIES
Air (typical room conditions)
1012
Lead
129
Aluminium
897
Mercury
139.5
Carbon dioxide
839
Methane
2191
Chromium
449
Nitrogen
1040
Copper
385
Neon
1030.1
Diamond
509.1
Oxygen
918
Ethanol
2440
Paraffin wax
2500
Gasoline
2220
Polyethylene
2302.7
Glass
840
Gold
129
Silica
703
Granite
790
Water at 100 °C (steam)
2080
Graphite
710
Water at 25 °C
4181.3
Helium
5193.2
Water at −10 °C (ice)
2050
Hydrogen
14300
Zinc
387
Iron
450
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
Q = m∆t
m=
c=
∆t=
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
A kettle contains 1.5kg of water at a
temperature of 18ºC. How much energy is
needed to bring the water to the boil?
specific heat capacity of water is
4200J/kgºC
Water has a very high specific heat capacity –
this is why it is used in radiators. The higher
the specific heat capacity – the more
energy the substance can store.
50kJ of energy was transferred to a material
with a mass of 5kg. The temperature
increased from 20ºC to 60ºC. What is the
specific heat capacity of the material?
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.