Thermal Energy
Thermal Energy
• Temperature & Heat
• Temperature is related to the
average kinetic energy of the
particles in a substance.
• SI unit for temp. is the Kelvin
• K = °C + 273 (10°C = 283K)
• °C = K – 273 (10K = -263°C)
• Thermal Energy –
the total of all the
kinetic and
potential energy
of all the particles
in a substance.
• Thermal energy relationships
• As temperature increases, so
does thermal energy (because the
kinetic energy of the particles
increased).
• Even if the temperature doesn’t
change, the thermal energy in a
more massive substance is higher
(because it is a total measure of
energy).
• Heat
Cup gets cooler while
hand gets warmer
• The flow of
thermal energy
from one object
to another.
• Heat always
flows from
warmer to
cooler objects.
Ice gets warmer
while hand gets
cooler
• Specific Heat
• Some things heat up or cool down
faster than others.
Land heats up and cools down faster than water
• Specific heat is the amount of heat
required to raise the temperature
of 1 kg of a material by one degree
(C or K).
C water = 4184 J / kg C
C sand = 664 J / kg C
This is why land heats up quickly
during the day and cools quickly at
night and why water takes longer.
Specific Heat
• Which take longer to
heat to 100°C?
50g Al 50g Cu
C
Water
Aluminum
Copper
Silver
Gold
C
(cal/g°C)
(J/g°C)
1.00
0.22
0.093
0.057
0.031
4.18
0.90
0.39
0.24
0.13
Aluminum has a higher specific heat, so it
will take longer to heat up. It will ALSO
take longer to cool down.
Why does water have such a
high specific heat?
water
metal
Water molecules form strong bonds
with each other; therefore it takes
more heat energy to break them.
Metals have weak bonds and do not
need as much energy to break them.
How to calculate changes
in thermal energy
q = mCpT
q = change in thermal energy (heat)
m = mass of substance
T = change in temperature (Tf – Ti)
Cp = specific heat of substance
-q means heat loss
+q = heat gain
Heat Transfer
A 32g silver spoon cools from 60°C to 20°C.
How much heat is lost by the spoon?
GIVEN
m = 32g
Ti = 60°C
Tf = 20°C
q = ??
C = 0.235 J/g°C
WORK
q = mCΔT
m = 32g
ΔT = Tf - Ti
ΔT = 20°C – 60°C = -40°C
q = (32g)(0.235J/g°C)(-40°C)
q = -300.8 J
Heat Transfer
How much heat is required to warm 230 g of
water from 12°C to 90°C?
GIVEN
m = 230g
Ti = 12°C
Tf = 90°C
Q = ??
C = 4.184 J/g°C
WORK
q = mCΔT
m = 230g
ΔT = Tf - Ti
ΔT = 90°C – 12°C = 78°C
q = (230g)(4.184J/g°C)(78°C)
q = 75,061 J
A piece of iron at a temperature of 145°C cools off
to 45°C. If the iron has a mass of 10 g and a
specific heat of 0.449 J/g°C, how much heat is
given up?
GIVEN
m = 10g
Ti = 145°C
Tf = 45°C
q = ??
C = 0.449 J/g°C
WORK
q = mCΔT
m = 10g
ΔT = Tf - Ti
ΔT = 45°C – 145°C = -100°C
q = (10g)(0.449J/g°C)(-100°C)
q = -449 J
A calorimeter is used to
help measure the
specific heat of a
substance.
Heat
gained
Heat lost
First,
mass= and
Knowing
its q value,
temperature
of
its mass,
and its
water
are measured
Then
heated
This
gives
the
T, its Cp can be
T is measured
sample
is
put
heat
lost
by
the
calculated
for water to help
inside
and heat
substance
get its heat gain
flows into water
Let’s Practice
A 55.1 g piece of metal is heated to a temp of 45.1°C, and placed
into a cup containing 359g of water at 20.0°C. The final temp of
the water and metal is 22.3°C.
• How much heat energy did the water absorb?
q = mcΔT
q = (359g)(4.18J/g°C)(22.3°C – 20.0°C) = 3.45 x 103J
• How much heat energy did the metal release to the water?
q lost = q gained
q lost by the metal = - 3.45 x 103J
The q is negative because heat was lost.
• What is the specific heat of the metal?
3.45 x 103J = (55.1g)(C)(22.3°C – 45.1°C)
2.75 J/g°C = C