Chapter 10
Ch 10-1: Temperature and
Thermal Equilibrium
• Determining an object’s temperature with
precision requires
• a definition of temperature
• established measurements that determine how
“hot” or “cold” objects are
Heat Energy
• Energy must be added to
or removed from a
substance to change its
temperature.
• Temperature is
PROPORTIONAL to the
kinetic energy of atoms.
Thermal Energy…
• Internal Energy – the energy of a
substance due to the random motions of
its component particles and equal to the
total energy of those particles.
Internal Energy
• For an ideal gas, the internal energy
depends only on the temperature of the
gas.
• For gases with 2 or more atoms per
molecule, as well as for liquids and solids,
other properties besides temperature
contribute to the internal energy.
INTERNAL ENERGY
• The symbol, U, stands for internal energy
• Thus, delta U, stands for the change in
internal energy
Heating and Cooling
• If an object has become hotter,
it means that it has gained heat energy.
• If an object cools down, it means it has
lost energy
Thermal Equilibrium
• The state in which two bodies in physical
contact with each other have identical
temperatures
Thermal Equilibrium
• Ex: When you put a warm coke can into a large
beaker of cold water, there is a noticeable
temperature difference between the two.
• After about 15 minutes, the can of soda will be
cooler and the water surrounding it will be slightly
warmer. Eventually, both the can and the water
will be at the same temperature.
• This temperature will not change as long
as conditions remain unchanged in
the beaker. This is Thermal Equilibrium.
• Thermal Equilibrium is the basis for
measuring temperature with
thermometers.
• This is because the thermometer is at the
same temperature, or is in thermal
equilibrium with, the object.
HEAT ENERGY
What is HEAT?
• Form of energy and measured in JOULES
• Particles move about more and take up
more room if heated – this is why things
expand if heated
• It is also why substances change from:
solids
liquids
gases
when heated
Matter and Temperature
• Matter Expands as its temperature
increases
• This is known as Thermal Expansion
Heating and Cooling cont…
• Heat energy always moves from:
HOT object
COOLER object
e.g. Cup of water at 20 °C in a room at 30°C gains heat energy and heats up – its
temperature rises
Cup of water at 20 °C in a room at 10°C
loses heat energy and cools down – its
temperature will fall.
Measuring Temperature
• Units depend on scale used:
– Fahrenheit, Celsius, and Kelvin (or absolute)
scales.
– Fahrenheit commonly used in US
– Fahrenheit and Celsius temps can be
converted…
The Fahrenheit temperature
scale is an English
measurement scale.
FREEZING POINT: Water
freezes at 32 oF and BOILING
POINT: boils at 2l2 oF.
The Fahrenheit degree is
smaller than the Celsius
degree or Kelvin.
On the Celsius temperature
scale, (the METRIC SCALE based on
the freezing and boiling point of water)
water’s freezing point is 0 oC
& water’s boiling point is l00 oC
The Kelvin (K) scale is based on
the boiling and freezing point of
water and absolute zero.
The Kelvin scale does not use
the degree symbol. The units are
Kelvins (K), not degrees Kelvin.
1 Kelvin unit = 1
oC
scale.
On this scale the lowest possible
temperature is absolute zero,
written ”0 K."
At absolute zero, the average
kinetic energy of particles is
zero.
Absolute zero on the Kelvin scale is
equal to -273 degrees Celsius.
Temperature Conversion
Formulas:
oC
=
K – 273
K = oC + 273
oC
= (oF-32)
1.8
oF
= (1.8oC) + 32
HOMEWORK:
Page 363: #1-3 all
THIS WILL BE FOR A GRADE
WHEN YOU WALK INTO
CLASS TOMORROW!!!
Must show all work 
HEAT and ENERGY
• HEAT: the energy transfer
between two objects
because of a difference in
their temperatures.
• Energy transferred as heat
ALWAYS transfers from
an object of a higher temp
to that of a lower temp.
Internal Energy
• If the can and the water
in the glass reach
thermal equilibrium,
which has greater
internal energy?
Internal Energy
• If the can and the water
in the glass reach
thermal equilibrium,
which has greater
internal energy?
• A: The glass of water
because of its greater
mass
Key Points to remember…
• Temperature – measures the average KE
of molecules in an object
• HEAT – energy transferred from one
object to another
• Internal Energy – sum of the energies of
the molecules.
The calorie – the
quantity of heat
needed to raise the
temperature of one
0
gram of water 1 C.
1 calorie = 4.18 joules
The calorie –
The Calorie (also
called a kilocalorie) –
often measures the
energy content in food
1 Calorie = 1000 calories
There is a difference!
1 Calorie = 1000 calories
There is a difference!
Example item: That means a
190 Calorie bowl of cereal
REALLY has 190,000
calories of energy!
Heat - units of Energy
• Heat Energy (heat flow): Q
• Internal energy: U
calorie (cal) = 4.186 J
kilocalorie (kcal) = 4186 J
Calorie (dietary Calorie) = 1 kcal = 4186 J
Thus, for every Calorie you consume you
actually obtain 4186 J of energy
Heat and Work
• Friction can increase a substances internal
energy
• For solids, internal energy can be
increased by deforming their structure.
• EX: Rubber band stretched or a piece of
metal bent.
Conservation of Energy
• If changes in internal energy are taken into
account along with changes in mechanical
energy, the total energy is a universally
conserved property
Delta PE + Delta KE + Delta U = 0
Ex: Conservation of Energy
• A 0.10 kg ball falls 10 m onto a hard floor
and then bounces back up 9.0 m. How
much of its mechanical energy is
transformed to the internal
energy of the
ball
ball and the floor?
Ex: Conservation of Energy
A 0.10 kg ball falls 10 m onto a hard floor and then
bounces back up 9.0 m. How much of its
mechanical energy is transformed to the internal
energy of the ball and the floor?
Givens:
m = 0.10 kg
h = 10.0 m
g = 9.81
PE = ?
Formula: Delta PE + Delta KE + Delta U = 0
Or
PEi + KEi + Ui = PEf + KEf + Uf
U=?
KE=?
Ex: Conservation of Energy
A 0.10 kg ball falls 10 m onto a hard floor and then
bounces back up 9.0 m. How much of its mechanical
energy is transformed to the internal energy of the ball and
the floor?
Givens:
m = 0.10 kg
h = 10.0 m
g = 9.81
PE = ?
Formula: Delta PE + Delta KE + Delta U = 0
or
PEi + KEi + Ui = PEf + KEf + Uf
mg(h2-h1) + 0 + Uf-Ui = 0
mg (h2-h1) = Uf-Ui
0.10 (9.81) (1) = Uf-Ui
0.98 J
U=?
KE= ?
HOMEWORK
• Page 370 Practice: # 1
• Page 370 Review: # 1 and 3
QOTD
1: A substance’s temperature increases as
a direct result of:
a. Energy being removed from the
particles of the substance
b. Kinetic Energy being added to the
particles of the substance
c: A change in the number of atoms
and molecules in a substance
QOTD CONT
2: What happens to the internal energy of an
ideal gas when it is heated from 0 degrees
Celsius to 4 degrees Celsius?
a: increases
b: decreases
c: stays the same
d: impossible to determine
QOTD cont
• Which of the following is proportional to
the KE of atoms and molecules?
a: Elastic Energy
b: Thermal Equilibrium
c: Potential Energy
d: Temperature
QOTD
• Heat Flow occurs between two bodies in
thermal contact when they differ in which
of the following properties?
A: mass
B: density
C: temperature
D: specific heat
QOTD
• Which of the following is equivalent to 88
degrees Fahrenheit?
A: 49 degrees C
B: 31 degrees C
C: 16 degrees C
D: 58 degrees C
10-3: Changes in temperature
and phase
• Specific Heat Capacity:
the quantity of energy needed to raise
the temperature of 1 kg of substance by
1 degree Celsius at constant pressure.
Specific Heat Capacity (c)
• Relates mass, temperature change, and
energy transferred as heat.
• Specific heat capacity equation:
cp = Q/m(T)
Where the subscript p represents heat
capacity measured at a constant pressure
Specific Heat Capacity
• The equation applies to both substances
that absorb energy AND those that
transfer energy to their surroundings.
• When temp increases; change in T and Q
are taken to be positive.
• When temp decreases; they are negative
and energy is transferred from the
substance.
Calorimetry
• An experimental procedure used to
measure the energy transferred from one
substance to another as heat.
See page 372
• Table 10-4 Specific Heat capacities
• **You must know cp for water =
4.186 x 103 J/(kg * C)
The higher the specific heat,
the more energy the material
contains when it is at the same
temperature as a material with a
lower specific heat.
Ex. 2 kg of Copper ( 385 J/kg*oC)
2 kg of Glass [664 J/kg*oC ]
If both are 20
more energy!
oC,
Glass has
Determining Specific Heat Capacity
• For simplicity, a subscript w will always stand
for “water” in problems involving specific heat
capacities.
Energy absorbed by water = energy released by the substance
Or
Qw = Qx
Where
Q = mc ΔT
Calculating Thermal Energy
Q = mc ΔT
Q = change in thermal energy
Joules ( J )
m = mass (kg )
o
c = specific heat ( J/kg C )
= change in temperature oC
ΔT
Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise
the temp. of 20 kg of iron from 35oC up to 50oC
if the specific heat for iron is 450 J/kgoC ?
given:
Q=
m=
c=
ΔT =
formula:
Q = mc ΔT
sub:
answer & unit
Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of
20 kg of iron from 35oC up to 50oC if the specific heat for
iron is 450 J/kgoC ?
given:
unit
formula:
Q=Q
Q = mc ΔT
m= 20 kg
c = 450 J/kgoC
ΔT = 50 - 35 = 15 oC
sub:
answer &
Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of
20 kg of iron from 35oC up to 50oC if the specific heat for
iron is 450 J/kgoC ?
given:
Q=Q
formula:
sub:
answer & unit
Q = m c ΔT
m= 20 kg
C = 450 J/kgoC
ΔT = 50 - 35 = 15 oC
BE SURE TO
WRITE THE
UNITS IN
YOUR NOTES !
Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of
20 kg of iron from 35oC up to 50oC if the specific heat for
iron is 450 J/kgoC ?
given:
unit
formula:
Q=Q
Q = mc ΔT
m= 20 kg
sub:
Q = 20 x 450 x 15
C = 450 J/kgoC
ΔT = 50 - 35 = 15 oC
answer &
Thermal Energy Problem A.
Ex. How much thermal energy is needed to raise the temp. of
20 kg of iron from 35oC up to 50oC if the specific heat for
iron is 450 J/kgoC ?
given:
unit
formula:
Q=Q
Q = mc ΔT
m= 20 kg
sub:
answer &
Q = 20 x 450 x 15 = 135,000 J
C = 450 J/kgoC
ΔT = 50 - 35 = 15 oC
A 0.05 kg metal bolt is heated to an unknown initial temperature.
It is then dropped into a beaker containing 0.15 kg of water with
an initial temperature of 21.0 C. The bolt and the water then
reach a final temperature of 25.0 C. If the metal has a specific
heat capacity of 899 J/kg *C, find the initial temperature of the
metal.
Givens: m metal = 0.05 kg
m water = 0.15 kg
T water = 21.0 C
Unknown T metal = ?
cp, m = 899 J/kg *C
cp, w = 4186 J/kg *C
T final = 25.00 C
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then
dropped into a beaker containing 0.15 kg of water with an initial temperature of
21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If
the metal has a specific heat capacity of 899 J/kg *C, find the initial
temperature of the metal.
Givens: m metal = 0.05 kg
m water = 0.15 kg
T water = 21.0 C
cp, m = 899 J/kg *C
cp, w = 4186 J/kg *C
T final = 25.00 C
Unknown T metal = ?
energy removed from the metal = energy absorbed by water
Qm = Qw
mmcp, m ΔTm = mw cp, w ΔTw
ΔTm = mw cp, w ΔTw
mmcp, m
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then
dropped into a beaker containing 0.15 kg of water with an initial temperature of
21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If
the metal has a specific heat capacity of 899 J/kg *C, find the initial
temperature of the metal.
Unknown T metal = ?
energy removed from the metal = energy absorbed by water
Qm = Qw
mmcp, m ΔTm = mw cp, w ΔTw
ΔTm = mw cp, w ΔTw
mmcp, m
Next step…
Plug and Chug!
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then
dropped into a beaker containing 0.15 kg of water with an initial temperature of
21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If
the metal has a specific heat capacity of 899 J/kg *C, find the initial
temperature of the metal.
Unknown T metal = ?
energy removed from the metal = energy absorbed by water
Qm = Qw
mmcp, m ΔTm = mw cp, w ΔTw
ΔTm = mw cp, w ΔTw
mmcp, m
ΔTm = (0.15) (4186) (4) = 56 oC
(.05) (899)
Not done
yet….looking for
INITIAL TEMP OF
THE METAL!!!
A 0.05 kg metal bolt is heated to an unknown initial temperature. It is then
dropped into a beaker containing 0.15 kg of water with an initial temperature of
21.0 C. The bolt and the water then reach a final temperature of 25.0 C. If
the metal has a specific heat capacity of 899 J/kg *C, find the initial
temperature of the metal.
ΔTm = (0.15) (4186) (4) = 56 oC
(.05) (899)
Δ Tm = Tf + Δ Tm
Δ Tm = 25.0 C + 56.0 C
Δ Tm =81.0 C
Homework Part 1
• Page 374 #1,2