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Temperature and Thermal
Equilibrium
Defining Temperature
• Temperature is a measure of the average kinetic
energy of the particles in a substance.
• Adding or removing energy usually changes
temperature.
• Internal energy is the energy of a substance due to
both the random motions of its particles and to the
potential energy that results from the distances and
alignments between the particles.
Temperature and Thermal
Equilibrium
Thermal Equilibrium
• Thermal equilibrium is the state in which two bodies
in physical contact with each other have identical
temperatures.
– By placing a thermometer in contact with an object and
waiting until the column of liquid in the thermometer stops
rising or falling, you can find the temperature of the object.
– The reason is that the thermometer is in thermal equilibrium
with the object.
• The temperature of any two objects in thermal
equilibrium always lies between their initial
temperatures.
Temperature and Thermal
Equilibrium
Thermal Expansion
• In general, if the temperature of a substance
increases, so does its volume. This phenomenon is
known as thermal expansion.
• Different substances undergo different amounts of
expansion for a given temperature change.
• The thermal expansion characteristics of a material
are indicated by a quantity called the coefficient of
volume expansion.
• Gases have the largest values for this coefficient.
Solids typically have the smallest values.
Temperature and Thermal
Equilibrium
Measuring Temperature
• The most common
thermometers use a glass
tube containing a thin column
of mercury, colored alcohol, or
colored mineral spirits.
• When the thermometer is
heated, the volume of the
liquid expands.
• The change in length of the
liquid column is proportional to
the temperature.
Temperature and Thermal
Equilibrium
Measuring Temperature, continued
• When a thermometer is in thermal equilibrium with a
mixture of water and ice at one atmosphere of
pressure, the temperature is called the ice point or
melting point of water. This is defined as zero
degrees Celsius, or 0°C.
• When the thermometer is in thermal equilibrium with
a mixture of steam and water at one atmosphere of
pressure, the temperature is called the steam point
or boiling point of water. This is defined as 100°C.
Temperature and Thermal
Equilibrium
Measuring Temperature, continued
• The temperature scales most widely used today are
the Fahrenheit, Celsius, and Kelvin scales.
• Celsius and Fahrenheit temperature measurements
can be converted to each other using this equation:
9
TF  TC  32.0
5
9

Fahrenheit temperature    Celsius temperature   32.0
5

• The number 32.0 indicates the difference between
the ice point value in each scale: 0.0ºC and 32.0ºF.
Temperature and Thermal
Equilibrium
Measuring Temperature, continued
• Temperature values in the Celsius and Fahrenheit
scales can have positive, negative, or zero values.
• But because the kinetic energy of the atoms in a
substance must be positive, the absolute
temperature that is proportional to that energy
should be positive also.
• A temperature scale with only positive values is
suggested by the graph on the next slide. This scale
is called the Kelvin scale.
Temperature and Thermal
Equilibrium
Measuring Temperature, continued
• The graph suggests that
if the temperature could
be lowered to –273.15°C,
the pressure would be
zero.
• This temperature is
designated in the Kelvin
scale as 0.00 K, where K
represents the
temperature unit called
the kelvin.
• Temperatures in the Kelvin scale are indicated by the
symbol T.
Temperature and Thermal
Equilibrium
Measuring Temperature, continued
• A temperature difference of one degree is the same
on the Celsius and Kelvin scales. The two scales
differ only in the choice of zero point.
• Thus, the ice point (0.00°C) equals 273.15 K, and the
steam point (100.00°C) equals 373.15 K.
• The Celsius temperature can therefore be converted
to the Kelvin temperature by adding 273.15:
T  TC  273.15
Kelvin temperature  Celsius temperature  273.15
Temperature and Thermal
Equilibrium
Temperature Scales and Their Uses
Defining Heat
Heat and Energy
• Heat is the energy transferred between objects
because of a difference in their temperatures.
• From a macroscopic viewpoint, energy transferred as
heat tends to move from an object at higher
temperature to an object at lower temperature.
• The direction in which energy travels as heat can be
explained at the atomic level, as shown on the next
slide.
Defining Heat
Transfer of Particles’ Kinetic Energy as Heat
Energy is transferred as heat from the higher-energy particles to
the lower-energy particles, as shown on the left. The net energy
transferred is zero when thermal equilibrium is reached, as
shown on the right.
Defining Heat
Heat and Energy, continued
• The atoms of all objects are in continuous motion, so
all objects have some internal energy.
– Because temperature is a measure of that energy,
all objects have some temperature.
• Heat, on the other hand, is the energy transferred
from one object to another because of the
temperature difference between them.
– When there is no temperature difference
between a substance and its surroundings, no net
energy is transferred as heat.
Defining Heat
Heat and Energy, continued
• Just as other forms of energy have a symbol that
identifies them (PE for potential energy, KE for kinetic
energy, U for internal energy, W for work), heat is
indicated by the symbol Q.
• Because heat, like work, is energy in transit, all heat
units can be converted to joules, the SI unit for
energy.
Defining Heat
Thermal Units and Their Values in Joules
Defining Heat
Thermal Conduction
• The type of energy transfer that
is due to atoms transferring
vibrations to neighboring atoms
is called thermal conduction.
• The rate of thermal
conduction depends on the
substance.
• Two other mechanisms for
transferring energy as heat are
convection and
electromagnetic radiation.
When this burner is
turned on, the skillet’s
handle heats up
because of conduction.
Defining Heat
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.
• In other words, the sum of the changes in
potential, kinetic, and internal energy is equal
to zero.
CONSERVATION OF ENERGY
DPE + DKE + DU = 0
the change in potential energy + the change in kinetic energy
+ the change in internal energy = 0
Defining Heat
Sample Problem
Conservation of Energy
An arrangement similar to the one
used to demonstrate energy
conservation is shown in the figure.
A vessel contains water. Paddles
that are propelled by falling masses
turn in the water. This agitation
warms the water and increases its
internal energy. The temperature of
the water is then measured, giving
an indication of the water’s internal
energy increase.
Defining Heat
Sample Problem, continued
Conservation of Energy, continued
If a total mass of 11.5 kg falls 1.3 m
and all of the mechanical energy is
converted to internal energy, by how
much will the internal energy of the
water increase? (Assume no energy
is transferred as heat out of the
vessel to the surroundings or from
the surroundings to the vessel’s
interior.)
Defining Heat
Sample Problem, continued
1. Define
Given:
m = 11.5 kg
h = 1.3 m
g = 9.81 m/s2
Unknown:
DU = ?
Defining Heat
Sample Problem, continued
2. Plan
Choose an equation or situation: Use the
conservation of energy, and solve for DU.
DPE + DKE + DU = 0
(PEf – PEi) + (KEf – KEi) + DU = 0
DU = –PEf + PEi – KEf + KEi
Tip: Don’t forget that a change in any quantity,
indicated by the symbol ∆, equals the final value
minus the initial value.
Defining Heat
Sample Problem, continued
Because the masses begin at rest, KEi equals
zero. If we assume that KEf is small compared to
the loss of PE, we can set KEf equal to zero also.
KEf = 0
KEi = 0
Because all of the potential energy is assumed to
be converted to internal energy, PEi can be set
equal to mgh if PEf is set equal to zero.
PEi = mgh
PEf = 0
Substitute each quantity into the equation for ∆U:
∆U = –PEf + PEi – KEf + KEi
∆U = 0 + mgh + 0 + 0 = mgh
Defining Heat
Sample Problem, continued
3. Calculate
Substitute the values into the equation and
solve:
DU = mgh
DU = (11.5 kg)(9.81 m/s2)(1.3 m)
DU = 1.5  102 J
4. Evaluate
The answer can be estimated using rounded
values. If m ≈ 10 kg and g ≈ 10 m/s2, then ∆U ≈
130 J, which is close to the actual value calculated.
Changes in Temperature and
Phase
Specific Heat Capacity
• The specific heat capacity of a substance is defined
as the energy required to change the temperature of
1 kg of that substance by 1°C.
• Every substance has a unique specific heat capacity.
• This value tells you how much the temperature of a
given mass of that substance will increase or
decrease, based on how much energy is added or
removed as heat.
Changes in Temperature and
Phase
Specific Heat Capacity, continued
• Specific heat capacity is expressed mathematically
as follows:
Q
cp 
mDT
energy transferred as heat
specific heat capacity =
mass  change in temperature
• The subscript p indicates that the specific heat capacity is
measured at constant pressure.
• In this equation, DT can be in degrees Celsius or in degrees
Kelvin.
Changes in Temperature and
Phase
Specific Heat Capacities
Changes in Temperature and
Phase
Calorimetry
• Calorimetry is used
to determine specific
heat capacity.
• Calorimetry is an
experimental
procedure used to
measure the energy
transferred from one
substance to another
as heat.
A simple
calorimeter
allows the
specific
heat
capacity of
a substance
to be
determined.
Changes in Temperature and
Phase
Calorimetry, continued
Because the specific heat capacity of water is well
known (cp,w= 4.186 kJ/kg•°C), the energy transferred as
heat between an object of unknown specific heat
capacity and a known quantity of water can be
measured.
energy absorbed by water = energy released by substance
Qw = –Qx
cp,wmw∆Tw = –cp,xmx∆Tx
Changes in Temperature and
Phase
Sample Problem
Calorimetry
A 0.050 kg metal bolt is heated to an unknown initial
temperature. It is then dropped into a calorimeter
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.
Changes in Temperature and
Phase
Sample Problem, continued
1. Define
Given:
Unknown:
Diagram:
mm = 0.050 kg
mw = 0.15 kg
Tw = 21.0°C
Tm = ?
cp,m = 899 J/kg•°C
cp,w = 4186 J/kg•°C
Tf = 25.0°C
Changes in Temperature and
Phase
Sample Problem, continued
2. Plan
Choose an equation or situation: The energy absorbed by the
water equals the energy removed from the bolt.
Qw  –Qm
c p,w mw DTw  –c p, m mm DTm
c p,w mw (T f  Tw )  –c p, m mm (T f  Tm )
Rearrange the equation to isolate the unknown:
Tm 
c p,w mw (T f  Tw )
c p,m mm
 Tf
Changes in Temperature and
Phase
Sample Problem, continued
3. Calculate
Substitute the values into the equation and solve:
Tm 
Tm 
c p , w mw (T f  Tw )
c p ,m mm
 Tf
(4186 J/kg•C)(0.15 kg)(25.0C  21.0C)
(899 J/kg  C)(0.050 kg)
Tm  81C
4. Evaluate
Tm is greater than Tf, as expected.
 25.0C
Tip: Because Tw is less
than Tf, you know that Tm
must be greater than Tf.
Changes in Temperature and
Phase
Latent Heat
• When substances melt, freeze, boil, condense, or
sublime, the energy added or removed changes the
internal energy of the substance without changing the
substance’s temperature.
• These changes in matter are called phase changes.
• The energy per unit mass that is added or removed
during a phase change is called latent heat,
abbreviated as L.
Q = mL
energy transferred as heat during phase change = mass  latent heat
Changes in Temperature and
Phase
Latent Heat, continued
• During melting, the energy that is added to a
substance equals the difference between the total
potential energies for particles in the solid and the
liquid phases. This type of latent heat is called the
heat of fusion, abbreviated as Lf.
• During vaporization, the energy that is added to a
substance equals the difference in the potential
energy of attraction between the liquid particles and
between the gas particles. In this case, the latent
heat is called the heat of vaporization, abbreviated
as Lv.
Multiple Choice
1. What must be true about two given objects for energy
to be transferred as heat between them?
A. The objects must be large.
B. The objects must be hot.
C. The objects must contain a large amount of
energy.
D. The objects must have different temperatures.
Multiple Choice
1. What must be true about two given objects for energy
to be transferred as heat between them?
A. The objects must be large.
B. The objects must be hot.
C. The objects must contain a large amount of
energy.
D. The objects must have different temperatures.
Multiple Choice, continued
2. A metal spoon is placed in one of two identical cups
of hot coffee. Why does the cup with the spoon have
a lower temperature after a few minutes?
F. Energy is removed from the coffee mostly by
conduction through the spoon.
G. Energy is removed from the coffee mostly by
convection through the spoon.
H. Energy is removed from the coffee mostly by radiation
through the spoon.
J. The metal in the spoon has an extremely large specific
heat capacity.
Multiple Choice, continued
2. A metal spoon is placed in one of two identical cups
of hot coffee. Why does the cup with the spoon have
a lower temperature after a few minutes?
F. Energy is removed from the coffee mostly by
conduction through the spoon.
G. Energy is removed from the coffee mostly by
convection through the spoon.
H. Energy is removed from the coffee mostly by radiation
through the spoon.
J. The metal in the spoon has an extremely large specific
heat capacity.
Multiple Choice, continued
Use the passage below to answer questions 3–4.
The boiling point of liquid hydrogen is –252.87°C.
3. What is the value of this temperature on the
Fahrenheit scale?
A. 20.28°F
B. –220.87°F
C. –423.2°F
D. 0°F
Multiple Choice, continued
Use the passage below to answer questions 3–4.
The boiling point of liquid hydrogen is –252.87°C.
3. What is the value of this temperature on the
Fahrenheit scale?
A. 20.28°F
B. –220.87°F
C. –423.2°F
D. 0°F
Multiple Choice, continued
Use the passage below to answer questions 3–4.
The boiling point of liquid hydrogen is –252.87°C.
4. What is the value of this temperature in kelvins?
F. 273 K
G. 20.28 K
H. –423.2 K
J. 0 K
Multiple Choice, continued
Use the passage below to answer questions 3–4.
The boiling point of liquid hydrogen is –252.87°C.
4. What is the value of this temperature in kelvins?
F. 273 K
G. 20.28 K
H. –423.2 K
J. 0 K
Multiple Choice, continued
5. A cup of hot chocolate with a temperature of 40°C is placed
inside a refrigerator at 5°C. An identical cup of hot chocolate at
90°C is placed on a table in a room at 25°C. A third identical cup
of hot chocolate at 80°C is placed on an outdoor table, where
the surrounding air has a temperature of 0°C. For which of the
three cups has the most energy been transferred as heat when
equilibrium has been reached?
A. The first cup has the largest energy transfer.
B. The second cup has the largest energy transfer.
C. The third cup has the largest energy transfer.
D. The same amount of energy is transferred as heat for all
three cups.
Multiple Choice, continued
5. A cup of hot chocolate with a temperature of 40°C is placed
inside a refrigerator at 5°C. An identical cup of hot chocolate at
90°C is placed on a table in a room at 25°C. A third identical cup
of hot chocolate at 80°C is placed on an outdoor table, where
the surrounding air has a temperature of 0°C. For which of the
three cups has the most energy been transferred as heat when
equilibrium has been reached?
A. The first cup has the largest energy transfer.
B. The second cup has the largest energy transfer.
C. The third cup has the largest energy transfer.
D. The same amount of energy is transferred as heat for all
three cups.
Multiple Choice, continued
6. What data are required in order to determine the
specific heat capacity of an unknown substance by
means of calorimetry?
F. cp,water, Twater, Tsubstance, Tfinal, Vwater, Vsubstance
G. cp,substance, Twater, Tsubstance, Tfinal, mwater, msubstance
H. cp,water, Tsubstance, mwater, msubstance
J. cp,water, Twater, Tsubstance, Tfinal, mwater, msubstance
Multiple Choice, continued
6. What data are required in order to determine the
specific heat capacity of an unknown substance by
means of calorimetry?
F. cp,water, Twater, Tsubstance, Tfinal, Vwater, Vsubstance
G. cp,substance, Twater, Tsubstance, Tfinal, mwater, msubstance
H. cp,water, Tsubstance, mwater, msubstance
J. cp,water, Twater, Tsubstance, Tfinal, mwater, msubstance
Multiple Choice, continued
7. During a cold spell, Florida orange growers often
spray a mist of water over their trees during the night.
Why is this done?
A. The large latent heat of vaporization for water keeps the trees
from freezing.
B. The large latent heat of fusion for water prevents it and thus
the trees from freezing.
C. The small latent heat of fusion for water prevents the water
and thus the trees from freezing.
D. The small heat capacity of water makes the water a good
insulator.
Multiple Choice, continued
7. During a cold spell, Florida orange growers often
spray a mist of water over their trees during the night.
Why is this done?
A. The large latent heat of vaporization for water keeps the trees
from freezing.
B. The large latent heat of fusion for water prevents it and thus
the trees from freezing.
C. The small latent heat of fusion for water prevents the water
and thus the trees from freezing.
D. The small heat capacity of water makes the water a good
insulator.
Multiple Choice, continued
Use the heating curve to answer questions 8–10. The
graph shows the change in temperature of a 23 g sample
as energy is added to the sample as heat.
8. What is the specific
heat capacity of the
liquid?
F. 4.4  105 J/kg•°C
G. 4.0  102 J/kg•°C
H. 5.0  102 J/kg•°C
J. 1.1  103 J/kg•°C
Multiple Choice, continued
Use the heating curve to answer questions 8–10. The
graph shows the change in temperature of a 23 g sample
as energy is added to the sample as heat.
8. What is the specific
heat capacity of the
liquid?
F. 4.4  105 J/kg•°C
G. 4.0  102 J/kg•°C
H. 5.0  102 J/kg•°C
J. 1.1  103 J/kg•°C
Multiple Choice, continued
Use the heating curve to answer questions 8–10. The
graph shows the change in temperature of a 23 g sample
as energy is added to the sample as heat.
9. What is the latent
heat of fusion?
A. 4.4  105 J/kg
B. 4.0  102 J/kg•°C
C. 10.15  103 J
D. 3.6  107 J/kg
Multiple Choice, continued
Use the heating curve to answer questions 8–10. The
graph shows the change in temperature of a 23 g sample
as energy is added to the sample as heat.
9. What is the latent
heat of fusion?
A. 4.4  105 J/kg
B. 4.0  102 J/kg•°C
C. 10.15  103 J
D. 3.6  107 J/kg
Multiple Choice, continued
Use the heating curve to answer questions 8–10. The
graph shows the change in temperature of a 23 g sample
as energy is added to the sample as heat.
10. What is the specific
heat capacity of the
solid?
F. 1.85  103 J/kg•°C
G. 4.0  102 J/kg•°C
H. 5.0  102 J/kg•°C
J. 1.1  103 J/kg•°C
Multiple Choice, continued
Use the heating curve to answer questions 8–10. The
graph shows the change in temperature of a 23 g sample
as energy is added to the sample as heat.
10. What is the specific
heat capacity of the
solid?
F. 1.85  103 J/kg•°C
G. 4.0  102 J/kg•°C
H. 5.0  102 J/kg•°C
J. 1.1  103 J/kg•°C
Short Response
Base your answers to questions 11–12 on the
information below.
The largest of the Great Lakes, Lake Superior, contains
1.20  1016 kg of fresh water, which has a specific heat
capacity of 4186 J/kg•°C and a latent heat of fusion of
3.33  105 J/kg.
11. How much energy would be needed to increase the
temperature of Lake Superior by 1.0°C?
Short Response
Base your answers to questions 11–12 on the
information below.
The largest of the Great Lakes, Lake Superior, contains
1.20  1016 kg of fresh water, which has a specific heat
capacity of 4186 J/kg•°C and a latent heat of fusion of
3.33  105 J/kg.
11. How much energy would be needed to increase the
temperature of Lake Superior by 1.0°C?
Answer: 5.0  1019 J
Short Response, continued
Base your answers to questions 11–12 on the
information below.
The largest of the Great Lakes, Lake Superior, contains
1.20  1016 kg of fresh water, which has a specific heat
capacity of 4186 J/kg•°C and a latent heat of fusion of
3.33  105 J/kg.
12. If Lake Superior were still liquid at 0°C, how much
energy would need to be removed from the lake for
it to become completely frozen?
Short Response, continued
Base your answers to questions 11–12 on the
information below.
The largest of the Great Lakes, Lake Superior, contains
1.20  1016 kg of fresh water, which has a specific heat
capacity of 4186 J/kg•°C and a latent heat of fusion of
3.33  105 J/kg.
12. If Lake Superior were still liquid at 0°C, how much
energy would need to be removed from the lake for
it to become completely frozen?
Answer: 5.00  1021 J
Short Response, continued
13. Ethyl alcohol has about one-half the specific heat
capacity of water. If equal masses of alcohol and
water in separate beakers at the same temperature
are supplied with the same amount of energy, which
will have the higher final temperature?
Short Response, continued
13. Ethyl alcohol has about one-half the specific heat
capacity of water. If equal masses of alcohol and
water in separate beakers at the same temperature
are supplied with the same amount of energy, which
will have the higher final temperature?
Answer: the ethyl alcohol
Short Response, continued
14. A 0.200 kg glass holds 0.300 kg of hot water, as
shown in the figure. The glass and water are set on
a table to cool. After the temperature has decreased
by 2.0°C, how much energy has been removed from
the water and glass?
(The specific heat capacity of
glass is 837 J/kg•°C, and that
of water is 4186 J/kg•°C.)
Short Response, continued
14. A 0.200 kg glass holds 0.300 kg of hot water, as
shown in the figure. The glass and water are set on
a table to cool. After the temperature has decreased
by 2.0°C, how much energy has been removed from
the water and glass?
(The specific heat capacity of
glass is 837 J/kg•°C, and that
of water is 4186 J/kg•°C.)
Answer: 2900 J
Extended Response
15. How is thermal energy transferred by the process
of convection?
Extended Response
15. How is thermal energy transferred by the process
of convection?
Answer: The increasing temperature of a liquid or gas
causes it to become less dense, so it rises above
colder liquid or gas, transferring thermal energy
with it.
Extended Response, continued
16. Show that the temperature –40.0° is unique in that it
has the same numerical value on the Celsius and
Fahrenheit scales. Show all of your work.
Extended Response, continued
16. Show that the temperature –40.0° is unique in that it
has the same numerical value on the Celsius and
Fahrenheit scales. Show all of your work.
Answer:
9
TF  (–40.0C)  32.0  (–72.0  32.0)F  –40.0F
5
Temperature and Thermal
Equilibrium
Measuring Temperature
Temperature and Thermal
Equilibrium
Determining Absolute Zero for an Ideal Gas
Changes in Temperature and
Phase
Calorimetry
Relationships Between Heat
and Work
Heat, Work, and Internal Energy
• Heat and work are energy transferred to or from a
system. An object never has “heat” or “work” in it; it
has only internal energy.
• A system is a set of particles or interacting
components considered to be a distinct physical
entity for the purpose of study.
• The environment the combination of conditions and
influences outside a system that affect the behavior
of the system.
Relationships Between Heat
and Work
Heat, Work, and Internal Energy, continued
• In thermodynamic systems, work is defined in terms
of pressure and volume change.
 A  F 
W  Fd  Fd      ( Ad )  PDV
 A  A
W  P DV
work = pressure  volume change
• This definition assumes that P is constant.
Relationships Between Heat
and Work
Heat, Work, and Internal Energy, continued
• If the gas expands, as
shown in the figure, DV is
positive, and the work done
by the gas on the piston is
positive.
• If the gas is compressed,
DV is negative, and the
work done by the gas on
the piston is negative. (In
other words, the piston
does work on the gas.)
Relationships Between Heat
and Work
Heat, Work, and Internal Energy, continued
• When the gas volume remains constant, there is no
displacement and no work is done on or by the
system.
• Although the pressure can change during a process,
work is done only if the volume changes.
• A situation in which pressure increases and volume
remains constant is comparable to one in which a
force does not displace a mass even as the force is
increased. Work is not done in either situation.
Relationships Between Heat
and Work
Thermodynamic Processes
• An isovolumetric process is a thermodynamic
process that takes place at constant volume so that
no work is done on or by the system.
• An isothermal process is a thermodynamic process
that takes place at constant temperature.
• An adiabatic process is a thermodynamic process
during which no energy is transferred to or from the
system as heat.
The First Law of
Thermodynamics
Energy Conservation
• If friction is taken into account, mechanical energy
is not conserved.
• Consider the example of a roller coaster:
– A steady decrease in the car’s total mechanical energy
occurs because of work being done against the friction
between the car’s axles and its bearings and between the
car’s wheels and the coaster track.
– If the internal energy for the roller coaster (the system) and
the energy dissipated to the surrounding air (the
environment) are taken into account, then the total energy
will be constant.
The First Law of
Thermodynamics
Energy Conservation
The First Law of
Thermodynamics
Energy Conservation, continued
• The principle of energy conservation that takes into
account a system’s internal energy as well as work
and heat is called the first law of thermodynamics.
• The first law of thermodynamics can be expressed
mathematically as follows:
DU = Q – W
Change in system’s internal energy = energy
transferred to or from system as heat – energy
transferred to or from system as work
The First Law of
Thermodynamics
Signs of Q and W for a system
The First Law of
Thermodynamics
Sample Problem
The First Law of Thermodynamics
A total of 135 J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases by 114 J during
the process, what is the total amount of energy
transferred as heat? Has energy been added to or
removed from the refrigerant as heat?
The First Law of
Thermodynamics
Sample Problem, continued
1. Define
Given:
W = –135 J Tip: Work is done
on the gas, so work
DU = 114 J (W) has a negative
Unknown:
Q=?
value. The internal
energy increases
during the process,
so the change in
internal energy
(DU) has a positive
value.
Diagram:
The First Law of
Thermodynamics
Sample Problem, continued
2. Plan
Choose an equation or situation:
Apply the first law of thermodynamics using the values
for DU and W in order to find the value for Q.
DU = Q – W
Rearrange the equation to isolate the unknown:
Q = DU + W
The First Law of
Thermodynamics
Sample Problem, continued
3. Calculate
Substitute the values into the equation and solve:
Q = 114 J + (–135 J)
Q = –21 J
Tip: The sign for the value of Q is negative. This
indicates that energy is transferred as heat from
the refrigerant.
The First Law of
Thermodynamics
Sample Problem, continued
4. Evaluate
Although the internal energy of the refrigerant
increases under compression, more energy is
added as work than can be accounted for by the
increase in the internal energy. This energy is
removed from the gas as heat, as indicated by the
minus sign preceding the value for Q.
The First Law of
Thermodynamics
Cyclic Processes
• A cyclic process is a thermodynamic process in
which a system returns to the same conditions under
which it started.
• Examples include heat engines and refrigerators.
• In a cyclic process, the final and initial values of
internal energy are the same, and the change in
internal energy is zero.
DUnet = 0 and Qnet = Wnet
The First Law of
Thermodynamics
Cyclic Processes, continued
• A heat engine uses heat to do
mechanical work.
• A heat engine is able to do
work (b) by transferring energy
from a high-temperature
substance (the boiler) at Th (a)
to a substance at a lower
temperature (the air around the
engine) at Tc (c).
• The internal-combustion engine found in most
vehicles is an example of a heat engine.
The First Law of
Thermodynamics
The Steps of a Gasoline Engine Cycle
The First Law of
Thermodynamics
The Steps of a Refrigeration Cycle
The First Law of
Thermodynamics
Thermodynamics of a Refrigerator
The Second Law of
Thermodynamics
Efficiency of Heat Engines
• The second law of thermodynamics can be stated
as follows:
No cyclic process that converts heat entirely
into work is possible.
• As seen in the last section, Wnet = Qnet = Qh – Qc.
– According to the second law of thermodynamics,
W can never be equal to Qh in a cyclic process.
– In other words, some energy must always be
transferred as heat to the system’s surroundings
(Qc > 0).
The Second Law of
Thermodynamics
Efficiency of Heat Engines, continued
• A measure of how well an engine operates is given
by the engine’s efficiency (eff ).
• In general, efficiency is a measure of the useful
energy taken out of a process relative to the total
energy that is put into the process.
Wnet Qh – Qc
Qc
eff 

 1
Qh
Qh
Qh
• Note that efficiency is a unitless quantity.
• Because of the second law of thermodynamics, the
efficiency of a real engine is always less than 1.
The Second Law of
Thermodynamics
Sample Problem
Heat-Engine Efficiency
Find the efficiency of a gasoline engine that, during
one cycle, receives 204 J of energy from combustion
and loses 153 J as heat to the exhaust.
1. Define
Given:
Diagram:
Qh = 204 J
Qc = 153 J
Unknown
eff = ?
The Second Law of
Thermodynamics
Sample Problem, continued
2. Plan
Choose an equation or situation: The efficiency of
a heat engine is the ratio of the work done by the
engine to the energy transferred to it as heat.
Wnet
Qc
eff 
 1
Qh
Qh
The Second Law of
Thermodynamics
Sample Problem, continued
3. Calculate
Substitute the values into the equation and
solve:
Qc
153 J
eff  1 
 1
Qh
204 J
eff  0.250
4. Evaluate
Only 25 percent of the energy added as heat is used
by the engine to do work. As expected, the efficiency
is less than 1.0.
The Second Law of
Thermodynamics
Entropy
• In thermodynamics, a system left to itself tends to go
from a state with a very ordered set of energies to
one in which there is less order.
• The measure of a system’s disorder or randomness
is called the entropy of the system. The greater the
entropy of a system is, the greater the system’s
disorder.
• The greater probability of a disordered arrangement
indicates that an ordered system is likely to
become disordered. Put another way, the entropy
of a system tends to increase.
The Second Law of
Thermodynamics
Entropy, continued
• Greater disorder means there is less energy to do
work.
• If all gas particles moved toward the piston, all of the
internal energy could be used to do work. This
extremely well ordered system is highly improbable.
The Second Law of
Thermodynamics
Entropy, continued
• Because of the connection between a system’s
entropy, its ability to do work, and the direction of
energy transfer, the second law of
thermodynamics can also be expressed in terms of
entropy change:
The entropy of the universe increases in all
natural processes.
• Entropy can decrease for parts of systems, provided
this decrease is offset by a greater increase in
entropy elsewhere in the universe.
The Second Law of
Thermodynamics
Energy Changes Produced by a Refrigerator
Freezing Water
Because of the refrigerator’s less-than-perfect efficiency, the entropy of
the outside air molecules increases more than the entropy of the
freezing water decreases.
Multiple Choice
1. If there is no change in the internal energy of a gas,
even though energy is transferred to the gas as heat
and work, what is the thermodynamic process that
the gas undergoes called?
A. adiabatic
B. isothermal
C. isovolumetric
D. isobaric
Multiple Choice
1. If there is no change in the internal energy of a gas,
even though energy is transferred to the gas as heat
and work, what is the thermodynamic process that
the gas undergoes called?
A. adiabatic
B. isothermal
C. isovolumetric
D. isobaric
Multiple Choice, continued
2. To calculate the efficiency of a heat engine, which
thermodynamic property does not need to be known?
F. the energy transferred as heat to the engine
G. the energy transferred as heat from the engine
H. the change in the internal energy of the engine
J. the work done by the engine
Multiple Choice, continued
2. To calculate the efficiency of a heat engine, which
thermodynamic property does not need to be known?
F. the energy transferred as heat to the engine
G. the energy transferred as heat from the engine
H. the change in the internal energy of the engine
J. the work done by the engine
Multiple Choice, continued
3. In which of the following processes is no work done?
A. Water is boiled in a pressure cooker.
B. A refrigerator is used to freeze water.
C. An automobile engine operates for several
minutes.
D. A tire is inflated with an air pump.
Multiple Choice, continued
3. In which of the following processes is no work done?
A. Water is boiled in a pressure cooker.
B. A refrigerator is used to freeze water.
C. An automobile engine operates for several
minutes.
D. A tire is inflated with an air pump.
Multiple Choice, continued
4. A thermodynamic process occurs in which the
entropy of a system decreases. From the second law
of thermodynamics, what can you conclude about the
entropy change of the environment?
F. The entropy of the environment decreases.
G. The entropy of the environment increases.
H. The entropy of the environment remains
unchanged.
J. There is not enough information to state what
happens to the environment’s entropy.
Multiple Choice, continued
4. A thermodynamic process occurs in which the
entropy of a system decreases. From the second law
of thermodynamics, what can you conclude about the
entropy change of the environment?
F. The entropy of the environment decreases.
G. The entropy of the environment increases.
H. The entropy of the environment remains
unchanged.
J. There is not enough information to state what
happens to the environment’s entropy.
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
5. Which of the figures
describes a situation in
which DU < 0, Q < 0, and
W = 0?
A. (a)
B. (b)
C. (c)
D. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
5. Which of the figures
describes a situation in
which DU < 0, Q < 0, and
W = 0?
A. (a)
B. (b)
C. (c)
D. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
6. Which of the figures
describes a situation in
which DU > 0, Q = 0, and
W < 0?
F. (a)
G. (b)
H. (c)
J. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
6. Which of the figures
describes a situation in
which DU > 0, Q = 0, and
W < 0?
F. (a)
G. (b)
H. (c)
J. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
7. Which of the figures
describes a situation in
which DU < 0, Q = 0, and
W > 0?
A. (a)
B. (b)
C. (c)
D. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
7. Which of the figures
describes a situation in
which DU < 0, Q = 0, and
W > 0?
A. (a)
B. (b)
C. (c)
D. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
8. Which of the figures
describes a situation in
which DU > 0, Q > 0, and
W = 0?
F. (a)
G. (b)
H. (c)
J. (d)
Multiple Choice, continued
Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
8. Which of the figures
describes a situation in
which DU > 0, Q > 0, and
W = 0?
F. (a)
G. (b)
H. (c)
J. (d)
Multiple Choice, continued
9. A power plant has a power output of 1055 MW and
operates with an efficiency of 0.330. Excess energy
is carried away as heat from the plant to a nearby
river. How much energy is transferred away from the
power plant as heat?
A. 0.348  109 J/s
B. 0.520  109 J/s
C. 0.707  109 J/s
D. 2.14  109 J/s
Multiple Choice, continued
9. A power plant has a power output of 1055 MW and
operates with an efficiency of 0.330. Excess energy
is carried away as heat from the plant to a nearby
river. How much energy is transferred away from the
power plant as heat?
A. 0.348  109 J/s
B. 0.520  109 J/s
C. 0.707  109 J/s
D. 2.14  109 J/s
Multiple Choice, continued
10. How much work must be done by air pumped into a
tire if the tire’s volume increases from 0.031 m3 to
0.041 m3 and the net, constant pressure of the air is
300.0 kPa?
F. 3.0  102 J
G. 3.0  103 J
H. 3.0  104 J
J. 3.0  105 J
Multiple Choice, continued
10. How much work must be done by air pumped into a
tire if the tire’s volume increases from 0.031 m3 to
0.041 m3 and the net, constant pressure of the air is
300.0 kPa?
F. 3.0  102 J
G. 3.0  103 J
H. 3.0  104 J
J. 3.0  105 J
Short Response
Use the passage below to answer questions 11–12.
An air conditioner is left running on a table in the middle
of the room, so none of the air that passes through the
air conditioner is transferred to outside the room.
11. Does passing air through the air conditioner affect
the temperature of the room? (Ignore the thermal
effects of the motor running the compressor.)
Short Response
Use the passage below to answer questions 11–12.
An air conditioner is left running on a table in the middle
of the room, so none of the air that passes through the
air conditioner is transferred to outside the room.
11. Does passing air through the air conditioner affect
the temperature of the room? (Ignore the thermal
effects of the motor running the compressor.)
Answer: No, because the energy removed from the
cooled air is returned to the room.
Short Response, continued
Use the passage below to answer questions 11–12.
An air conditioner is left running on a table in the middle
of the room, so none of the air that passes through the
air conditioner is transferred to outside the room.
12. Taking the compressor motor into account, what
would happen to the temperature of the room?
Short Response, continued
Use the passage below to answer questions 11–12.
An air conditioner is left running on a table in the middle
of the room, so none of the air that passes through the
air conditioner is transferred to outside the room.
12. Taking the compressor motor into account, what
would happen to the temperature of the room?
Answer: The temperature increases.
Short Response, continued
13. If 1600 J of energy are transferred as heat to an
engine and 1200 J are transferred as heat away
from the engine to the surrounding air, what is the
efficiency of the engine?
Short Response, continued
13. If 1600 J of energy are transferred as heat to an
engine and 1200 J are transferred as heat away
from the engine to the surrounding air, what is the
efficiency of the engine?
Answer: 0.25
Extended Response
14. How do the temperature of combustion and the
temperatures of coolant and exhaust affect the
efficiency of automobile engines?
Extended Response
14. How do the temperature of combustion and the
temperatures of coolant and exhaust affect the
efficiency of automobile engines?
Answer: The greater the temperature difference is, the
greater is the amount of energy transferred as heat. For
efficiency to increase, the heat transferred between the
combustion reaction and the engine (Qh) should be
made to increase, whereas the energy given up as
waste heat to the coolant and exhaust (Qc) should be
made to decrease.
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
15. How much work is done by
the steam shovel in lifting the
dirt?
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
15. How much work is done by
the steam shovel in lifting the
dirt?
Answer: 3.8  104 J
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
16. What is the efficiency of the
steam shovel?
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
16. What is the efficiency of the
steam shovel?
Answer: 0.19
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
17. Assuming there is no change
in the internal energy of the
steam shovel’s engine, how
much energy is given up by
the shovel as waste heat?
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
17. Assuming there is no change
in the internal energy of the
steam shovel’s engine, how
much energy is given up by
the shovel as waste heat?
Answer: 1.62  105 J
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
18. Suppose the internal energy
of the steam shovel’s engine
increases by 5.0  103 J. How
much energy is given up now
as waste heat?
Extended Response, continued
Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00  105 J of energy as heat
for the steam shovel to lift the dirt.
18. Suppose the internal energy
of the steam shovel’s engine
increases by 5.0  103 J. How
much energy is given up now
as waste heat?
Answer: 1.57  105 J
Extended Response, continued
19. One way to look at heat and work is to think of
energy transferred as heat as a “disorganized” form
of energy and energy transferred as work as an
“organized” form. Use this interpretation to show
that the increased order obtained by freezing
water is less than the total disorder that results
from the freezer used to form the ice.
Extended Response, continued
19. One way to look at heat and work is to think of
energy transferred as heat as a “disorganized” form
of energy and energy transferred as work as an
“organized” form. Use this interpretation to show
that the increased order obtained by freezing
water is less than the total disorder that results
from the freezer used to form the ice.
Answer: Disorganized energy is removed from water to
form ice, but a greater amount of organized energy
must become disorganized to operate the freezer.
The Second Law of
Thermodynamics
Entropy
The Second Law of
Thermodynamics
Energy Changes Produced by a Refrigerator
Freezing Water