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Resources
Chapter Presentation
Transparencies
Visual Concepts
Sample Problems
Standardized Test Prep
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Chapter 9
Heat
Table of Contents
Section 1 Temperature and Thermal Equilibrium
Section 2 Defining Heat
Section 3 Changes in Temperature and Phase
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Objectives
• Relate temperature to the kinetic energy of atoms
and molecules.
• Describe the changes in the temperatures of two
objects reaching thermal equilibrium.
• Identify the various temperature scales, and convert
from one scale to another.
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Forms of Internal Energy
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Thermal Equilibrium
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Thermal Expansion
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 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.
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Chapter 9
Section 1 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
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Temperature Scales and Their Uses
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Chapter 9
Section 2 Defining Heat
Objectives
• Explain heat as the energy transferred between
substances that are at different temperatures.
• Relate heat and temperature change on the
macroscopic level to particle motion on the
microscopic level.
• Apply the principle of energy conservation to
calculate changes in potential, kinetic, and internal
energy.
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 Defining Heat
Temperature and Heat
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 Defining Heat
Thermal Units and Their Values in Joules
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 Defining Heat
Convection, Conduction, and Radiation
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Chapter 9
Section 2 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
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Chapter 9
Section 2 Defining Heat
Conservation of Energy
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 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.)
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Chapter 9
Section 2 Defining Heat
Sample Problem, continued
1. Define
Given:
m = 11.5 kg
h = 1.3 m
g = 9.81 m/s2
Unknown:
DU = ?
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Chapter 9
Section 2 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.
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Chapter 9
Section 2 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
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Chapter 9
Section 2 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.
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Chapter 9
Section 3 Changes in
Temperature and Phase
Objectives
• Perform calculations with specific heat capacity.
• Interpret the various sections of a heating curve.
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Chapter 9
Section 3 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.
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Chapter 9
Section 3 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.
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Chapter 9
Section 3 Changes in
Temperature and Phase
Specific Heat Capacities
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Chapter 9
Section 3 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.
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Chapter 9
Section 3 Changes in
Temperature and Phase
Calorimetry
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Chapter 9
Section 3 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
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Chapter 9
Section 3 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.
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Chapter 9
Section 3 Changes in
Temperature and Phase
Sample Problem, continued
1. Define
Given:
Unknown:
mm = 0.050 kg
mw = 0.15 kg
Tw = 21.0°C
cp,m = 899 J/kg•°C
cp,w = 4186 J/kg•°C
Tf = 25.0°C
Tm = ?
Diagram:
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Chapter 9
Section 3 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
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Chapter 9
Section 3 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
 25.0C
Tip: Because Tw is less
than Tf, you know that Tm
must be greater than Tf.
Tm is greater than Tf, as expected.
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Chapter 9
Section 3 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
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Chapter 9
Section 3 Changes in
Temperature and Phase
Latent Heat
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Chapter 9
Section 3 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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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?
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 9
Standardized Test Prep
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
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 9
Standardized Test Prep
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?
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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?
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
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.)
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Standardized Test Prep
Extended Response
15. How is thermal energy transferred by the process
of convection?
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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.
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Chapter 9
Standardized Test Prep
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
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Measuring Temperature
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Chapter 9
Section 1 Temperature and
Thermal Equilibrium
Determining Absolute Zero for an Ideal Gas
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Chapter 9
Section 3 Changes in
Temperature and Phase
Calorimetry
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