Ch12. Temperature and Heat
Common Temperature Scales
A number of
different
temperature
scales have been
devised, two
popular choices
being the Celsius
(formerly,
centigrade) and
Fahrenheit scales.
1
On the Celsius scale, an ice point of 0 °C (0 degrees Celsius)
and a steam point of 100 °C were selected. On the Fahrenheit
scale, an ice point of 32 °F (32 degrees Fahrenheit) and a
steam point of 212 °F were chosen. The Celsius scale is used
worldwide, while the Fahrenheit scale is used mostly in the
United States.
The temperature of the human body is about 37 °C, where the
symbol °C stands for “degrees Celsius.” However, the change
between two temperatures is specified in “Celsius
degrees ”(C°)—not in “degrees Celsius”.
The separation between the ice and steam points on the Celsius
scale is divided into 100 Celsius degrees, while on the
Fahrenheit scale the separation is divided into 180 Fahrenheit
degrees. Therefore, the size of the Celsius degree is larger than
that of the Fahrenheit degree by a factor of 180 , or 9 .
2
100
5
Example 1. Converting from a
Fahrenheit to a Celsius Temperature
A healthy person has an oral temperature of 98.6 °F. What
would this reading be on the Celsius scale?
Reasoning and Solution A temperature of 98.6 °F is 66.6
Fahrenheit degrees above the ice point of 32.0 °F.
Since
, the difference of 66.6 F° is
equivalent to
Thus, the person’s temperature is 37.0 Celsius degrees above the
ice point. Adding 37.0 Celsius degrees to the ice point of 0 °C
on the Celsius scale gives a Celsius temperature of
3
Example 2. Converting from a Celsius
to a Fahrenheit Temperature
A time and temperature sign on a bank indicates that the
outdoor temperature is –20.0 °C. Find the corresponding
temperature on the Fahrenheit scale .
The temperature, then, is 36.0 Fahrenheit degrees below
the ice point. Subtracting 36.0 Fahrenheit degrees from the
ice point of 32.0 °F on the Fahrenheit scale gives a
Fahrenheit temperature of
4
Reasoning Strategy
Converting Between Different Temperature Scales
1. Determine the magnitude of the difference between
the stated temperature and the ice point on the initial
scale.
2. Convert this number of degrees from one scale to the
other scale by using the fact that.
3.Add or subtract the number of degrees on the new
scale to or from the ice point on the new scale.
5
Check Your Understanding 1
On a new temperature scale the steam point is 348 °X,
and the ice point is 112 °X. What is the temperature on
this scale that corresponds to 28.0 °C?
178 °X
6
The Kelvin Temperature Scale
Kelvin temperature scale was introduced by the Scottish
physicist William Thompson (Lord Kelvin, 1824–1907),
and in his honor each degree on the scale is called a kelvin
(K). By international agreement, the symbol K is not
written with a degree sign (°), nor is the word “degrees”
used when quoting temperatures. For example, a
temperature of 300 K (not 300 °K) is read as “three
hundred kelvins,” not “three hundred degrees kelvin.” The
kelvin is the SI base unit for temperature.
7
The ice point (0 °C) occurs
at 273.15 K on the Kelvin
scale.
When a gas confined to a
fixed volume is heated, its
pressure increases.
Conversely, when the gas is
cooled, its pressure decreases.
The change in gas pressure
with temperature is the basis
for the constant-volume gas
thermometer.
8
A constant-volume gas
thermometer.
9
A plot of absolute pressure
versus temperature for a
low-density gas at constant
volume. The graph is a
straight line and, when
extrapolated (dashed line),
crosses the temperature axis
at –273.15 °C.
“Absolute zero” means that
temperatures lower than –
273.15 °C cannot be
reached by continually
cooling a gas or any other
substance.
10
Thermometers
A property that changes with temperature is called a
thermometric property.
The thermocouple is a thermometer used extensively in
scientific laboratories. It consists of thin wires of different
metals, welded together at the ends to form two junctions.
11
12
One of the junctions, called the “hot” junction, is placed in
thermal contact with the object whose temperature is being
measured. The other junction, termed the “reference”
junction, is kept at a known constant temperature (usually
an ice–water mixture at 0 °C). The thermocouple
generates a voltage that depends on the difference in
temperature between the two junctions. This voltage is the
thermometric property and is measured by a voltmeter.
13
Because this electrical resistance changes with temperature,
electrical resistance is another thermometric property.
Electrical resistance thermometers are often made from
platinum wire, because platinum has excellent mechanical
and electrical properties in the temperature range from –270
°C to +700 °C. The electrical resistance of platinum wire is
known as a function of temperature. Thus, the temperature
of a substance can be determined by placing the resistance
thermometer in thermal contact with the substance and
measuring the resistance of the platinum wire.
14
Radiation emitted by an object can also be used to indicate
temperature. At low to moderate temperatures, the
predominant radiation emitted is infrared. As the
temperature is raised, the intensity of the radiation
increases substantially.
“Thermal painting” is called a thermograph or
thermogram. Thermography is an important diagnostic
tool in medicine.
15
Linear Thermal Expansion
NORMAL SOLIDS
The increase in any one dimension of a solid is called
linear expansion .
When the temperature of a
rod is raised by  T, the length
of the rod increases by L .
16
For modest temperature
changes, experiments show
that the change in length is
directly proportional to the
change in temperature L  T
In addition, the change in
length is proportional to the
initial length of the rod. L
is proportional to both L0 and
T ( L  L0 T ) by using a
proportionality constant  ,
which is called the coefficient
of linear expansion.
17
LINEAR THERMAL EXPANSION OF A SOLID
The length L0 of an object changes by an amount  L
when its temperature changes by an amount  T:
L  L0 T
where

is the coefficient of linear expansion.
Common Unit for the Coefficient of Linear Expansion:
1
1
 C 
C
18
Coefficient of Thermal
Expansion (C°)–1
Substance
Linear (a)
Volumetric
(b)
Solids
23 × 10–6
69 × 10–6
Brass
19 × 10–6
57 × 10–6
Concrete
12 × 10–6
36 × 10–6
Copper
17 × 10–6
51 × 10–6
Glass
(common)
8.5 × 10–6
26 × 10–6
Aluminium
3.3 × 10–6
9.9 × 10–6
Gold
14 × 10–6
42 × 10–6
Iron or steel
12 × 10–6
36 × 10–6
Lead
29 × 10–6
87 × 10–6
Nickel
13 × 10–6
39 × 10–6
Quartz (fused)
0.50 × 10–6
1.5 × 10–6
Silver
19 × 10–6
Glass (Pyrex)
57 × 10–6
19
Liquidsb
Benzene
—
1240 × 10–6
Carbon
tetrachloride
—
1240 × 10–6
Ethyl alcohol
—
1120 × 10–6
Gasoline
—
950 × 10–6
Mercury
—
182 × 10–6
Methyl alcohol
—
1200 × 10–6
Water
—
207 × 10–6
20
Example 3. Buckling of a Sidewalk
21
A concrete sidewalk is constructed between two buildings on a
day when the temperature is 25 °C. The sidewalk consists of
two slabs, each three meters in length and of negligible
thickness . As the temperature rises to 38 °C, the slabs
expand, but no space is provided for thermal expansion. The
buildings do not move, so the slabs buckle upward. Determine
the vertical distance y in part b of the drawing.
22
Antiscalding device screws onto the end of a faucet and quickly
shuts off the flow of water when it becomes too hot. As the water
temperature rises, the actuator spring expands and pushes the
plunger forward, shutting off the flow. When the water cools,
the spring contracts and the water flow resumes.
23
THERMAL STRESS: Example 4.
The Stress on a Steel Beam
A steel beam is used in the
roadbed of a bridge. The beam
is mounted between two
concrete supports when the
temperature is 23 °C, with no
room provided for thermal
expansion. What compressional
stress must the concrete
supports apply to each end of
the beam, if they are to keep the
beam from expanding when the
temperature rises to 42 °C?
24
F
L
Stress   Y
A
L0
L0 T
L
Stress  Y
Y
 YT
L0
L0
Y = 2.0 × 1011 N/m2
 = 12 × 10–6 (C°) –1
T = 19 C°
25
THE BIMETALLIC STRIP
A bimetallic strip is made from two thin strips of metal that
have different coefficients of linear expansion.
Bass
Steel
26
Bimetallic strips are frequently used as adjustable
automatic switches in electrical appliances.
27
28
THE EXPANSION OF HOLES:
Conceptual Example 5.
Do Holes Expand or Contract When
the Temperature Increases?
Eight square tiles that are arranged to form a square
pattern with a hole in the center. If the tiles are heated, what
happens to the size of the hole?
29
The hole expands just as if it were made of the
material of the surrounding tiles.
30
Example 6.
A Heated Engagement Ring
A gold engagement ring has an inner diameter of 1.5 × 10–2
m and a temperature of 27 °C. The ring falls into a sink of
hot water whose temperature is 49 °C. What is the change
in the diameter of the hole in the ring?
 = 14 × 10–6 (C°)–1
31
Conceptual Example 7.
Expanding Cylinders
In a cross-sectional view of three cylinders, A, B, and C,
each is made from a different material: one is lead, one is
brass, and one is steel. All three have the same temperature,
and they barely fit inside each other. As the cylinders are
heated to the same, but higher, temperature, cylinder C
falls off, while cylinder A becomes tightly wedged to
cylinder B. Which cylinder is made from which material?
32
A = brass, B = steel, and C = lead
A = lead, B = steel, and C = brass
33
Check Your Understanding 2
A metal ball has a diameter that is slightly greater than the
diameter of a hole that has been cut into a metal plate. The
coefficient of linear thermal expansion for the metal from which
the ball is made is greater than that for the metal of the plate.
Which one or more of the following procedures can be used to
make the ball pass through the hole?
(a)Raise the temperatures of the ball and the plate by the same
amount.
(b)Lower the temperatures of the ball and the plate by the same
amount.
(c) Heat the ball and cool the plate.
(d)Cool the ball and heat the plate.
b&d
34
Volume Thermal Expansion
VOLUME THERMAL EXPANSION
The volume V0 of an object changes by an amount  V
when its temperature changes by an amount  T:
V  V0 T
where

is the coefficient of volume expansion.
Common Unit for the Coefficient of Volume
Expansion: (C°) –1
 = 3.
35
Example 8. An Automobile Radiator
A small plastic container, called the
coolant reservoir, catches the
radiator fluid that overflows when
an automobile engine becomes hot .
The radiator is made of copper, and
the coolant has a coefficient of
volume expansion of . If the
radiator is filled to its 15-quart
capacity when the engine is cold (6.0
°C), how much overflow from the
radiator will spill into the reservoir
when the coolant reaches its
operating temperature of 92 °C?
36

V  V0T  4.10 10 C
1
4
15quarts86C
 0.53quarts
  5110

6
C 
1
V  V0T  5110 C
6
1
15quarts86C
 0.066quarts
The overflow volume is 0.53 quarts – 0.066 quarts = 0.46 quarts.
37
The fact that water has its greatest density at 4 °C, rather
than at 0 °C, has important consequences for the way in
which a lake freezes.
38
The fact that the density of ice is smaller than the
density of water has an important consequence for
home owners, who have to contend with the
possibility of bursting water pipes during severe
winters.
39
Heat and Internal Energy
Heat is energy in transit from
hot to cold.
(a)Heat flows from the hotter
coffee cup to the colder
hand.
(b)Heat flows from the warmer
hand to the colder glass of
ice water.
40
DEFINITION OF HEAT
Heat is energy that flows from a higher-temperature object
to a lower-temperature object because of the difference in
temperatures.
SI Unit of Heat: joule (J)
The internal energy of a substance is the sum of the molecular
kinetic energy (due to the random motion of the molecules),
the molecular potential energy (due to forces that act between
the atoms of a molecule and between molecules), and other
kinds of molecular energy. When heat flows in circumstances
where the work done is negligible, the internal energy of the
hot substance decreases and the internal energy of the cold
substance increases.
41
42
Heat and Temperature Change:
Specific Heat Capacity
SOLIDS AND LIQUIDS
HEAT SUPPLIED OR REMOVED IN CHANGING THE
TEMPERATURE OF A SUBSTANCE
The heat Q that must be supplied or removed to change the
temperature of a substance of mass m by an amount  T is
Q  cmT
where c is the specific heat capacity of the substance.
Common Unit for Specific Heat Capacity: J/(kg·C°)
43
Substance
Specific Heat Capacity, c
J/(kg·C°)
Solids
Aluminum
9.00 × 102
Copper
387
Glass
840
Human body (37 °C,
average)
3500
Ice (–15 °C)
2.00 × 103
Iron or steel
452
Lead
128
Silver
235
44
Liquids
Benzene
1740
Ethyl alcohol
2450
Glycerin
2410
Mercury
139
Water (15 °C)
4186
45
Example 9. A Hot Jogger
In a half hour, a 65-kg jogger can generate 8.0 × 105 J of
heat. This heat is removed from the jogger’s body by a
variety of means, including the body’s own temperatureregulating mechanisms. If the heat were not removed,
how much would the body temperature increase?
46
Example 10. Taking a Hot Shower
Cold water at a temperature of 15 °C enters a heater, and
the resulting hot water has a temperature of 61 °C. A
person uses 120 kg of hot water in taking a shower. (a) Find
the energy needed to heat the water. (b) Assuming that the
utility company charges $0.10 per kilowatt·hour for
electrical energy, determine the cost of heating the water.
(a )
47
(b)
At a cost of $0.10 per kWh, the bill for the heat is $0.64
or 64 cents.
48
GASES
The value of the specific heat capacity depends on
whether the pressure or volume is held constant while
energy in the form of heat is added to or removed
from a substance. The distinction between constant
pressure and constant volume is usually not
important for solids and liquids but is significant for
gases.
49
HEAT UNITS OTHER THAN THE
JOULE
There are three heat units other than the joule in common use.
One kilocalorie (1 kcal) was defined historically as the amount
of heat needed to raise the temperature of one kilogram of
water by one Celsius degree.
one calorie (1 cal) was defined as the amount of heat needed to
raise the temperature of one gram of water by one Celsius
degree
The British thermal unit (Btu) is the other commonly used
heat unit and was defined historically as the amount of heat
needed to raise the temperature of one pound of water by one
Fahrenheit degree.
50
Joule’s experiments revealed that the performance
of mechanical work, like rubbing your hands
together, can make the temperature of a substance
rise, just as the absorption of heat can.
This conversion factor is known as the
mechanical equivalent of heat.
51
CALORIMETRY
The kind of heat transfer that occurs within a thermos
of iced tea also occurs within a calorimeter, which is the
experimental apparatus used in a technique known as
calorimetry.
52
Example 11. Measuring the Specific
Heat Capacity
A calorimeter cup is made from 0.15 kg of aluminum and
contains 0.20 kg of water. Initially, the water and the cup
have a common temperature of 18.0 °C. A 0.040-kg mass
of unknown material is heated to a temperature of 97.0 °C
and then added to the water. The temperature of the water,
the cup, and the unknown material is 22.0 °C after
thermal equilibrium is reestablished. Ignoring the small
amount of heat gained by the thermometer, find the specific
heat capacity of the unknown material.
53
(cmT ) Al  (cmT ) water  (cmT )unknown
Heat gained by aluminum and water
Heat lost by unknown
material
c Al m Al TAl  cwater mwater Twater
cunknown 
munknownTunknown
54
TAl = Twater = 22.0 °C – 18.0 °C = 4.0 C°
Tunknown = 97.0 °C – 22.0 °C = 75.0 C°.
55
Check Your Understanding 3
Consider a mass m of a material and a change  T in its
temperature. Various possibilities for these variables are
listed in the table below. Rank these possibilities in
descending order (largest first), according to how much heat
is needed to bring about the change in temperature.
m (kg)
c, b, d, a
 T (C°)
(a)
2.0
15
(b)
1.5
40
(c)
3.0
25
(d)
2.5
20
56
Heat and Phase Change: Latent Heat
Three familiar phases of
matter—solid, liquid,
and gas—and the phase
changes that can occur
between any two of
them.
57
The graph shows the way the temperature of water
changes as heat is added, starting with ice at –30
°C. The pressure is atmospheric pressure.
58
Conceptual Example 12.
Saving Energy
Suppose you are cooking spaghetti for dinner, and
the instructions say “boil the pasta in water for ten
minutes.” To cook spaghetti in an open pot with the
least amount of energy, should you turn up the
burner to its fullest so the water vigorously boils, or
should you turn down the burner so the water barely
boils?
Turn down the heat, because the least
amount of energy is expended when the
water barely boils.
59
HEAT SUPPLIED OR REMOVED IN CHANGING THE
PHASE OF A SUBSTANCE
The heat Q that must be supplied or removed to change the
phase of a mass m of a substance is
where L is the latent heat of the substance.
SI Unit of Latent Heat: J/kg
The latent heat of fusion Lf refers to the change between solid
and liquid phases, the latent heat of vaporization Lv applies to
the change between liquid and gas phases, and the latent heat
of sublimation Ls refers to the change between solid and gas
phases.
60
Substance
Melting Point
(°C)
Latent Heat of Fusion,
Lf (J/kg)
Boiling Point
(°C)
Latent Heat of Vaporization,
Lv (J/kg)
Ammonia
–77.8
33.2 × 104
–33.4
13.7 × 105
Benzene
5.5
12.6 × 104
80.1
3.94 × 105
Copper
1083
20.7 × 104
2566
47.3 × 105
Ethyl
alcohol
–114.4
10.8 × 104
78.3
8.55 × 105
Gold
1063
6.28 × 104
2808
17.2 × 105
Lead
327.3
2.32 × 104
1750
8.59 × 105
Mercury
–38.9
1.14 × 104
356.6
2.96 × 105
Nitrogen
–210.0
2.57 × 104
–195.8
2.00 × 105
Oxygen
–218.8
1.39 × 104
–183.0
2.13 × 105
Water
0.0
33.5 × 104
100.0
22.6 × 105
61
Example 13. Ice-cold Lemonade
Ice at 0 °C is placed in a Styrofoam cup containing 0.32
kg of lemonade at 27 °C. The specific heat capacity of
lemonade is virtually the same as that of water; that is, c
= 4186 J/(kg·C°). After the ice and lemonade reach an
equilibrium temperature, some ice still remains. The
latent heat of fusion for water is Lf = 3.35 × 105 J/kg.
Assume that the mass of the cup is so small that it
absorbs a negligible amount of heat, and ignore any heat
lost to the surroundings. Determine the mass of ice that
has melted.
62
Q = cm T
Q = mLf
mL 
f ice

 cm T
Heat gained by ice

lemonade
Heat lost by lemonade
63
Example 14. Getting Ready for a Party
A 7.00-kg glass bowl [c = 840 J/(kg·C°)] contains 16.0 kg
of punch at 25.0 °C. Two-and-a-half kilograms of ice [c =
2.00 × 103 J/(kg·C°)] are added to the punch. The ice
has an initial temperature of –20.0 °C, having been kept
in a very cold freezer. The punch may be treated as if it
were water [c = 4186 J/(kg·C°)], and it may be assumed
that there is no heat flow between the punch bowl and the
external environment. The latent heat of fusion for water
is 3.35 × 105 J/kg. When thermal equilibrium is reached,
all the ice has melted, and the final temperature of the
mixture is above 0 °C. Determine this temperature.
64
(a)
(b)
(c)
65
(d)
(e)
.
66
A dye-sublimation printer.
As the plastic film passes in
front of the print head, the
heat from a given heating
element causes one of three
pigments or dyes on the
film to sublime from a solid
to a gas. The gaseous dye is
absorbed onto the coated
paper as a dot of color. The
size of the dots on the paper
has been exaggerated for
clarity.
67
Check Your Understanding 4
When ice cubes are used to cool a drink, both their mass
and temperature are important in how effective they are.
The table below lists several possibilities for the mass and
temperature of the ice cubes used to cool a particular
drink. Rank the possibilities in descending order (best
first), according to their cooling effectiveness.
Mass of ice Temperature of
cubes
ice cubes
c, a, b
(a)
m
–6.0 °C
(b)
½m
–12 °C
(c)
2m
–3.0 °C
68
Equilibrium Between Phases of Matter
The pressure of the vapor that coexists in equilibrium with the
liquid is called the equilibrium vapor pressure of the liquid. 69
The equilibrium vapor pressure does not depend on the
volume of space above the liquid. Only when the
temperature and vapor pressure correspond to a point on
the curved line, which is called the vapor pressure curve or
the vaporization curve, do liquid and vapor phases coexist
70
in equilibrium.
Conceptual Example 15.
How to Boil Water That Is Cooling Down
71
Water is boiling in an open flask. Shortly after the
flask is removed from the burner, the boiling stops.
A cork is then placed in the neck of the flask to seal
it. To restart the boiling, should you pour hot (but
not boiling) water or cold water over the neck of
the flask, as in part b of the drawing?
It is possible to restart the boiling by pouring
cold water over the neck of the flask.
72
The operation of spray cans is
based on the equilibrium
between a liquid and its vapor.
73
As is the case for
liquid/vapor equilibrium, a
solid can be in equilibrium
with its liquid phase only at
specific conditions of
temperature and pressure.
For each temperature,
there is a single pressure at
which the two phases can
coexist in equilibrium. A
plot of the equilibrium
pressure versus
equilibrium temperature is
referred to as the fusion
curve.
74
Humidity
The partial pressure of a gas is the pressure it would exert
if it alone occupied the entire volume at the same
temperature as the mixture.
When the partial pressure of the water vapor equals the equilibrium
vapor pressure of water at a given temperature, the relative
humidity is 100%. In such a situation, the vapor is said to be
saturated because it is present in the maximum amount, as it would
be above a pool of liquid at equilibrium in a closed container.
75
Example 16. Relative Humidities
One day, the partial
pressure of water vapor in
the air is 2.0 × 103 Pa. Using
the vaporization curve for
water, determine the relative
humidity if the temperature
is (a) 32 °C and (b) 21 °C.
76
(a)
(b)
77
When air containing a
given amount of water
vapor is cooled, a
temperature is reached
in which the partial
pressure of the vapor
equals the equilibrium
vapor pressure. This
temperature is known as
the dew point.
78
79
Concepts & Calculations Example 17.
Linear and Volume Thermal Expansion
Three rectangular blocks are made from the same
material. The initial dimensions of each are
expressed as multiples of D, where D = 2.00 cm.
They are heated and their temperatures increase by
35.0 C°.
80
The coefficients of linear and volume expansion are
and
respectively. Determine the change in their (a) vertical heights
and (b) volumes.
81
(b)
82
Concepts & Calculations Example 18.
Heat and Temperature Changes
83
Objects A and B are made from copper, but the mass of
B is three times that of A. Object C is made from glass
and has the same mass as B. The same amount of heat Q
is supplied to each one: Q = 14 J. Determine the rise in
temperature for each.
84
Conceptual Question 3
REASONING AND SOLUTION The plate is made of
aluminum; the spherical ball is made of brass. The
coefficient of linear expansion of aluminum is greater than
the coefficient of linear expansion of brass. Therefore, if the
plate and the ball are heated, both will expand; however,
the diameter of the hole in the aluminum plate will expand
more than the diameter of the brass ball. In order to
prevent the ball from falling through the hole, the plate and
the ball must be cooled. Both the diameter of the hole in the
plate and the diameter of the ball will contract. The
diameter of the hole will decrease more than the diameter of
the ball, thereby preventing the ball from falling through
the hole.
85
Problem 12
REASONING AND SOLUTION
a. The radius of the hole will be larger when the plate is
heated, because the hole expands as if it were made of copper.

b. The expansion of the radius is  r =
r0 T . Using
the value for the coefficient of thermal expansion of copper
given in Table 12.1, we find that the fractional change in
the radius is
r/r0 = T = (17 * 10–6 C°–1)(110 °C  11 °C)
 0.0017
86
Problem 18
REASONING AND SOLUTION The initial diameter of the
sphere, ds, is
ds = (5.0 * 10–4)dr + dr
where dr is the initial diameter of the ring. Applying
to the diameter of the sphere gives
L  L0 T
d s   s d s T
and to the ring gives
dr   r dr T
87
If the sphere is just to fit inside the ring, we must have
d s  d s  d r  d r
T 
d r  ds
 sd s   r d r
Substituting Equation (1) in this result and taking values for the
coefficients of thermal expansion of steel and lead from Table 10.1
yield
4
T 


5.0  10

29  10 6 C 1 5.0  10 4 1  12 10 6 C 1
Tf = 70.0 °C - 29 C° =
 29 C
41 C
88
Problem 30
REASONING AND SOLUTION Both the water and pipe
expand as the temperature increases.
VP  CV0 T
VW  W V0 T
The initial volume of the pipe and water is
V0  r L
2
V  VW  VP  ( W  C )V0 T
V  (207 10 C  5110 C ) (9.5 10 m) (76m)(54C)
6
1
4
 1.8 10 m
6
3
1
3
2
89
Problem 32
REASONING AND SOLUTION Both the coffee and
beaker expand as the temperature increases.
VC  WV0 T
Vb  bV0 T
V  VC  Vb  (  w   b )V0 T
Taking the coefficients of volumetric expansion  w and c
for coffee (water) and glass (Pyrex) from Table 12.1, we find

6
1
6
1

3
V  207 10 (C)  9.9 10 (C) (0.5 10 m )(92C  18C )
=7.3*10-6m3
3
90
Problem 46
REASONING AND SOLUTION We wish to convert
2.0% of the heat Q into gravitational potential energy,
i.e., (0.020)Q = mgh. Thus
mg 
0.020 Q
h

 4186 J 
 0.020 110 Calories 

1 Calorie 
2.1 m
 4.4  10 3 N
91