Transcript Ch13 The Transfer of Heat

Ch13. The Transfer of Heat
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Convection
Convection is the process in which heat is carried from
place to place by the bulk movement of a fluid.
Convection currents are set up when a pan of water is
heated.
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Conceptual Example 1. Hot Water
Baseboard Heating and Refrigerators
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Hot water baseboard heating units are frequently used in
homes, where they are mounted on the wall next to the
floor. In contrast, the cooling coil in a refrigerator is
mounted near the top of the refrigerator. The locations for
these heating and cooling devices are different, yet each
location is designed to maximize the production of
convection currents. Explain how.
(a) The air above the baseboard unit is heated, like the air
above a fire. Buoyant forces from the surrounding cooler
air push the warm air upward. Cooler air near the ceiling is
displaced downward and then warmed by the baseboard
heating unit, leading to the convection current. Had the
heating unit been located near the ceiling, the warm air
would have remained there, with very little convection to
distribute the heat.
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(b) The air in contact with the top-mounted coil is cooled,
its volume decreases, and its density increases. The
surrounding warmer and less dense air cannot provide
sufficient buoyant force to support the colder air, which
sinks downward. In the process, warmer air near the
bottom is displaced upward and is then cooled by the coil,
establishing the convection current. Had the cooling coil
been placed at the bottom of the refrigerator, stagnant,
cool air would have collected there, with little convection to
carry the heat from other parts of the refrigerator to the
coil for removal.
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Updrafts, or thermals, are caused by the convective
movement of air that the ground has warmed .
Sometimes, meteorological conditions cause a layer to
form in the atmosphere where the temperature increases
with increasing altitude. Such a layer is called an
inversion layer.
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Natural convection, in
which a temperature
difference causes the
density at one place in
a fluid to be different
from that at another.
The forced convection generated by a pump circulates
radiator fluid through an automobile engine to
remove excess heat.
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Conduction
Conduction is the process whereby heat is transferred
directly through a material, with any bulk motion of the
material playing no role in the transfer.
Those materials that conduct heat well are called thermal
conductors, and those that conduct heat poorly are
known as thermal insulators.
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The amount of heat Q conducted through the bar from the
warmer end to the cooler end depends on a number of
factors:
1. Q is proportional to the time t during which conduction
takes place (Q  t).
2. 2. Q is proportional to the temperature difference T
between the ends of the bar (Q   T).
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3. Q is proportional to the cross-sectional area A of the
bar (Q  A).
4. Q is inversely proportional to the length L of the bar
(Q  1/L).
Q
(A T)t/L.
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CONDUCTION OF HEAT THROUGH A MATERIAL
The heat Q conducted during a time t through a bar of
length L and cross-sectional area A is
(kA T )t
Q
L
where  T is the temperature difference between the ends
of the bar and k is the thermal conductivity of the
material.
SI Unit of Thermal Conductivity: J/(s·m·C°)
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Since k = QL/(tA T), the SI unit for thermal
conductivity is J·m/(s·m2·C°) or J/(s·m·C°). The
SI unit of power is the joule per second (J/s) or watt
(W), so the thermal conductivity is also given in units
of W/(m·C°).
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Substance
Thermal Conductivity, k
[J/(s·m·C°)]
Metals
Aluminum
240
Brass
110
Copper
390
Iron
79
Lead
35
Silver
420
Steel (stainless)
14
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Other Materials
Gases
Air
Hydrogen (H2)
Asbestos
0.090
Body fat
0.20
Concrete
1.1
Diamond
2450
Glass
0.80
Goose down
0.025
Ice (0 °C)
2.2
Styrofoam
0.010
Water
0.60
Wood (oak)
0.15
0.0256
0.180
Nitrogen (N2)
0.0258
Oxygen (O2)
0.0265
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Styrofoam is an excellent
thermal insulator because it
contains many small, deadair spaces. These small
spaces inhibit heat transfer
by convection currents, and
air itself has a very low
thermal conductivity.
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Example 2.
Heat Transfer in the Human Body
When excessive heat is produced within the body, it must be
transferred to the skin and dispersed if the temperature at the
body interior is to be maintained at the normal value of 37.0
°C. One possible mechanism for transfer is conduction
through body fat. Suppose that heat travels through 0.030 m of
fat in reaching the skin, which has a total surface area of 1.7
m2 and a temperature of 34.0 °C. Find the amount of heat
that reaches the skin in half an hour (1800 s).
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Example 3. Layered Insulation
One wall of a house consists
of 0.019-m-thick plywood
backed by 0.076-m-thick
insulation. The temperature
at the inside surface is 25.0
°C, while the temperature
at the outside surface is 4.0
°C, both being constant.
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The thermal conductivities of the insulation and the
plywood are, respectively, 0.030 and 0.080 J/(s·m·C°),
and the area of the wall is 35 m2. Find the heat
conducted through the wall in one hour (a) with the
insulation and (b) without the insulation.
(a)
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(b)
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Conceptual Example 4.
An Iced-up Refrigerator
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In a refrigerator, heat is removed by a cold refrigerant
fluid that circulates within a tubular space embedded
within a metal plate. A good refrigerator cools food as
quickly as possible. Decide whether the plate should be
made from aluminum or stainless steel and whether the
arrangement works better or worse when it becomes
coated with a layer of ice.
The plate should be made from aluminum .
When covered with ice, the cooling plate
works less well.
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Check Your Understanding 1
Two bars are placed between plates whose temperatures are Thot
and Tcold (see the drawing). The thermal conductivity of bar 1 is
six times that of bar 2 (k1 = 6k2), but bar 1 has only one-third the
cross-sectional area (
). Ignore any heat loss through
the sides of the bars. Which statement below correctly describes
the heat conducted by the bars in a given amount of time?
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a. Bar 1 conducts 1/4 the heat as does bar 2;
b. Bar 1 conducts 1/8 the heat as does bar 2;
c. Bar 1 conducts twice the heat as does bar 2;
d. Bar 1 conducts four times the heat as does bar 2;
e. Both bars conduct the same amount of heat;
c
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Radiation
Radiation is the process in which energy is transferred by
means of electromagnetic waves.
In the transfer of energy by radiation, the absorption
of electromagnetic waves is just as important as their
emission. The surface of an object plays a significant
role in determining how much radiant energy the
object will absorb or emit.
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Since the color black is
associated with nearly
complete absorption of
visible light, the term
perfect blackbody or,
simply, blackbody is used
when referring to an
object that absorbs all the
electromagnetic waves
falling on it.
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Since absorption and emission are balanced, a material that
is a good absorber, like lampblack, is also a good emitter, and
a material that is a poor absorber, like polished silver, is also
a poor emitter.
THE STEFAN-BOLTZMANN LAW OF RADIATION
The radiant energy Q, emitted in a time t by an object that
has a Kelvin temperature T, a surface area A, and an
emissivity e, is given by

where
is the Stefan-Boltzmann constant and has a
value of 5.67 × 10–8 J/(s·m2·K4).
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Example 5. A Supergiant Star
The supergiant star Betelgeuse has a surface temperature of
about 2900 K (about one-half that of our sun) and emits a
radiant power (in joules per second, or watts) of approximately 4
× 1030 W (about 10 000 times as great as that of our sun).
Assuming that Betelgeuse is a perfect emitter (emissivity e = 1)
and spherical, find its radius.
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Example 6. A Wood-Burning Stove
A wood-burning stove stands unused in a room where the
temperature is 18 °C (291 K). A fire is started inside the stove.
Eventually, the temperature of the stove surface reaches a
constant 198 °C (471 K), and the room warms to a constant 29
°C (302 K). The stove has an emissivity of 0.900 and a surface
area of 3.50 m2. Determine the net radiant power generated by
the stove when the stove (a) is unheated and has a temperature
equal to room temperature and (b) has a temperature of 198 °C.
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(a)
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(b)
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Check Your Understanding 2
Two identical cubes have the same temperature. One of them,
however, is cut in two and the pieces are separated (see the
drawing). What is true about the radiant energy emitted in a
given time?
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a.The cube cut into two pieces emits twice as much radiant
energy as does the uncut cube;
b.The cube cut into two pieces emits more radiant energy than
does the uncut cube according to
c.The cube cut into two pieces emits the same amount of radiant
energy as does the uncut cube;
d.The cube cut into two pieces emits one-half the radiant energy
emitted by the uncut cube;
e.The cube cut into two pieces emits less radiant energy than does
the uncut cube according to
(b)
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Applications
Insulation inhibits convection between inner and outer walls
and minimizes heat transfer by conduction. With respect to
conduction, the logic behind home insulation ratings comes
directly from Equation
Q AT

t
L/k
The term L/k in the denominator is called the R value of
the insulation. Larger R values reduce the heat per unit
time flowing through the material and, therefore, mean
better insulation.
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A thermos bottle minimizes energy transfer due to
convection, conduction, and radiation.
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In a halogen cooktop, quartz-iodine lamps emit a large
amount of electromagnetic energy that is absorbed
directly by a pot or pan.
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Concepts & Calculations
Example 7. Boiling Water
Two pots are identical, except that in one case the flat
bottom is aluminum and in the other it is copper. Each pot
contains the same amount of boiling water and sits on a
heating element that has a temperature of 155 °C. In the
aluminum-bottom pot, the water boils away completely in
360 s. How long does it take the water in the copper-bottom
pot to boil away completely?
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Concepts & Calculations
Example 8. Freezing Water
One half of a kilogram of liquid water at 273 K (0 °C) is
placed outside on a day when the temperature is 261 K (–
12 °C). Assume that heat is lost from the water only by
means of radiation and that the emissivity of the radiating
surface is 0.60. How long does it take for the water to freeze
into ice at 0 °C when the surface area from which the
radiation occurs is (a) 0.035 m2 (as it could be in a cup) and
(b) 1.5 m2 (as it could be if the water were spilled out to
form a thin sheet)?
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(a) Smaller area
(b) Larger area
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Problem 1
REASONING AND SOLUTION According to Equation
13.1, the heat per second lost is
Q k A T [0.040 J/(sm  C o )] (1.6 m2 )(25 Co )
2



8.0

10
J/s
–3
t
L
2.0 10 m
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Problem 8
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REASONING To find the total heat conducted, we will apply Equation
13.1 to the steel portion and the iron portion of the rod. In so doing, we
use the area of a square for the cross section of the steel. The area of the
iron is the area of the circle minus the area of the square. The radius of
the circle is one half the length of the diagonal of the square.
SOLUTION In preparation for applying Equation 13.1, we need the area of
the steel and the area of the iron. For the steel, the area is simply ASteel = L2,
where L is the length of a side of the square. For the iron, the area is
AIron = R2 – L2. To find the radius R, we use the Pythagorean theorem,
which indicates that the length D of the diagonal is related to the length of
the sides according to D2 = L2 + L2. Therefore, the radius of the circle
is R  D / 2  2 L / 2. For the iron, then, the area is
2
AIron
 2L 



2
  L    1 L2
 R  L   

2
2 


2
2
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Taking values for the thermal conductivities of steel and iron from
Table 13.1 and applying Equation 13.1, we find
QTotal  QSteel  QIron

 kAT t 
 kAT t 
   2  T t
2


 ksteel L  k Iron   1 L 


 L  Steel  L  Iron 
2   L

J
J


  
2
2
 14
0.01m   79
  10.01m 
 s  m  C  2 
 s  m  C 


78C  18C 120 s 

 85 J
0.5m
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Problem 22
REASONING AND SOLUTION
a. The radiant power lost by the body is
PL = e T4A = (0.80)[5.67 *10–8 J/(sm2K4)](307 K)4(1.5 m2) = 604 W
The radiant power gained by the body from the room is
Pg = (0.80)[5.67 * 10–8 J/(sm2K4)](298 K)4(1.5 m2) = 537 W
The net loss of radiant power is P = PL - Pg =
67 W
b. The net energy lost by the body is
 1 Calorie 
Q  Pt  (67 W)(3600 s) 
  58 Calories
 4186 J 
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Problem 28
REASONING AND SOLUTION The rate at which energy is
gained through the refrigerator walls is
Q kAT 0.03J /( s  m  C )(5.3m 2 )(2.0 101 C )


 42J / s
t
L
0.075m
Therefore, the amount of heat per second that must be
removed from the unit to keep it cool is 42 J / s
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Problem 34
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REASONING AND SOLUTION Using Equation 13.1,
Q  ( kA T ) t / L
 Q  kT
 
L
 At 
Before Equation (1) can be applied to the ice-aluminum combination, the
temperature T at the interface must be determined. We find the
temperature at the interface by noting that the heat conducted through
the ice must be equal to the heat conducted through the aluminum: Qice
= Qaluminum. Applying Equation 13.1 to this condition, we have
 kATt 
 kATt 

 

 L ice  L  alu minum
2.2 J / (s  m  C ) A ( 10.0  C)  T t
0.0050 m

240 J / (s  m  C ) A T  ( 25.0  C) t
0.0015 m
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The factors A and t can be eliminated algebraically. Solving for T
gives T = –24.959 °C for the temperature at the interface
a. Applying Equation (1) to the ice leads to
QQI  [ 2.2 J / (s  m  C )] ( 10.0  C)  ( 24.959  C )
F
 J 
G
HAtAtK
0.0050 m
 6.58  10 3 J / (s  m 2 )
iceice
Since heat is not building up in the materials, the rate of heat transfer per
unit area is the same throughout the ice-aluminum combination. Thus,
this must be the heat per second per square meter that is conducted
through the ice-aluminum combination.
b. Applying Equation (1) to the aluminum in the absence of any ice
gives:
QQI  [240 J / (s  m  C )] ( 10.0  C)  ( 25.0  C)
F
 J 
G
H
0.0015 m
AtAtK

 2.40  10 6 J / (s  m 2 )
Al Al
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