Ch13. The Transfer of Heat

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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
• Q is proportional to the time t during which
  conduction takes place (Q t).
• 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 DT)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
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/(smC)
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k QL/(tA DT),
Since the SI
unit for thermal conductivity is Jm/(sm2C) or
J/(smC). 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/(mC).
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Substance   Thermal Conductivity, k J/(smC) 
 Metals 
   Aluminum   240 
   Brass   110 
   Copper   390 
   Iron    79 
   Lead    35 
   Silver   420 
   Steel (stainless)    14 
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Other Materials 
   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 
Gases 
   Air   0.0256 
   Hydrogen (H2)   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, dead-air 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/(smC), 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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1. Bar 1 conducts 1/4 the heat as does bar 2  
2. Bar 1 conducts 1/8 the heat as does bar 2   
3. Bar 1 conducts twice the heat as does bar 2   
4. Bar 1 conducts four times the heat as does bar 2
     
5. 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 5.67 108 J/(sm2K4).
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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
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
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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
. For the iron, then, the
area is
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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
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Problem 22
REASONING AND SOLUTION
a. The radiant power lost by the body is
PL e T4A (0.80)5.67 108 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)5.67 108 J/(s?m2?K4)(298 K)4(1.5
m2) 537 W
The net loss of radiant power is P PL - Pg
b. The net energy lost by the body is
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Problem 28
REASONING AND SOLUTION The rate at which energy
is gained through the refrigerator walls is
Therefore, the amount of heat per second that
must be removed from the unit to keep it cool is
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Problem 34
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REASONING AND SOLUTION Using Equation 13.1,
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
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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
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