Unformatted text preview: at a rate that is given by
(Equation 13.1), where the bottom of
t
L
a pot has a thermal conductivity k, a crosssectional area A, and a thickness L. The thermal
conductivities of copper and aluminum (kCu, kAl) are different (see Table 13.1), but the two
pot bottoms are identical in every other respect. Further, because both pots are boiling away
Q
at the same rate, the flow of heat must occur at the same rate
through both bottoms. We
t
will use Equation 13.1 and the temperature TAl of the heating element under the aluminum 716 THE TRANSFER OF HEAT bottomed pot to determine the temperature TCu of the heating element under the copperbottomed pot.
Q
of heat flow, as well as equal crosst
Q kA ∆T
=
sectional areas A and thicknesses L. Therefore,
(Equation 13.1) yields
t
L SOLUTION Both pot bottoms have identical rates Q kCu A ( ∆T )Cu kAl A ( ∆T )Al
=
=
t
L
L kCu ( ∆T )Cu = kAl ( ∆T )Al or (1) Solving Equation (1) for ( ∆T )Cu yields ( ∆T )Cu = kAl ( ∆T )Al
kCu (2) Substituting ( ∆T )Cu = TCu − T0 into Equation (2) and solving for TCu, we obtain TCu − T0 = kAl ( ∆T )Al
kCu or TCu = kAl ( ∆T )Al
+ T0
kCu Therefore, the temperature of the heating element underneath the copperbottomed pot is TCu ( )(
( ) 240 J/ s ⋅ m ⋅ Co 155.0 oC − 100.0 oC =
+ 100.0 oC = 134 oC
o
390 J/ s ⋅ m ⋅ C ) 43. SSM WWW REASONING Heat flows along the rods via conduction, so that Equation
( k A∆T ) t , where Q is the amount of heat that flows in a time t, k is the
13.1 applies: Q =
L
thermal conductivity of the material from which a rod is made, A is the crosssectional area
of the rod, and ∆T is the difference in temperature between the ends of a rod. In arrangement
a, this expression applies to each rod and ∆T has the same value of ∆T = TW − TC . The total
heat Q ′ is the sum of the heats through each rod. In arrangement b, the situation is more
complicated. We will use the fact that the same heat flows through each rod to determine the
temperature at the interface between the rods and then use this temperature to determine ∆T
and the heat flow through either rod. Chapter 13 Problems 717 SOLUTION For arrangement a, we apply Equation 13.1 to each rod and obtain for the total
heat that
k A ( TW − TC ) t k2 A ( TW − TC ) t ( k1 + k2 ) A ( TW − TC ) t
Q′ = Q1 + Q2 = 1
+
=
(1)
L
L
L
For arrangement b, we use T to denote the temperature at the interface between the rods and
note that the same heat flows through each rod. Thus, using Equation 13.1 to express the
heat flowing in each rod, we have k1 A ( TW − T ) t
144L
244
3 = k2 A ( T − TC ) t
144L
244
3 Heat flowing
through rod 1 or k1 ( TW − T ) = k2 ( T − TC ) Heat flowing
through rod 2 Solving this expression for the temperature T gives T= k1TW + k2TC
k1 + k2 (2) Applying Equation 13.1 to either rod in arrangement b and using Equation (2) for the
interface temperature, we can determine the heat Q that is flowing. Choosing rod 2, we find
that kT +k T k2 A 1 W 2 C − TC t k +k k A ( T − TC ) t
1
2 Q= 2
=
L
L k T −k T k2 A 1 W 1 C t k +k 1
2 = k2 Ak1 ( TW − TC ) t
=
L
L ( k1 + k2 )
Using Equations (1) and (3), we obtain for the desired ratio that Q′
=
Q ( k1 + k2 ) A ( TW − TC ) t
L
k2 Ak1 ( TW − TC ) t
L ( k1 + k2 ) ( k1 + k2 ) A ( TW − TC ) t L ( k1 + k2 ) ( k1 + k2 )2
=
=
k2 k1
L k2 A k1 ( TW − TC ) t Using the fact that k2 = 2k1 , we obtain (3) 718 THE TRANSFER OF HEAT ( k + 2k1 ) = 4.5
Q′ ( k1 + k2 )
=
=1
Q
k2 k1
2k1 k1
2 2 44. REASONING The drawing shows a crosssectional view of the small sphere inside the larger
spherical asbestos shell. The small sphere produces
a net radiant energy, because its temperature
(800.0 °C) is greater than that of its environment
(600.0 °C). This energy is then conducted through
the thin asbestos shell (thickness = L). By setting
the net radiant energy produced by the small
sphere equal to the energy conducted through the
asbestos shell, we will be able to obtain the
temperature T2 of the outer surface of the shell. T2
600.0 °C
L r2 r1
800.0 °C SOLUTION The heat Q conducted during a time through the thin asbestos shell is given
A ∆T ) t
(k
by Equation 13.1 as Q = asbestos 2
, where kasbestos is the thermal conductivity of
L
2
asbestos (see Table 13.1), A2 is the surface area of the spherical shell A2 = 4π r2 , ∆T is the ( ) temperature difference between the inner and outer surfaces of the shell
(∆T = 600.0 °C − T2), and L is the thickness of the shell. Solving this equation for the T2
yields
QL
T2 = 600.0 °C −
kasbestos 4π r22 t ( ) The heat Q is produced by the net radiant energy generated by the small sphere inside the
asbestos shell. According to Equation 13.3, the net radiant energy is ( ) Q = Pnet t = eσ A1 T 4 − T04 t , where e is the emissivity, σ is the StefanBoltzmann constant,
A1 is the surface area of the sphere ( A1 = 4π r12 ) , T is the temperature of the sphere (T = 800.0 °C = 1073.2 K) and T0 is the temperature of the environment that surrounds the
sphere (T0 = 600.0 °C = 873.2 K). Substit...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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