This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ctivity kss of stainless steel from Table 13.1, we find that the
temperature difference ∆Tss between the stainless steel surfaces is
∆Tss ( mLv ) L = ( 0.15 kg ) ( 22.6 ×105 J/kg ) (1.4 ×10−3 m ) = 9.4 C°
=
kss At ( ) 14 J/ ( s ⋅ m ⋅ C° ) 0.015 m 2 ( 240 s ) 700 THE TRANSFER OF HEAT The temperature at the steelburner interface is T = 101.2 °C + ∆Tss = 110.6 °C . 18. REASONING If the cylindrical rod were made of solid copper, the amount of heat it would
conduct in a time t is, according to Equation 13.1, Qcopper = (kcopper A2 ∆T / L)t . Similarly,
the amount of heat conducted by the leadcopper combination is the sum of the heat
conducted through the copper portion of the rod and the heat conducted through the lead
portion:
Qcombination = kcopper ( A2 − A1 ) ∆T / L + klead A1∆T / L t . Since the leadcopper combination conducts onehalf the amount of heat than does the solid
copper rod, Qcombination = 1 Qcopper , or
2 kcopper ( A2 − A1 ) ∆T
L + klead A1∆T
L 1 kcopper A2 ∆T = L
2 This expression can be solved for A1 / A2 , the ratio of the crosssectional areas. Since the
crosssectional area of a cylinder is circular, A = π r 2 . Thus, once the ratio of the areas is
known, the ratio of the radii can be determined. SOLUTION Solving for the ratio of the areas, we have kcopper
A1
=
A2 2 ( kcopper − klead )
The crosssectional areas are circular so that A1 / A2 = (π r12 ) /(π r22 ) = (r1 / r2 )2 ; therefore, r1
r2 = kcopper
2(kcopper − klead ) = 390 J/(s ⋅ m ⋅ C°)
= 0.74
2[390 J/(s ⋅ m ⋅ C°) − 35 J/(s ⋅ m ⋅ C°)] where we have taken the thermal conductivities of copper and lead from Table 13.1. Chapter 13 Problems 701 19. SSM WWW REASONING The rate at which heat is conducted along either rod is
given by Equation 13.1, Q / t = ( k A ∆T ) / L . Since both rods conduct the same amount of
heat per second, we have ks As ∆T
Ls = ki Ai ∆T (1) Li Since the same temperature difference is maintained across both rods, we can algebraically
cancel the ∆T terms. Because both rods have the same mass, ms = mi ; in terms of the
densities of silver and iron, the statement about the equality of the masses becomes
ρs ( Ls As ) = ρi ( Li Ai ) , or
As
Ai = ρi Li
ρs Ls (2) Equations (1) and (2) may be combined to find the ratio of the lengths of the rods. Once the
ratio of the lengths is known, Equation (2) can be used to find the ratio of the crosssectional
areas of the rods. If we assume that the rods have circular cross sections, then each has an
area of A = π r 2 . Hence, the ratio of the crosssectional areas can be used to find the ratio of
the radii of the rods.
SOLUTION
a. Solving Equation (1) for the ratio of the lengths and substituting the right hand side of
Equation (2) for the ratio of the areas, we have Ls
Li = ks As
ki Ai = ks ( ρi Li ) ki ( ρs Ls ) 2 or Ls ks ρi =
L ki ρs i Solving for the ratio of the lengths, we have
Ls
Li = ks ρi ki ρs = [420 J/(s ⋅ m ⋅ C°)](7860 kg/m3 )
= 2.0
[79 J/(s ⋅ m ⋅ C°)](10 500 kg/m3 ) b. From Equation (2) we have π rs2 ρL
= ii
2
ρs Ls
π ri Solving for the ratio of the radii, we have 2 or rs ρi Li =
r ρs Ls i 702 THE TRANSFER OF HEAT rs
ri = ρi Li 7860 kg/m3 1 = = 0.61 ρs Ls 10 500 kg/m3 2.0 20. REASONING According to Equation 6.10b, power P is the change in energy Q divided by
the time t during which the change occurs, or P = Q/t. The power radiated by a filament is
given by the StefanBoltzmann law as
P= Q
= eσ T 4 A
t (13.2) where e is the emissivity, σ is the StefanBoltzmann constant, T is the temperature (in
kelvins), and A is the surface area. This expression will be used to find the ratio of the
filament areas of the bulbs. SOLUTION Solving Equation (13.2) for the area, we have
A= P
eσ T 4 Taking the ratio of the areas gives P
1 A1
e1 σ T14
=
P2
A2
e2 σ T24
Setting e2 = e1, and P2 = P1, we have that
P
1
4
e1 σ T14
A1
T24 ( 2100 K )
=
= 4=
= 0.37
A2
P
T1
( 2700 K )4
1
e1 σ T24 21. SSM WWW REASONING AND SOLUTION Solving the StefanBoltzmann law,
Equation 13.2, for the time t, and using the fact that Qblackbody = Qbulb , we have
tblackbody = Qblackbody σ T4A = Qbulb σ T4A = Pbulb tbulb σ T4A Chapter 13 Problems 703 where Pbulb is the power rating of the light bulb. Therefore, tblackbody = 5.67 ×10 –8 (100.0 J/s) (3600 s)
J/(s ⋅ m ⋅ K ) (303 K) 4 (6 sides)(0.0100 m)2 / side 2 4 1 h 1 d × = 14.5 d 3600 s 24 h 22. REASONING According to the StefanBoltzmann law, the radiant power emitted by the
Q
“radiator” is
= e σ T 4 A (Equation 13.2), where Q is the energy radiated in a time t, e is
t
the emissivity of the surface, σ is the StefanBoltzmann constant, T is the temperature in
Kelvins, and A is the area of the surface from which the radiant energy is emitted. We will
apply this law to the “radiator” before and after it is painted. In either case, the same radiant
power is emitted.
SOLUTION Applying the StefanBoltzmann law, we obtain the following: Q
4
= eafter σ Tafter A t after and Q
4
= ebefore...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details