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Unformatted text preview: σ A ( T 4 − T04 ) . Solving for T gives FP
I
T = G +T J
e
Hσ A K
R
=S
(0.900)[5.67 × 10
T
1/ 4 net 4
0 7300 W
+ (302 K) 4
–8
J / (s ⋅ m 2 ⋅ K 4 )](2.00 m 2 ) U=
V
W
1/4 532 K 28. REASONING According to the StefanBoltzmann law, the power radiated by an object is
Q/t = eσT A (Equation 13.2), where e is the emissivity of the object, σ is the
StefanBoltzmann constant, T is the temperature (in kelvins) of the object, and A is the
2
surface area of the object. The surface area of a sphere is A = 4πR , where R is the radius.
Substituting this expression for A into Equation 13.2, and solving for the radius yields
4 Q
t
R=
4π e σ T 4 (1) This expression will be used to find the radius of Sirius B. SOLUTION Writing Equation (1) for both stars, we have RSirius Q t Sirius
=
4
4π eSirius σ TSirius and RSun Q t Sun
=
4
4π eSun σ TSun Chapter 13 Problems 707 Dividing RSirius by RSun and remembering that (Q/t)Siuris = 0.040(Q/t)Sun and eSirius = eSun,
we obtain RSirius
=
RSun Q t Sirius
4
4π eSirius σ TSirius = Q t Sun
4
4π eSun σ TSun Q
0.040 t Sun
4
4π eSun σ TSirius
Q t Sun
4
4π eSun σ TSun 4
0.040TSun = 4
TSirius Solving for the radius of Sirius B, and noting that TSirius = 4TSun, gives
2 RSirius T T
= 0.040 Sun RSun = 0.040 Sun
T 4T Sirius Sun 2 ( 6.96 ×108 m ) = 8.7 × 106 m 29. SSM REASONING AND SOLUTION The power radiated per square meter by the car
when it has reached a temperature T is given by the StefanBoltzmann law, Equation 13.2,
Pradiated / A = eσ T 4 , where Pradiated = Q / t . Solving for T we have
1/ 4 / A) (P
T = radiated eσ 1/4 560 W/m 2 =
–8
2
4 (1.00) 5.67 ×10 J/(s ⋅ m ⋅ K ) = 320 K 30. REASONING AND SOLUTION
a. The radiant power lost by the body is
PL = eσ T 4A = (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 2 The net loss of radiant power is P = PL − Pg =
b. The net energy lost by the body is 4 4 2 J/(s⋅m ⋅K )](298 K) (1.5 m ) = 537 W
67 W 708 THE TRANSFER OF HEAT 1 Calorie Q = P t = (67 W)(3600 s) = 4186 J 58 Calories 31. REASONING The liquid helium is at its boiling point, so its temperature does not rise as it
absorbs heat from the radiating shield. Instead, the net heat Q absorbed in a time t converts a
mass m of liquid helium into helium gas, according to Q = mLv (Equation 12.5), where
Lv = 2.1×104 J/kg is the latent heat of vaporization of helium. The ratio of the net heat Q
absorbed by the helium to the elapsed time t is equal to the net power Pnet absorbed by
Q
helium: Pnet = (Equation 6.10b). The net power absorbed by the helium depends upon the
t
temperature T = 77 K maintained by the radiating shield and the temperature T0 = 4.2 K of ( the boiling helium, as we see from Pnet = eσ A T 4 − T04 ) (Equation 13.3), where σ = 5.67×10−8 J/(s·m2·K4) is the StefanBoltzmann constant, e = 1 is the emissivity of the
container (a perfect blackbody radiator), and A is the surface area of the container. Because
the container is a sphere of radius R, its surface area is given by A = 4πR2. Therefore, the net
power absorbed by the helium can be expressed as ( ) ( Pnet = eσ A T 4 − T04 = 4π R 2eσ T 4 − T04 ) (1) SOLUTION Solving Q = mLv (Equation 12.5) for m, we obtain m= Solving Pnet = Q
Lv (2) Q
(Equation 6.10b) for Q yields Q = Pnet t . Substituting this into Equation (2),
t we find that m= Q Pnet t
=
Lv
Lv (3) Substituting Equation (1) into Equation (3) gives
m= Pnet t
Lv = ( ) 4π R 2 eσ T 4 − T04 t
Lv Since 1 hour is equivalent to 3600 seconds, the mass of helium that boils away in one hour
is Chapter 13 Problems m= 709 2
4
4
4π ( 0.30 m ) (1) 5.67 ×10−8 J/ ( s ⋅ m 2 ⋅ K 4 ) ( 77 K ) − ( 4.2 K ) ( 3600 s ) = 0.39 kg
4
2.1×10 J/kg 32. REASONING
a. According to Equation 13.2, the radiant power P (or energy per unit time) emitted by an
object is P = Q / t = e σ T 4 A , where e is the emissivity of the object, σ is the
StefanBoltzmann constant, T is the temperature (in kelvins) of the object, and A is its
surface area. This expression will allow us to find the ratio of the two absorbed powers. The
reason is that the object and the room have the same constant temperature. Since the
object’s temperature is constant, it must be absorbing the same power that it is emitting.
b. The temperature of the two bars in part (b) of the text drawing can be obtained directly
P
from Equation 13.2: T = 4
.
eσ A
SOLUTION
a. The power absorbed by the two bars in part (b) of the text drawing is given by
Q / t = e σ T 4 A2 (Equation 13.2), where A2 is the total surface area of the two bars that is
exposed to the room: A2 = 28 L2 . The power absorbed by the single bar in part (a) of the
0 text drawing is Q / t = e σ T 4 A1 , where A1 is the total surface area of the single bar:
A1 = 22 L2 . The ratio of the power P2 absorbed by the two bars to the power P1 absorbed by
0
the single bar is (
( )
) 4
2
P2 eσ T 28 L0
=
= 1.27
P eσ T 4 22 L2
1
0 b. The temperature T2 of the two bars in part (b) of the text drawing and the temperature T1
of the single bar in part (a) are
T2 = 4 P2
e σ A2 and T1 = 4 Dividing T2 by T1 , and...
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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

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