This preview shows page 1. Sign up to view the full content.
Unformatted text preview: σ Tbefore A t before Since the same radiant power is emitted before and after the “radiator” is painted, we have Q
Q
= t after t before or 4
4
eafter σ Tafter A = ebefore σ Tbefore A The terms σ and A can be eliminated algebraically, so this result becomes
4
4
eafter σ Tafter A = ebefore σ Tbefore A or 4
4
eafter Tafter = ebefore Tbefore Remembering that the temperature in the StefanBoltzmann law must be expressed in
Kelvins, so that Tbefore = 62 °C +273 = 335 K (see Section 12.2), we find that
4
Tafter = 4
ebefore Tbefore eafter or Tafter = 4 ebefore
eafter 0.75
(Tbefore ) = 4 0.50 ( 335 K ) = 371 K On the Celsius scale, this temperature is 371 K − 273 = 98 °C . 704 THE TRANSFER OF HEAT 23. REASONING The radiant energy Q absorbed by the person’s head is given by
Q = e σ T 4 At (Equation 13.2), where e is the emissivity, σ is the StefanBoltzmann
constant, T is the Kelvin temperature of the environment surrounding the person
(T = 28 °C + 273 = 301 K), A is the area of the head that is absorbing the energy, and t is the
time. The radiant energy absorbed per second is Q/t = e σ T 4 A.
SOLUTION
a. The radiant energy absorbed per second by the person’s head when it is covered with hair
(e = 0.85) is ( ) ( ) Q
4
= e σ T 4 A = ( 0.85 ) 5.67 ×10−8 J/ s ⋅ m 2 ⋅ K 4 ( 301 K ) 160 × 10−4 m 2 = 6.3 J/s t
b. The radiant energy absorbed per second by a bald person’s head (e = 0.65) is ( ) ( ) Q
4
= e σ T 4 A = ( 0.65 ) 5.67 ×10−8 J/ s ⋅ m 2 ⋅ K 4 ( 301 K ) 160 × 10−4 m 2 = 4.8 J/s t ( ) 24. REASONING The net radiant power of the baking dish is given by Pnet = eσ A T 4 − T04 ,
(Equation 13.3), where e is the emissivity of the dish, σ is the StefanBoltzmann constant, A
is the total surface area of the dish, T is the Kelvin temperature of the dish, and T0 is the
temperature of the kitchen. As the dish cools down, both its net radiant power Pnet and
temperature T decrease, but its emissivity e and surface area A remain constant. We will
therefore solve Equation 13.3 for the product of these constant quantities and the StefanBoltzmann constant:
Pnet
eσ A =
(1)
4
T − T04 ( ) Because the left side of Equation (1) is constant, the ratio on the right hand side cannot
change as the temperature T decreases. We will use this fact to determine the radiant power
Pnet,1 of the baking dish when its temperature is 175 °C.
SOLUTION From Equation (1), we have that Pnet,1 = Pnet,2 (T14 − T04 ) (T24 − T04 ) or T 4 − T 4 Pnet,1 = Pnet,2 14 04 T2 − T0 (2) In Equation (2), Pnet,2 = 12.0 W is the net radiant power when the temperature is 35 °C. We
first convert temperatures to the Kelvin scale by adding 273 to each temperature given in
degrees Celsius (see Equation 12.1). The initial temperature of the baking dish is Chapter 13 Problems 705 T1 = 175 °C + 273 = 448 K, its final temperature is T2 = 35 °C + 273 = 308 K, and the room
temperature is T0 = 22 °C + 273 = 295 K. Equation (2), then, gives the net radiant power
when the dish is first brought out of the oven: ( 448 K )4 − ( 295 K )4 = 275 W
Pnet,1 = (12.0 W ) ( 308 K )4 − ( 295 K )4 25. REASONING According to the discussion in Section 13.3, the net power Pnet radiated by ( ) the person is Pnet = eσ A T 4 − T04 , where e is the emissivity, σ is the StefanBoltzmann
constant, A is the surface area, and T and T0 are the temperatures of the person and the
environment, respectively. Since power is the change in energy per unit time (see
Equation 6.10b), the time t required for the person to emit the energy Q contained in the
dessert is t = Q/Pnet.
SOLUTION The time required to emit the energy from the dessert is
t= Q
Q
=
Pnet eσ A T 4 − T 4
0 ( ) 4186 J The energy is Q = ( 260 Calories ) , and the Kelvin temperatures are 1 Calorie T = 36 °C + 273 = 309 K and T0 = 21 °C + 273 = 294 K. The time is t= ( 260 Calories ) ( 0.75 ) 5.67 ×10−8 4186 J 1 Calorie = 1.2 ×104 s
4
4
2
4
2
J/ s ⋅ m ⋅ K
1.3 m ( 309 K ) − ( 294 K ) ( )( ) 26. REASONING The power radiated by an object is given by Q / t = eσ T 4 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 its surface area.
The power that the object absorbs from the room is given by Q/t = eσT0 A. Except for the
temperature T0 of the room, this expression has the same form as that for the power radiated
by the object. Note especially that the area A is the surface area of the object, not the room.
Review Example 8 in the text to understand this important point.
4 706 THE TRANSFER OF HEAT SOLUTION The object emits three times as much power as it absorbs from the room, so it
follows that (Q/t)emitted = 3(Q/t)absorbed. Using the StefanBoltzmann law for each of the
powers, we find
eσ T 43
3e 4 A
1 24 = 1 σ T0 3
4A
4
24
Power emitted Power absorbed Solving for the temperature T of the object gives T = 4 3T0 = 4 3 ( 293 K ) = 386 K 27. SSM REASONING AND SOLUTION The net power generated by the stove is given by
Equation 13.3, Pnet = e...
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