Unformatted text preview: °C âˆ’ 5 Â°C )
=
=
= 42 J/s
t
L
0.075 m Therefore, the amount of heat per second that must be removed from the unit to keep it cool
is 42 J / s . 37. REASONING The radiant energy Q radiated by the sun is given by Q = e Ïƒ T 4 At
(Equation 13.2), where e is the emissivity, Ïƒ is the StefanBoltzmann constant, T is its
temperature (in Kelvins), A is the surface area of the sun, and t is the time. The radiant
energy emitted per second is Q/t = e Ïƒ T 4 A. Solving this equation for T gives the surface
temperature of the sun. Chapter 13 Problems 713 26 SOLUTION The radiant power produced by the sun is Q/t = 3.9 Ã— 10 W. The surface area
2
of a sphere of radius r is A = 4Ï€r . Since the sun is a perfect blackbody, e = 1. Solving
Equation 13.2 for the surface temperature of the sun gives
T= 4 Q/t
3.9 Ã—1026 W
=4
e Ïƒ 4Ï€ r 2
(1) 5.67 Ã—10âˆ’8 J/ s â‹… m 2 â‹… K 4 4Ï€ 6.96 Ã—108 m ( ) ( ) 2 = 5800 K 38. REASONING The net rate at which energy is being lost via radiation cannot exceed the
production rate of 115 J/s, if the body temperature is to remain constant. The net rate at
which an object at temperature T radiates energy in a room where the temperature is T0 is
given by Equation 13.3 as Pnet = eÏƒA(T4 â€“ T04). Pnet is the net energy per second radiated.
We need only set Pnet equal to 115 J/s and solve for T0. We note that the temperatures in this
equation must be expressed in Kelvins, not degrees Celsius. SOLUTION According to Equation 13.3, we have ( Pnet = eÏƒ A T 4 âˆ’ T04 ) or T04 = T 4 âˆ’ Pnet eÏƒ A Using Equation 12.1 to convert from degrees Celsius to Kelvins, we have T = 34 + 273 =
307 K. Using this value, it follows that T0 = 4 T 4 âˆ’ Pnet eÏƒ A = 4 ( 307 K ) âˆ’
4 115 J/s ( )( 0.700 5.67 Ã—10âˆ’8 J/ s â‹… m 2 â‹… K 4 1.40 m 2 ) = 287 K (14 Â°C) 39. REASONING AND SOLUTION The heat Q conducted during a time t through a wall of
thickness L and cross sectional area A is given by Equation 13.1:
Q= kA âˆ†T t
L The radiant energy Q, emitted in a time t by a wall that has a Kelvin temperature T, surface
area A, and emissivity e is given by Equation (13.2): 714 THE TRANSFER OF HEAT Q = eÏƒ T 4 At If the amount of radiant energy emitted per second per square meter at 0 Â°C is the same as
the heat lost per second per square meter due to conduction, then
Q
Q
= t A conduction t A radiation Making use of Equations 13.1 and 13.2, the equation above becomes
k âˆ†T
= eÏƒ T 4
L Solving for the emissivity e gives: e= k âˆ†T
[1.1 J/(s â‹… m â‹… K)](293.0 K âˆ’ 273.0 K)
=
= 0.70
4
LÏƒ T
(0.10 m)[5.67 Ã—10âˆ’8 J/(s â‹… m 2 â‹… K 4 )] (273.0 K) 4 Remark on units: Notice that the units for the thermal conductivity were expressed as
J/(s.m.K) even though they are given in Table 13.1 as J/(s.m.CÂ°). The two units are
equivalent since the "size" of a Celsius degree is the same as the "size" of a Kelvin; that is,
1 CÂ° = 1 K. Kelvins were used, rather than Celsius degrees, to ensure consistency of units.
However, Kelvins must be used in Equation 13.2 or any equation that is derived from it. 40. REASONING AND SOLUTION According to Equation 13.2, for the sphere we have
Q/t = eÏƒAsTs4, and for the cube Q/t = eÏƒAcTc4. Equating and solving we get
4 4 Tc = (As/Ac)Ts
Now As/Ac = (4Ï€ R2)/(6L2)
1/ 3 3
3
3
The volume of the sphere and the cube are the same, (4/3) Ï€ R = L , so R = 4Ï€ The ratio of the areas is As
Ac = 4Ï€ R 2 4Ï€ 3 = 6 4Ï€ 6 L2 2/3 = 0.806 . The temperature of the cube is, then
1/4 L. A 1/ 4
Tc = s Ts = ( 0.806 ) ( 773 K ) = 732 K
A c Chapter 13 Problems 715 41. REASONING The heat lost per second due to conduction through the glass is given by
Equation 13.1 as Q/t = (kAâˆ†T)/L. In this expression, we have no information for the
thermal conductivity k, the crosssectional area A, or the length L. Nevertheless, we can
apply the equation to the initial situation and again to the situation where the outside
temperature has fallen. This will allow us to eliminate the unknown variables from the
calculation.
SOLUTION Applying Equation 13.1 to the initial situation and to the situation after the
outside temperature has fallen, we obtain ( kA TIn âˆ’ TOut, initial
Q
= L t Initial ) ( kA TIn âˆ’ TOut, colder
Q
= L t Colder and ) Dividing these two equations to eliminate the common variables gives ( Q / t )Colder
( Q / t )Initial ( ) ( ) kA TIn âˆ’ TOut, colder
= TIn âˆ’ TOut, colder
L
=
TIn âˆ’ TOut, initial
kA TIn âˆ’ TOut, initial
L Remembering that twice as much heat is lost per second when the outside is colder, we find
2 ( Q / t )Initial ( Q / t )Initial =2= TIn âˆ’ TOut, colder
TIn âˆ’ TOut, initial Solving for the colder outside temperature gives TOut, colder = 2TOut, initial âˆ’ TIn = 2 ( 5.0 Â°C ) âˆ’ ( 25 Â°C ) = âˆ’15 Â°C 42. REASONING The flow of heat from the heating elements through the pot bottoms and into
the boiling water occurs because the temperature T of each burner is greater than the
temperature T0 = 100.0 Â°C of the boiling water. The temperature difference âˆ†T = T â€“ T0
Q kA âˆ†T
=
drives heat flow...
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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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