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Unformatted text preview: these expressions for Q and P will allow us to determine the time t.
SOLUTION Substituting Equations 12.4 and 20.6c into the expression for the time and
recognizing that the normal boiling point of water is 100.0 ºC, we find that Chapter 20 Problems t= = 1071 Q cm∆T Rcm∆T
=
=
P V2 /R
V2 (15 Ω ) 4186 J/ ( kg ⋅ C° ) ( 0.50 kg )(100.0 °C − 13 °C ) = 190 s
(120 V )2 30. REASONING The total volume of ice melted is the product of the thickness h and the area
A of the layer that melts, which permits us to determine the thickness in terms of the volume
and the area:
hA = Volume h= or Volume
A Given the density ρ = 917 kg/m3, we can find the volume of the ice from ρ = (1)
m
Volume (Equation 11.1), where m is the mass of the ice.
Volume = m ρ (2) The mass m of the ice determines the heat Q required to melt it, according to Equation 12.5
Q = mL (12.5) In Equation 12.5, L = 33.5 ×104 J/kg is the latent heat of fusion for water (see Table 12.3).
In order to find the maximum thickness of ice the defroster can melt, we will assume that all
the heat generated by the defroster goes into melting the ice.
The power output P of the defroster is the rate at which it converts electrical energy to heat,
so we have that
P= Q
t or Q = Pt (6.10b) where t is the elapsed time (3.0 minutes). The power output P is, in turn, found from the
operating voltage V and current I: P = IV
SOLUTION Substituting Equation (2) into Equation (1) yields (20.6a) 1072 ELECTRIC CIRCUITS h= Volume ( m ρ ) m
=
=
A
A
ρA Solving Q = mL (Equation 12.5) for m, we find that m =
Equation (3), we obtain
h= (3)
Q
. Substituting this result into
L m (Q L)
Q
=
=
ρA
ρA
Lρ A (4) Substituting Equation (6.10b) into Equation (4) yields
h= Q
Pt
=
Lρ A Lρ A (5) We note that the elapsed time t is given in minutes, which must be converted to seconds.
Substituting Equation 20.6a into Equation (5), we obtain the maximum thickness of the ice
that the defroster can melt: 60 s min Pt
IVt 1 min = 3.1×10−4 m
h=
=
=
L ρ A L ρ A 33.5 ×104 J/kg 917 kg/m3 0.52 m2 ( 23 A )(12 V ) ( 3.0 ( )( ) )( ) ______________________________________________________________________________
31. SSM REASONING AND SOLUTION As a function of temperature, the resistance of the wire is given by Equation 20.5: R = R0 1 + α (T − T0 ) , where α is the temperature 2 coefficient of resistivity. From Equation 20.6c, we have P = V / R . Combining these two
equations, we have
2
P0
V
P=
=
R0 1 + α (T – T0 ) 1 + α (T – T0 ) 1 2 where P0 = V /R0, since the voltage is constant. But P = P0 , so we find
2 P0
2 = P0 1 + α (T – T0 ) or 2 = 1 + α (T − T0 ) Solving for T, we find
1
+ 28° = 250 °C
–1
α
0.0045 (C°)
______________________________________________________________________________
T= 1 + T0 = Chapter 20 Problems 1073 32. REASONING Substituting V = 1 V0 into Equation 20.7 gives a result that can be solved
2
directly for the desired time.
SOLUTION From Equation 20.7 we have V = 1 V0 = V0 sin 2π f t
2 or 1
2 = sin 2π f t Using the inverse trigonometric sine function, we find 2π f t = sin −1 ( 1 ) = 0.524
2 In this result, the value of 0.524 is in radians and corresponds to an angle of 30.0º. Thus we
find that the smallest value of t is t= 0.524
0.524
=
= 1.39 × 10−3 s
2π f
2π ( 60.0 Hz ) 33. REASONING
a. The average power P delivered to the copy machine is equal to the square of the
2
rmscurrent Irms times the resistance R, or P = I rms R (Equation 20.15b). Both Irms and R
are known.
b. According to the discussion in Section 20.5, the peak power Ppeak is twice the average
power, or Ppeak = 2 P .
SOLUTION
a. The average power is
2
2
P = I rms R = ( 6.50 A ) (18.6 Ω ) = 786 W (20.15b) b. The peak power is twice the average power, so
Ppeak = 2 P = 2 ( 786 W ) = 1572 W
______________________________________________________________________________ 1074 ELECTRIC CIRCUITS 34. REASONING AND SOLUTION The expression relating the peak current, I0, to the rmscurrent, Irms, is
I
2.50 A
I rms = 0 =
= 1.77 A
2
2
______________________________________________________________________________
35. 2
SSM REASONING The average power is given by Equation 20.15c as P = Vrms / R . In
this expression the rms voltage Vrms appears. However, we seek the peak voltage V0. The relation between the two types of voltage is given by Equation 20.13 as Vrms = V0 / 2 , so
we can obtain the peak voltage by using Equation 20.13 to substitute into Equation 20.15c. SOLUTION Substituting Vrms from Equation 20.13 into Equation 20.15c gives P= 2
Vrms R (V / 2 )
=
0 R 2 = V02
2R Solving for the peak voltage V0 gives V0 = 2 RP = 2 ( 4.0 Ω ) ( 55 W ) = 21 V
______________________________________________________________________________
36. REASONING Because we are ignoring the effects of temperature on the heater, the
resistance R of the heater is the same whether it is plugged into a 120V outlet or a 230V
outlet. The average power output P of the heater is related to the rms outlet voltage Vrms and
the heater’s resistance R by P =...
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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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