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Unformatted text preview: Ω) = 0.135 W. Assuming a steady rate, the amount of thermal energy generated in 30 minutes is found to
be (0.135 J/s)(30 × 60 s) = 2.43 × 102 J.
47. (a) From P = V 2/R = AV 2 / ρL, we solve for the length:
L= (b) Since L ∝ V 2 AV 2 (2.60 × 10−6 m2 )(75.0 V) 2
.
=
= 585 m.
ρP (5.00 × 10−7 Ω ⋅ m)(500 W) FV ′I
the new length should be L ′ = LG J
HV K 2 F 100 V IJ
= (585 m) G
.
H 75.0 V K 48. The mass of the water over the length is
m = ρ AL = (1000 kg/m3 )(15 × 10−5 m 2 )(0.12 m) = 0.018 kg , and the energy required to vaporize the water is
Q = Lm = (2256 kJ / kg)(0.018 kg) = 4.06 × 104 J . The thermal energy is supplied by joule heating of the resistor:
Q = PΔt = I 2 RΔt . Since the resistance over the length of water is
R= ρw L
A = (150 Ω ⋅ m ) ( 0.120 m ) = 1.2 ×105 Ω ,
15 × 10−5 m 2 the average current required to vaporize water is 2 = 10.4 m. 1052 CHAPTER 26 I= Q
4.06 × 104 J
=
= 13.0 A .
RΔt
(1.2 × 105 Ω)(2.0 ×10−3 s) 49. (a) Assuming a 31day month, the monthly cost is
(100 W)(24 h/day)(31days/month) (6 cents/kW ⋅ h) = 446 cents = US$4.46 .
(b) R = V 2/P = (120 V)2/100 W = 144 Ω.
(c) i = P/V = 100 W/120 V = 0.833 A...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.
 Fall '08
 schuller
 Magnetism, Work

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