Homework 6 Solutions

46 b r v 2p 120 v2100 w 144 c i pv 100

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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 31-day 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.

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