775_Physics ProblemsTechnical Physics

775_Physics ProblemsTechnical Physics - Chapter 27 P27.33 R...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 27 P27.33 R = R 0 1 + αT R − R0 = R0α∆T R − R0 = α∆T = 5.00 × 10 −3 25.0 = 0.125 R0 e P27.34 j a Assuming linear change of resistance with temperature, R = R0 1 + α∆T a ja fe f f R77 K = 1.00 Ω 1 + 3.92 × 10 −3 −216° C = 0.153 Ω . a f 1 FG ρ Hρ W IJ K ρ = ρ 0 1 + α∆T or ∆TW = Require that ρ W = 4ρ 0 Cu so that ∆TW = Therefore, P27.35 TW = 47.6° C + T0 = 67.6° C . Section 27.5 αW −1 0W F1 GH 4.50 × 10 I FG 4e1.70 × 10 j − 1IJ = 47.6° C . JK ° C J G 5.60 × 10 KH −8 −3 Superconductors Problem 48 in Chapter 43 can be assigned with this section. Section 27.6 P27.36 I= Electric Power P ∆V = and R = *P27.37 P27.38 600 W = 5.00 A 120 V ∆V 120 V = = 24.0 Ω . I 5.00 A e b gb g P = 0.800 1 500 hp 746 W hp = 8.95 × 10 5 W b 8.95 × 10 5 = I 2 000 P = I∆V P27.39 j P = I∆V = 500 × 10 −6 A 15 × 10 3 V = 7.50 W g I = 448 A The heat that must be added to the water is b gb f ga Q = mc∆T = 1.50 kg 4 186 J kg ° C 40.0° C = 2.51 × 10 5 J . Thus, the power supplied by the heater is P= W Q 2.51 × 10 5 J = = = 419 W ∆t ∆t 600 s a∆V f = a110 V f 2 and the resistance is R = P 2 419 W = 28.9 Ω . −8 115 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online