Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - Problems - Page 522: 22

Answer

280°C

Work Step by Step

$ρ_{Al} [1+α(T-20°C)] = ρ_{W}$ divide both sides by $ρ_{Al}$ $1+α(T-20°C)= \frac{ρ_{W}}{ρ_{Al}}$ subtract 1 from both sides $α(T-20°C)= \frac{ρ_{W}-ρ_{Al}}{ρ_{Al}}$ divide out $\alpha$ $T-20°C= \frac{ρ_{W}-ρ_{Al}}{ρ_{Al}{α}}$ add $20°C$ $T= \frac{ρ_{W}-ρ_{Al}}{ρ_{Al}{α}} +20°C = 279.485°C \approx 280°C$
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