79. (a) In Eq. 26-17, we let ρ= 2ρ0where ρ0is the resistivity at T0= 20°C: ρρρρρ α−=−=−000002TTbg,and solve for the temperature T: TT=+=°+×≈°−0312014 310250αCKC../(b) Since a change in Celsius is equivalent to a change on the Kelvin temperature scale,
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