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Sagot :
Take into account that the resistance in the wire, for a certain temperature T, can be written as follow:
[tex]R=R_o(1+\alpha T)[/tex]where,
Ro: initial resistance
T: temperature
α: coefficient of resistivity
R: resistance
Based on the given information, you can write for each temperature:
[tex]\begin{gathered} 10.0\Omega=R_o(1+\alpha\cdot20C) \\ 10.5\Omega=R_o(1+\alpha\cdot90C) \end{gathered}[/tex]If you divide the first equation over the second one, you have:
[tex]\frac{10.0\Omega}{10.5\Omega}=\frac{R_o(1+\alpha\cdot20C)}{R_o(1+\alpha\cdot90C)}[/tex]By solving for α, you obtain:
[tex]\begin{gathered} 10.0\Omega(1+\alpha\cdot90C)=10.5\Omega(1+\alpha\cdot20C) \\ 10.0+\alpha\cdot900C=10.5+\alpha\cdot210C \\ \alpha(900C-210C)=10.5-10.0 \\ \alpha(690C)=0.5 \\ \alpha=\frac{0.5}{690C}=7.24\cdot10^{-4}C^{-1} \end{gathered}[/tex]Hence, the coefficient of the resistivity of the given metal is approximately
7.24*10^-4°C^-1
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