Answer:
[tex]3.34\Omega[/tex]
Explanation:
The resistance of a metal rod is given by
[tex]R=\frac{\rho L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity
L is the length of the rod
A is the cross-sectional area
The resistivity changes with the temperature as:
[tex]\rho(T)=\rho_0 (1+\alpha (T-T_0))[/tex]
where in this case:
[tex]\rho_0[/tex] is the resistivity of silver at [tex]T_0=21.0^{\circ}C[/tex]
[tex]\alpha=6.1\cdot 10^{-3} ^{\circ}C^{-1}[/tex] is the temperature coefficient for silver
[tex]T=180.0^{\circ}C[/tex] is the current temperature
Substituting,
[tex]\rho(180^{\circ}C)=\rho_0 (1+6.1\cdot 10^{-3}(180-21))=1.970\rho_0[/tex]
The length of the rod changes as
[tex]L(T)=L_0 (1+\alpha_L(T-T_0))[/tex]
where:
[tex]L_0[/tex] is the initial length at [tex]21.0^{\circ}C[/tex]
[tex]\alpha_L = 18\cdot 10^{-6} ^{\circ}C^{-1}[/tex] is the coefficient of linear expansion
Substituting,
[tex]L(180^{\circ}C)=L_0(1+18\cdot 10^{-6}(180-21))=1.00286L_0[/tex]
The cross-sectional area of the rod changes as
[tex]A(T)=A_0(1+2\alpha_L(T-T_0))[/tex]
So, substituting,
[tex]A(180^{\circ}C)=A_0(1+2\cdot 18\cdot 10^{-6}(180-21))=1.00572A_0[/tex]
Therefore, if the initial resistance at 21.0°C is
[tex]R_0 = \frac{\rho_0 L_0}{A_0}=1.70\Omega[/tex]
Then the resistance at 180.0°C is:
[tex]R(180^{\circ}C)=\frac{\rho(180)L(180)}{A(180)}=\frac{(1.970\rho_0)(1.00285L_0)}{1.00572A_0}=1.9644\frac{\rho_0 L_0}{A_0}=1.9644 R_0=\\=(1.9644)(1.70\Omega)=3.34\Omega[/tex]
