The resistivity of a metal increases slightly with increased temperature. This can be expressed as $\rho ={\rho }_0\left[1+\alpha \left(T-T_0\right)\right]$, where $T_0$is a reference temperature, usually $20°\mathrm{C}$, and $\alpha$is the temperature coefficient of resistivity. For copper, $\alpha =3.9\times {10}^{-3}{{}^{\circ }\mathrm{C}}^{-1}$. Suppose a long. thin copper wire has a resistance of $0.25\Omega$at $20°\mathrm{C}$. At what temperature, in ${}^{\circ }\mathrm{C}$, will its resistance be $0.30\Omega$?

Resistivity of a metal increasing $\rho ={\rho }_0\left[1+\alpha \left(T-T_0\right)\right]$

Temperature$T_0={20}^{\circ }C$

Copper's temperature coefficient of resistivity $\alpha =3.9\times {10}^{-3}{{}^{\circ }\mathrm{C}}^{-1}$

Resistancerole="math" localid="1649139996990" $=0.25\Omega$

Temperature atrole="math" localid="1649140003581" $0.30\Omega =?$

Given that,
$\rho ={\rho }_0\left[1+\alpha \left(T-T_0\right)\right],$

It gives the length-based resistivity of a unit area resistor. Neither the length nor the cross-section area of the resistor - that is, the wire - will change, so the given expression can simply be rewritten as follows:
$R=R_0\left[1+\alpha \left(T-T_0\right)\right]$

Solve for $T$:

$\frac{R}{R_0}=1+\alpha \left(T-T_0\right)$

Modify the expression,

$\frac{R-R_0}{R_0}=\alpha \left(T-T_0\right)$

Find the temperature,

$T-T_0=\frac{R-R_0}{R_0\alpha }$

Hence, the expression will be,

$T=T_0+\frac{R-R_0}{R_0\alpha }$

Substitute the values,

role="math" localid="1649140489954" $T=\left({20}^{\circ }C+\frac{0.3\Omega -0.25\Omega }{0.3\Omega \times 0.{0039}^{\circ }{C}^{-1}}\right)$

$=63°\mathrm{C}.$

Hence, the temperature will be${63}^{\circ }C.$