Question

To what temperature must you raise a silver wire (c = 0.0038), originally at 20.0°C, to double its resistance, neglecting any changes in dimensions? (You can assume that resistivity varies with temperature linearly)

Answer 1

Answer:

$T=283^{\circ}C$

Explanation:

Given a material with temperature coefficient of resistance c, the equation that relates the resistance $R_0$ at temperature $T_0$ and the resistance $R$ at temperature $T$ is

$\frac{R-R_0}{R_0}=c(T-T_0)$

We want to double our resistance, so $R=2R_0$, thus having:

$\frac{2R_0-R_0}{R_0}=\frac{R_0}{R_0}=1=c(T-T_0)$

For this T must be:

$1=cT-cT_0$

$T=\frac{1+cT_0}{c}$

which for our values means (with $T=20^{\circ}C=293^{\circ}K$, remember to write temperature in S.I., and that for silver $c=0.0038^{\circ}K^{-1}$):

$T=\frac{1+(0.0038^{\circ}K^{-1})(293^{\circ}K)}{(0.0038^{\circ}K^{-1})}=556^{\circ}K=283^{\circ}C$