Question

a 100 - w lightbulb has a resistance of about 12 ω when cold (0 °c) and 140 ω when on (hot). part a estimate the temperature of the filament when hot assuming an average temperature coefficient of resistivity α = 0.0045 (c)^(-1). express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall resistance - temperature formula

$R = R_0(1+\alpha(T - T_0))$, where $R$ is the resistance at temperature $T$, $R_0$ is the resistance at temperature $T_0$, $\alpha$ is the temperature - coefficient of resistivity.
We know that $R_0 = 12\Omega$, $T_0=0^{\circ}C$, $R = 140\Omega$ and $\alpha=0.0045(^{\circ}C)^{-1}$.

Step2: Rearrange the formula to solve for $T$

First, expand the formula: $R = R_0+R_0\alpha(T - T_0)$. Then, $R - R_0=R_0\alpha(T - T_0)$. Since $T_0 = 0^{\circ}C$, we have $T=\frac{R - R_0}{R_0\alpha}$.

Step3: Substitute the values

$T=\frac{140 - 12}{12\times0.0045}=\frac{128}{12\times0.0045}=\frac{128}{0.054}\approx2407.41^{\circ}C$.

Step4: Round to two significant figures

Rounding $2407.41^{\circ}C$ to two significant figures gives $T = 2400^{\circ}C$.

Answer:

$2400^{\circ}C$