Question

At room temperature $27^\circ\text{C} ,$ the resistance of a heating element is $50 \, \Omega$. The temperature coefficient of the material is $2.4 \times 10^{-4}$ $^\circ\text{C}^{-1}$. The temperature of the element, when its resistance is $62 \, \Omega$ , is _________$\degree C$.

Answer 1

The relationship between resistance and temperature is given by:
$R = R_0 \left( 1 + \alpha \Delta T \right),$
where: \begin{itemize} \item $R_0 = 50 \, \Omega$ (resistance at room temperature), $R = 62 \, \Omega$ (resistance at the higher temperature), \alpha = 2.4 \times 10^{-4} \degree{C}^{-1} (temperature coefficient), $\Delta T = T - T_0$ (change in temperature), $T_0 = 27^\circ \mathrm{C}$ (initial temperature).
Rearrange to solve for $\Delta T$:
$\Delta T = \frac{R - R_0}{\alpha R_0}.$

Substitute the given values:
$\Delta T = \frac{62 - 50}{(2.4 \times 10^{-4}) \cdot 50}.$

Simplify: $\Delta T = \frac{12}{(2.4 \times 10^{-4}) \cdot 50} = \frac{12}{0.012} = 1000^\circ \mathrm{C}.$
The final temperature $T$ is: $T = T_0 + \Delta T = 27 + 1000 = 1027^\circ \mathrm{C}.$

Answer 2

The relationship between resistance and temperature is given by:
$R = R_0 \left( 1 + \alpha \Delta T \right),$
where: \begin{itemize} \item $R_0 = 50 \, \Omega$ (resistance at room temperature), $R = 62 \, \Omega$ (resistance at the higher temperature), \alpha = 2.4 \times 10^{-4} \degree{C}^{-1} (temperature coefficient), $\Delta T = T - T_0$ (change in temperature), $T_0 = 27^\circ \mathrm{C}$ (initial temperature).
Rearrange to solve for $\Delta T$:
$\Delta T = \frac{R - R_0}{\alpha R_0}.$

Substitute the given values:
$\Delta T = \frac{62 - 50}{(2.4 \times 10^{-4}) \cdot 50}.$

Simplify: $\Delta T = \frac{12}{(2.4 \times 10^{-4}) \cdot 50} = \frac{12}{0.012} = 1000^\circ \mathrm{C}.$
The final temperature $T$ is: $T = T_0 + \Delta T = 27 + 1000 = 1027^\circ \mathrm{C}.$