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 _________\(\degree C\).
Correct Answer: 1027
Approach Solution - 1
To find the temperature at which the resistance of the heating element is \(62 \, \Omega\), we use the formula for resistance as a function of temperature: \(R = R_0 (1 + \alpha \Delta T)\), where \(R_0\) is the initial resistance, \(\alpha\) is the temperature coefficient, and \(\Delta T\) is the change in temperature.
Given:
- Initial resistance, \( R_0 = 50 \, \Omega \)
- Final resistance, \( R = 62 \, \Omega \)
- Temperature coefficient, \(\alpha = 2.4 \times 10^{-4} \, ^\circ\text{C}^{-1}\)
- Initial temperature, \( T_0 = 27^\circ\text{C} \)
First, solve for \(\Delta T\):
\[\begin{align*} 62 &= 50(1 + 2.4 \times 10^{-4} \Delta T) \\ \frac{62}{50} &= 1 + 2.4 \times 10^{-4} \Delta T \\ 1.24 - 1 &= 2.4 \times 10^{-4} \Delta T \\ 0.24 &= 2.4 \times 10^{-4} \Delta T \\ \Delta T &= \frac{0.24}{2.4 \times 10^{-4}} = 1000^\circ\text{C} \end{align*}\]Next, calculate the final temperature:
\[T = T_0 + \Delta T = 27 + 1000 = 1027^\circ\text{C}\]Confirming the solution, \(1027^\circ\text{C}\) is within the expected range \(1027, 1027\). Therefore, the temperature of the element when its resistance is \(62 \, \Omega\) is \(\boxed{1027^\circ\text{C}}\).
Approach Solution -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: \[ T = T_0 + \Delta T = 27 + 1000 = 1027^\circ \mathrm{C}. \]
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