Question

current attempt in progress a coil of wire has a resistance of 47.6 ω at 28.0 °c and 45.2 ω at 69.0 °c. what is the temperature coefficient of resistivity? number units etextbook and media hint assistance the resistance of a piece of material depends on temperature as described by equation 20.5. save for later attempts: 0 of 3 used submit a using multiple attempts will impact your score. 20% score reduction after attempt 1

Explanation:

Step1: Recall the temperature-resistance formula

The formula relating resistance and temperature is \( R = R_0(1 + \alpha (T - T_0)) \), where \( R \) is the resistance at temperature \( T \), \( R_0 \) is the resistance at reference temperature \( T_0 \), and \( \alpha \) is the temperature coefficient of resistivity. We can rearrange this formula to solve for \( \alpha \): \( \alpha=\frac{R - R_0}{R_0(T - T_0)} \).

Let \( R_0 = 47.6\,\Omega \) (resistance at \( T_0 = 28.0^\circ\text{C} \)) and \( R = 45.2\,\Omega \) (resistance at \( T = 69.0^\circ\text{C} \)).

Step2: Calculate the temperature difference

First, find the difference in temperature \( \Delta T=T - T_0 \).
\( \Delta T = 69.0^\circ\text{C}- 28.0^\circ\text{C}=41.0^\circ\text{C} \)

Step3: Substitute values into the formula for \( \alpha \)

Substitute \( R = 45.2\,\Omega \), \( R_0 = 47.6\,\Omega \), and \( \Delta T = 41.0^\circ\text{C} \) into the formula for \( \alpha \):
\( \alpha=\frac{45.2 - 47.6}{47.6\times(69.0 - 28.0)} \)
First, calculate the numerator: \( 45.2 - 47.6=- 2.4 \)
Then, calculate the denominator: \( 47.6\times41.0 = 47.6\times40+47.6\times1=1904 + 47.6 = 1951.6 \)
Now, divide the numerator by the denominator: \( \alpha=\frac{-2.4}{1951.6}\approx - 1.23\times 10^{-3}\,^\circ\text{C}^{-1} \)

Answer:

The temperature coefficient of resistivity is approximately \(-1.23\times 10^{-3}\,^\circ\text{C}^{-1}\) (or \(-0.00123\,^\circ\text{C}^{-1}\)). The units are \(^\circ\text{C}^{-1}\) (per degree Celsius).