In: Physics
Concepts and reason
The concept required to solve this problem is the variation of resistance with respect to temperature. Rearrange the equation of variation of resistance with respect to temperature to solve for temperature and plug all values in the equation to calculate the temperature.
Fundamentals
Variation of resistance with temperature is given by the following formula. \(R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\)
Here, \(R\) is the temperature of resistance at temperature \(T, R_{0}\) is the resistance of resistance at some reference temperature \(T_{0}\), and \(\alpha\) is the coefficient of resistivity.
Variation of resistance with temperature is given by the following formula. \(R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\)
Rearrange, variation of temperature equation for temperature. \(T=\frac{1}{\alpha}\left[\frac{R}{R_{0}}-1\right]+T_{0}\)
Resistance of the resistor changes with temperature and that variation is given by the following expression. \(R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\)
The expression for temperature is, \(T=\frac{1}{\alpha}\left[\frac{R}{R_{0}}-1\right]+T_{0}\)
Substitute, \(-5.00 \times 10^{-4 \circ} \mathrm{C}^{-1}\) for \(\alpha, 216.0 \Omega\) for \(R, 217.3 \Omega\) for \(R_{0}\) and \(4.00^{\circ} \mathrm{C}\) for \(T_{0}\).
$$ \begin{array}{c} T=\frac{1}{-5.00 \times 10^{-4} \mathrm{C}^{-1}}\left[\frac{216.0 \Omega}{217.3 \Omega}-1\right]+4.00^{\circ} \mathrm{C} \\ =15.96^{\circ} \mathrm{C} \end{array} $$
The reference temperature is \(4.00^{\circ} \mathrm{C}\) and resistance at this temperature is \(217.3 \Omega\) which is \(R_{0}\).
The temperature was \(15.96^{\circ} \mathrm{C}\) when resistance was \(216.0 \Omega\).