At room temperature (25.0 °C) the resistance of a heating element is 100 . The temperature of the element at which resistance becomes 125.5 Ω is: (Given that the temperature coefficient of the material of the resistor is $1.70 × 10^{-4} °C^{-1}$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1525 °C
Rechecking the calculation with the given temperature coefficient:
Given: $R_0 = 100\ \Omega$, $R = 125.5\ \Omega$, $\alpha = 1.70 \times 10^{-4}\ {}^\circ\text{C}^{-1}$, $T_0 = 25\ {}^\circ\text{C}$
Resistance-temperature relation: $R = R_0 [1 + \alpha (T - T_0)]$
Substitute values:
$125.5 = 100 [1 + 1.70 \times 10^{-4} (T - 25)]$
$1.255 = 1 + 1.70 \times 10^{-4} (T - 25)$
$0.255 = 1.70 \times 10^{-4} (T - 25)$
$T - 25 = \frac{0.255}{1.70 \times 10^{-4}} \approx 1500\ {}^\circ\text{C}$
$T \approx 1525\ {}^\circ\text{C}$
∴ Correct temperature of the element = 1525 °C