# 3.5 At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is $1.70\times 10^{-4}\degree C^{-1}$.

Given,

Temperature coefficient of filament, $\alpha$ = $1.70\times 10^{-4}\degree C^{-1}$

$\dpi{100} T_1= 27\degree C$ ;  $\dpi{100} R_1 = 100 \Omega$

Let $\dpi{100} T_2$ be the temperature of element, $\dpi{100} R_2= 117\Omega$

(Positive $\alpha$ means that the resistance increases with temperature. Hence we can deduce that $\dpi{100} T_2$ will be greater than  $\dpi{100} T_1$)

We know,

$\dpi{100} R_2 = R_1[1 + \alpha \Delta T]$

$\dpi{80} \implies$ $117 = 100[1 +(1.70\times 10^{-4})(T_2 - 27) ]$

$\\ \Rightarrow T_2-27=\frac{117-100}{1.7\times10^{-4}}\\\Rightarrow T_2-27=1000\\\Rightarrow T_2=1027^\circ C$

Hence, the temperature of the element is 1027 °C.

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