Question

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 × 10⁻⁴ °C⁻¹)?

Answer 1

Answer:

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Answer 2

Explanation:

$\Large{\pink{\underline{\underline{\sf{\orange{Solution:}}}}}}$

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$\hookrightarrow$ Room temperature, $\sf T\:=\:27^{\circ}C$

$\hookrightarrow$ Resistance of the heating elememt at T, $\sf R\:=\:100\Omega$

$\hookrightarrow$ Let $\sf T_1$ is the increased temperature of the filament

$\hookrightarrow$ Resistance of the heating elememt at $\sf T_1$, $\sf R_1\:=\:100\Omega$

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Temperature co-efficient of the material of the filament,

$\sf \alpha\:=\:1.70\times 10^{-4}°C^{-1}$

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$\sf \alpha$ is given by the relation,

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$\boxed{\sf{\pink{\alpha\:=\: \dfrac{R_1-R}{R(T_1-T)}}}}$

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$\leadsto$ $\sf T_1-T\:=\: \dfrac{R_1-R}{R\alpha}$

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$\leadsto$ $\sf T_1-27\:=\: \dfrac{117-100}{100(1.7\times 10^{-4})}$

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$\leadsto$ $\sf T_1-27\:=\:1000$

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$\leadsto$ $\sf T_1\:=\:1027^{\circ}C$

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$\longrightarrow$ Hence at $\sf{\green{T_1\:=\:1027^{\circ}C}}$, the resistance of the element is $\sf{\pink{117\Omega}}$