At room temperature `(27.0^@C)` the resistance of a heating element is `100 Omega`. What is the temperature of the element if the resistance is found to be `117 Omega`, given that the temperature coefficient of the material of the resistor is `(1.70 xx 10^(-4))^@C^(-1)`.

Explanation

Room temperature, `T = 27^(@)C`
Resistance of the heating element at T, `R = 100 Omega`
Let `T_(1)` is the increased temperature of the filament. Resistanc of the heating element at `T_(1),R_(1)=117 Omega`
Temperature co-efficient of the material of the filament,
`alpha = 1.70 xx 10^(-1) .^(@)C^(-1)`
`alpha` is given by the relation,
`alpha = (R_(1)-R)/(R(T_(1)-T))`
`T_(1)-T = (R_(1)-R)/(R alpha)`
`T_(1)-27 = (117 xx 100)/(100(1.7 xx 10^(-4)))`
`T_(1)-27 = 1000`
`T_(1)=1027^(@)C`
Therefore, at `1027^(@)C`, the resistance of the element is `117 Omega`.