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 coefficeint of the material of the resistor is `1.70 xx 10^(-4) .^(@)C^(-1)`

Explanation

Here `R_(27) = 100 Omega, R_(1) = 117 Omega, t = ? alpha = 1.70 xx 10^(-4) .^(@C^(-1)`
We know that
`alpha = (R_(t) - R_(27))/(R_(27) (t - 27))` or `t - 27 = R_(t) - R_(27)`
`t = (R_(1) - R_(2))/(R_(27) xx alpha) + 27 = (117 - 100)/(100 xx 1.7 xx 10^(-4)) + 27`
`= 1000 + 27 = 1027^(@)C`.