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

If `R_(2)` is the resistance of a material at temperature `t_(2) and R_(1)` is the resistance at temperature `t_(1)`, then the temperature coefficient of resistivity of the material is given by: `alpha = (R_(2) - R_(1))/(R_(1) (t_(2)-t_(1)))` …(i)
Let the desired temperature of the heating element be t adn the resistance of heating element at this temperature be `R_(1)`. Also, resistance of the heating element at `27.0 ""^(@)C " is " R_(27)`. Substituting `R_(2) = R_(t), R_(1) = R_(27), t_(2) = t and t_(1) = 27` in (i), we get
`alpha = (R_(t)-R_(27))/(R_(27) (t-27))`
`rArr t-27 = (R_(t)-R_(27))/(R_(27)xx alpha)`
Here, `R_(t) = 117 Omega, R_(27) = 100 Omega and alpha= 1.70 xx 10^(-4) ""^(@)C^(-1)`
`:. t= 27 = (117 - 100)/(100 xx 1.7 xx10^(-4))`
`=1,000 ""^(@)C`
`rArr t= 1,027 ""^(@)C`