#### Question

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

*R *= *R*_{o} [1 + α (*T *– *T*_{o})]

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

#### Solution 1

Here, R_{0} = 101.6 Ω; T_{0} = 273.16 K Case (i) R_{1}= 165.5 Ω; T_{1 }= 600.5 K, Case (ii) R_{2} = 123.4 , T_{2} = ?

Using the relation R = R_{0}[1 + α (T – T_{0})]

Case (i) 165.5 = 101.6 [1 + α (600.5 – 273.16)]

`alpha= (165.5 - 101.6)/(101.6xx(600.5-273.16)) = 63.9/(101.6xx327xx37)`

Case II `123.4 = 101.6 [1 + alpha(T_2 - 273.16)]`

or `123.4 = 101.6 [1+ 63.9/(101.6xx327.34)(T_2-273.16)]`

`= 101.6 + 63.9/327.37 (T_2 - 273.16)`

or `T_2 = ((123.4-101.6)xx327.34)/63.9+ 273.16 = 111.67 + 273.16`

= 384.83 K

#### Solution 2

It is given that:

*R *= *R*_{0} [1 + α (*T *– *T*_{0})] … (*i*)

Where,

*R*_{0} and *T*_{0} are the initial resistance and temperature respectively

*R* and *T* are the final resistance and temperature respectively

α is a constant

At the triple point of water, *T*_{0} = 273.15 K

Resistance of lead, *R*_{0} = 101.6 Ω

At normal melting point of lead, *T* = 600.5 K

Resistance of lead, *R* = 165.5 Ω

Substituting these values in equation (*i*), we get:

`R = R_0[1+alpha(T - T_0)]`

`165.5=101.6 [1+alpha(600.5 - 273.15)]`

`1.629 = 1 + alpha(327.35)`

`:.alpha = 0.629/327.35 = 1.92 xx 10^(-3) K^(-1)`

For resistance, `R_1= 123.4 Omega`

`R_1 = R_0[1+alpha(T-T_0)]`

Where T is the temperrature when the resistance of lead is `123.4 Omega`

`123.4 = 101.6[]1+1.92 xx 10^(-3)(T-273.15)]`

`1.214 = 1 + 1.92 xx 10^(-3)(T - 273.15)`

`0.214/(1.92xx10^(-3)) = T - 273.15`

:. T = 384.61 K