Question

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = \(R_0\) [1 + α (T – \(T_0\))]
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 ?

Answer 1

It is given that:
R = \(R_0\) [1 + α (T – \(T_0\))] … (i) 
Where,
\(R_0 \space and \space T_0\) 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+α (T -\(\space T_0\))]
165.65 = 101.6 [1 + α (600.5 - 273.15)]
1.629 = 1 + α (327.35)
∴ α = \(\frac{0.629}{327.35}\) = 1.92 x \(10^{-3}\) \(K^{-1}\)

For resistance, \(R_1\)= 123.4 Ω
\(R_1\) = \(R_0\)[1+α (T - \(T_0\))]
Where, T is the temperature when the resistance of lead is 123.4 Ω
123.4 = 101.6[1 + 1.92 x \(10^{-3}\)(T - 273.15)]
1.214 = 1 + 1.92 x \(10^{-3}\)(T - 273.15)
\(\frac{0.214}{1.92}\) x \(10^{-3}\) = T - 273.15

∴ T = 384.61 K