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