The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = Ro [1 + α(T – To)]
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?
Given,
Triple point of water, T0 = 273.16 K
Resistance at triple point, R0 = 101.6 Ω
Normal melting point of lead, T = 600.5 K
Resistance at normal melting point of lead, R = 165.5 Ω
Also,
R = R0[1+α(T-T0)]
Where, R0 is the initial resistance
Ris the final resistance
T0 is the initial temperature
T is the final temperature
∴ 165.5 Ω = (101.6 Ω)[1 + α(600.5 K - 273.16 K)]
⇒ 1+α(327.34) = 165.5/101.6
⇒ 1+327.34α = 1.629
⇒ 327.34α = 0.629
⇒ α = 0.629/327.34
⇒ α = 1.92 × 10-3 K-1
For R = 123.4 Ω,
123.4 Ω = (101.6 Ω)[1 + (1.92 × 10-3 K-1)(T - 273.16 K)]
⇒ 1 + (1.92 × 10-3 K-1)(T - 273.16 K) = 123.4/101.6
⇒ 1 + (1.92 × 10-3 K-1)(T - 273.16 K) = 1.214
⇒ (1.92 × 10-3 K-1)(T - 273.16 K) = 0.214
⇒ T - 273.16 K = 0.214/(1.92 × 10-3 K-1)
⇒ T – 273.16 K = 111.75 K
⇒ T = 111.75 K + 273.16 K
⇒ T = 384.91 K