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?

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