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


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