Question

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?

Answer 1

It is given that:

R = R₀ [1 + α (T – T₀)] … (i)

Where,

R₀ and T₀ 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, T₀ = 273.15 K

Resistance of lead, R₀ = 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: