Question

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : $R = R_0 [1 + \alpha (T - T_0)]$. 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 = 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:

R = Ro [1 + α (T – To)]

165.5 = 101.6 [ 1 + α(600.5 – 273.15) ]

1.629 = 1 + α (327.35)

∴ α = 0.629 / 327.35 = 1.92 × 10-3 K-1

For resistance, R1 = 123.4 Ω

R1= R0 [1 + α (T – T0)]

where,

T is the temperature when the resistance of lead is 123.4 Ω

123.4 = 101.6 [ 1 + 1.92 × 10-3( T – 273.15) ]

Solving for T, we get

T = 384.61 K.