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Question

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?

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Solution

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:

For resistance, R1 = 123.4 Ω


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