1.

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:R = R0[1 + α(T - T0)]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»

At the triple point of water,

T = 273.16 K and R = 101.6 Ω (given)

∴ R = R0[1 + 5 x 10-3(T - T0)]

⇒ 101.6 = R0[1 + 5 x 10-3 (273.16 - T0)]   ...(i)

At temperature, T = 600.5 K, R = 165.5 Ω

∴ 165.5 = R0[1 + 5 x 10-3(600.5 - T0)]  ...(ii)

Dividing (ii) by (i), we have,

165.5/101.6 = {R0[1 + 5 x 10-3(600.5 - T0)]}/{[R0[1 + 5 x 10-3(273.16 - T0)]]}

⇒ 1.63[1 + 5 x 10-3(273.16 - T0)]

= [1 + 5 x 10-3(600.5 - T0)]

⇒ 1.63 x 5 x 10-3(273.16 - T0)] + 1.63

= 1 + 5 x 10-3(600.5 - T0)

⇒ 5 x 10-3(600.5 - T0 - 1.63 x 273.16 + 1.63 T0) = 1.63 - 1

∴ 0.63 T0 + 155.25 = {0.63}/{5 x 10-3}

⇒ T0 = -46.42 K   ....(iii)

∴ From equation (i),

101.6 = R0[1 + 5 x 10-3{273.16 - (-46.42)}]

⇒ R0 = 101.6/2.6 = 39.1 Ω   ...(iv)

Now, when R = 123.4 Ω, T = ?

∴ Using, R = R0[1 + 5 x 10-3(T - T0)]

Taking value of R0 and T0 from equations (iii) and (iv), we get

123.4 = 39.1 x [1 + 5 x 10-3{T - (-46.42)}]

or, 1 + 5 x 10-3(T + 46.42) = 123.4/39.1 = 3.156

or, T + 46.42 = {3.156 - 1}/{5 x 10-3} = {2.156 x 1000}/{5} = 431.2

∴ T = 431.2 - 46.42 = 384.78 K


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