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?

Answer 1

Here,
Ro = 101.6 ohm , To = 273.16K
R1 = 165.5 ohm , T1 = 600.5 K
R2 = 123.4 ohm , T2 = ?

we know,
R = Ro [1 +a( T - To)]
So, R1 = Ro [ 1 + a(T1-To)]
165.5 = 101.6[ 1 + a( 600.5 - 273.16)]
a = 63.9/( 101.6 × 327.34 )

Now,
R2 = Ro [1 + a(T2 - To)]
123.4 = 101.6 [ 1 + 63.9/101.6×327.34 ( T2 - 273.16)
T2 = ( 123.4 - 101.6)×327.34/63.9 + 273.16
= 384.83 K