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 W at the triple-point of water 273.16 K, and 165.5 W at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 W

Answer 1

Given: The resistance is 101.6 Ω at the triple point of water 273.15 K, the resistance is 165.5 Ω at the normal melting point of lead 600.5 K.

The equation for the resistance of thermometer is given as,

R= R 0 [ 1+α( T− T 0 ) ](1)

Where, R 0 and T 0 are the resistance and temperature at triple point of water, R and T are the resistance and temperature at the normal melting point of lead and α is a constant.

By substituting the given values in the above equation, we get

165.5=101.6[ 1+α( 600.5−273.16 ) ] 165.5 101.6 =[ 1+327.34α ] α= 1.63−1 327.34 α=1.924× 10 −3   K −1

Again, substituting the values for R=123.4 Ω in equation (1), we get

123.4=101.6[ 1+1.924× 10 −3 ( T−273.16 ) ] 123.4 101.6 −1=1.924× 10 −3 ( T−273.16 ) T= 0.2146 1.924× 10 −3 +273.16 =384.7 K

Thus, the value of the temperature is 384.7 K.