Question 14 of 20intermediate🔧 ApplyNumerical3 marks

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R=R0[1+α(T−T0)]R = R_0 [ 1 + \alpha (T - T_0) ]. The resistance is 101.6 Ω\Omega at the triple-point of water 273.16 K, and 165.5 Ω\Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω\Omega?

Correct Answer

The temperature when the resistance is 123.4 Ω is approximately 384.9 K.

Exercise: EXERCISES | Q: 10.3 | (Chapter: 22)
For More Understanding

Explanation

This problem applies the linear relationship between electrical resistance and temperature. The triple-point of water serves as the reference point (R₀, T₀), and we first determine the temperature coefficient α using the known resistance at lead's melting point. Then, we substitute the given resistance value to find the corresponding temperature.

Solution Steps

  1. Step 1: Identify given values At triple-point of water: T₀ = 273.16 K, R₀ = 101.6 Ω At normal melting point of lead: T₁ = 600.5 K, R₁ = 165.5 Ω Given resistance: R = 123.4 Ω, Find T

  2. Step 2: Calculate temperature coefficient α Using R = R₀[1 + α(T - T₀)]: 165.5 = 101.6[1 + α(600.5 - 273.16)] 165.5 = 101.6[1 + α(327.34)] 165.5/101.6 = 1 + 327.34α 1.629 = 1 + 327.34α α = 0.629/327.34 = 0.00192 K⁻¹

  3. Step 3: Find temperature T when R = 123.4 Ω 123.4 = 101.6[1 + 0.00192(T - 273.16)] 123.4/101.6 = 1 + 0.00192(T - 273.16) 1.215 = 1 + 0.00192(T - 273.16) 0.215 = 0.00192(T - 273.16) T - 273.16 = 111.98 T = 384.9 K

PreviousNext