Question

A resistance thermometer reads R = 20.0 Ω, 27.5 Ω, and 50.0 Ω at the ice point (0°C), the steam point (100°C) and the zinc point (420°C), respectively. Assuming that the resistance varies with temperature as R_θ = R₀ (1 + αθ + βθ²), find the values of R₀, α and β. Here θ represents the temperature on the Celsius scale.

Answer 1

Given:
Reading on resistance thermometer at ice point, R₀ = 20 Ω
Reading on resistance thermometer at steam point, R₁₀₀ = 27.5 Ω
Reading on resistance thermometer at zinc point, R₄₂₀ = 50 Ω
The variation of resistance with temperature in Celsius scale,θ, is given as:

\[R_{100} = R_0 \left( 1 + \alpha \theta + \beta \theta^2 \right)\]

\[ \Rightarrow R_{100} = R_0 + R_0 \alpha\theta + R_0 \beta \theta^2 \]

\[ \Rightarrow R_{100} = R_0 + R_0 \alpha\theta + R_0 \beta \theta^2 \]

\[ \Rightarrow \frac{\left( R_{100} - R_0 \right)}{R_0} = \alpha\theta + \beta \theta^2 \]

\[ \Rightarrow \frac{27 . 5 - 20}{20} = \alpha\theta + \beta \theta^2 \]

\[ \Rightarrow \frac{7 . 5}{20} = \alpha \times 100 + \beta \times 10000 . . . \left( i \right)\]

\[Also, R_{420} = R_0 \left( 1 + \alpha\theta + \beta \theta^2 \right)\]

\[ \Rightarrow R_{420} = R_0 + R_0 \alpha\theta + R_0 \beta \theta^2 \]

\[ \Rightarrow \frac{R_{420} - R_0}{R_0} = \alpha\theta + \beta \theta^2 \]

\[ \Rightarrow \frac{50 - 20}{20} = 420\alpha + 176400 \beta\]

\[ \Rightarrow \frac{3}{2} = 420\alpha + 176400 \beta . . . \left( ii \right)\]

Solving (i) and (ii), we get:
α = 3.8 ×10^–3°C^​-1
β = –5.6 ×10^–7°C^-1
Therefore, resistance R₀ is 20 Ω and the value of α is 3.8 ×10^–3°C^​-1  and that of β is –5.6 ×10^–7°C^-1.