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} = R_{0} (1 + αθ + βθ^{2}), find the values of R_{0}, α and β. Here θ represents the temperature on the Celsius scale.

#### Solution

Given:

Reading on resistance thermometer at ice point, *R*_{0} = 20 Ω

Reading on resistance thermometer at steam point, *R*_{100} = 27.5 Ω

Reading on resistance thermometer at zinc point, *R*_{420} = 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 *R*_{0 }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}.