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.

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Solution

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:
$R_{100} = R_{0} (1 + α θ + β θ^{2}) \Rightarrow R_{100} = R_{0} + R_{0} α θ + R_{0} β θ^{2} \Rightarrow R_{100} = R_{0} + R_{0} α θ + R_{0} β θ^{2} \Rightarrow \frac{(R_{100} - R_{0})}{R_{0}} = α θ + β θ^{2} \Rightarrow \frac{27.5 - 20}{20} = α θ + β θ^{2} \Rightarrow \frac{7.5}{20} = α \times 100 + β \times 10000 . . . (i) Also, R_{420} = R_{0} (1 + α θ + β θ^{2}) \Rightarrow R_{420} = R_{0} + R_{0} α θ + R_{0} β θ^{2} \Rightarrow \frac{R_{420} - R_{0}}{R_{0}} = α θ + β θ^{2} \Rightarrow \frac{50 - 20}{20} = 420 α + 176400 β \Rightarrow \frac{3}{2} = 420 α + 176400 β . . . (i i)$
Solving (i) and (ii), we get:
α = 3.8 ×10^–3°C^-1
β = –5.6 ×10^–7°C^-1
Therefore, resistance R₀is 20 Ω and the value of α is 3.8 ×10^–3°C^-1and that of β is –5.6 ×10^–7°C^-1.

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A resistance thermometer reads R = 20.0 $Ω$ , 27.5 $Ω$ , and 50.0 $Ω$ at the ice point ( $0^{\circ} C$ ), the steam point ( $100^{\circ} C$ ) and the zinc point ( $420^{\circ} C$ ) respectively. Assuming that the resistance varies with temperature as $R_{θ} = R_{0} (1 + α θ + β θ^{2})$ , find the values of $R_{0}, α$ and $β$ . Here $θ$ represents the temperature on Celsius scale.