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 Rθ = R0 (1 + αθ + βθ2), find the values of R0, α and β. Here θ represents the temperature on the Celsius scale.

Answer 1

$\alpha=3.8 \times 10^{-3}$$^{\circ} \mathrm{C}^{-2}$, $\beta=-5.6 \times 10^{-7}$$^{\circ} \mathrm{C}^{-2}$, $\mathrm{R}_{0}=20 \Omega$.

Explanation:

resistance at 0°C is , $\mathrm{R}_{0}=20 \Omega$

resistance at 100°C is, $R_{100}=27.5 \Omega$

resistance at 420°C is , $R_{420}=50.0 \Omega$

Given equation is    

$R_{\theta}=R_{0}\left(1+\alpha \theta+\beta \theta^{2}\right)$

Take $R_{0}$ inside the bracket        

$R_{\theta}=R_{0}+R_{0} \alpha \theta+R_{0} \beta \theta^{2}$

Case 2:

When $\theta=100^{\circ} \mathrm{C}$

$R_{100}-R_{0}=R_{0} \alpha 100+R_{0} \beta(100)^{2}$

$\frac{R_{100}-R_{0}}{R_{0}}=\alpha 100+\beta 10000$ ………………(1)

$\frac{27.5-20}{20}=\alpha 100+\beta 10000$

$800 \alpha+80000 \beta=3$   ………………..(2)

Case 2:

When $\theta=420^{\circ} \mathrm{C}$

From equation (1)

$\frac{R_{420}-R_{0}}{R_{0}}=\alpha 420+\beta 176400$  

$840 \alpha+352800 \beta=3$    ……………….(3)

By solving equations (2) and (3)

$\alpha=-6820 \beta$

=>  $\beta=-5.6 \times 10^{-7}$$^{\circ} \mathrm{C}^{-2}$

=>  $\alpha=3.8 \times 10^{-3}$$^{\circ} \mathrm{C}^{-2}$

Answer 2

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