Answer to Question #103567 in Molecular Physics | Thermodynamics for Ajay

Question

For a platinum wire, the coefficients of variation of resistance with temperature are α=3.94×10^(–3) °C^(–1)
and β=–5.82×10^(–7) °C^(–2).
A thermometer is fabricated using
this wire with
R0=12.00 Ω
where
R0 signifies the resistance of the material of wire
at ice point. When the thermometer is placed in contact with a heat bath, the
resistance is found to be
15.20Ω.
Calculate the temperature of the bath.

Answer 1

Solution. The temperature dependence of resistance on temperature can be represented by the formula

$$R_t=R_0(1+\alpha t+\beta t^2)$$

where R_t is resistance at temperature t⁰C ; R₀ is resistance at temperature 0⁰C;  α=3.94×10^(–3) °C^(–1) and β=–5.82×10^(–7) °C^(–2) are constants. (the formula is valid in the temperature range 0⁰С-850⁰С) According to the conditions of the problem R_t=15.20Ω,

R₀=12.00 Ω. As result get quadratic equation

$$15.20=12.00(1+3.94×10^{–3} t - 5.82×10^{–7} t^2)$$ $$- 5.82×10^{–7} t^2+3.94×10^{–3} t - 0.267=0$$

Find the roots of the quadratic equation

$$D=(3.94×10^{–3})^2-4( - 5.82×10^{–7} )(-0.267) \approx (3.86×10^{–3})^2$$

$$t_1=\frac {-3.94\times10^{-3}-3.86\times10^{-3}}{-2\times 5.82\times 10^{-7}}\approx 6692^0C$$

The temperature t₁ does not belong to the interval 0⁰С-850⁰С.

$$t_2=\frac {-3.94\times10^{-3}+3.86\times10^{-3}}{-2\times 5.82\times 10^{-7}}\approx 69^0C$$

Answer. 69⁰C