# Theoretical justification for the range of validity of the approximation $R(T)\approx R(T_0)[1+\alpha (T-T_0)]$

In the experiment for calibrating a platinum resistance thermometer, we are always approximating the resistance of the platinum thermometer by $$R(T) \approx R_0 (1+\alpha T),$$ taking the reference at $$0^o C$$ and using Celsius scale.

We measure temperature by a thermometer of least count $$0.5^o C$$, and resistance of the platinum thermometer by a Carey Foster bridge which is sensitive up to $$0.005\Omega$$.

We work in the range of $$0^o C-100^o C$$.

Qusetion: How can we justify (without saying that, experimentally it is observed...) the usage of the approximation for $$R(T)$$ for this temperature range given the precision of our apparatus?

What I’m basically asking is the justification for ignoring the higher order terms in the Taylor expansion of $$R(T)$$ at $$T_0$$ for ranges as large as $$100^o C$$?

• Well, you could look at the tables of resistance vs temperature for your particular resistor and see how linear it is. – Jon Custer Sep 20 at 13:59
• I asked on theoretical grounds... – Atom Sep 20 at 14:00
• Well, the theory is in the Taylor expansion. The check if it is reasonable is the actual data for you particular material. Deriving resistivity vs temperature for an arbitrary material across a large temperature range is, well, quite difficult. – Jon Custer Sep 20 at 14:02
• @JonCuster So the range of validity comes purely from experiment? – Atom Sep 20 at 14:04
• Indeed. You might be interested in tsapps.nist.gov/publication/get_pdf.cfm?pub_id=904572 - it would suggest that the range should be based on the particular calibration requested. – Jon Custer Sep 20 at 14:07

The response of a platinum resistance thermometer (PRT) is better approximated with the so-called Callendar-Van Dusen equation (this is defined in the standard IEC EN 60751):

$$R(t) = \begin{cases} R_0\left[1+At+Bt^2+C(t-100\,{}^\circ\mathrm{C})t^3\right]& t<0 \\[5mm] R_0(1+At+Bt^2)& t\geqslant 0 \end{cases}$$

where $$t$$ is the Celsius temperature and the nominal values of the coefficients are (a calibration can give more accurate values)

\begin{aligned} &R_0 = 100\,\Omega \\ &A = 3.9083\times 10^{-3}/{}^\circ\mathrm{C}, \\ &B = -5.775\times 10^{-7}/{}^\circ\mathrm{C}^2, \\ &C = -4.183\times 10^{-12}/{}^\circ\mathrm{C}^4, \end{aligned}

For $$t$$ from $$0\,{}^\circ \mathrm{C}$$ to $$100\,{}^\circ \mathrm{C}$$, the response is

$$R(t) = R_0(1+At+Bt^2)$$

If you use instead the linear approximation $$R(t) = R_0(1+\alpha t)$$, you will measure the temperature $$t'$$ such that (assuming that $$R_0$$ is the same for the two equations; otherwise, modify accordingly)

$$R_0(1+At+Bt^2) = R_0(1+\alpha t')$$

The difference $$\Delta t = t'-t$$ is the measurement error that you get from using the linear approximation. Substituting $$t' = t+\Delta t$$ in the above equation and simplifying, you get

$$At+Bt^2 = \alpha(t+\Delta t)$$

from which

$$\Delta t = \dfrac{1}{\alpha}\left[(A-\alpha)t+B t^2\right].$$

The above equation represents a parabola, and substituting the value of $$\alpha$$ that you use you can find the maximum error along the whole measurement range. For instance, if you use $$\alpha = 3.85\times 10^{-3}/{}^\circ \mathrm{C}$$, the nominal average temperature coefficient from $$0\,{}^\circ \mathrm{C}$$ to $$100\,{}^\circ \mathrm{C}$$ (according to the EU standards, for a US PRT you may get $$\alpha = 3.92\times 10^{-3}/{}^\circ \mathrm{C}$$), the error $$\Delta t$$ is maximum at $$t = 50{}^\circ \mathrm{C}$$ and $$(\Delta t)_\mathrm{max}\approx 0.38{}^\circ \mathrm{C}$$. This specific case is represented in the figure below (own picture):

Depending on your application and the accuracy class of the PRT, you can then evaluate whether you can consider the above error negligible or not.