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I have plotted data of temperature against time for a cooling cup of coffee and gotten a nice curve. I have then linearised this data by taking the log of the temperature for the purposes of gatting a straight line. My question now is how do I deal with the uncertainty in the temperature measurement? If the absolute uncertainty was +/-0.5 degrees C what is it now that I've taken the log of the temperature? I found something else here that MIGHT be what I'm looking for but I'm afraid I don't understand that either :(

Many thanks

David

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This question is a little bit on the homework-like side of things for us, but there's so many heuristic rules about propagating uncertainties floating around that it seems worth stating correctly again.

The right way to propagate uncertainties in a single variable is to use calculus to decide what a small variation in the input does to the output.

In your case, you have $y = \ln x$, so

\begin{align} \delta y &= \frac{\partial y}{\partial x} \delta x = \frac{\delta x}{x} \end{align}

So the absolute error in $\ln x$ is the same as the fractional error in $x$. Beware of temperature units (like centigrade) with non-physical zeros.

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  • $\begingroup$ Note that this is only an approximate (first-order) formula, as many others that involve uncertainties. $\endgroup$ – Federico Poloni Nov 23 '18 at 13:28
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    $\begingroup$ To account for second-order corrections, you can either (a) Taylor-expand around the central value, or (b) repeat your experiment with better precision. $\endgroup$ – rob Nov 23 '18 at 15:08
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The general rule is that when you have a value $g$ that depends of another value $f$ then if you write $u(g)$ for the incertitude of $g$ then : $$ u(g(f)) = \frac{\partial g}{\partial f}(f) \times u(f) $$ With here $ g = \mathrm{log}(T) $ and so : $$ u(\mathrm{log}(T)) = \frac{u(T)}{T} $$

So for a temperature $T$ the incertitude on its log will be of $\frac{0.5}{T}$

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