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I have not been able to find an answer to what may be a silly question so I am asking it on here as a last resort.

If I am to plug in say a number like 14.0 Celsius into the formula that solves for temperature in Fahrenheit. So I would do 9/5 times 14.0 to get 25.2. When I then add the 25.2 to 32 does the rule for addition matter here?

Would the answer be 57.2 or 57?

Is there a general rule for when I am plugging in a measured value into a formula that is given to me? I believe that only the measured values are significant and all numbers in the formula are exact with infinite significant digits.

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If the conversion is exact, all uncertainties carry over. Arguing the same for significant digits is very tricky and can lead you on erroneous paths. It is thus easier for you to convert the uncertainty that is present in the originally measured value into a new uncertainty for the converted value and then infer from that the number of significant digits.

Let's assume that your measurement is $\vartheta = 14.0 \pm 0.1 ~ \mathrm{^\circ C}$.

The conversion to Fahrenheit is $$ \phi(\vartheta) = \left(\frac{9}{5} \frac{\vartheta}{\mathrm{^\circ C}} + 32 \right) ~ \mathrm{^\circ F} \,,$$ which we can straightforwardly apply to the mean of the measurement which gives us $57.2~\mathrm{^\circ F}$.

To do the error propagation under the assumption that the errors are independent, we do the following: $$\sigma_\phi = \sqrt{\left( \frac{\partial\phi}{\partial\vartheta} \right)^2 \sigma_\vartheta^2} = \frac{\partial\phi}{\partial\vartheta} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \sigma_\vartheta = \frac{9}{5}\frac{\mathrm{^\circ F}}{\mathrm{^\circ C}} \cdot 0.1~\mathrm{^\circ C} = 0.18~\mathrm{^\circ F}$$ The term under the root would in principle be a sum, but here we only have one variable with an uncertainty associated with it. For more details, look up this Wikipedia article on error propagation.

Anyway, our newly converted value is correctly expressed as $$ \phi = 57.20 \pm 0.18 ~ \mathrm{^\circ F} \,.$$

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  • $\begingroup$ This mostly answers the question. Since 14.0 Celsius has three significant figures then when converted to Fahrenheit we'd generally write three significant figures also, so 57.2 F would be the better conversion. But that obviously belies the exact situation. The 14.0 Celsius temperatures value is measured +/- 0.05 degree which is not exactly the same as +/- 0.05 F which is indicated by 57.2 F value. You'd have to make a judgement call if the error conversion is necessary too. $\endgroup$ – MaxW Jan 28 '16 at 20:52
  • $\begingroup$ Point being the rule of thumb is to use same number of significant figures in the answer as the smallest number of significant figures used in the any one of values used in the calculation. $\endgroup$ – MaxW Jan 28 '16 at 20:58
  • $\begingroup$ This is new to me. How would I choose between the error conversion and not using it? So the error being +/- 0.1 why do you put +/- o.o5 degrees MaxW? $\endgroup$ – theRealChainman Jan 29 '16 at 6:15
  • $\begingroup$ So the error must be converted to the new units as well always? What does it mean to assume that the errors are independent? $\endgroup$ – theRealChainman Jan 29 '16 at 6:21
  • $\begingroup$ Usually for our purposes the errors are independent. If they are not (and that can sometimes be the case) then you need to use more advanced modelling of the error than the one I gave here. $\endgroup$ – tschoppi Jan 29 '16 at 6:53

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