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Why is it not possible to know the ratio of 1°C to how many Fahrenheit and use that to convert from or to Celsius/Fahrenheit?

What is really happening? Fahrenheit increases linearly and so does Celsius, so why is there no ratio?

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    $\begingroup$ Fahrenheit and Celsius do not scale linearly with the amount of heat an object contains. $\endgroup$ – DumpsterDoofus Nov 16 '13 at 5:08
  • $\begingroup$ @DumpsterDoofus: That should be an answer; it's much better than the existing ones, which seem to not get the question. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 16 '13 at 6:56
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    $\begingroup$ @DumpsterDoofus -- ??? How do they not scale linearly?? When you look at the charts below the lines are straight. If they did not scale linearly then Alfred Centauri's chart below would show a curve. $\endgroup$ – Hot Licks Jan 29 '15 at 21:55
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There is a conversion factor, but it isn't just a multiplication because the two scales do not have their zero at the same point.

The celcius scale is defined as having its zero at the freezing point of water. The fahrenheit scale, on the other hand, appears simply arbitrary

To convert from celcius to fahrenheit, you multiply by $9/5$ (the ratio to account for the difference in slope) and then add $32$ to account for the difference in zeros.

Graphing the celcius and fahrenheit scales

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A picture, as they say, is worth a thousand words:

enter image description here

See that, in fact, the ratio of Fahrenheit to Celsius is:

$$ \dfrac{32 + \frac{9}{5}x}{x}$$

where $x$ is the temperature in degrees Celsius.

Clearly, the reason the ratio is not constant is the presence of the constant offset 32. Only in the limit of arbitrarily high temperature does the ratio approach the constant $\frac{9}{5}$.

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    $\begingroup$ As can be seen, there is a "linear" relationship between the two scales, it's just that they have different origins. $\endgroup$ – Hot Licks Jan 29 '15 at 21:57
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Well there IS a 1.8 x ratio between Fahrenheit degrees and Celsius degrees, but the origin is at -40 degrees instead of at zero.

So F+40 = 1.8 x (C+40)

For the boiling point of water F+40 = 1.8 x (100+40) = 252 = 212 + 40.

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Linearity does not necessarily mean a ratio exists. Linearity in this case means that a given change in one variable causes the same change in the other variable no matter what the value of the variable was before the change. As explained by the other answers, $F=32+\frac95 C$. $F$ is a linear polynomial in $C$ (i.e, $F(c)=aC+b$), but $F$ is not a linear map ($F(C_1+C_2)\neq F(C_1)+F(C_2)$ )

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Fahrenheit and Celsius scales begin at different arbitrary points, so there cannot be a simple ratio between the two. Imagine if we asked two identical twins how many years it's been since they reached the legal drinking age - one lives in the US, where the drinking age is 21, and the other in Germany where the drinking age is 18. At the age of 21, the ratio is 0:3, at 22 it's 1:4, at 23 it's 2:5...

Clearly, both values are linear, yet there is not a simple ratio between them.

Note that if you convert both temperatures to absolute scales (Kelvin and Rankine), then there is a simple ratio between them, because they now have a common starting point.

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Well, there is a linear transformation from Fahrenheit to Celcius and vice versa (http://en.wikipedia.org/wiki/Fahrenheit#Definition_and_conversions ), but it is not just a coefficient, you need to shift the origin as well, as zero degree Fahrenheit is not the same as zero degree Celcius.

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Celsius has 100 degrees between the freezing point of water at 0°C, and boiling point at 100°C.

Fahrenheit has 180 degrees between freezing point a 32°F, and boiling point at 212°F (212-32=180).

Since 100 degrees C = 180 degrees F, then 1 degree C = 1.8 degree F

And since the Fahrenheit scales define the freezing point of water at 32°F, it can be seen that a temperature in Fahrenheit = 1.8 x (temp in Centigrade) + 32, or $T_F = 1.8T_C + 32$ or $T_F = {^9/_5}T_C + 32$

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protected by Qmechanic Aug 1 '17 at 16:46

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