You're on the phone with a friend in the USA on a freezing cold winter day. "It's minus \(40\) degrees here!" you both exclaim simultaneously. Normally, this would be a matter of clarifying who means Celsius and who means Fahrenheit—but not at this particular temperature. Why is that? This point is the only temperature at which the Celsius and Fahrenheit scales agree!
\(−40\) degrees Fahrenheit is exactly \(−40\) degrees Celsius. This is no coincidence, but a direct consequence of the linear relationship between the two scales. Both temperature scales are affine transformations (linear + shift) of the same physical quantity, "temperature." Converting between these two scales is often tedious. However, there is an interesting point at which both scales have the same numerical value.
- Celsius scale (°C):
\(0^\circ\mathrm{C}\) Freezing point of water
\(100^\circ\mathrm{C}\) boiling point of water
Distance between these fixed points: \(100\) degrees. - Fahrenheit scale (°F):
\(32^\circ\mathrm{F}\) Freezing point of water
\(212^\circ\mathrm{F}\) Boiling point of water
Distance between these fixed points: \(212-32=180\) degrees.
This determines the ratio (slope) between the scales:
\[
\frac{180}{100}=\frac{9}{5}
\]
The zero point (offset) is also different: \(0^\circ\mathrm{C}\) corresponds to \(32^\circ\mathrm{F}\) .
To derive the standard formula, we look for an affine mapping of the form
\[
T_\mathrm{F}=a T_\mathrm{C}+b,
\]
where \(a\) the slope (scale factor) and \(b\) is the offset.
The following two conditions are sufficient because an affine mapping through two points is uniquely determined:
- \(T_\mathrm{C}=0 \Rightarrow T_\mathrm{F}=32 \Rightarrow 32 = a\cdot 0 + b \Rightarrow b=32.\)
- \(T_\mathrm{C}=100 \Rightarrow T_\mathrm{F}=212 \Rightarrow 212 = a\cdot 100 + 32 \Rightarrow a=\frac{212-32}{100}=\frac{180}{100}=\frac{9}{5}.\)
Substituting yields the standard formula:
\[
T_\mathrm{F}=\frac{9}{5} T_\mathrm{C}+32
\]
The inverse (from Fahrenheit to Celsius) is obtained by solving for \(T_\mathrm{C}\) :
\[
T_\mathrm{C}=\frac{5}{9}\left(T_\mathrm{F}-32\right)
\]
Now we are looking for the temperature \(T\) at which the identical numerical value appears in both scales:
\[
T_\mathrm{F}=T_\mathrm{C}\equiv T
\]
Now insert \(T_\mathrm{F}\) into the standard formula:
\[
T=\frac{9}{5}T+32 \Leftrightarrow T-\frac{9}{5}T=32
\]
and finally
\[
\left(1-\frac{9}{5}\right)T=32 \quad\Rightarrow\quad \left(\frac{5}{5}-\frac{9}{5}\right)T=32 \quad\Rightarrow\quad -\frac{4}{5}T=32.
\]
This results in \(T\)
\[
T=-32\cdot\frac{5}{4}=-8\cdot5=-40
\]
and thus
\[
-40^\circ\mathrm{F} = -40^\circ\mathrm{C}.
\]
For positive Celsius values, \(T_\mathrm{F}=\tfrac{9}{5}T_\mathrm{C}+32\) always a larger numerical value than \(T_\mathrm{C}\) (e.g. \(0^\circ\mathrm{C} \rightarrow 32^\circ\mathrm{F}\), \(20^\circ\mathrm{C}\rightarrow68^\circ\mathrm{F})\). For sufficiently negative Celsius values, the \(32\) Degrees at the start of the Fahrenheit scale is actually below zero. At some point, this compensates for the scale factor \(\frac{9}{5}\). This balancing point is exactly \(−40\): there is the additional shift \(+32\) just large enough so that both numerical values are identical. Graphically, \(T_\mathrm{F}= \tfrac{9}{5}T_\mathrm{C}+32\) (a straight line) and \(T_\mathrm{F}=T_\mathrm{C}\) (diagonal) – the intersection point of their lines is at \((-40,-40)\).
In contrast, absolute temperatures (e.g., for thermodynamic calculations) are given in Kelvin or Rankine, where there is no offset in the scale conversion (only a pure scale factor). For example, between Celsius and Kelvin \(T_\mathrm{K} = T_\mathrm{C} + 273{,}15\) applies. The existence of this offset is precisely the reason why the Celsius-Fahrenheit mapping is affine and not purely linear. The equality \(-40^\circ\mathrm{F}=-40^\circ\mathrm{C}\) follows directly from the affine relationship between Fahrenheit and Celsius.
If you substitute \(T_\mathrm{F}=T_\mathrm{C}\) into \(T_\mathrm{F}=\tfrac{9}{5}T_\mathrm{C}+32\) and solve, you clearly get \(T=-40\) . This is exactly where the two scales intersect. This intersection point at \(-40\) is the only point at which the numerical values of both scales are identical. This is due to the linear nature of the conversion: two non-parallel lines always intersect at exactly one point. So the next time someone mentions \(-40\) degrees, you don't have to explicitly ask which scale they mean.