Celsius and Fahrenheit are two of the common temperature scales used today. Here I explain how to derive the formula for converting between the two temperature scales.

Conversion from Celsius to Fahrenheit

Figure 1: “Side-by-side diagrams of Celsius and Fahrenheit thermometers, with boiling/freezing point of water labelled.” by Gringer is released into the public domain.

First, let’s examine the two scales. The freezing point of water is set at \(0\degree C\) on the Celsius scale and \(32\degree F\) on the Fahrenheit scale. Similarly, the boiling point of water is marked as \(100\degree C\) on the Celsius scale and \(212\degree F\) on the Fahrenheit scale.

Thus, the difference between the boiling and freezing points of water is \(100\) units on the Celsius scale \((100\degree C - 0\degree C)\) and \(180\) units on the Fahrenheit scale \((212\degree F - 32\degree F)\). In other words, a difference of \(100\) units on the Celsius scale is equivalent to a difference of \(180\) units on the Fahrenheit scale.

From the above, we can conclude that \(1\) unit change on the Celsius scale corresponds to \(1.8\) (which is \(\frac{180}{100}\) or \(\frac{9}{5}\)) unit change on the Fahrenheit scale.

So, if \(0\degree C\) is equal to \(32\degree F\), then \(1\degree C\) is equal to \(32\degree F + 1.8\degree F\). This is because we are adding \(1\degree C\) to the Celsius scale and adding the equivalent \(1.8\degree F\) to the Fahrenheit scale.

Similarly, to find the equivalent of \(5\degree C\) in Fahrenheit, we add \(1.8\degree F\) five times to \(32\degree F\). This calculation is:

\[32 + 5 \times 1.8 = 32 + 5 \times \frac{9}{5} = 32 + 9 = 41\degree F\]

To find the equivalent of \(-5\degree C\) in Fahrenheit, we subtract \(1.8\degree F\) five times from \(32\degree F\). This calculation is:

\[32 - 5 \times 1.8 = 32 - 5 \times \frac{9}{5} = 32 - 9 = 23\degree F\]

Note

From the above, we can then see that to convert a Celsius temperature, C, to its equivalent Fahrenheit temperature, F, we can use the formula:

\[F = \frac{9}{5}\times C + 32\]

Conversion from Fahrenheit to Celsius

We can derive the formula for converting a Fahrenheit temperature, F, to its equivalent Celsius temperature, C, by solving for C from following formula:

\[F = \frac{9}{5}\times C + 32\]

Subtracting \(32\) from both side, we get:

\[F - 32 = \frac{9}{5}\times C\]

Multiplying both side by \(\frac{5}{9}\), we get:

\[C = \frac{5}{9} \times (F-32)\]

Note

So to convert a Fahrenheit temperature, F, to its equivalent Celsius temperature, C, we can use the formula:

\[C = \frac{5}{9} \times (F-32)\]

In summary

Let us focus on a particular temperature, measured as C on the Celsius scale and as F on the Fahrenheit scale. Assume it falls between the freezing and boiling points of water, only to make it easier to picture (see Figure 1). Let us also denote freezing point of water as FP and boiling point of water as BP.

Now the change in temperate to C from (FP)_Celsius is proportionate to the change in temperature to F from (FP)_Fahrenheit.

This proportion can be stated in the equation:

\[\frac{C - \text{(FP)}_{\text{Celsius}}}{\text{(BP)}_\text{Celsius} - \text{(FP)}_{\text{Celsius}}} = \frac{F - \text{F}_{\text{Fahrenheit}}}{\text{(BP)}_\text{Fahrenheit} - \text{(FP)}_{\text{Fahrenheit}}}\]

Substituting the values for the freezing and boiling points of water in the respective scales, we get:

\[\frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32} \implies \frac{C}{100} = \frac{F - 32}{180}\]

Taking the cross product, we get:

\[180 \times C = 100 \times (F-32) \implies 9 \times C = 5 \times (F-32)\]

Solving for F in the above equation, we can see that:

\[F = \frac{9}{5}\times C + 32\]

and, solving for C in the above equation, we can see that:

\[C = \frac{5}{9} \times (F-32)\]