Two of the most common temperature scales are the Celsius and Fahrenheit scales. One common problem is converting between these two. We might have encountered these two formulas used to convert values:
- \(T_F=\frac{9}{5}T_C+32\)
- \(T_C=\frac{5}{9}(T_F-32)\)
If you want to convert 18ºC to Fahrenheit, substitute 18ºC to the formula:
\(T_F=\frac{9}{5}T_C+32\)
\(T_F=\frac{9}{5}\times{18}+32\)
At the end of your computation, you should get 64.4ºF. Conversely, if I want to convert something in Fahrenheit to Celsius, I would use the other formula.
A common question is, what if I forgot the formula?
We Don't Need to Memorize the Formula
There is a way to convert temperature accurately without memorizing the formula. This method will require two things:
Celsius Versus Fahrenheit Graph
First, we must create a graph between the Celsius and Fahrenheit scales. We let the x-axis be the Celsius and the y-axis be the Fahrenheit scale.
Next, we need to plot the coordinates of the freezing and boiling points in both the Celsius and Fahrenheit scales.
Celsius Scale (Water)
- 0ºC = Freezing Point
- 100ºC = Boiling Point
Fahrenheit Scale (Water)
- 32ºF = Freezing Point
- 212ºF = Boiling Point
In our graph, these would be points (0, 32) and (100, 212). After plotting, we connect these two points with a straight line.
Our graph should look something like this:
This illustration is the only thing we need to convert from one temperature scale to another.
To demonstrate, let's convert 18ºC to Fahrenheit using the illustration. If we try to plot this point, it will be a third coordinate (18, \(T_F\)). Our objective is to find the missing coordinate of this third point. The question is, how?
How To Use The Graph?
Since we have a straight line, we can use linear interpolation. It may seem an intimidating term for the non-technical person, but it's simple.
The idea is that the slope between points along a straight line should be equal. Now, what is the slope? In Geometry, the slope \(m\) is rise-over-run. Expressing it between two coordinates \((x_1, y_1)\) and \((x_2, y_2)\), \(m\) is equal to:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
Let's compute the slope between the boiling and freezing points of water.
\(m=\frac{212-32}{100-0}\)
\(m=1.8=\frac{9}{5}\)
According to interpolation, this computed \(m\) must be the same between ANY two points along the line. So, if we consider the slope between the boiling point AND the third point, it must equal to \(m=1.8=\frac{9}{5}\). From there, we can compute for the missing temperature:
\(1.8=\frac{9}{5}=\frac{212-T_F}{100-18}\)
\(T_F=64.4\)
Congratulations! We have converted 18ºC to Fahrenheit (64.4ºF) without a formula!
We can also convert Fahrenheit to Celsius using the graph above. Just follow the same steps as we did. To check your answer, you can interact with the graphic above and look for the coordinates along the line.
It's good to note that this strategy still works if you want to convert beyond the freezing and boiling points - say you would like to know what is 300ºF in Celsius.
Why This Method is Better?
Now that we've illustrated how to convert using a graph, why should we use this method instead?
One good reason is that this strategy of converting temperatures is how we created those formulas.
Consider the graph again. We want to find the Fahrenheit value \(T_F\) for any Celsius value \(T_C\):
- The slope \(m\) between the freezing and boiling points is: \(m=\frac{212-32}{100-0}=\frac{9}{5}\)
- Let's consider \((T_C, T_F)\) and the freezing point. From interpolation, this slope must equal to: \(\frac{9}{5}=\frac{T_F-32}{T_C-0}\)
Isolating \(T_F\) in one side of the equation:
\(\frac{9}{5}=\frac{T_F-32}{T_C-0}\)
\(T_F=\frac{9}{5}T_C+32\)
This expression is the equation of the line in the graph in its slope-intercept form. It's the same as the well-known formula: \(T_F=\frac{9}{5}T_C+32\). From this equation, we can transpose the variables so that \(T_C\) is isolated. If we do, we'll get the other equation: \(T_C=\frac{5}{9}(T_F-32)\)
Another reason why converting temperature this way is better is because it isn't limited to just between Celsius and Fahrenheit. You can repeat the same procedure we did between any two scales (Celsius, Fahrenheit, Kelvin, Newton, Rankine, etc.). We need only two pieces of information - water's freezing and boiling point in both scales.
