# Why does this temperature conversion procedure work?

Recently while playing with conversion between different temperature scales I found a quite interesting and simple procedure for conversion from one scale to another.

This is as follows:

Suppose we have two temperature scales in which melting and boiling point of water has been labeled differently (it is not necessary to take melting and boiling point of water rather any other phenomena which occurs at a given temperature).

• In scale A the melting point is numerically labeled as $$a$$ whereas boiling point is labeled as $$b$$.

• In scale Z the melting point is labeled as $$\alpha$$ and boiling point as $$\beta$$.

Now in scale A we measure a body's temperature as $$t_A$$ and in scale Z we measure the same body's temperature as $$t_z$$. Then it happens so that:

$$\frac {t_A-a}{b-a} = \frac {t_Z-\alpha}{\beta-\alpha}$$

So my question is:

• Why does this procedure work?

• Is there a physical meaning attached to this? If so, what?

Sorry I don't know what this is called so I wasn't able to search for it on the internet.

An example of if this method really works is the conversion between Celsius ($$t_C$$) and Fahrenheit($$t_F$$):

$$\frac {t_F-32}{180} = \frac {t_C}{100}$$

• You are converting both temperatures to a common scale in which the melting point is 0 and the boiling point is 1.
– user856
Feb 1, 2020 at 13:36

Temperature scales are designed based on one assumption that there is a linear relationship between Heat (h) and temperature (t). $$h = mt + c$$ For different scales you may get different values of constants m and c.
Consider for scale A, $$\frac {t_A-a}{b-a} =\frac {\frac {h_A-c}{m}-\frac {h_a-c}{m}}{b-a}=\frac {h_A-h_a}{m(b-a)}=1$$ The term $$\frac {h_A-h_a}{b-a}$$ is nothing but m.
Similarly, for any other scale Z, you will get $$\frac {t_Z-\alpha}{\beta-\alpha} =1$$
Hence, $$\frac {t_A-a}{b-a} = \frac {t_Z-\alpha}{\beta-\alpha}$$