Question: Suppose that on a linear temperature scale X, water boils at $-53.5°X$and freezes at$-170°X$. What is a temperature of$340K$on the X scale? (Approximate water’s boiling point as 373K.)

• i) Linear X temperature scale water boils at$-53.5°X$
• ii) Linear X temperature scale water freezes at $-170°X$
• iii) Linear Y scale =340K
• iv) Approximate water’s boiling point = 373K

Find a temperature of 340Kon the Xscaleby using linear dependence of temperature.

The linear dependence relation of temperature of X- and Y-axis is given as;

$Y=mX+b$

Consider scales X and Y as linearly related; therefore, by using the linear relationship of equation (i).

The boiling temperature relation in both scales is given as:

373.15 = m(-53.5)+b ……. (ii)

Determine freezing temperature relation in both scales is given as:

273.15 = m( -170.0) +b

…… (iii)

Subtracting (iii) from (ii),determine value of the slope, m as:

100.00 = 116. 5m
m = 0.858

Substitute the value of m in equation (ii), and solve for the value of constant, b as:
$$\begin{gathered} 373.15=(0.858\times (-53.5\left)\right)+b \\ 373=-45.92+b \\ b=419 \end{gathered}$$

From equation (i) and the above values, solve for the value of X-scale as:

$X=\frac{Y-b}{m}$

Substitute the value and solve as:

$$\begin{gathered} X=\frac{340-419}{0.858} \\ =-92.07 \\ ~-92.1° \end{gathered}$$

Therefore, the temperature of 340 K on the X scale is $-92.1°X$.