Question

The normal boiling point of water is ${100}^{\circ }C$ or ${212}^{\circ }F$ and the freezing point of water is $0^{\circ }C$ or ${32}^{\circ }F$.
(i) Find the linear relationship between $C$ anf $F$
Find (ii) the value of $C$ for ${98.6}^{\circ }F$ and (iii) the value of $F$ for ${38}^{\circ }C$

Answer 1

Let the x-axis represent the temperature in Celsius

and y-axis represent the temperature is Fahrenheit.

It is given that, the normal boiling point of water is (100, 212)

and the freezing point of water is (0,32)

(1) the equation of line between (100, 212) and (0, 32) is,

$y=mx+c...\left(1\right)$

where $m=\frac{y_2-y_1}{x_2-x_1}=\frac{32-212}{0-100}=\frac{-180}{-100}=\frac{9}{5}$

Therefore equation (1) becomes

$y=\frac{9}{5}x+c_1$

which can also be written as,

$F=\frac{9}{5}C+C_1...\left(ii\right)$

Initially at C = 0, F = 32

Substituting in (ii) we get,

$32=\frac{9}{5}\left(0\right)+c_1$

$\Rightarrow c_1=32.$

Hence, $F=\frac{9}{5}C+32.$ or $C=(F-32)\frac{5}{9}...\left(iii\right)$

(ii) Given, $F=\mathrm{98.6.}C=(\mathrm{9.8.6}-32)\frac{5}{9}=66.6\times \frac{5}{9}$

$\therefore C=37$ (using equation (ii))

(iii) Given,$C=38,F=\frac{9}{5}\left(38\right)+32=68.4+32=100.4$

$\therefore F=100.4$ (using equation (iii))