Question

Question 4
The linear equation that converts Fahrenheit (F) to Celsius (C) if is given by the relation, $C=\frac{5F-160}{9}$

(i) If the temperature is ${86}^{\circ }$F, what is the temperature in Celsius?

(ii) If the temperature is ${35}^{\circ }$C, what is the temperature in Fahrenheit?

(iii) If the temperature is $0^{\circ }$C, what is the temperature in Fahrenheit and if the temperature is $0^{\circ }$F, what is the temperature in Celsius?

(iv) What is the numerical value of the temperature which is same on both the scales?

Answer 1

Given relation is $C=\frac{5F-160}{9}$ ....(i)
$\Rightarrow 9C=5F-160\Rightarrow 5F=9C+160$
$\Rightarrow F=\frac{9C+160}{5}$ ....(ii)

(i) Given F = ${86}^{\circ }$F, then from Eq(i); we get,
$C=\frac{5\times 86-160}{9}=\frac{430-160}{9}=\frac{270}{9}={30}^{\circ }C$

(ii) Given C = ${35}^{\circ }$C, then from eq(ii); we get,
$F=\frac{9\times 35+160}{5}=\frac{315+160}{5}=\frac{475}{5}={95}^{\circ }F$

(iii) Given, C = $0^{\circ }$C, then from eq(ii); we get,
$F=\frac{9\times 0+160}{5}=\frac{+160}{5}={32}^{\circ }F$
F = $0^{\circ }$F, then from Eq.(i); we get,
$C=\frac{5\times 0-160}{9}=\frac{-160}{9}={(-\frac{160}{9})}^{\circ }C$

(iv) By the given condition; C = F.
Put this value in Eq (i); we get,
$C=\frac{5C-160}{9}\Rightarrow 9C=5C-160$
$\Rightarrow 9C-5C=-160$
$\Rightarrow 4C=-160\Rightarrow C=\frac{-160}{4}\Rightarrow C=-40=F$
[Numerical value of the temperature]