Question

The Fahrenheit temperature $F$ and absolute temperature $K$ satisfy a linear equation. Given that $K=273$ when $F=32$ and that $K=373$ when $F=212$. Express $K$ in terms of $F$ and find the value of $F$ when $K=0$.

Answer 1

Step $1:$ Simplification of given data.
Assuming $F$ along $x$-axis and $K$ along $y$-axis.
Given that $K=273$ when $F=32$ and
$K=373$ when $F=212$
Now, we have two points
$(32,273)$ and $(212,373)$ in $XY$ plane.

Step $2:$ Required line
We know that equation of line through two points$(x_1,y_2)$ and $(x_2,y_2)$ is
$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
The point $(F,K)$ satisfies the equation of line and the line passes through points $(32,273)and(212,373)$
So, required equation
$\Rightarrow (K-273)=\frac{373-273}{212-32}(F-32)$
$\Rightarrow K-273=\frac{100}{180}(F-32)$
$\Rightarrow K-273=\frac{5}{9}(F-32)$
$\Rightarrow K=\frac{5}{9}(F-32)+273\cdots \left(i\right)$

Step $3:$ Solve for value of $F$
Now, finding the value of $F$, when $K=0$.
Putting $K=0$ in the equation $\left(i\right)$
$\Rightarrow K=\frac{5}{9}(F-32)+273$
$\Rightarrow 0=\frac{5}{9}(F-32)+273$
$\Rightarrow -273=\frac{5}{9}(F-32)$
$\Rightarrow \frac{-273\times 9}{5}+32=F$
$\Rightarrow \frac{-273\times 9+32\times 5}{5}=F$
$\Rightarrow \frac{-2457+160}{5}+32=F$
$\Rightarrow F=-459.4$

Therefore, the required equation is
$K=\frac{5}{9}(F-32)+273$ and when $K=0,F=-459.4$.