# A wood fire burns at 260 °C. What is the temperature in Fahrenheit?

Jan 8, 2018

color(red)(F=9/5color(blue)((C))+32-> "the derived formula"
where:
$C = {260}^{o}$

#### Explanation:

It is confusing to memorize formulas. To start with, just remember the standard values; freezing and boiling points of water expressed in $C$ and $F$. Knowing these data provide the important conversion formula that can be derived from it.

$\text{Freezing Point of water in } C \mathmr{and} F = {0}^{o} \mathmr{and} {32}^{o}$
$\text{Boiling Point of water in } C \mathmr{and} F = {100}^{o} \mathmr{and} {212}^{o}$

Now, formulate the ratios that relate the two temperature scales where the temperature units and zero points are different; hence,

$\frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32}$, simplify

$\frac{C}{100} = \frac{F - 32}{180}$, cross multiply

$100 \left(F - 32\right) = 180 \left(C\right)$, divide both sides by 100 and simplify

(cancel(100)(F-32))/cancel(100)=(cancel(180)9(C))/(cancel(100)5, simplify

$F - 32 = \frac{9}{5} \left(C\right)$, add 32 both sides of the equation
$F \cancel{- 32 + 32} = \frac{9}{5} \left(C\right) + 32$, simplify
color(red)(F=9/5color(blue)((C))+32-> "the derived formula"

Then, given the value $\textcolor{b l u e}{C = {260}^{o}}$, plug in the value to the formula to convert $C$ to $F$; hence

$F = \frac{9}{5} \left(C\right) + 32$

$F = \frac{9}{5} \left(260\right) + 32$

$F = 468 + 32$

$F = {500}^{o}$