# Question #8d154

Mar 13, 2016

#### Explanation:

1. This problem is merely conversion from $F$ to $C$;
2. I don't recommend memorizing the formula, but always remember the following data to accurately write the conversion formula for these temperature scales:

## Boiling Pt of Water

$F = {212}^{o}$
$C = {100}^{o}$

## Freezing Pt of Water

$F = {32}^{o}$
$C = {0}^{o}$
3. Write the ratio of the two temperatures with the zero points and the size units;

$\frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32}$
4. Simplify the equation

$\frac{C}{100} = \frac{F - 32}{180}$
5. Isolate $C$ to derive the formula

$C = \frac{100}{180} \left(F - 32\right)$
6. Reduce fraction to lowest form
$C = \frac{\cancel{20} .5}{\cancel{20} .9} \left(F - 32\right)$
7. The derived formula is
$C = \frac{5}{9} \left(F - 32\right)$
8. Plug in the given value $F = {230}^{o}$ to find the value in $C$
9. The answer is $C = {110}^{o}$
10. Don't really know if there is a significant difference between the two temperatures. Some say, it has approximately ${1.5}^{o}$ difference between alcohol and mercury, maybe because alcohol has low boiling point compared to mercury... If so, this ${230}^{o} F$ = ${110}^{o} C$ for mercury-type thermometer. For alcohol-type thermometer this value is equal to ${111.5}^{o} C$.

Moreover, manufacturers probably have all types of thermometers properly calibrated such that all can give the same results...