Question

Convert each of the following temperatures in ${}^ \circ F$ to the Celsius and kelvin scale.
(i) $68{}^ \circ F$
(ii) $5{}^ \circ F$
(iii) $176{}^ \circ F$

Answer 1

Hint: We can convert the given temperatures in ${}^ \circ F$ into Celsius and kelvin scale by using the conversion factors and formulae of ${}^ \circ F$ into ${}^ \circ C$ and ${}^ \circ F$ into ${}^ \circ K$ . Then, we discuss why we use such conversion factors for given conversions.

Complete step by step answer:
To convert from ${}^ \circ F$ into ${}^ \circ C$ , we have to subtract $32$ from ${}^ \circ F$ and then multiply it by $\dfrac{5}{9}$ , to get the value in ${}^ \circ C$ i.e. $\left( {{}^ \circ F - 32} \right) \times \dfrac{5}{9} = {}^ \circ C$. To convert ${}^ \circ F$ into ${}^ \circ K$ , we have to subtract $32$ from ${}^ \circ F$ and then multiply it by $\dfrac{5}{9}$ and finally add $273.15$, to get the value in ${}^ \circ K$ i.e. $\left( {{}^ \circ F - 32} \right) \times \dfrac{5}{9} + 273.15 = {}^ \circ K$.

(i) $68{}^ \circ F$
Celsius : $\left( {68{}^ \circ F - 32} \right) \times \dfrac{5}{9} = 36 \times \dfrac{5}{9} = 20{}^ \circ C$
Kelvin : $\left( {68{}^ \circ F - 32} \right) \times \dfrac{5}{9} + 273.15 = 20 + 273.15 = 293.15{}^ \circ K$

(ii) $5{}^ \circ F$
Celsius : $\left( {5{}^ \circ F - 32} \right) \times \dfrac{5}{9} = - 27 \times \dfrac{5}{9} = - 15{}^ \circ C$
Kelvin : $\left( {5{}^ \circ F - 32} \right) \times \dfrac{5}{9} + 273.15 = - 27 + 273.15 = 258.15{}^ \circ K$

(iii) $176{}^ \circ F$
Celsius : $\left( {176{}^ \circ F - 32} \right) \times \dfrac{5}{9} = 144 \times \dfrac{5}{9} = 80{}^ \circ C$.
Kelvin : $\left( {176{}^ \circ F - 32} \right) \times \dfrac{5}{9} + 273.15 = 144 + 273.15 = 353.15{}^ \circ K$

Note: The freezing point is 0° in the Celsius scale and 32° on the Fahrenheit scale and therefore, we subtract 32 when converting from Fahrenheit to Celsius, and add 32 when converting from Celsius to Fahrenheit. Also, a change of 5 degree Celsius is equal to a change of 9 degrees Fahrenheit. For this reason, we use the factor $\dfrac{5}{9}$ when converting from ${}^ \circ F$ into ${}^ \circ C$.