At what temperature is the Fahrenheit scale reading equal to

(a) twice that of the Celsius scale and

(b) half that of the Celsius scale?

• i)The Fahrenheit scale equal to twice that of the Celsius scale:$T_F=2T_C$
• ii) The Fahrenheit scale equal to half that of the Celsius scale:$T_F=\frac{1}{2}{T}_C$

By using the relation between the Fahrenheit and Celsius scales, we will find all the answers.

Formula:

Using the relation between the Fahrenheit and Celsius scales as follows:
$T_F=\frac{9}{5}{T}_C+32.......\left(i\right)$

In the first case, it is given that the Fahrenheit scale reading is equal to twice that of the Celsius scale that is given as:

$T_F=2T_C$ ….. (ii)

Using condition (ii) and equation (i), and solve for the Celsius temperature as:
$$\begin{gathered} 2T_C=\frac{9}{5}{T}_C+321 \\ 2T_C-\frac{9}{5}{T}_C=32 \\ \left(2-\frac{9}{5}\right)T_C=32 \\ {T}_C=160° \end{gathered}$$

Hence, the Fahrenheit scale using equation (ii) is given as:
$$\begin{gathered} T_F=2\times 160 \\ =320°F \end{gathered}$$

At $T_C=160°C$and $T_F=320°F$the Fahrenheit scale reading is equal to twice that of the Celsius scale.

In the second case, it is given that the Fahrenheit scale reading is equal to half of the Celsius scale that is

$T_F=\frac{1}{2}{T}_C$ …. (iii)

Using condition (iii) and equation (i), we get the Celsius temperature as:

$$\begin{gathered} \frac{1}{2}{T}_C=\frac{9}{5}{T}_C+32 \\ \frac{1}{2}{T}_C-\frac{9}{5}{T}_C=32 \\ \left(\frac{1}{2}-\frac{9}{5}\right)T_C=32 \\ {T}_C=-24.61°F \end{gathered}$$

Hence, the Fahrenheit scale using condition (iii) is given as:
$$\begin{gathered} T_F=\frac{1}{2}\times (-24.61) \\ =-12.3°F \end{gathered}$$

At $T_C=-24.61°C$and $T_F=-12.3°F$the Fahrenheit scale reading is equal to half of the Celsius scale.