Chemistry: Molecular Approach (4th Edition)

$\underline{40{}^\circ }$
The relation between the temperature in $^{\circ }\text{F}$ with Celsius can be expressed as follows: $^{\circ }\text{C}=\frac{\left( ^{\circ }\text{F}-\text{32} \right)}{1.8}$ Here, $^{\circ }\text{C}$ is the temperature unit in Celsius and $^{\circ }\text{F}$ is the temperature unit in Fahrenheit. Let the temperature at which the Celsius scale is equal to Fahrenheit scale be $x{}^\circ$. Substitute $^{\circ }\text{F}$ and $^{\circ }\text{C}$ with $x{}^\circ$ in the above expression: \begin{align} & \,\,\,\,\,\,\,x{}^\circ =\frac{\left( x{}^\circ -\text{32} \right)}{1.8} \\ & \Rightarrow \,1.8x{}^\circ =x{}^\circ -\text{32} \\ & \Rightarrow \,1.8x{}^\circ -x{}^\circ =\text{32} \\ & \Rightarrow \,-0.8x{}^\circ =\text{32} \\ & \Rightarrow \,x{}^\circ =-\frac{\text{32}}{0.8} \\ & \Rightarrow \,x{}^\circ =-40 \\ \end{align} At $‒40^{\circ }$, Celsius and Fahrenheit will be same.