Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 1 - Exercises - Page 40: 108


\[\underline{40{}^\circ }\]

Work Step by Step

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.
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