Question

A Fahrenheit thermometer indicates a temperature of \[{41^\circ }F\] . What is the corresponding reading in the Centigrade scale? At what temperature will the Centigrade thermometer read twice as much as the Fahrenheit thermometer?

Answer 1

Hint:Using the equation that relates Celsius scale with Fahrenheit scale, we can link the values in the equation to reveal the answer. In case to relate Celsius as double of Fahrenheit we need to assume an imaginary variable and resolve using the same equation.

Formula used:
$^\circ C = \dfrac{5}{9} \times \left( {^\circ F - 32} \right)$

Complete step by step answer:
According to question temperature on Fahrenheit scale reads ${41^\circ }F$
Use the equation relating Fahrenheit and Centigrade scale:
$^\circ C = \dfrac{5}{9} \times \left( {^\circ F - 32} \right)$
Where,
$^\circ C = {\text{Temp}}{\text{. on centigrade scale}}$
$^\circ F = {\text{Temp}}{\text{. on fahrenheit scale}}{\text{.}}$
Putting value of Fahrenheit scale in the equation will give,
$^\circ C = \dfrac{5}{9} \times \left( {41 - 32} \right)$
$\Rightarrow ^\circ C = {5^\circ }C$

To calculate temperature at which Celsius scale read twice as much as Fahrenheit reading,Let that temperature on Fahrenheit scale read $2{x^\circ }F$.According to question temperature on Celsius scale is half of that on Fahrenheit scale, hence Celsius scale will read temperature ${x^\circ }C$.
Putting both values in equation we will get,
$x = \dfrac{5}{9} \times \left( {2x - 32} \right)$
Evaluating equation will give,
$\therefore x = 160$
As $x$ is assumed to be variable for Celsius scale so we can say ${160^\circ }C$ is the temperature at which the Centigrade thermometer reads twice as much as the Fahrenheit thermometer.

Note:A Fahrenheit scale ranges from ${32^\circ }F$ to ${212^\circ }F$ and is divided into 180 parts with each part corresponding to one degree. The Celsius scale ranges from ${0^\circ }C$ to ${100^\circ }C$ and is divided into 100 parts with each part corresponding to one degree. It is important to use the equation correctly. This is the only spot where there are chances of the result getting tempered or wrong.