Courses
Courses for Kids
Free study material
Offline Centres
More
Store

At what temperature are the readings of Celsius scale and Fahrenheit scale will be equal?

Last updated date: 13th Jun 2024
Total views: 401.7k
Views today: 5.01k
Verified
401.7k+ views
Hint: The value of temperature at Fahrenheit scale is equivalent to the product of value of the same temperature in Celsius scale with $\dfrac{9}{5}$ and then add a $32$ into it. This way we can convert the Celsius scale reading into Fahrenheit scale. When they are equal substitute the value of Celsius scale equation in the equation of Fahrenheit scale. These all will definitely help you in solving this question.

First of all let us take a look at how we can convert a degree Celsius scale into Fahrenheit scale reading. This can be expressed by the equation given as,
${}^\circ F=\left( {}^\circ C\times \dfrac{9}{5} \right)+32$……… (1)
By rearranging the same equation in terms of degree Celsius, we can see that the value given in Fahrenheit scale can be converted into Celsius scale. This can be written by the equation,
$\left( {}^\circ F-32 \right)\times \dfrac{5}{9}={}^\circ C$……… (2)
As per the question is concerned, it has been mentioned that they are equal. Therefore we can write that,
${}^\circ C={}^\circ F$
Substituting the values of degree Fahrenheit in it, we can write that,
${}^\circ C=\left( {}^\circ C\times \dfrac{9}{5} \right)+32$
Rearranging the equation will give,
${}^\circ C-\left( {}^\circ C\times \dfrac{9}{5} \right)=32$
Thus we can simplify the equation as,
$\left( {}^\circ C\times \dfrac{-4}{5} \right)=32$
That is,
${}^\circ C=-40{}^\circ C$
Therefore the question has been answered.

Note: Let us do it in the opposite way also. It is also possible. That means by substituting the values of the degree Celsius scale reading in the equation which can be given as,
${}^\circ F={}^\circ C$
Substituting the equation (2) in it will give,
${}^\circ F=\left( {}^\circ F-32 \right)\times \dfrac{5}{9}$
Rearranging this equation can be written as,
$\dfrac{5}{9}{}^\circ F-{}^\circ F=32\times \dfrac{5}{9}$
Simplifying this equation further will give,
$\dfrac{-4}{9}{}^\circ F=32\times \dfrac{5}{9}$
Cancelling the denominator can be possible, as they are equal. That is,
${}^\circ F=32\times \dfrac{-5}{4}$
That is,
${}^\circ F=-40{}^\circ C$
Therefore the temperature if both the Celsius and Fahrenheit scales are the same is $-40{}^\circ C$.