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Question: What is the temperature at which we get the same reading on both the Centigrade and Fahrenheit scale...

What is the temperature at which we get the same reading on both the Centigrade and Fahrenheit scales?

Explanation

Solution

To find the temperature, we need to state the formulae for converting between Celsius and Fahrenheit scales. Then on equating the two values we will get the temperature at which we will get the same reading for both the Centigrade and Fahrenheit scale.

Formula Used: The following formulas are used to solve this question.
C=(F32)×59^ \circ C = \left( {^ \circ F - 32} \right) \times \dfrac{5}{9} the temperature in Celsius is C^ \circ C and the temperature in Fahrenheit is F^ \circ F and
F=(C×95)+32^ \circ F = \left( {^ \circ C \times \dfrac{9}{5}} \right) + 32 the temperature in Celsius is C^ \circ C and the temperature in Fahrenheit is F^ \circ F.

Complete step by step answer
The formulae for converting between Celsius and Fahrenheit scales have been given below as,
C=(F32)×59^ \circ C = \left( {^ \circ F - 32} \right) \times \dfrac{5}{9} and
F=(C×95)+32^ \circ F = \left( {^ \circ C \times \dfrac{9}{5}} \right) + 32
To find the temperature when both are equal, we set F=C^ \circ F{ = ^ \circ }C where the temperature in Celsius is C^ \circ C and the temperature in Fahrenheit is F^ \circ F.
F=C^ \circ F{ = ^ \circ }C
  C=  (C×95)+32\Rightarrow {\;^ \circ }C = \;\left( {^ \circ C \times \dfrac{9}{5}} \right) + 32
This is derived from the equations given above.
C32=(C×95)^ \circ C - 32 = \left( {^ \circ C \times \dfrac{9}{5}} \right)
Simplifying the equation further we get,
45×C=32- \dfrac{4}{5}{ \times ^ \circ }C = 32
C=32×54{ \Rightarrow ^ \circ }C = - 32 \times \dfrac{5}{4}
Thus, we have, C=40C^ \circ C = - {40^ \circ }C
Similarly for the Fahrenheit scale, we get
F=(F32)×59^ \circ F = \left( {^ \circ F - 32} \right) \times \dfrac{5}{9}
F×95=(F32){ \Rightarrow ^ \circ }F \times \dfrac{9}{5} = \left( {^ \circ F - 32} \right)
Simplifying the equation further,
F95×F=32^ \circ F - \dfrac{9}{5}{ \times ^ \circ }F = 32
45F=32\Rightarrow - {\dfrac{4}{5}^ \circ }F = 32
Hence we get,
F=40{ \Rightarrow ^ \circ }F = - 40
So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

Note
Celsius scale, or centigrade scale, introduced by Anders Celsius is a temperature scale that is based on the freezing point of water at 0C0^\circ C and the boiling point of water at 100C100^\circ C. This scale uses the symbol C^\circ C.
In this scale, the lower fixed point is considered 0C0^\circ C, and the upper fixed point is considered 100C100^\circ C. Thus, in the Celsius scale, the freezing point of water is 0C0^\circ C, and the boiling point of water is 100C100^\circ C.The region between these two temperatures is divided into 100 equal parts so that each part equals to one degree Celsius (1C)\left( {1^\circ C} \right).
Fahrenheit temperature scale, introduced by Daniel Gabriel Fahrenheit scale is based on 3232^\circ for the freezing point of water and 212212^\circ for the boiling point of water, the interval between the two being divided into 180 equal parts.
The scale was originally taken as: the temperature of an equal ice-salt mixture was the zero of his scale and the values of 3030^\circ and 9090^\circ were selected for the freezing point of water and normal body temperature, respectively.