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Question: At what temperature does the temperature in Celsius and Fahrenheit equalize? A. -40 B. 40 C. 3...

At what temperature does the temperature in Celsius and Fahrenheit equalize?
A. -40
B. 40
C. 36.6
D. 38

Explanation

Solution

In order to answer the given question, firstly we have to recall the relation that is used to convert the temperatures from one of the given scales to the other. Then, you could assume a temperature T for which both scales show the same reading. Now rearranging the above relation will find you the value of T and hence the answer.

Formula used: Relation between degree Celsius and degree Fahrenheit,
TC=(TF32)59T{}^\circ C=\left( T{}^\circ F-32 \right)\dfrac{5}{9}

Complete step by step answer:
Degree Celsius and degree Fahrenheit are both units of temperature on the Celsius and Fahrenheit scale respectively.
The Celsius scale is based on 0C0{}^\circ C as the freezing point and 100C100{}^\circ Cas the boiling point of water at 1atm pressure. But in the case of Fahrenheit, the scaling is very different from that of Celsius. Here, the freezing point of water is at 32F32{}^\circ F and its boiling point is212F212{}^\circ F under standard atmospheric pressure.
In the given question, we are asked to find the temperature at which both these scales show the same value. In order for us to find that, we have to recall the relation used for converting reading inC{}^\circ C toF{}^\circ F and vice versa. The relation is given by,
TC=(TF32)59T{}^\circ C=\left( T{}^\circ F-32 \right)\dfrac{5}{9}
Let T be the temperature at which both scales show the same reading, then,
T=(T32)59\Rightarrow T=\left( T-32 \right)\dfrac{5}{9}
95T=T32\Rightarrow \dfrac{9}{5}T=T-32
5T9T5=32\Rightarrow \dfrac{5T-9T}{5}=32
4T=160\Rightarrow -4T=160
T=40\therefore T=-40{}^\circ

So, the correct answer is “Option A”.

Note: Though both the given units are that of temperature, none of them is its SI unit. The SI unit of temperature is Kelvin. The conversion of Fahrenheit and Celsius to Kelvin is given by the following relations:
T(F)=95T(K)459.67T\left( {}^\circ F \right)=\dfrac{9}{5}T\left( K \right)-459.67
T(C)+273.15=T(K)T\left( {}^\circ C \right)+273.15=T\left( K \right)