Question

If the temperature of an object is $140^\circ F$ , then its temperature in centigrade is
A) $105^\circ \,C$
B) $32^\circ \,C$
C) $140^\circ C$
D) $60^\circ C$

Answer 1

Hint: The relation between the Fahrenheit and centigrade is linear in nature. Use the relation between the two scales and place the value of temperature given in the question.
Formula used: $C = \dfrac{5}{9}(F - 32)$ where $C$ is the temperature in the centigrade scale and $F$ is the temperature in Fahrenheit scale.

Complete step by step solution:
The relation between the temperature of an object in centigrade and Fahrenheit is given as:
$\Rightarrow C = \dfrac{5}{9}(F - 32)$
Since we’ve been given the temperature of the object as $140^\circ F$ and we want to find the temperature in centigrade, we place $F = 140^\circ $ in the formula and get
$\Rightarrow C = \dfrac{5}{9}(140 - 32) $
$\Rightarrow C = \dfrac{5}{9}(108) $
On further simplifying the RHS of the equation, we get
$\Rightarrow C = 60^\circ $
So, $140^\circ \,F$ is equivalent to $60^\circ \,C$ in the centigrade option which corresponds to option (D).

Hence, the correct option is option (D).

Additional Information:
In the Fahrenheit scale, the reference temperatures are $32\;^\circ F$ as a freezing point of water and $212\;^\circ F$ as a boiling point of water. Whereas in the Celsius scale $0\;^\circ C$ is the freezing point of water and $100\;^\circ C$ is the boiling point of water. Both scales are related directly and linearly which means that if the temperature of an object rises, its value rises both in the centigrade and in the Fahrenheit scale. The common point of both scales is $ - 40^\circ C$ which means that both the centigrade and the Fahrenheit scales will have the same value i.e. $ - 40^\circ C = - 40^\circ F$.

Note:
We must remember different temperature conversion formulae to solve such questions. Alternatively, if you remember the relation in the following manner:
$\Rightarrow F = \dfrac{9}{5}C + 32$
Then place the value of $F = 140^\circ $
$\Rightarrow 140 = \dfrac{9}{5}C + 32$
On subtracting both sides by $32^\circ C$ , we get
$\Rightarrow 108 = \dfrac{9}{5}C$
Multiplying both sides by $\dfrac{5}{9}$, we again get
$\Rightarrow C = 60^\circ $