
At what temperature on the Fahrenheit scale will the reading be double the reading on Celsius scale?
$ A.\;{340^\circ }F \\
B.\;{240.6^\circ }F \\
C.\;{330^\circ }F \\
D.\;{320^\circ }F $
Answer
405.3k+ views
Hint :To find the answer, we need to know the relation between Celsius and Fahrenheit scale. The given condition is that the temperature on Fahrenheit is double to that of Celsius. So, let us assume that the temperature on Celsius scale as ‘x’ and the temperature on Fahrenheit is taken as ‘2x’ as mentioned above.
Complete Step By Step Answer:
There is a simple formula that explains the relation between Fahrenheit and Celsius scales. The formula we need to use here is,
$ {T_c} = \left( {{T_f} - 32} \right) \times \dfrac{5}{9} $
Where
$ {T_c} $ = temperature in Celsius
$ {T_f} $ = temperature in Fahrenheit
Now, let us assume that Celsius temperature $ \left( {{T_c}} \right) $ as ‘x’ and Fahrenheit temperature $ \left( {{T_f}} \right) $ as ‘2x’. By substituting the values in the above formula, we get
$ \Rightarrow x = \left( {2x - 32} \right) \times \dfrac{5}{9} \\
\Rightarrow 9x = \left( {2x - 32} \right) \times 5 \\
\Rightarrow 9x = 10x - 160 \\
\Rightarrow 9x - 10x = - 160 \\
\Rightarrow - 1x = - 160 \\
\Rightarrow x = 160 \\
\Rightarrow {T_c} = 160 $
We have got the value of ‘x’, which is the temperature in Celsius. When you look at the options, the temperature is given in Fahrenheit, so we need to convert it into Fahrenheit. We assumed that the Fahrenheit temperature was ‘2x’. Now substituting the value of ‘x’ we get
$ \Rightarrow {T_f} = 2x \\
\Rightarrow 2\left( {160} \right) = 320 \\
\Rightarrow {T_f} = {320^\circ }F $
So at the temperature of $ {320^\circ }F $ , the reading will be double that of the Celsius scale.
By the above explanation, we can conclude that the correct option is D.
Note :
The Celsius scale has two fixed points namely freezing point $ \left( {{0^\circ }C} \right) $ and boiling point $ \left( {{{100}^\circ }C} \right) $ . In the similar way, the Fahrenheit scale has freezing point at $ {32^\circ }F $ and boiling point at $ {212^\circ }F $ . There is a temperature where both Celsius and Fahrenheit scale will give the same reading. This is at $ - {40^\circ } $ .
Complete Step By Step Answer:
There is a simple formula that explains the relation between Fahrenheit and Celsius scales. The formula we need to use here is,
$ {T_c} = \left( {{T_f} - 32} \right) \times \dfrac{5}{9} $
Where
$ {T_c} $ = temperature in Celsius
$ {T_f} $ = temperature in Fahrenheit
Now, let us assume that Celsius temperature $ \left( {{T_c}} \right) $ as ‘x’ and Fahrenheit temperature $ \left( {{T_f}} \right) $ as ‘2x’. By substituting the values in the above formula, we get
$ \Rightarrow x = \left( {2x - 32} \right) \times \dfrac{5}{9} \\
\Rightarrow 9x = \left( {2x - 32} \right) \times 5 \\
\Rightarrow 9x = 10x - 160 \\
\Rightarrow 9x - 10x = - 160 \\
\Rightarrow - 1x = - 160 \\
\Rightarrow x = 160 \\
\Rightarrow {T_c} = 160 $
We have got the value of ‘x’, which is the temperature in Celsius. When you look at the options, the temperature is given in Fahrenheit, so we need to convert it into Fahrenheit. We assumed that the Fahrenheit temperature was ‘2x’. Now substituting the value of ‘x’ we get
$ \Rightarrow {T_f} = 2x \\
\Rightarrow 2\left( {160} \right) = 320 \\
\Rightarrow {T_f} = {320^\circ }F $
So at the temperature of $ {320^\circ }F $ , the reading will be double that of the Celsius scale.
By the above explanation, we can conclude that the correct option is D.
Note :
The Celsius scale has two fixed points namely freezing point $ \left( {{0^\circ }C} \right) $ and boiling point $ \left( {{{100}^\circ }C} \right) $ . In the similar way, the Fahrenheit scale has freezing point at $ {32^\circ }F $ and boiling point at $ {212^\circ }F $ . There is a temperature where both Celsius and Fahrenheit scale will give the same reading. This is at $ - {40^\circ } $ .
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