Answer
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Hint: We know that the two scales used for temperature measurements are Celsius and Fahrenheit. We also know that the temperature in the centigrade scale will be expressed in degrees Celsius. So, we find the relationship and then derive the above statement to correct two separate scales.
Formula used:
Celsius to Fahrenheit,
$F = \left( {\dfrac{9}{5} \times C} \right) + 32$
Fahrenheit to Celsius,
$C = \dfrac{5}{9}\left( {F - 32} \right)$
Where,
$F$ is the temperature in Fahrenheit,
$C$ is the temperature on the centigrade scale.
Complete step-by-step solution:
Given by,
The relation between the Celsius and Fahrenheit scales,
The Celsius-Fahrenheit relationship is direct. Celsius is specifically linked to Fahrenheit.
Which suggests that
As the temperature rises on the Celsius scale, the equivalent temperature of Fahrenheit would also be high.
As the temperature falls on the Celsius scale, the temperature equal to Fahrenheit would also be low.
$\dfrac{{C - L.F.P}}{{U.F.P - L.F.P}} = \dfrac{{F - L.F.P}}{{U.F.P - L.F.P}}$
Here, we define,
$L.F.P$ is Lower Fixed Point
$U.F.P$ is Upper Fixed Point
According to that,
$\dfrac{{C - 0}}{{100 - 0}} = \dfrac{{F - 32}}{{212 - 32}}$
On simplifying,
We get,
$\dfrac{C}{{100}} = \dfrac{{F - 32}}{{180}}$
Again, we resolve the above equation,
We get,
$\dfrac{C}{5} = \dfrac{{\left( {F - 32} \right)}}{9}$
A temperature scale based on the freezing point of water at ${0^ \circ }C$ and the boiling point of water at ${100^ \circ }C$ is the Celsius scale, or centigrade scale. The Fahrenheit scale is a temperature scale based on the ${32^ \circ }F$ water freezing point and the ${220^ \circ }F$ water boiling point.
Hence,
Thus, the Celsius and Fahrenheit scales of temperature using the fixed point in two different scales $\dfrac{C}{5} = \dfrac{{\left( {F - 32} \right)}}{9}$
Note: The above derivation is evaluating the relation between Celsius and Fahrenheit is proportional. Both have different freezing points of water and both follow the varied unit difference between each scale. That is because of the Celsius temperature scale, part of the metric system that denotes the temperature.
Formula used:
Celsius to Fahrenheit,
$F = \left( {\dfrac{9}{5} \times C} \right) + 32$
Fahrenheit to Celsius,
$C = \dfrac{5}{9}\left( {F - 32} \right)$
Where,
$F$ is the temperature in Fahrenheit,
$C$ is the temperature on the centigrade scale.
Complete step-by-step solution:
Given by,
The relation between the Celsius and Fahrenheit scales,
The Celsius-Fahrenheit relationship is direct. Celsius is specifically linked to Fahrenheit.
Which suggests that
As the temperature rises on the Celsius scale, the equivalent temperature of Fahrenheit would also be high.
As the temperature falls on the Celsius scale, the temperature equal to Fahrenheit would also be low.
$\dfrac{{C - L.F.P}}{{U.F.P - L.F.P}} = \dfrac{{F - L.F.P}}{{U.F.P - L.F.P}}$
Here, we define,
$L.F.P$ is Lower Fixed Point
$U.F.P$ is Upper Fixed Point
According to that,
$\dfrac{{C - 0}}{{100 - 0}} = \dfrac{{F - 32}}{{212 - 32}}$
On simplifying,
We get,
$\dfrac{C}{{100}} = \dfrac{{F - 32}}{{180}}$
Again, we resolve the above equation,
We get,
$\dfrac{C}{5} = \dfrac{{\left( {F - 32} \right)}}{9}$
A temperature scale based on the freezing point of water at ${0^ \circ }C$ and the boiling point of water at ${100^ \circ }C$ is the Celsius scale, or centigrade scale. The Fahrenheit scale is a temperature scale based on the ${32^ \circ }F$ water freezing point and the ${220^ \circ }F$ water boiling point.
Hence,
Thus, the Celsius and Fahrenheit scales of temperature using the fixed point in two different scales $\dfrac{C}{5} = \dfrac{{\left( {F - 32} \right)}}{9}$
Note: The above derivation is evaluating the relation between Celsius and Fahrenheit is proportional. Both have different freezing points of water and both follow the varied unit difference between each scale. That is because of the Celsius temperature scale, part of the metric system that denotes the temperature.
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