Question

How do you convert $45^\circ C$ to Fahrenheit?

Answer 1

Hint:
As we know the relation ${T_F} = \left( {\dfrac{9}{5} \times {T_C}} \right) + 32$ , where the ${T_F}$ is the temperature in Fahrenheit scale and ${T_C}$ is the temperature in Celsius scale. You can substitute the value of ${T_C}$ in this relation to convert that temperature from Celsius to Fahrenheit scale. The relation is between the constants, not the units $(^\circ C{\text{ and }}^\circ F)$.

Complete Step by step Solution:
Here in this question, we are given a temperature in Celsius scale, i.e. $45^\circ C$ . And we need to convert this temperature given in Celsius scale into Fahrenheit scale.
Before starting with a solution to the problem we should understand a few concepts of the Celsius and Fahrenheit temperature scale. The Celsius scale, also known as the centigrade scale, is a temperature scale based on $0^\circ $ for the freezing point of water and $100^\circ $ for the boiling point of water. Initially, the Celsius scale was used $0^\circ $ to denote the boiling point of water and $100^\circ $ to denote the freezing point of water.
Fahrenheit temperature scale, scale based on $32^\circ $ for the freezing point of water and $212^\circ $ for the boiling point of water, the interval between the two being divided into $180$ equal parts. Daniel Gabriel Fahrenheit originally took as the zero of his scale the temperature of an equal ice-salt mixture and selected the values of $30^\circ $ and $90^\circ $ for the freezing point of water and normal body temperature, respectively; these later were revised to $32^\circ $ and $96^\circ $ , but the final scale required an adjustment to $98.6^\circ $ for the latter value.
As we know the relation between Fahrenheit and Celsius temperature as:
$ \Rightarrow {T_F} = \left( {\dfrac{9}{5} \times {T_C}} \right) + 32$ , where ${T_F}$ represents the temperature in Fahrenheit and ${T_C}$ represents the temperature in Celsius
Here in this case we have temperature given in Celsius. So let us substitute ${T_C} = 45$ in the above relation, we get:
$ \Rightarrow {T_F} = \left( {\dfrac{9}{5} \times 45} \right) + 32$
This relation will give us the value of the same temperature in Fahrenheit. Solving the parenthesis, we get:
$ \Rightarrow {T_F} = \left( {\dfrac{9}{5} \times 45} \right) + 32 = \left( {9 \times 9} \right) + 32 = 81 + 32$
Now, this can be easily solved as:
$ \Rightarrow {T_F} = 81 + 32 = 113$

Thus, we get the temperature in Fahrenheit as $113^\circ F$.

Note:
In this question, the use of the fundamental concepts of temperature scales played a crucial role. An alternate approach to solve this problem is to use the relationship ${T_C} = \dfrac{5}{9} \times \left( {{T_F} - 32} \right)$ , where we can substitute the value ${T_C} = 45^\circ C$ in this equation to get the required value of temperature in the Fahrenheit scale.