Question

What is $${42}^{\circ }$$ Celsius in Fahrenheit?

Answer 1

Hint: We are given a temperature in Celsius scale and we have to convert the temperature to Fahrenheit scale. To convert the temperature from Celsius to Fahrenheit, we will use the formula, which is, $$F={\displaystyle \frac{9}{5}}C+32$$ where ‘F’ is the temperature in Fahrenheit and ‘C’ is the temperature in Celsius. We will put the given temperature in Celsius in the above formula and solve further to find the temperature in Fahrenheit. Hence, we will have the Fahrenheit equivalent temperature for the given Celsius temperature.

Complete step by step answer:
According to the given question, we have a temperature given and it is in Celsius scale. We are asked to convert the given temperature to Fahrenheit scale.
The given temperature we have is,
$${42}^{\circ }$$ Celsius ----(1)
We can carry out the conversion by using a formula and we have,
$$F={\displaystyle \frac{9}{5}}C+32$$-----(2)
where ‘F’ is the temperature in Fahrenheit and ‘C’ is the temperature in Celsius.
We will substitute the given temperature in Celsius in the equation and solve the equation to obtain the temperature in Fahrenheit.
we will put $$C={42}^{\circ }$$ in the equation (2) and we get,
$$\Rightarrow F={\displaystyle \frac{9}{5}}\left({42}^{\circ }\right)+32$$
In the above expression, we will divide $${42}^{\circ }$$ by 5 and we get,
$$\Rightarrow F=(9\times 8.4)+32$$
Now, we will multiply the components within the brackets and get the value as,
$$\Rightarrow F=75.6+32$$
Now, we will add up the above terms and we get the value as,
$$\Rightarrow F={107.6}^{\circ }$$

Therefore, the Fahrenheit equivalent of the given temperature is $${107.6}^{\circ }F$$.

Note: The formula that we used above, which is, $$F={\displaystyle \frac{9}{5}}C+32$$ can also be used to convert the temperature in Fahrenheit to Celsius and for that our formula will become, $$C={\displaystyle \frac{5}{9}}(F-32)$$. So, when asked, modify the formula as per the requirement. Also, read the question well and on what scale the temperature is given, if this part is understood wrongly the entire solution gets wrong.