Question

Convert ${38^ \circ }C$ to Fahrenheit.

Answer 1

Before we get into the question, let's go over the definitions of Celsius and Fahrenheit. Temperatures are measured in Celsius, or centigrade, across the majority of the world. Water freezes at ${0^ \circ }C$ and boils at ${100^ \circ }C$. Fahrenheit is a temperature scale that is extensively used in the United States. Now, in order to answer the given question we need to multiply the $^\circ C$ temperature by $1.8$ . After that we will add $32$ to this number and our answer will come out in $^\circ F$ .

Complete step by step answer:
When you ask to convert ${38^ \circ }C$ to $F$ , you mean to convert ${38^ \circ }C$ to ${}^ \circ F$. We'll show you how to convert ${38^ \circ }C$ to Fahrenheit so you can see how hot or cold ${38^ \circ }C$ is in Fahrenheit. The $C$ to $F$ formula is,
$\left( {C{\text{ }} \times {\text{ }}\dfrac{9}{5}} \right){\text{ }} + {\text{ }}32{\text{ }} = {\text{ }}F$
When we enter $38$ for $C$ in the formula we get,
$\left( {38{\text{ }} \times {\text{ }}\dfrac{9}{5}} \right){\text{ }} + {\text{ }}32{\text{ }} = {\text{ }}F$

To solve $\left( {38{\text{ }} \times {\text{ }}\dfrac{9}{5}} \right){\text{ }} + {\text{ }}32{\text{ }} = {\text{ }}F$ , we first multiply $9$ by $\,38$ , then we divide the product by $5$ , and then finally we add $32$ to the quotient to get the answer. Here is the math to illustrate:

\Rightarrow \dfrac{{342{\text{ }}}}{5} = 68.4 \\\ \Rightarrow 68.4 + {\text{ }}32 = 100.4$$ **Therefore, the answer to $${38^ \circ }C$$ to $$F$$ is $$100.4$$ which can be written as follows: $$38{\text{ }}^\circ C{\text{ }} = {\text{ }}100.4{\text{ }}^\circ F$$.** **Note:** Because the Celsius and Fahrenheit scales are both offset–that is, none is defined as beginning at zero. Furthermore, with each additional unit of heat energy, the Celsius and Fahrenheit scales add a different value. Because of this configuration, it is impossible to assert that doubling the $$^\circ C$$ or $$^\circ F$$ value doubles the quantity of heat energy, making it difficult to have an intuitive understanding of how much energy one degree Fahrenheit or Celsius actually is.