Q. 74 - Precalculus Enhanced with Graphing Utilities

Question

Temperature Conversion The function $C\left(F\right)=\frac{5}{9}\left(F-32\right)$

converts a temperature in degrees Fahrenheit, F, to a temperature in degrees Celsius, C. The function $K\left(C\right) =C + 273$ converts a temperature in degrees Celsius to a temperature in kelvins, K.

(a) Find a function that converts a temperature in degrees Fahrenheit to a temperature in kelvins.

(b) Determine 80 degrees Fahrenheit in kelvins.

Step-by-Step Solution

Answer

(a)$K\circ C=\frac{5F+2297}{9}$

(b) $K\left(80\right)=299.67$

Step 1. Given Information

The function $C\left(F\right)=\frac{5}{9}\left(F-32\right)$ and $K\left(C\right)=C+273$

We need to find a function that converts a temperature in degrees Fahrenheit to a temperature in kelvins.

Part (a) Step 1. Finding a function converts a temperature in degrees Fahrenheit to a temperature in kelvins

$$\begin{gathered} K\circ C=K\left(C\left(F\right)\right)=K\left(\frac{5}{9}\left(F-32\right)\right) \\ K\circ C=\frac{5}{9}\left(F-32\right)+273 \\ K\circ C=\frac{5}{9}F-\frac{160}{9}+273 \\ K\circ C=\frac{5}{9}F-\frac{160}{9}+\frac{2457}{9} \\ K\circ C=\frac{5}{9}F+\frac{2297}{9} \\ K\circ C=\frac{5F+2297}{9} \end{gathered}$$

Part (b) Step 1. Determine 80 degrees Fahrenheit in Kelvins

$$\begin{gathered} K\left(F\right)=\frac{5}{9}\left(F-32\right)+273 \\ K\left(80\right)=\frac{5}{9}\left(80-32\right)+273 \\ K\left(80\right)=\frac{5}{9}\left(48\right)+273 \\ K\left(80\right)=299.67 \end{gathered}$$