Chapter 1.2, Problem 10CFU

a.

To determine

To find: functions to relate degree Fahrenheit and Kelvin by using composition function.

Answer to Problem 10CFU

  $K\left(F\right)=\frac{5}{9}\left(F-32\right)+273.5$ .

Explanation of Solution

Given information: The functions for degree Celsius and Kelvin are as follows:

  $\begin{array}{l}C\left(F\right)=\frac{5}{9}\left(F-32\right)\\ K\left(C\right)=C+273.5\end{array}$

Calculation:

Use composition functions to write Fahrenheit $\left(F\right)$ as a function of Kelvin as shown below.

  $\left(K\circ C\right)\left(F\right)=K\left(C\left(F\right)\right)$

Substitute $C\left(F\right)$ into $K\left(C\right)$ , to get

  $\begin{array}{l}K\left(C\left(F\right)\right)=C\left(F\right)+273.15\\ K\left(\frac{5}{9}\left(F-32\right)\right)=\frac{5}{9}\left(F-32\right)+273.15\left[\because C\left(F\right)=\frac{5}{9}\left(F-32\right)\right]\\ K\left(F\right)=\frac{5}{9}\left(F-32\right)+273.15\end{array}$

Thus, the function $K\left(F\right)=\frac{5}{9}\left(F-32\right)+273.15$

a.

To determine

To convert: the given degree Fahrenheit temperatures in Kelvin.

Answer to Problem 10CFU

  $\begin{array}{c}K\left(-40\right)=233.15K\\ K\left(-12\right)=248.71K\\ K\left(0\right)=255.37K\\ K\left(32\right)=273.15K\\ K\left(212\right)=373.15K\end{array}$ .

Explanation of Solution

Given information: The function is as follows:

  $K\left(F\right)=\frac{5}{9}\left(F-32\right)+273.5$

The temperatures that need to be converted into Kelvin are as follows:

  $\begin{array}{l}{F}_1=-40°F\\ F_2=-12°F\\ F_3=0°F\\ F_4=32°F\\ F_5=\text{212}°F\end{array}$

Calculation:

Substitute $F_1$ into $K\left(F\right)$ and simplify it as follows:

  $\begin{array}{c}K\left(-40\right)=\frac{5}{9}\left(-40-32\right)+273.5\\ =\frac{5}{9}\left(-72\right)+273.5\\ =-40+273.5\\ =233.15K\end{array}$

Thus, $F_1$ is $233.15K$ .

Substitute $F_2$ into $K\left(F\right)$ and simplify it as follows:

  $\begin{array}{c}K\left(-12\right)=\frac{5}{9}\left(-12-32\right)+273.5\\ =\frac{5}{9}\left(-44\right)+273.5\\ =-24.44+273.5\\ =248.71K\end{array}$

Thus, $F_2$ is $248.71K$ .

Substitute $F_3$ into $K\left(F\right)$ and simplify it as follows:

  $\begin{array}{c}K\left(0\right)=\frac{5}{9}\left(0-32\right)+273.5\\ =\frac{5}{9}\left(-32\right)+273.5\\ =-17.78+273.5\\ =255.37K\end{array}$

Thus, $F_3$ is $255.37K$ .

Substitute $F_4$ into $K\left(F\right)$ and simplify it as follows:

  $\begin{array}{c}K\left(32\right)=\frac{5}{9}\left(32-32\right)+273.5\\ =\frac{5}{9}\left(0\right)+273.5\\ =0+273.5\\ =273.5K\end{array}$

Thus, $F_4$ is $273.5K$ .