Chapter 6.5, Problem 56CCR

To determine

To find: How much would the temperature changes from sea level to the highest point on each mountain.

Answer to Problem 56CCR

For Highest point of Mt. McKinley temperature decreases by ${101.6}^{0}F$ .

For Highest point of Mt. St. Elias temperature decreases by ${90.04}^{0}F$ .

For Highest point of Mt. Foraker temperature decreases by ${85}^{0}F$ .

Explanation of Solution

Given information:

Highest point of Mt. McKinley $=20,320$

Highest point of Mt. St. Elias $=18,008$

Highest point of Mt. Foraker $=17,000$

For every $1000m$ temperature decreases by $5^0$ F.

For Mt. McKinley

Highest point of Mt. McKinley $=20,320$

  $1000m\to 5^{0}F$

Dividing with $1000$ on both sides

We get

  $\begin{array}{l}\Rightarrow \frac{1000}{1000}\to \frac{5}{1000}\\ \Rightarrow 1\to \frac{1}{200}\end{array}$

  $\Rightarrow 1\to 0.005$

Therefore,

For $1$ m temperature decreases by ${0.005}^{0}F$ .

  $\Rightarrow 1\to 0.005$

Multiplying with $20,320$ on both sides

We get

  $\begin{array}{l}\Rightarrow 1\times 20,320\to 0.005\times 20,320\\ \Rightarrow 20,320m\to {101.6}^{0}F\end{array}$

Therefore,

For Highest point of Mt. McKinley temperature decreases by ${101.6}^{0}F$ .

For Mt. St. Elias

Highest point ofMt. St. Elias $=18,008$

For $1$ m temperature decreases by ${0.005}^{0}F$ .

  $\Rightarrow 1\to 0.005$

Multiplying with $18,008$ on both sides

We get

  $\begin{array}{l}\Rightarrow 1\times 18,008\to 0.005\times 18,008\\ \Rightarrow 18,008m\to {90.04}^{0}F\end{array}$

Therefore,

For Highest point of Mt. St. Elias temperature decreases by ${90.04}^{0}F$ .

For Mt. Foraker

Highest point of Mt. Foraker $=17,000$

For $1$ m temperature decreases by ${0.005}^{0}F$ .

  $\Rightarrow 1\to 0.005$

Multiplying with $17,000$ on both sides

We get

  $\begin{array}{l}\Rightarrow 1\times 17,000\to 0.005\times 17,000\\ \Rightarrow 17,000m\to {85}^{0}F\end{array}$

Therefore,

For Highest point of Mt. Foraker temperature decreases by ${85}^{0}F$ .