Question

Cooling Air As a parcel of air rises (for example, as it is pushed over a mountain), it cools at the dry adiabatic lapse rate of 5.5°F per 1000 feet until it reaches its dew point. If the ground temperature is 67°F, write a formula for the sequence of temperatures, {Tn}, of a parcel of air that has risen n thousand feet. What is

the temperature of a parcel ofair if it has risen 5000 feet?

Short answer

Formula for the sequence of temperatures

$T_n=-5.5n+67$

the temperature of a parcel of air if it has risen 5000 feet

$T_5=39.5°F$

Step-by-step solution

Given Information 

The ground temperature is 67°F. The difference in temperature per 1000 feet is -5.5°F .

Formula for temperature 

The ground temperature is 67°F. The difference in temperature per 1000 feet is -5.5°F .

The temperature of parcel of air at 1000 feet is T1= 67-5.5=62.5

The temperature of parcel of air at 2000 feet is$T_2=67-2(5.5)$

The temperature of parcel of air at n thousand feet

$$\begin{gathered} T_n=67-5.5n \\ {T}_n=-5.5n+67 \end{gathered}$$

Step 3:  Temperature at 5000 feet

Now we find the temperature of a parcel of air if it has risen 5000 feet

n=5

$$\begin{gathered} T_n=-5.5n+67 \\ n=5 \\ {T}_5=-5.5\left(5\right)+67=39.5°F \end{gathered}$$