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 \(5.5^{\circ} \mathrm{F}\) per 1000 feet until it reaches its dew point. If the ground temperature is \(67^{\circ} \mathrm{F},\) write a formula for the sequence of temperatures, \(\left\\{T_{n}\right\\},\) of a parcel of air that has risen \(n\) thousand feet. What is the temperature of a parcel of air if it has risen 5000 feet?

Short answer

The temperature of the air parcel at 5000 feet is 39.5°F.

Step-by-step solution

Understand the dry adiabatic lapse rate

When air rises, it cools at a consistent rate known as the dry adiabatic lapse rate. This rate is given as 5.5°F per 1000 feet. Therefore, for every 1000 feet the parcel rises, its temperature decreases by 5.5°F.

Determine the ground temperature

The ground temperature is given as 67°F. This is the starting temperature, denoted as \(T_0\).

Create the formula for temperature at any height

We need a formula to represent the temperature, \(T_n\), of the parcel of air after it has risen \(n\) thousand feet. Using the lapse rate, the temperature decrease per 1000 feet is 5.5°F. Thus, the temperature at \(n\) thousand feet can be represented as: \[T_n = T_0 - 5.5n.\]\

Apply the formula to find the temperature at 5000 feet

Substitute \(T_0 = 67^{\circ}F\) and \(n = 5\) (since 5000 feet is 5 thousand feet) into the formula \[T_n = 67 - 5.5 \times 5.\]\

Calculate the temperature

Compute the value: \[T_5 = 67 - 5.5 \times 5 = 67 - 27.5 = 39.5^{\circ}F.\]\

Key concepts

temperature decrease

When air rises, it experiences a temperature decrease due to lower air pressure at higher elevations. This cooling process follows a specific rate known as the dry adiabatic lapse rate. Essentially, for every 1000 feet that a parcel of air ascends, its temperature drops by 5.5°F. Understanding this rate helps us predict temperature changes at various altitudes with precision. This rate of change is consistent in dry air and forms the basis for calculating how much cooler the air gets as it moves upward.
This is why mountaintops are generally cooler than valleys. The higher you go, the colder it gets!

ground temperature

Ground temperature is crucial as a starting point for understanding how temperature changes with altitude. It's the initial condition of our problem and is denoted as \(T_0\). In our exercise, the ground temperature is given as 67°F.
Starting from this temperature point, we apply the adiabatic lapse rate to determine the temperature at various elevations.

• Ground temperature: 67°F
• First 1000 feet: 61.5°F
• Second 1000 feet: 56°F

This consistent rate helps us derive a formula that predicts temperatures at different heights.

formula derivation

To find the temperature of rising air at any height, we must derive a formula using the ground temperature and the adiabatic lapse rate. We start with the known ground temperature, \(T_0\), which is 67°F in this case. Given the lapse rate, the temperature decreases by 5.5°F for each 1000 feet, or 5.5° for each unit of \(n\) thousand feet.
Thus, the formula for the temperature at \(n\) thousand feet is: \[T_n = T_0 - 5.5n\] Here, \(n\) represents the number of thousand feet the air rises. This allows easy calculation of temperature at any point by substituting \(n\) with the desired height in thousand feet.

height in thousand feet

In this problem, height is measured in units of thousand feet denoted by \(n\). This makes it simpler to apply the dry adiabatic lapse rate, which is defined per 1000 feet.

• 1 thousand feet = 1 unit of \(n\)
• 2 thousand feet = 2 units of \(n\)
• 5 thousand feet (as given in the problem) = 5 units of \(n\)

Using this convention, we can easily slot in values for \(n\) into our formula to find temperatures at corresponding heights.

sequence of temperatures

Once we have our formula \(T_n = T_0 - 5.5n\), we can determine the sequence of temperatures as the parcel of air rises. Let's compute the temperatures at various heights:

• At the ground (0 thousand feet): \(T_0 = 67\)
• At 1000 feet: \(T_1 = 67 - 5.5 \times 1 = 61.5\)
• At 2000 feet: \(T_2 = 67 - 5.5 \times 2 = 56\)
• At 5000 feet: \(T_5 = 67 - 5.5 \times 5 = 39.5\)

Thus, by following the formula, we find that the air temperature at 5000 feet is 39.5°F. This simple sequence helps visualize how temperature drops incrementally with elevation.