How much does the density altitude increase when the temperature rises from 25 to 50°F at a pressure altitude of 5,000 feet?

How much does the density altitude increase when the temperature rises from 25 to 50°F at a pressure altitude of 5,000 feet?

• A. 850 feet
• B. 1,650 feet
• C. 2,000 feet
• D. 2,500 feet

Answer: (B)

To determine how much the density altitude increases when the temperature rises from 25°F to 50°F at a pressure altitude of 5,000 feet, it's important to understand the relationship between temperature, pressure altitude, and density altitude. The density altitude can be significantly affected by changes in temperature. The standard atmosphere assumes a certain lapse rate, typically around 3.5°F per 1,000 feet. Given that the temperature is increasing by 25°F (from 25°F to 50°F), it's essential to calculate how this translates to an increase in density altitude. First, let's find out how many thousands of feet correspond to the temperature change. By dividing the total temperature change (25°F) by the standard lapse rate (3.5°F/1,000 feet), the calculation becomes: 25°F / (3.5°F / 1,000 feet) = approximately 7,143 feet. This indicates an increase in density altitude. However, this value needs to be adjusted according to the base pressure altitude of 5,000 feet, leading to a final adjustment for how much the density altitude has increased overall. To find this more directly, we can also use a simpler approximation often used in flight training, where each

To determine how much the density altitude increases when the temperature rises from 25°F to 50°F at a pressure altitude of 5,000 feet, it's important to understand the relationship between temperature, pressure altitude, and density altitude.

The density altitude can be significantly affected by changes in temperature. The standard atmosphere assumes a certain lapse rate, typically around 3.5°F per 1,000 feet. Given that the temperature is increasing by 25°F (from 25°F to 50°F), it's essential to calculate how this translates to an increase in density altitude.

First, let's find out how many thousands of feet correspond to the temperature change. By dividing the total temperature change (25°F) by the standard lapse rate (3.5°F/1,000 feet), the calculation becomes:

25°F / (3.5°F / 1,000 feet) = approximately 7,143 feet.

This indicates an increase in density altitude. However, this value needs to be adjusted according to the base pressure altitude of 5,000 feet, leading to a final adjustment for how much the density altitude has increased overall.

To find this more directly, we can also use a simpler approximation often used in flight training, where each