Answer:
This is a problem of lapse rate. The lapse rate is the rate at which an atmospheric variable, normally temperature in Earth’s atmosphere, falls with altitude. To answer the questions, we need to plot the given data and find the equation of the line that best fits the data. Using a graphing calculator or an online tool, we can obtain the following equation:
T = -4.8E + 40
where T is the temperature in °F and E is the elevation in 1000 ft. Using this equation, we can answer the questions as follows:
The temperature at sea level is T = -4.8(0) + 40 = 40°F. The temperature at 20,000 ft is T = -4.8(20) + 40 = -56°F.
The rate of change in temperature is the slope of the line, which is -4.8°F/1000 ft.
The elevation at which the temperature is 0°F is found by setting T = 0 and solving for E. We get E = (0 - 40)/(-4.8) = 8.33 (1000 ft) or 8330 ft.
The temperature outside a jet flying at 40,000 ft is T = -4.8(40) + 40 = -152°F.
Explanation: