Normal Lapse Rate The air temperature at 200 ft above the sea level is 40ºF for

Question

Normal Lapse Rate
The air temperature at 200 ft above the sea level is 40ºF for a stratified column of air.
A. What will its temperature be at level B (2,000 ft). Show your work. _____________ º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
B. What will its temperature be at the sea level? _______________º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
C. At what elevation will the temperature reach the freezing point (32ºF)?
Show your work. Round your answer to the nearest whole number. [Hint: First find the temperature difference and then use normal lapse rate of 3.5º F/1,000 ft to find the new elevation].
_____________________________ ft.
Normal Lapse Rate
The air temperature at 200 ft above the sea level is 40ºF for a stratified column of air.
A. What will its temperature be at level B (2,000 ft). Show your work. _____________ º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
B. What will its temperature be at the sea level? _______________º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
C. At what elevation will the temperature reach the freezing point (32ºF)?
Show your work. Round your answer to the nearest whole number. [Hint: First find the temperature difference and then use normal lapse rate of 3.5º F/1,000 ft to find the new elevation].
_____________________________ ft.
Normal Lapse Rate
The air temperature at 200 ft above the sea level is 40ºF for a stratified column of air.
A. What will its temperature be at level B (2,000 ft). Show your work. _____________ º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
B. What will its temperature be at the sea level? _______________º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
C. At what elevation will the temperature reach the freezing point (32ºF)?
Show your work. Round your answer to the nearest whole number. [Hint: First find the temperature difference and then use normal lapse rate of 3.5º F/1,000 ft to find the new elevation].
_____________________________ ft. [Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
B. What will its temperature be at the sea level? _______________º F
[Hint: First, find the elevation difference, and then use normal lapse rate of 3.5º F/1,000 ft to find the new temperature].
C. At what elevation will the temperature reach the freezing point (32ºF)?
Show your work. Round your answer to the nearest whole number. [Hint: First find the temperature difference and then use normal lapse rate of 3.5º F/1,000 ft to find the new elevation].
_____________________________ ft.

Explanation / Answer

A. Temperature at 200ft above sea level =40oF

Now with every 1000ft elevation we gain 3.5oF we lose.(Normal lapse rate)

elevation at B is 2000ft.

So, the difference in elevation = 2000-200 ft = 1800ft

So,loss in temperature = (1800x3.5) / 1000 = 6.3oF

Temperature at B=40o - 6.3o F = 33.7oF

B. For sea level elevation difference = 200ft

So, the temperature difference =(200x 3.5) / 1000 oF =0.7oF

So temperature at sea level = 40o + 0.7oF = 40.7oF

C. Here, we have temperature = 32oF

So,Temperature difference= 40o-32oF = 8oF

Now with every 1000ft elevation we gain 3.5oF we lose.(Normal lapse rate)

So for 8oF loss gain in elevation =(8x 1000) / 3.5ft=2285.7 ft

So, the required elevation = 200+ 2285.7 ft = 2485.7ft = 2486ft (rounded off).

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