What does isothermal mean? Identify all the isothermal layers of the atmosphere.
After solving on your own, check the
What is an inversion? Identify all the inversion layers of the atmosphere.
After solving on your own, check the
Why is the stratosphere warmer than the troposphere?
After solving on your own, check the
If you jump in a hot air balloon when the surface air temperature is \(25^\circ C (77^\circ F)\), what would be the temperature at an altitude of 3 km (about 10,000 ft)? Assume we have an average lapse rate.
After solving on your own, check the
## Question 1.7.1
1. What does isothermal mean? Identify all the isothermal layers of the atmosphere.
Isothermal = Regions of equal temperature.
The isothermal layers are the tropopause, stratopause, and mesopause.
## Question 1.7.2
2. What is an inversion? Identify all the inversion layers of the atmosphere.
The temperature normally decreases when going up in height. An inversion is a region where the temperature actually *increases* with height.
The inversion layers of the atmosphere are the stratosphere and the thermosphere.
## Question 1.7.3
3. Why is the stratosphere warmer than the troposphere?
The ozone in the stratosphere absorbs large amounts of UV radiation. More energy means higher temperatures.
## Question 1.7.4
4. If you jump in a hot air balloon when the surface air temperature is $$25^\circ C (77^\circ F)$$, what would be the temperature at an altitude of 3 km (about 10,000 ft)? Assume we have an average lapse rate.
The average lapse rate is $$6.5^\circ C/km$$. So rising 1 km, the temperature drops $$6.5\circ C$$. Rising 2 km, the temperature drops another 6.5, so a total drop of $$2*6.5\circ C = 13.0\circ C$. Rising 3 km, the temperature drops another 6.5, so a total drop of $$3*6.5\circ C = 19.5\circ C$$.
$$25^\circ C - 19.5\circ C = 5.5^\circ C$$
A shortcut: $$6.5\tfrac{^\circ C}{km}\cdot 3~km = 19.5^\circ C$$. Then $$25^\circ C - 19.5\circ C = 5.5^\circ C$$.
If you want to do this in Fahrenheit and feet, the average lapse rate is $$3.5^\circ F/1,000~ft$$.
$$\frac{3.5^\circ F}{1,000~ft}\cdot 10,000~ft = 35^\circ F$$
The temperature then drops to $$77^\circ F - 35^\circ F = 42^\circ F$$.