Question

On a spring day, a middle-latitude city (about 40∘ north latitude) has a surface (sea-level) temperature of 10 ∘c. if vertical soundings reveal a nearly constant environmental lapse rate of 6.5 ∘c per kilometer and a temperature at the tropopause of –55 ∘c, what is the height of the tropopause?

Answer 1

10 km

Step-by-step explanation:

We are told that the temperature at the surface is

`T_0 = 10^(\circ)`

and that the rate of drop of the temperature vs height is

`k=-6.5^(\circ)/km`

Therefore we can write the temperature at a generic altitude h as

`T(h) = T_0 + kh`

If we call h the height of the tropopause, we have

`T(h) = -55^(\circ)`

Therefore we can solve the equation to find h, the height of the tropopause:

`h=(T(h)-T_0)/(k)=(-55-10)/(-6.5)=10 km`