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?

Respuesta :

Answer:

10 km

Explanation:

We are told that the temperature at the surface is

[tex]T_0 = 10^{\circ}[/tex]

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

[tex]k=-6.5^{\circ}/km[/tex]

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

[tex]T(h) = T_0 + kh[/tex]

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

[tex]T(h) = -55^{\circ}[/tex]

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

[tex]h=\frac{T(h)-T_0}{k}=\frac{-55-10}{-6.5}=10 km[/tex]