Question

If the air temperature is +8 °C at an elevation of 1,350 feet and a standard (average) temperature lapse rate exists, what will be the approximate freezing level?

Answer 1

Approximate freezing level is at 5350 feet.

Step-by-step explanation:

We know that the standard lapse rate equals
`-2^(o)C` per 1000 feet

and freezing occurs at a height where the temperature becomes equal to

`0^(o)C`

Thus the standard equation of temperature at any height becomes

`t=mh+c\\\\\therefore t=(-2)/(1000)h+c`

The constant 'c' can be obtained by using the boundary condition that at

h=1350 feet temperature equals
`8^(o)C`

`8=(-2)/(1000)* 1350+c\\\\\therefore c=8+2.7=10.7\\\\\therefore t=(-2)/(1000)h+10.7`

Thus the height at which temperature becomes zero equals

`0=-0.002h+10.7\\\\h=(10.7)/(0.002)=5350feet`