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

Answer:

Approximate freezing level is at 5350 feet.

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

$t=mh+c\\\\\therefore t=\frac{-2}{1000}h+c$

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

h=1350 feet  temperature equals $8^{o}C$

$8=\frac{-2}{1000}\times 1350+c\\\\\therefore c=8+2.7=10.7\\\\\therefore t=\frac{-2}{1000}h+10.7$

Thus the height at which temperature becomes zero equals

$0=-0.002h+10.7\\\\h=\frac{10.7}{0.002}=5350feet$