I'm having a bit of trouble figuring out a solution to this problem:

There is an isothermal layer of air which has a temperature of 0 °C extending from the surface to 1 km height. A parcel is 0.5 °C is released at the surface in the isothermal layer. How high does the parcel rise before it descends back?

I know what an isothermal layer is and I know that the parcel will rise until it reaches the surrounding air temperature. I am just not sure how or which formula I would use to solve the problem.


1 Answer 1


Isothermal, as you are likely aware means constant temperature. Your problem has the givens:

  • The 0 - 1 km layer has a constant temperature of 0 C
  • A surface parcel initially has a temperature of 0.5 C

We can likely assume the parcel is not saturated and you are aware that that a parcel with a positive relative temperature perturbation is buoyant and will rise.

To question you need to look at next is: what happens to a parcel of air as it rises? Your 0 - 1 km layer has a constant temperature but not a constant pressure so the parcel will expand as it rises. If the parcel expands then its temperature will decrease.

At what rate does the temperature of an unsaturated parcel of air decrease as it rises? This number will allow you to calculate the height that the parcels temperature decreases by 0.5 C and neutral buoyancy is attained.

  • $\begingroup$ So would I use the dry adiabatic lapse rate somehow? Or poisson's equation? $\endgroup$
    – Sofia June
    Oct 22, 2014 at 3:57
  • $\begingroup$ Poisson's equation would yield the dry adiabatic lapse rate, so yes, you are on the right track. Use the dry adiabatic lapse rate to determine the height needed for a 0.5 C temperature decrease and you'll find your answer. $\endgroup$
    – casey
    Oct 22, 2014 at 4:17
  • $\begingroup$ So is the dry adiabatic lapse rate is 0.98 degrees Celsius / 100 meters then in this case? $\endgroup$
    – Sofia June
    Oct 22, 2014 at 4:41
  • 2
    $\begingroup$ Nice "look in this direction" answer to a homework problem IMO :-) $\endgroup$ Oct 22, 2014 at 10:03

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