Two identical spheres of dry-ice are left to sublimate in two different chambers.
In chamber 1, which is exposed to air, the rate of sublimation is proportional to the instantaneous surface area of the sphere.
In chamber 2, which is in a microwave oven, the rate of sublimation is proportional to the instantaneous volume of the sphere.
For the sake of simplicity, it is assumed that the sublimation rate is uniform spatially (i.e., the sphere retains its shape at all times),
Which sphere will be the first to evaporate completely?
Note : Here the rate of sublimation means rate of change of volume.
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In Chamber 1
Let the volume be V , then surface area is A = ( 4 π ) 1 / 3 ( 3 V ) 2 / 3 , if the rate constant for sublimation is taken as ( 4 π ) 1 / 3 3 2 / 3 α , then the dynamics of the volume is governed by
d t d V = − α V 2 / 3 , V ( 0 ) = V 0
On solving this, we have V ( t ) = 2 7 1 ( 3 3 V 0 − α t ) 3
There is complete evaporation at time t = α 3 3 V 0 .
In Chamber 2
If a rate constant of β is assumed in this case, the dynamics would be governed by
d t d V = − β V , V ( 0 ) = V 0
which has a solution, V ( t ) = e − β t V 0 , the volume tends to zero, but never quite reaches zero in this case.
Therefore irrespective of the relative magnitudes of α and β , as long as α > 0 , the sphere in chamber 1 evaporates completely at some finite time and the one in chamber 2 takes infinite amount of time.