 Solution

The target liquid-water mass isThe thin-plate ice-growth time uses the prescribed ice mass law, which therefore requires $r_{i\ast }=(m_{\ast }/\alpha {)}^{1/3}=45.31\mu \mathrm{m}$, not the same radius as the water droplet. With $C=2r_i/\pi$ and $F_i=F_k^i+F_d^i$,Hence, for negligible seed mass and constant conditions,This is about $8.24$ times faster than the liquid estimate to the same mass. Both laws have $dm/dt\propto m^{1/3}$ under the given shape assumptions. Ice benefits here from five times the atmospheric supersaturation and a larger geometrical capacitance for the same mass, partly offset by its slightly greater transport resistance. The mass coefficient $\alpha$ encodes the assumed plate shape; replacing it by bulk ice density would change the supplied model.

In a mixed-phase cloud, saturation vapour pressure over ice is lower than over supercooled liquid water. Air can therefore be subsaturated with respect to liquid yet supersaturated with respect to ice: droplets evaporate while ice crystals grow, transferring water through the vapour phase. This is the Wegener–Bergeron–Findeisen mechanism. Subsequent aggregation, riming and melting can produce atmospheric precipitation much faster than condensation alone.

The large difference is not a universal fixed ice/liquid time ratio. Atmospheric supersaturation evolves, latent heating increases thermal resistance, crystal habits and capacitance change, and surface attachment or ventilation may matter. Ice nucleation and the availability of freezing nuclei also precede the calculated growth. The boxed time describes the specified already-present crystals under maintained conditions, not a complete prediction of the time until rain begins.