Question

If a drop a liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop radius. Let a drop of radius R, break into N small droplets each of radius r. Estimate the drop in temperature.

Answer 1

When a big drop having radius R breaks into N droplets each of radius r, the volume remains constant.

∴ Volume of big drop = N × volume of each small drop

\(\frac{4}{3}πR^3\) = N × \(\frac{4}{3}πr_3\)

Or R³ = Nr³ or N = \(\frac{R^3}{r^3}\)

Change in surface area = 4\(πR^3\)− N4\(πr^2\)

= 4π(R² − Nr²)

Energy released = T × ∆A

= T × 4π(R² − Nr²)

Released energy lowers the temperature by ∆θ, then

Energy released = ms∆θ

T × 4π(R² − Nr²) = \((\frac{4}{3} × R^3 × ρ)\)s∆θ

\([^{s\, =\,\text {specific heat of liquid}}_{ρ\, =\, \text {density}}]\)

or ∆θ = \(\frac{T×4π(R^2−Nr^2)}{\frac{4}{3}πR^3ρ×s}\) = \(\frac{3T}{ρS} [\frac{R^2}{R^3} − \frac{Nr^2}{R^3}]\)

∆θ = \(\frac{3T}{ρS} [\frac{1}{R} − \frac{1}{r}]\)

∵ (N = \(\frac{R^3}{r^3}\))