If a drop of liquid breaks into smaller droplets, it results in lowering of the temperature of the droplets. Let a drop of radius R, break into N small droplets each of radius r. Estimate the drop in temperature
ΔE=σ(Final area−initial area of surface) ΔE=msΔt, By the law of conservation of mass, Final volume = Initial volume of one drop of rasius R, splitted in N drops of radius r ∴34πR3=N.34πr3 or R3=Nr3 r=N31R ΔE=σΔA=σ[Are of N drops of radius r−Area of big drop] msΔt=σ[N.4πr2−4πR2] VρsΔt=4πσ[Nr2−R2] N.(34πr3)ρsΔt=4πσ[Nr2−R2] M = mass of smaller drops ρ = density of liquid s = specific heat of liquid Δt = change in temperature Δt=N.4πr3ρs4πσ×3[Nr2−R2] Δt=Nρs3σ[r3Nr2−r2R2] (∵r3=NR3) Δt=ρNs3σN[r1−R1] Δt=ρs3σ[r1−R1] asR>r Δt will be positive i.e. R1<r1 Hence, formation of smaller drops require the temperature of drops to increase. This energy is taken from surrounding whose temperature decreases.
