If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius R , break into N small droplets each of radius r , then decrease (drop) in temperature Q (given, specific heat of liquid drop = S and surface tension = T )

Step-by-step solution

since, volume remains unchanged, during this phenomenon, so 4 3 π R 3 = N × 4 3 π r 3 N = R 3 r 3 Now, change in surface area = 4 π R 2 - N 4 π r 2 = 4 π R 2 - N r 2 Energy released ( Δ U ) = T × change in surface area = T × 4 π R 2 - N r 2 Here, all this energy released is al the cost of lowering the temperature and mass of the big drop of liquid = 4 3 π R 2 ρ Now, change in temperature, Δ θ = Δ U m s = T × 4 π R 2 - N r 2 4 3 &#9