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 )

Question

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 ), 3 T ρ S 1 r - 1 R, - 2 T ρ S 1 r - 1 R, 2 R ρ S 1 R - 1 r, 3 T ρ S 1 R - 1 r

MARKS App · Accepted Answer

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 π R 3 ρ S = 3 T μ S 1 R - N r 2 R 3 = 3 T ρ S 1 R - R 3 × r 2 r 3 × R 3 = 3 T ρ S 1 R - 1 r

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