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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. Estimate the drop in temperature.
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Solution
When a big drop of radius R, breaks into N droplets each of radius r, the volume remains constant.
∴ Volume of big drop = N × Volume of each small drop
`4/3 πR^3 = N xx 4/3 πR^3`
or `R^3 = Nr^3`
or `N = R^3/r^3`
Now, the change in surface area = `4πR^2 - N4πr^2`
= `4π(R^2 - Nr^2)`
The energy released = T × ΔA
= `S xx 4π(R^2 - Nr^2)` .....[T = Surface tension]
Due to releasing of this energy. the temperature is lowered.
If ρ is the density and s is the specific heat of liquid and its temperature is lowered by Δθ then the energy released = msΔθ ......[s = specific heat Δθ = change in temperature]
`T xx 4π(R^2 - Nr^2) = (4/3 xx R^3 xx ρ)sΔθ` ......`[∴m = vρ = 4/3 πR^3ρ]`
⇒ Δθ = `(T xx 4π(R^2 - Nr^2))/(4/3 πR^3 ρ xx s)`
= `(3T)/(ρs) [R^2/R^3 - (Nr^2)/R^3]`
= `(3T)/(ρs) [1/R - ((R^3-r^3) xx r^2)/R^3]`
= `(3T)/(ρs) [1/R - 1/r]`
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