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.

Answer 1

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]`