If a drop of liquid breaks into smaller droplets, it results in...

since, volume remains unchanged, during this phenomenon, so

$\dfrac{4}{3}\piR^{3} = N \times \dfrac{4}{3}\pir^{3}$

$N = \dfrac{R3}{r3}$

Now, change in surface area $= 4\piR^{2} - N4\pir^{2}$

$= 4\pi \left(R^{2} - Nr^{2}\right)$

Energy released $( \Delta U ) = T \timeschange in surface area</p><p>$= T \$times4\pi \left[R^{2} - Nr^{2}\right]$

Here, all this energy released is al the \cos t of lowering the temperature and mass of the big drop of liquid $= \dfrac{4}{3}\piR^{2}$ρ</p><p>Now, change in temperature, \Delta\theta = \dfrac{\Delta U}{ms}</p><p>=\dfrac{T \times 4\pi\left(R^{2} - Nr^{2}\right)}{ \left(\dfrac{4}{3}\piR^{3}ρ\right)S}

$= \dfrac{3T}{\muS}\left(\dfrac{1}{R} - \dfrac{Nr^{2}}{R3}\right)$

$= \dfrac{3T}{ρS} \left(\dfrac{1}{R} - \dfrac{R^{3} \times r^{2}}{r^{3} \times R^{3}}\right)$

= \dfrac{3T}{ρS} \left(\dfrac{1}{R} - \dfrac{1}{r}\right)$</p>$$