Question 25: Properties of Matter

Question

Two water drops each of radius ' $r$ ' coalesce to form a bigger drop. If ' $T$ ' is the surface tension, the surface energy released in this process is :

Accepted Answer

To determine the surface energy released when two identical water drops coalesce to form a larger drop, let's go through the process step by step:
Volume Conservation
:
When two droplets, each with a radius $ R $, merge to form a larger droplet, the total volume is conserved. Therefore:
$    2 	imes \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3    $
Solving for $ r $, the radius of the larger drop, we find:
$    r = 2^{1/3} R    $
Initial Surface Energy ($ U_i $)
:
The initial surface energy of the two smaller droplets is given by the surface area of each multiplied by the surface tension $ T $:
$    U_i = 2 	imes 4 \pi R^2 T    $
Final Surface Energy ($ U_f $)
:
The surface energy of the larger droplet is given by:
$    U_f = 4 \pi r^2 T = 4 \pi R^2 T (2^{2/3})    $
Energy Released
:
The energy released in the process is the difference between the initial and final surface energy:
$    	ext{Heat lost} = U_i - U_f = 4 \pi R^2 T \left[2 - 2^{2/3}ight]    $
This expression gives us the amount of surface energy that is released when two identical water drops combine to form a single larger drop.

Answer

$4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{1 / 3}\right]$

Answer

$4 \pi \mathrm{r}^2 \mathrm{~T}[1+\sqrt{2}]$

Answer

$4 \pi \mathrm{r}^2 \mathrm{~T}\left[2-2^{2 / 3}\right]$

Answer

$4 \pi \mathrm{r}^2 \mathrm{~T}[\sqrt{2}-1]$

Properties of Matter

Two water drops each of radius ' r ' coalesce to form a bigger drop. If ' T ' is the surface tension, the surface energy released in this process is :

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