Calculate the amount of energy evolved when $8$ droplets of water of radius $0.5 m m$ each combine into one. Surface tension of water is $72 d y n e c m^{- 1}$ .

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Solution

The volume surface of the droplet which spherical in shape is given as

$V = \frac{4}{3} π R^{3}$

The volume of the eight droplet is

$V = 8 \times \frac{4}{3} π {(0.5)}^{3}$

On equation the above two equation we get the radius of the combined droplet

The radius is $1 m m$

The surface energy of the droplet will be

$Δ E = 8 \times 4 π {(\frac{1}{2} \times \frac{1}{1000})}^{2} \times 0.072 - 4 π {(\frac{1}{1000})}^{2} \times 0.072$

The energy evolved will be $9 \times 1 0^{- 7} J$

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Calculate the amount of energy evolved when 8 droplets of water of radius 0.5 mm each combine into one. Surface tension of water is 72 dyne cm−1.

Calculate the amount of energy evolved when $8$ droplets of water of radius $0.5 m m$ each combine into one. Surface tension of water is $72 d y n e c m^{- 1}$ .