Calculate the energy evolved when 8 droplets of water (surface tension

1.	Calculate the energy evolved when 8 droplets of water (surface tension 0.72 N/m) of radius 0.5 mm combine into one.
Answer» Here, s = 0.072 N/m r = 0.5 mm = 0.5 × 10^-3 m. Let R be the radius of big drop formed. Volume of the big drop = volume of 8 small drops. \(\frac{4}{3}\)π R³ = 8 × \(\frac{4}{3}\)πr³ or R = 2 r = 2 × 0.5 × 10^-3= 10^-3 m Surface area of big drop : = 4π R² = 4π × (10^-3 )² = 4π × 10^-6m² Surface area of 8 small drops: = 8 × 4π × r² = 8 × 4π × (0.5 × 10^-3 )² = 8π × 10^-6m² Energy evolved = S.T × decrease in area. = 0.072 × 4π × 10^-6 = 9.05 × 10⁷J.

Calculate the energy evolved when 8 droplets of water (surface tension 0.72 N/m) of radius 0.5 mm combine into one.

Answer»

Here, s = 0.072 N/m

r = 0.5 mm = 0.5 × 10^-3 m.

Let R be the radius of big drop formed.

Volume of the big drop = volume of 8 small drops.

\(\frac{4}{3}\)π R³ = 8 × \(\frac{4}{3}\)πr³

or R = 2 r = 2 × 0.5 × 10^-3= 10^-3 m

Surface area of big drop :

= 4π R² = 4π × (10^-3 )²

= 4π × 10^-6m²

Surface area of 8 small drops:

= 8 × 4π × r²

= 8 × 4π × (0.5 × 10^-3 )²

= 8π × 10^-6m²

Energy evolved = S.T × decrease in area.

= 0.072 × 4π × 10^-6 = 9.05 × 10⁷J.