Question

What amount of energy is required to change a spherical drop of water with a diameter of 2.00 mm to two smaller spheric drops of equal size? The surface tension, y. of water at room temperature is 72.0 mJ/m². AE | 0,00012868

Answer 1

20 2.00 mon goa 1 mm 0.1 cm = = 0.1 x 10 m Area of the drop - 4802 - 4x3-14X(0-1x10²) * = 1125x100 m2 volume occupied - 4793 - 430144 (0.1102) = 401x10-6 m 3 If it is divided into two equal size drops and to be the native of new small drops, 2x Lagos = 401X 10 -6 on, 7o3 4.1X10**3 2X4K 3.14 on, v' = 7.8 x 10-3 m -= 4.8 x 10

Ewrfall. energy of one. Spherical drop - 4x2x (0-18 10^2) *8 Surface, energy of two new u n3 = 4878/7-2x16)X2XJ Energy required = 487x (7.8x10-3)*8x2 - 48x10.1X1052)x8 = 48 7x8 { 2x/7.8x10+)? (0-18 10-1);} = 4x9x8 {{1•2x1074) – 10.01x1024) = 4X 3• 12x 72 m3/m2 x 1:19X10-4 m2 = 0.1069 MI = 10.1069X 0.001) 11= 1000 m] -10.1069X 0.001)xröfengs 1m3 = 3 = 0.1069x 10^eng. = 0.001 3 : 1069 eng 1 J = 10 7 ergs 1000