Question

What amount of energy is required to change a spherical drop of water with a diameter of 1.60 mm to five smaller spherical drops of equal size? The surface tension, y, of water at room temperature is 72.0 mJ/m2 Number 0.34 mJ

Answer 1

Surface Area in 4 pi gamma^2 volume = 4/3 pi gamma^3 and gamma = 1.60 mm/2 = 0.8 mm S_A = 8042 mm^2 Volume = 2.145 mm^3 Total surface energy of original = 8.042 times 10^-6 times 72 = 0.579 times 10^3 mj Now for smaller drops: - 5V = 2.145 mm^3 v = 0.429 mm^3 gamma_small = 3 squareroot 3/4 pi times 0.429 = 0.467 mm surface area of each drop = 4 pi gamma_small^2 4 times pi times 0.467^2 mm^2 2.74 mm^2 therefore Total surface energy of the 5 smaller droplets = 5 * 2.74 times 10^6 times 72 = 9.864 times 10^-4 mJ \r\n The difference in energy = (9.864 times 10^-4 - 0.579 times 10^-3) mJ = 4.074 times 10^-4 Mj Number 4.074 times 10^-4 mJ or Number 0.0004074 mJ