Question

I've literally asked this question 4 times now... its pretty sad people can't read as smart as they are to be able to answer these questions....

What amount of energy is required to change a spherical drop of water with a diameter of 1.60 mm to SIX (6) smaller spherical drops of equal size? The surface tension, γ, of water at room temperature is 72.0 mJ/m2. PLEASE DO NOT GIVE ME ANSWER FOR FIVE (5) ITS SIX (6)!

Answer 1

You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³.

With a diameter of 1.6 mm you have an original droplet with a radius of 0.6 mm so the surface area is roughly 8.04 mm² (0.00000804 m²) and the volume is roughly 2.145 mm³. ( i have calculated using surface area and volume formula mentioned above)

The total surface energy of the original droplet is 0.00000804 * 72 ~ 0.00058 mJ

The six smaller droplets need to have the same volume as the original. Therefore

6 V = 2.145 mm³ so the volume of one of the smaller spheres is 2.145/6 = 0.357 mm³.

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.357/(4/3) π) = 0.44 mm.

Each of the smaller droplets has a surface area of 4π r² = 2.4 mm² or 0.0000024 m².

The surface energy of the 6 smaller droplets is then 6 * 0.0000024 * 72.0 = 0.000103 mJ
From this radius the surface energy of all smaller droplets is 0.000103 and the difference in energy is 0.000103- 0.00058 mJ = 0.00045 mJ.

Therefore you need roughly 0.00045 mJ or 0.45 µJ of energy to change a spherical droplet of water of diameter 1.6 mm into 6 identical smaller droplets.

this not may be the exact answer but this is approxmate solution.

hope im correct. comment below if its wrong, il check the calculations and il post it again

all the best for your examzz.