What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drops of equal size? The surface tension, γ, of water at room temperature is 72.0 mJ/m2.

Question

Chemistry

Blanche

6 December, 22:03

What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drops of equal size? The surface tension, γ, of water at room temperature is 72.0 mJ/m2.

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Amari Le · Answer 1

This is a straightforward question related to the surface energy of the droplet.

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

With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³.

The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ

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

5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³.

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root (0.287 / (4/3) π) = cube_root (4.39) = 0.4 mm.

Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m².

The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ

From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072 - 0.00044 mJ = 0.00028 mJ.

Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets.