The surface tension of water is 72 mJ/m–2. Calculate the amount of energy required to disperse one spherical drop of radius 3.0 mm into spherical drops of radius 3.0 × 10–3 mm. The surface area of a sphere of radius r is 4(pi)r2 and the volume is  4/3 (pi) r3 Please show step by step, the answer is 8.1mJ

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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The surface tension of water is 72 mJ/m–2. Calculate the amount of energy required to disperse one spherical drop of radius 3.0 mm into spherical drops of radius 3.0 × 10–3 mm. The surface area of a sphere of radius r is 4(pi)r2 and the volume is  4/3 (pi) r3

Please show step by step, the answer is 8.1mJ 

Expert Solution
Step 1

Surface area of sphere = 4πr2

Surface volume of sphere = 4/3 πr3

Surface tension of water = 72 mJ/m2

We have to calculate .,

The amount of energy required to disperse one spherical drop of radius 3.0 mm into spherical drops of radius 3.0 × 103 mm.

 

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