What amount of energy is required to change a spherical drop of water with a dia

Question

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

__mJ ?

I have tried finding the surface area and volume of the original drop by using the formulas: 4pi r^2 and (4)pi r^3 and then dividing the volume by two the get the smaller volume drops...then using the volume of the smaller drops to find the radius to find the surface area of the smaller drops....and then multiplying the surface area by two to get the total surface area of both drops...which i got 10.13... that answer was incorrect so then I multiplied that answer by 72 to get 729.57 and that was incorrect as well....

PLEASE HELP!

Explanation / Answer

Here is how to solve

4/3 pi x (2mm x 10^-3)^3 = volume of big drop

Take the volume of the big drop and divide it by 2 drops = the volume of each small drop

take that volume and calculate the radius of the smaller drop using 4/3 pi R^3 to get radius of each small drop

then take the radius of the small drops and multiply it by 4 pi to get the surface area of each smaller drop

then take that surface area of the small drops and multiply it by 2 drops to get the total surface area of both drops

then take the total surface of both drops and multiply it by 72mJ/ 1m^2 = Energy required

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