Question

At room temperature, the surface tension of water is 72.0 mJ·m–2. What is the energy required to change a spherical drop of water with a diameter of 1.20 mm to five smaller spherical drops of equal size? The surface area of a sphere of radius r is 4π r2 and the volume is 4π r3/3.

Answer 1

Since you need 0.072 Nm per meter squared of water, and you have a spherical drop of water of 0.0012 meter, this means that the area of the spherical drop of water is 0.0000180864 meter squared. Thus the total area of all the 5 spherical drop of water is 0.000090432 of water per meter squared. Multiplying this by 0.072 Nm per meter squared, you get 0,000006511 JOULES.

1)
\[72.0 mJ·m–2= 0.072Nm m ^{2}\]
Area of one drop of sphere:
\[1.2mm=0.0012m \rightarrow 4\times3.14\times 0.0012\times0.0012=0.0000180864 m ^{2}\ \]
Since there are 5 drops, thus:
\[0.0000180864\times5=0.000090432 m ^{2}\]
And we need about 0.072 joules for one meter squared of water, therefor for for spherical water drop we need:
\[0.000090432m ^{2}\times 0.072Nmm ^{2}= 0.000006511 joule\]
so the answer is 0,000006511joules