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

I'm not sure but guess the solution is :

the volume of the big drop :
\[\frac{ 4}{ 3 } \pi R^3 = \frac{ 4 }{ 3 } \pi (1.2 \times 10 ^-3)^3 = 7.24 \times 10 ^-9 m^3\]
the volume of each smaller drop :
\[7.24 \times 10 ^ -9 \div 5 = 1.45 \times 10^-9 m^3\]
now calculate the radius of each smaller drop :
\[\frac{ 4 }{ 3 } \pi R^3 = 1.45 \times 10 ^-9 \rightarrow R = 7.02 \times 10 ^-4 m\]
the surface of each smaller drop is :
\[4 \pi R^2 = 4\pi (7.02 \times 10 ^-4 )^2 = 6.19\times 10^-6 m^2\]
the total surface of the drops is :
\[6.19\times10^-6 \times 5 = 3.10 \times 10^-5 m^2\]
and the energy required is :
\[3.10\times 10^-5m^2 \times \frac{ 72 mJ }{ 1m^2 } = 2.23 \times 10^-3 mJ\]