Calculate the energy evolved when 8 droplets of water of radius 1/2mm each combine into one where s=0.072N/m.
To calculate the energy evolved when water droplets combine, we can use the equation for the surface energy of a liquid droplet. The surface energy (E) of a droplet is given by:
E = 4πr^2s
Where:
E = Surface energy
r = Radius of the droplet
s = Surface tension of the liquid
We are given that the radius of each droplet is 1/2 mm, which is equal to 0.0005 m. The surface tension (s) is given as 0.072 N/m.
First, let's calculate the surface energy of each droplet before they combine:
E₁ = 4π(0.0005)^2 * 0.072
E₁ = 0.000091783 Nm
Now, since we have 8 droplets, the total energy (E_total) for all droplets before combining is:
E_total = 8 * E₁
E_total = 8 * 0.000091783
E_total = 0.00073427 Nm
When the droplets combine, they form a single droplet. The radius of the new droplet will be the combined radius of the individual droplets, which is 8 times the original radius:
r_new = 8 * 0.0005
r_new = 0.004 m
Now, we can calculate the surface energy of the new droplet:
E_new = 4π(0.004)^2 * 0.072
E_new = 0.000723823 Nm
The energy evolved when the droplets combine is equal to the difference in energy before and after combining:
Energy_evolved = E_total - E_new
Energy_evolved = 0.00073427 - 0.000723823
Energy_evolved = 0.000010447 Nm
Therefore, the energy evolved when 8 droplets of water of radius 1/2mm each combine into one is approximately 0.000010447 Nm.