What amount of energy is released if two mercury droplets of radii 0.1 & 0.2cm collapse into one single drop. The surface tension of mercury is 500x10^-3 Nm^-1?
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Expert's answer
2013-01-25T07:03:11-0500
Energy of surface tension equal: E=sigma*S sigma - surface tension S - area Amount of energy dE = (E1+E2) - E E1 - energy of first drop E2 - energy of second drop E - energy of drop after collapse E1 = sigma*4*pi*R1^2 R1 - radius of first drop E2 = sigma*4*pi*R2^2 R1 - radius of second drop Mass is uniform: M = M1+ M2 4/3*pi*R^3*density = 4/3*pi*R1^3*density + 4/3*pi*R2^3*density R^3 = R1^3 + R2^3 E = sigma*4*pi*(R1^3 + R2^3)^(2/3) dE = (E1+E2) - E = sigma*4*pi*(R1^2+R2^2-(R1^3 + R2^3)^(2/3)) = = 500*10^-3Hm^-1*4*Pi*(0.1^2+0.2^2-(0.1^3 + 0.2^3)^(2/3))*10^-4m^2 =& 4.23*10^-6 J Answer: dE = 4.23*10^-6 J
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