Determine the surface energy of water contained in a cylindrical vessel with a radius of 6.0 cm . The surface tension of water is 0.075 J/m ^2 .

Step-by-step solution

The surface energy of a liquid is given by the product of its surface tension and the area of its free surface. The free surface of water in a cylindrical vessel is circular. The area of this surface is: A = r^2 Given the radius r = 6.0 cm = 6 10⁻² m , the area is: A = (6 10⁻²)^2 = 36 10⁻⁴ m ^2 The surface tension of water is T = 0.075 J/m ^2 . The surface energy E is: E = T A E = 0.075 36 10⁻⁴ E = 2.7 10⁻⁴ J Using 3.14 : E 2.7 3.14 10⁻⁴ J = 8.478 10⁻⁴ J Rounding off, we get: E 8.5 10⁻⁴ J