Determine the surface energy of water contained in a cylindrical...

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:

$A = \pi r^2$

Given the radius $r = 6.0 \text{ cm} = 6 \times 10^{-2} \text{ m}$, the area is:

$A = \pi \times (6 \times 10^{-2})^2 = 36\pi \times 10^{-4} \text{ m}^2$

The surface tension of water is $T = 0.075 \text{ J/m}^2$.

The surface energy $E$ is:

$E = T \times A$

$E = 0.075 \times 36\pi \times 10^{-4}$

$E = 2.7\pi \times 10^{-4} \text{ J}$

Using $\pi \approx 3.14$:

$E \approx 2.7 \times 3.14 \times 10^{-4} \text{ J} = 8.478 \times 10^{-4} \text{ J}$

Rounding off, we get:

$E \approx 8.5 \times 10^{-4} \text{ J}$