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Physics 13 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

At constant temperature the volume of water remains constant. So the new radius of the 5 identical small drops can be calculated by equating the initial and final volume. The new surface area can thus be calculated. The energy required is the change in surface area (5*4\[\pi\]*r^2)-4\[\pi\]R^2 times the surface tension.

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