Sphere, Tank and Water

Basic Thermodynamics ~ J. Pohl ©

www.THERMOspokenhere.com (60-B035)

The diameter of the sphere suspended above the tank containing water is one foot (D = 1 ft). When the sphere is lowered into the tank the water contained rises to a final depth of two feet. No water spills over.

Calculate the mass of water in the tank.
♦ Since the final depth of water will be 2 feet, we conclude the sphere will be completely submerged. Neither volume, that of the sphere nor water changes. Hence their final volume, taken together, will equal the sum of their initial volumes.

(1) The final volume equals the sum
of the initial volumes.

(2)Here, “h” is the initial depth
of the water in the tank.

Thus the mass of water initially present is: m = ρ V:

(3)Mass equals density times volume.

Though we tend to take geometry for granted, it can become complicated for the novice.

Suppose the initial depth of water were 0.5 feet (that is, h = 0.5 ft). Were the sphere lowered into the tank, what final depth of the water would occur?
♦ Hoping this calculation is simple, we assume the sphere becomes completely submerged. That way the sum of the spherical and rectangular volumes will become a rectangular total. With that assumption, we calculate the second depth of water as a sum of the water in the tank initially and volume of the sphere (subject to our assumption).

(4) CONTRADICTION: the calculated final depth, h,
is less that the diameter of the sphere.
Hence the sphere is not submerged.

A problem has arisen. We assumed complete submergence. The second depth determined (0.67ft) is less that the diameter of the sphere. Thus the sphere will not be completely submerged; our assumption was wrong. Our calculation must be recast to include a "cap" of the sphere extending above the water level. Do you know the volume of a spherical cap? Some tedious algebra will obtain an answer. We leave it here.