# A spherical vessel (diameter "= 2.00 cm") when empty has a mass of "2.00 g". What is the greatest volume of water that can be placed in the vessel and still have the vessel float at the surface of the water?

## Given: density of water $= {\text{1.00 g/cm}}^{3}$

Mar 17, 2018

Here's what I got.

#### Explanation:

For starters, you know that in order for an object to float on water, i.e. at the surface of the water, its density must be lower than the density of water.

In other words, the ratio that exists between the mass of the object and the volume it occupies must be lower than ${\text{1.00 g cm}}^{- 3}$, the given density of water, in order for the object to float on the surface of the water.

So, start by calculating the total volume of the sphere. You know

${V}_{\text{sphere}} = \frac{4}{3} \cdot \pi \cdot {r}^{3}$

Here

• $r$ is the radius of the sphere, equal to half its diameter, $d$

$d = \text{2.00 cm}$

which means that

$r = \text{2.00 cm"/2 = "1.00 cm}$

The volume of the sphere will thus be equal to

${V}_{\text{sphere" = 4/3 * pi * "1.00 cm}}^{3}$

${V}_{\text{sphere" = "4.189 cm}}^{3}$

Now, you know that the empty sphere has a mass of $\text{2.00 g}$, so you can say that its density when empty is equal to

${\rho}_{\text{empty" = "2.00 g"/"4.189 cm"^3 = "0.477 g cm}}^{- 3}$

Since the density of the empty sphere is lower than ${\text{1.00 g cm}}^{- 3}$, you can say that the empty sphere will float.

At this point, your goal is to figure out how much water will increase the total mass of the sphere to the point

${\text{mass sphere"/"4.189 cm"^3 <= "1.00 g cm}}^{- 3}$

Since you want the sphere to float at the surface of the water, you can take the maximum value of that expression and say that you need

${\text{mass sphere"/"4.189 cm"^3 = "1.00 g cm}}^{- 3}$

Keep in mind that technically, this is not the maximum value that the density of the sphere can take in order for it to float on the surface of the water because when the two densities are equal, the sphere will be suspended in the water.

This is equivalent to saying that the mass of the sphere must be equal to

"mass sphere" = 4.189 color(red)(cancel(color(black)("cm"^3))) * "1.00 g" color(red)(cancel(color(black)("cm"^(-3))))

$\text{mass sphere = 4.189 g}$

But remember, this includes the mass of the added water and the mass of the sphere itself, so you will have

$\text{mass added water" = "4.189 g " - " 2.00 g}$

$\text{mass added water = 2.189 g}$

Finally, to convert the mass of water to cubic centimeters, use the given density of water.

$2.189 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * "1 cm"^3/(1.00color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("2.19 cm}}^{3}}}}$

The answer is rounded to three sig figs.

So one could argue that something like ${\text{2.18 cm}}^{3}$ could be a more accurate answer here because ${\text{2.19 cm}}^{3}$ will cause the overall density of the sphere to be equal to that of the water, which like I said before means that the sphere will be suspended in the water, i.e. it will not float on the surface.

For example, in a multiple-choice problem that has ${\text{2.18 cm}}^{3}$ and ${\text{2.19 cm}}^{3}$ as options, I would go with ${\text{2.18 cm}}^{3}$. But if the only option close to these two is ${\text{2.19 cm}}^{3}$, then this is the answer you're looking for.