Question

For a material to float on the surface of water, the material must have a density less than that of water \((1.0 \mathrm{~g} / \mathrm{mL})\) and must not react with the water or dissolve in it. A spherical ball has a radius of \(0.50 \mathrm{~cm}\) and weighs \(2.0 \mathrm{~g}\). Will this ball float or sink when placed in water? (Note: Volume of a sphere \(=\frac{4}{3} \pi r^{3}\).)

Short answer

The volume of the sphere is calculated using the formula \(V = \frac{4}{3} \pi (0.50\mathrm{~cm})^{3}\), which yields approximately \(0.52 \mathrm{~mL}\). Then, we calculate the density of the sphere using the formula \(Density = \frac{2.0\mathrm{~g}}{0.52 \mathrm{~mL}} \approx 3.85 \mathrm{~g} / \mathrm{mL}\). Since the density of the sphere (3.85 \(\mathrm{g/mL}\)) is greater than the density of water (1.0 \(\mathrm{g/mL}\)), the ball will sink when placed in water.

Step-by-step solution

Calculate the volume of the sphere

The formula to calculate the volume of a sphere is given: \[ V = \frac{4}{3} \pi r^{3} \] With the given radius (\(r = 0.50\mathrm{~cm}\)), we can calculate the volume: \[ V = \frac{4}{3} \pi (0.50\mathrm{~cm})^{3} \]

Calculate the density of the sphere

The density formula is: \[ Density = \frac{Weight}{Volume} \] Using the sphere's given weight of \(2.0\mathrm{~g}\) and the volume calculated in Step 1 (assuming you already calculated it), we can calculate the density: \[ Density = \frac{2.0\mathrm{~g}}{V} \]

Compare the density of the sphere to the density of water

If the calculated density of the sphere is less than \(1.0\mathrm{~g} / \mathrm{mL}\) (the density of water), the sphere will float. If it is greater than \(1.0\mathrm{~g} / \mathrm{mL}\), the sphere will sink.

Key concepts

Volume of a Sphere

The volume of a sphere is a three-dimensional measure of the amount of space it occupies. The formula to calculate this is given by \[ V = \frac{4}{3} \pi r^{3} \]Here, \( V \) is the volume and \( r \) is the radius of the sphere. For example, if you have a sphere with a radius of \(0.50\mathrm{~cm}\), you substitute this into the formula: \[ V = \frac{4}{3} \pi (0.50)^3 \]This will give you the volume of the sphere. The volume is a crucial piece of information as it is used to determine other properties like density. Knowing how to calculate the volume helps in understanding how much space is inside a spherical object.

Floatation

Floatation is the concept that determines whether an object sinks or floats in a fluid like water. This process is governed by the relative densities of the object and the fluid. - An object will float if its density is less than the fluid it is placed in.- Conversely, if the object's density is greater than the fluid's density, the object will sink.In our specific example, if a ball has a density less than the density of water \((1.0 \mathrm{~g} / \mathrm{mL})\), it will float. This principle is often used in designing boats and other vessels that need to float by ensuring they are less dense than water. Understanding floatation can be critical for applications in engineering and daily life.

Density of Water

Density is a measure of mass per unit volume, and it is critical in understanding how substances interact. For water, the standard density is \(1.0 \mathrm{~g} / \mathrm{mL}\). This value is a benchmark for determining floatation because any object with a density less than this value is expected to float. - When dealing with problems involving water, always remember this density benchmark.- Also, be aware that temperature can slightly affect water's density.Essentials in aquatic projects or any study involving fluids include knowing how various densities compare with water. This basic knowledge is crucial when predicting how different materials will behave when submerged.