Question

Calculate the mass of each of these: (a) a sphere of gold of radius \(10.0 \mathrm{~cm}\) [the volume of a sphere of radius \(r\) is \(V=\left(\frac{4}{3}\right) \pi r^{3} ;\) the density of gold \(\left.=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right]\) (b) a cube of platinum of edge length \(0.040 \mathrm{~mm}\) (the density of platinum \(\left.=21.4 \mathrm{~g} / \mathrm{cm}^{3}\right),\) (c) \(50.0 \mathrm{~mL}\) of ethanol (the density of ethanol \(=0.798 \mathrm{~g} / \mathrm{mL})\)

Short answer

The mass of (a) the gold sphere, (b) the platinum cube, and (c) the ethanol are calculated respectively based on their volumes and densities. Apply the provided formulas and appropriate unit conversions where necessary.

Step-by-step solution

Calculate the mass of the sphere of gold

Firstly, we have to find the volume of the sphere. The formula for the volume of a sphere is \(V= \left( \frac{4}{3}\right) \pi r^{3}\). Replace \(r\) with 10.0 cm and calculate the volume. Then, to get the mass, multiply the volume by the density of gold, which is 19.3 g/cm³.

Calculate the mass of the cube of platinum

The volume of a cube is given by \(V= a³\), where 'a' is the length of an edge. Replace \(a\) by the given value of the edge length, which is 0.040 mm. Convert mm to cm by dividing by 10, as 1 cm = 10 mm. Then, multiply this volume by the density of platinum, which is 21.4 g/cm³, to get the mass.

Calculate the mass of 50.0 mL ethanol

This one is simpler, as the volume is already given. The mass is given by the product of density and volume. Simply multiply the volume of ethanol, which is 50.0 mL, by its density, which is 0.798 g/mL.

Key concepts

Volume of a Sphere

Understanding how to calculate the volume of a sphere is essential in various physics and engineering problems. The formula to find the volume of a sphere is \( V = \left( \frac{4}{3} \right) \pi r^3 \). Here, \( V \) represents the volume, and \( r \) is the radius of the sphere.
For example, if the radius \( r = 10.0 \) cm, you simply substitute this value into the formula, getting:

• \( V = \left( \frac{4}{3} \right) \pi (10.0)^3 \)
• Calculate \( (10.0)^3 \) to get 1000.
• Then multiply by \( \frac{4}{3} \pi \) to find the volume.

This gives you the volume in cubic centimeters (cm³), which you can then use to find the mass if the density is known.

Density

Density is a concept that measures how much mass exists in a given volume. It’s defined as mass per unit volume and is commonly expressed in grams per cubic centimeter \( \text{g/cm}^3 \) or kilograms per cubic meter \( \text{kg/m}^3 \).
The formula for density is:

• \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)

Knowing the density of a material allows you to calculate its mass, provided you know its volume. For example, if you know the density of gold \( = 19.3 \text{g/cm}^3 \) and have calculated the volume, you multiply these values to find the mass:

• Mass of gold \( = \text{Volume of the sphere} \times 19.3 \text{g/cm}^3 \).

This relationship helps you in converting descriptive information into quantitative measurements that are needed in scientific calculations.

Conversion of Units

When solving physics problems, it's crucial to ensure that all units are consistent. Often, you'll need to convert from one unit to another to maintain this consistency, particularly when dealing with measurements like length, volume, and mass.
For example, when dealing with a cube with edge length given in millimeters (mm), converting to centimeters (cm) is necessary since most density values are given in \( \text{g/cm}^3 \). Remember:

• 1 cm = 10 mm
  
• Therefore, to convert from mm to cm, divide by 10.

Being comfortable with these conversions helps ensure that you're working with the correct numbers, making it easier to derive accurate results when calculating volumes, masses, and other physical quantities. Proper unit conversion is essential for a seamless application of formulas in any scientific calculation.