Problem 77 - Chemistry: The Central Science

Question

A thief plans to steal a gold sphere with a radius of \(28.9 \mathrm{~cm}\) from a museum. If the gold has a density of \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass of the sphere in pounds? [The volume of a sphere is \(V=(4 / 3) \pi r^{3}\).] Is the thief likely to be able to walk off with the gold sphere unassisted?

Step-by-Step Solution

Answer

Calculating the volume of the sphere using the given formula and radius, \(V = \frac{4}{3} \pi (28.9)^{3}\), we get the mass of the gold sphere in grams using the density, \(m_\text{grams} = V \times 19.3\). Converting the mass to pounds, \(m_\text{pounds} = \frac{m_\text{grams}}{453.592}\). After calculating the mass of the gold sphere in pounds, we can compare it to a reasonable weight that an average person can carry (around 50 pounds) to determine if the thief can walk off with the gold sphere unassisted.

Step 1: Calculate the volume of the sphere

To find the volume of the sphere, we can use the given formula: \[V = \frac{4}{3} \pi r^{3}\] where \(r = 28.9 \mathrm{~cm}\), the radius of the gold sphere. Let's calculate the volume: \[V = \frac{4}{3} \pi (28.9)^{3}\]

Step 2: Calculate the mass of the gold in grams

Now that we have found the volume of the sphere, we can use the density of gold to find the mass of the gold sphere. The density of gold is given as 19.3 g/cm³. So, the mass of the gold sphere in grams can be calculated using: \[m_\text{grams} = V \times \rho\] where \(\rho = 19.3 \mathrm{~g} / \mathrm{cm}^{3}\), the density of gold. Let's calculate the mass: \[m_\text{grams} = V \times 19.3\]

Step 3: Convert the mass from grams to pounds

To convert the mass of the gold sphere from grams to pounds, we can use the conversion factor: \(1 \mathrm{~lb} = 453.592 \mathrm{~g}\) So, the mass of the gold sphere in pounds can be calculated using: \[m_\text{pounds} = \frac{m_\text{grams}}{453.592}\]

Step 4: Determine if the thief can carry the gold sphere

We can compare the mass of the gold sphere in pounds to a reasonable weight that an average person can carry with difficulty (around 50 pounds). If the mass of the gold sphere is significantly larger than this value, the thief is unlikely to be able to walk off with it unassisted. Let's analyze the situation after calculating the mass of the gold sphere in pounds. Now let's apply these steps to find the mass of the gold sphere in pounds and determine if the thief can carry it unassisted.

Key Concepts

Density

Density is a fundamental concept in both physics and material science, and it plays a critical role in our understanding of how different materials behave. Simply put, density (\rho) measures how much mass is contained in a given volume. It is commonly expressed as grams per cubic centimeter (g/cm³) for solids, such as the gold in our problem.

The formula for density is: \[\rho = \frac{m}{V}\] where \(m\) represents mass and \(V\) represents volume. When we know the density of a material and the volume it occupies, we can easily calculate the mass by rearranging the formula: \[m = V \times \rho\]
Understanding density is not only important for scientific calculations but also for practical applications, such as designing objects that need to float on water or for materials that must withstand certain forces.

Volume of a Sphere

When dealing with three-dimensional objects, such as a sphere, calculating volume allows us to understand the amount of space it occupies. The volume of a sphere, in particular, is determined by its radius, which is the distance from the center of the sphere to its surface.

To calculate the volume of a sphere, we use the formula: \[V = \frac{4}{3} \pi r^{3}\] In this formula, \(r\) is the radius of the sphere, and \(\pi\) is the constant (\(\approx 3.14159\)) that is the ratio of the circumference of a circle to its diameter.

For a gold sphere with a radius given in our problem, we would plug in the value of \(r\) into this formula to find its volume. The concept of a sphere's volume is essential in various fields such as architecture, engineering, and even space sciences, where it helps in the calculation of capacities, designing round structures, and understanding celestial bodies, respectively.

Unit Conversion

Unit conversion is a crucial skill in science, mathematics, and everyday life, as it allows us to translate measurements from one set of units to another. In physics and engineering, it is often necessary to convert between units of mass, length, and volume to solve problems effectively.

For instance, if we calculate the mass of an object in grams but need to understand its weight in pounds, we need to use a conversion factor. This conversion factor is a ratio that represents how many of one unit equals another unit. For the mass conversion, the factor from grams to pounds is: \[1 \mathrm{lb} = 453.592 \mathrm{g}\] To convert from grams to pounds, we divide the mass in grams by the conversion factor:
\[m_\text{pounds} = \frac{m_\text{grams}}{453.592}\]
Unit conversion ensures compatibility of measurements across different systems and countries, enabling consistent and accurate communication of quantities in science, industry, and trade.