Calculating Mass from Density and Volume

This problem asks us to find the expected mass of nuclear matter that would fill a matchbox of a given size. We are provided with the dimensions of the matchbox, which allows us to calculate its volume, and the density of nuclear matter. The relationship between mass, density, and volume is fundamental in physics.

Understanding Density, Mass, and Volume

Density ($\rho$) is defined as the mass ($M$) of a substance per unit volume ($V$). The formula relating these quantities is:

\[ \rho = \frac{M}{V} \]

To find the mass, we can rearrange this formula:

\[ M = \rho \times V \]

We are given the density ($\rho$) of nuclear matter and the dimensions of the matchbox, from which we can calculate the volume ($V$).

Calculating Matchbox Volume

The matchbox is rectangular with dimensions 5 cm, 4 cm, and 1 cm. The volume of a rectangular box is calculated by multiplying its length, width, and height.

\[ V = \text{length} \times \text{width} \times \text{height} \]

The dimensions are given in centimeters (cm), but the density is given in kilograms per cubic meter (kg m^-3). To ensure our units are consistent for the calculation, we must convert the dimensions from centimeters to meters (m).

• 1 cm = 0.01 m or $10^{-2}$ m

Converting the dimensions:

• Length = 5 cm = $5 \times 10^{-2}$ m
• Width = 4 cm = $4 \times 10^{-2}$ m
• Height = 1 cm = $1 \times 10^{-2}$ m

Now, calculate the volume in cubic meters (m³):

\[ V = (5 \times 10^{-2} \text{ m}) \times (4 \times 10^{-2} \text{ m}) \times (1 \times 10^{-2} \text{ m}) \]

\[ V = (5 \times 4 \times 1) \times (10^{-2} \times 10^{-2} \times 10^{-2}) \text{ m}^3 \]

\[ V = 20 \times 10^{-2-2-2} \text{ m}^3 \]

\[ V = 20 \times 10^{-6} \text{ m}^3 \]

We can also write this in scientific notation as:

\[ V = 2 \times 10^1 \times 10^{-6} \text{ m}^3 \]

\[ V = 2 \times 10^{1-6} \text{ m}^3 \]

\[ V = 2 \times 10^{-5} \text{ m}^3 \]

Calculating Mass of Nuclear Matter

Now that we have the volume ($V = 2 \times 10^{-5}$ m³) and the density ($\rho = 2.3 \times 10^{17}$ kg m^-3), we can calculate the mass ($M$) using the formula $M = \rho \times V$.

\[ M = (2.3 \times 10^{17} \text{ kg m}^{-3}) \times (2 \times 10^{-5} \text{ m}^3) \]

\[ M = (2.3 \times 2) \times (10^{17} \times 10^{-5}) \text{ kg} \]

\[ M = 4.6 \times 10^{17-5} \text{ kg} \]

\[ M = 4.6 \times 10^{12} \text{ kg} \]

The expected mass of nuclear matter that fills the matchbox is $4.6 \times 10^{12}$ kilograms.

Comparing with Options

Let's compare our calculated mass with the given options:

• 4.6 × 10¹² mg (milligrams)
• 4.6 × 10¹² μg (micrograms)
• 4.6 × 10¹² g (grams)
• 4.6 × 10¹² kg (kilograms)

Our calculated mass is $4.6 \times 10^{12}$ kg, which matches the fourth option.

Revision Table: Key Physics Concepts

Additional Information on Nuclear Matter Density

Nuclear matter is the extremely dense material that makes up the nucleus of an atom, consisting primarily of protons and neutrons. Its density is incredibly high compared to everyday materials. For example, the density of water is about $10^3$ kg m^-3. The density of nuclear matter ($2.3 \times 10^{17}$ kg m^-3) is about $2.3 \times 10^{14}$ times greater than water.

This extreme density means that even a small volume of nuclear matter has an enormous mass. The calculated mass of $4.6 \times 10^{12}$ kg is equivalent to about 4.6 trillion kilograms, which is roughly the mass of thousands of large cargo ships! This highlights just how incredibly dense nuclear matter is and why understanding nuclear density is important in physics and astrophysics (for example, in studying neutron stars, which are essentially giant nuclei).