Eighth Grade Science: Mass, Volume, Density & Buoyancy
Why does a massive 100,000-ton steel aircraft carrier float effortlessly on the ocean's surface, while a tiny pebble sinks straight to the bottom? The answer lies in the interplay between mass, volume, density, and the upward buoyant force described by Archimedes' Principle. Mastering these concepts allows you to predict how fluids interact, design floating vessels, and understand planetary layering.
Mass, Volume, and the Derivation of Density
To understand why objects sink or float, you must first differentiate between how much matter an object contains and how much physical space it occupies.
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| Mass (m) | Volume (V) | Density (D) |
+-----------------------+-----------------------+-----------------------+
| The total quantity of | The total amount of | The ratio of mass to |
| matter in an object. | three-dimensional | volume; matter packed |
| Measured in grams (g) | space occupied. | per unit volume. |
| using a balance. | Measured in cm^3 or mL| D = m / V |
+-----------------------+-----------------------+-----------------------+
The Density Equation
Density is a characteristic intensive physical property defined as mass per unit volume. It quantifies how tightly matter is packed inside an object. The mathematical relationship is expressed as:
Density = Mass / Volume ( D = m / V )
Mass = Density * Volume ( m = D * V )
Volume = Mass / Density ( V = m / D )
For solid objects, density is typically expressed in grams per cubic centimeter ($g/cm^3$). For liquids, density is expressed in grams per milliliter ($g/mL$). Because one cubic centimeter is defined as exactly equal to one milliliter ($1\ cm^3 = 1\ mL$), the numerical values are directly interchangeable.
Measuring Volume: Geometric Calculation versus Water Displacement
For regular geometric solids, volume is determined using mathematical formulas—such as length times width times height ($V = l * w * h$) for a rectangular prism. For irregular solids, such as a jagged rock or metal bolt, volume is measured using the water displacement method in a graduated cylinder.
Water Displacement Method
Initial Level: 50.0 mL Submerged Object: 65.5 mL
+-------+ +-------+
| | |~~~~~~~| <-- 65.5 mL
|~~~~~~~| <-- 50.0 mL | |
| | | (O) | <-- Submerged Object
+-------+ +-------+
Volume of Object = 65.5 mL - 50.0 mL = 15.5 mL (or 15.5 cm^3)
The Density Column and Relative Density
When immiscible liquids of differing densities are poured into the same column, they arrange themselves into distinct layers based on relative density. The liquid with the highest density sinks to the bottom, while the liquid with the lowest density floats to the top. Solid objects placed into the column sink through liquids denser than themselves and float on liquids less dense than themselves.
The Density Column Stratification
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| Lamp Oil (D = 0.80 g/mL) | <-- Lowest Density
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| Rubbing Alcohol (D = 0.87 g/mL) |
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| Pure Water (D = 1.00 g/mL) | <-- Baseline Standard
+------------------------------------+
| Dish Soap (D = 1.06 g/mL) |
+------------------------------------+
| Corn Syrup (D = 1.38 g/mL) |
+------------------------------------+
| Honey (D = 1.42 g/mL) | <-- Highest Density
+------------------------------------+
Pure liquid water at 4 degrees Celsius has a density of exactly 1.00 gram per milliliter ($1.00\ g/mL$). Any substance with a density less than $1.00\ g/mL$ floats in water, while any substance with a density greater than $1.00\ g/mL$ sinks.
Buoyancy and Archimedes' Principle
Buoyancy is the net upward force exerted by a fluid on any object placed in it. Buoyant force arises from fluid pressure: because fluid pressure increases with depth, the upward pressure against the bottom of a submerged object is always greater than the downward pressure on its top.
Fluid Pressure & Buoyant Force
Downward Force
|
v
+-----------+
| OBJECT |
+-----------+
^
|
Upward Force (Greater pressure at depth!)
Net Result = Upward Buoyant Force
Archimedes' Principle
Articulated over two thousand years ago by the Greek mathematician Archimedes, Archimedes' Principle states that the upward buoyant force exerted on an object submerged in a fluid is exactly equal to the weight of the fluid that the object displaces.
Buoyant Force (F_b) = Weight of Displaced Fluid (W_fluid)
If a solid block displaces 50 Newtons of water when pushed beneath the surface, the water pushes back upward with exactly 50 Newtons of buoyant force.
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| Sinking / Floating | Force & Density Relationship |
+-----------------------+---------------------------------------------+
| Sinking Object | Gravitational Weight > Buoyant Force |
| | Object Density > Fluid Density |
| Neutrally Buoyant | Gravitational Weight = Buoyant Force |
| | Object Density = Fluid Density (Suspended) |
| Floating Object | Gravitational Weight = Buoyant Force |
| | Average Density < Fluid Density |
+-----------------------+---------------------------------------------+
How Steel Ships Float: Average Density
Solid steel has a density of approximately $7.85\ g/cm^3$, which is nearly eight times greater than water. If dropped in water, a solid block of steel sinks immediately. However, an aircraft carrier or cargo ship is not a solid block of steel; it is constructed as an expansive hollow hull filled with air ($D pprox 0.0012\ g/cm^3$). The *average density* of the ship—calculated by dividing the total mass of the steel, cargo, and air by the massive volume enclosed by the hull—is significantly less than the density of water ($1.00\ g/mL$). As a result, the ship floats stably, displacing an amount of water whose weight matches the total weight of the vessel.
Chapter Exercises
Exercise 1: An irregularly shaped metal artifact has a mass of 189.0 grams. When placed into a graduated cylinder containing 40.0 milliliters of water, the water level rises to 65.0 milliliters. Calculate the density of the metal artifact in grams per cubic centimeter and determine whether it would sink or float in pure water.
Exercise 2: A sealed plastic cube with a side length of 5.0 centimeters has a total mass of 100.0 grams. Calculate the volume of the cube, find its density, and explain what happens when the cube is placed into a tank of water with a density of 1.00 g/cm^3.
Exercise 3: Explain Archimedes' Principle in your own words. Use it to explain why a swimmer floats much more easily in the hypersaline Great Salt Lake or Dead Sea (density approximately 1.24 g/mL) than in a freshwater swimming pool (density 1.00 g/mL).
Solutions and Explanations
Solution 1: To find the density, first calculate the artifact's volume using the water displacement method. Subtract the initial volume from the final volume: 65.0 mL - 40.0 mL = 25.0 mL. Because one milliliter equals one cubic centimeter, the artifact's volume is 25.0 cubic centimeters. Next, calculate density by dividing mass by volume: 189.0 grams divided by 25.0 cubic centimeters = 7.56 grams per cubic centimeter. Because the density of the metal artifact ($7.56\ g/cm^3$) is significantly greater than the density of pure water ($1.00\ g/cm^3$), the artifact will sink to the bottom.
Solution 2: The volume of the cube is calculated using the geometric formula for a cube: length times width times height ($5.0\ cm * 5.0\ cm * 5.0\ cm = 125.0\ cm^3$). Next, calculate the density: 100.0 grams divided by 125.0 cubic centimeters = 0.80 grams per cubic centimeter. When placed in water ($1.00\ g/cm^3$), the cube will float because its density ($0.80\ g/cm^3$) is less than that of water. Specifically, because its density is 80% that of water, the cube will float with exactly 80% of its volume (100 cm^3) submerged beneath the water surface and 20% of its volume (25 cm^3) protruding above the surface.
Solution 3: Archimedes' Principle states that any object placed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. When a swimmer enters water, their body displaces a volume of fluid equal to the submerged portion of their body. In a freshwater pool with a density of 1.00 g/mL, the weight of the displaced water is relatively low, requiring the swimmer to submerge nearly their entire body (or tread water) to generate enough buoyant force to balance their weight. In the Dead Sea or Great Salt Lake, where the water is densely packed with dissolved salts ($D pprox 1.24\ g/mL$), each milliliter of displaced water weighs 24% more than freshwater. Consequently, displacing a smaller volume of saltwater produces an upward buoyant force equal to the swimmer's total weight, causing the swimmer to float higher in the water with minimal physical effort.