Mass, Volume, Specific Gravity Calculator • 2026
\( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
\( \text{Mass} = \text{Density} \times \text{Volume} \)
\( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \)
\( \text{Specific Gravity} = \frac{\text{Density of Substance}}{\text{Density of Water}} \)
\( \text{Buoyant Force} = \text{Volume} \times \text{Density of Fluid} \times g \)
Where:
Density calculations are fundamental in physics, chemistry, engineering, and geology. These formulas help scientists determine material properties, identify substances, and predict behavior in fluids.
Example: For a 50g object with volume 20 cm³: \( \text{Density} = \frac{50}{20} = 2.5 \) g/cm³.
| Parameter | Value | Unit | Formula |
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| Buoyancy Parameter | Value | Unit | Formula |
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| Specific Gravity Parameter | Value | Unit | Description |
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Density is the mass per unit volume of a substance. It's an intensive property that remains constant regardless of the amount of material. Density is crucial for identifying materials, predicting whether objects will sink or float, and understanding how materials behave under different conditions.
Density (ρ): Mass per unit volume
Mass (m): Amount of matter in an object
Volume (V): Space occupied by an object
Specific Gravity: Density relative to water
Buoyant Force: Upward force on submerged objects
Density calculations are essential for material identification, quality control, engineering design, geological surveys, and scientific research. They're used in industries ranging from mining to aerospace to determine material properties and predict behavior.
A metal cube has a mass of 78.75g and a volume of 10 cm³. What is its density?
The answer is B) 7.88 g/cm³. Using the formula: Density = Mass / Volume = 78.75g / 10cm³ = 7.875 g/cm³ ≈ 7.88 g/cm³. This density is consistent with iron.
This question tests the fundamental density calculation. Understanding that density is mass per unit volume is crucial. This property helps identify materials and predict their behavior in different environments.
Density: Mass per unit volume
Mass: Amount of matter in an object
Volume: Space occupied by an object
• Density = Mass / Volume
• Units must be consistent
• Density is an intensive property
• Use precise measurements
• Consider temperature effects
• Compare to known values
• Using wrong formula (V/m instead of m/V)
• Inconsistent units
• Forgetting to round appropriately
A 200g object with a volume of 80 cm³ is placed in water. Will it sink or float? Calculate the buoyant force acting on it. (Density of water = 1.0 g/cm³, g = 9.81 m/s²)
Step 1: Calculate object density
ρ_object = Mass / Volume = 200g / 80cm³ = 2.5 g/cm³
Step 2: Compare to water density
Since 2.5 g/cm³ > 1.0 g/cm³, the object will sink.
Step 3: Calculate buoyant force
First, convert volume to m³: 80 cm³ = 80 × 10⁻⁶ m³
ρ_water = 1.0 g/cm³ = 1000 kg/m³
F_b = Volume × ρ_water × g
F_b = (80 × 10⁻⁶ m³) × (1000 kg/m³) × (9.81 m/s²) = 0.785 N
Therefore, the object will sink and experience a buoyant force of 0.785 N.
This problem demonstrates Archimedes' principle, which states that the buoyant force on an object equals the weight of fluid displaced. Objects denser than the fluid sink, while less dense objects float. The buoyant force calculation requires unit conversions.
Buoyant Force: Upward force on submerged objects
Archimedes' Principle: Force equals weight of displaced fluid
Displacement: Volume of fluid moved by object
• Objects sink if ρ_object > ρ_fluid
• Objects float if ρ_object < ρ_fluid
• F_b = V_displaced × ρ_fluid × g
• Compare densities directly
• Convert units consistently
• Consider shape effects on buoyancy
• Confusing mass and density
• Forgetting unit conversions
• Using wrong volume in buoyancy calculation
A liquid has a density of 0.85 g/cm³. What is its specific gravity? If a 50g piece of cork (density = 0.24 g/cm³) is placed in this liquid, will it sink or float?
Step 1: Calculate specific gravity
SG = ρ_liquid / ρ_water = 0.85 g/cm³ / 1.0 g/cm³ = 0.85
Step 2: Compare cork density to liquid density
Cork density = 0.24 g/cm³
Liquid density = 0.85 g/cm³
Since 0.24 g/cm³ < 0.85 g/cm³, the cork will float in the liquid.
Step 3: Verify with specific gravity approach
SG_cork = 0.24 / 1.0 = 0.24
Since SG_cork (0.24) < SG_liquid (0.85), cork floats.
Therefore, the liquid's specific gravity is 0.85, and the cork will float.
This problem demonstrates how specific gravity provides a unitless comparison to water. It also shows that an object floats in a liquid if its density is less than the liquid's density, regardless of whether it's less than water.
Specific Gravity: Density relative to water
Unitless Ratio: No units since it's a comparison
Floatation: When object density < fluid density
• SG = ρ_substance / ρ_water
• SG < 1 means substance is less dense than water
• Floatation depends on relative densities
• Specific gravity is easier to work with
• Compare SG values directly
• Water density = 1.0 g/cm³ (reference)
• Forgetting that specific gravity is unitless
• Comparing absolute densities incorrectly
• Not using water as reference point
A 100g aluminum block at 25°C has a density of 2.70 g/cm³. If heated to 100°C, calculate its new density. (Linear expansion coefficient for aluminum = 23×10⁻⁶ /°C)
Step 1: Calculate initial volume
V_initial = Mass / Density = 100g / 2.70 g/cm³ = 37.04 cm³
Step 2: Calculate volume expansion
For volume expansion: β ≈ 3α = 3 × (23×10⁻⁶) = 69×10⁻⁶ /°C
ΔV = V_initial × β × ΔT = 37.04 cm³ × (69×10⁻⁶) × (100-25)°C
ΔV = 37.04 × 69×10⁻⁶ × 75 = 0.192 cm³
Step 3: Calculate final volume
V_final = V_initial + ΔV = 37.04 + 0.192 = 37.23 cm³
Step 4: Calculate final density
Since mass remains constant: ρ_final = Mass / V_final = 100g / 37.23 cm³ = 2.69 g/cm³
Therefore, the new density is 2.69 g/cm³.
This problem demonstrates how temperature affects density through thermal expansion. As temperature increases, most materials expand, increasing volume while keeping mass constant, resulting in decreased density. This is important in engineering applications.
Thermal Expansion: Increase in size with temperature
Linear Coefficient: Expansion per degree per unit length
Volume Coefficient: Approx. 3 times linear coefficient
• β ≈ 3α (volume ≈ 3 × linear expansion)
• ΔV = V₀ × β × ΔT
• Mass remains constant during heating
• Most materials expand when heated
• Gases show much larger expansion than solids
• Consider thermal effects in precision work
• Forgetting that mass remains constant
• Using linear instead of volume coefficient
• Not accounting for temperature change correctly
How does pressure affect the density of a gas at constant temperature?
The answer is A) Density increases proportionally with pressure. According to the ideal gas law (PV = nRT), at constant temperature: PV = constant, so P₁V₁ = P₂V₂. Since density ρ = m/V and mass is constant, ρ₁V₁ = ρ₂V₂. Combining: ρ₂/ρ₁ = V₁/V₂ = P₂/P₁. Therefore, density is directly proportional to pressure at constant temperature.
This question explores the relationship between pressure and density for gases. At constant temperature, increasing pressure compresses the gas, reducing volume and increasing density. This relationship is described by Boyle's law and is crucial for understanding gas behavior.
Boyle's Law: P₁V₁ = P₂V₂ at constant temperature
Ideal Gas Law: PV = nRT
Compressibility: Ability to reduce volume under pressure
• For gases: ρ ∝ P (at constant T)
• Liquids and solids show minimal compression
• Gases are highly compressible
• Gases compress significantly under pressure
• Liquids are essentially incompressible
• Solids compress minimally
• Applying gas laws to liquids/solids
• Forgetting that temperature must be constant
• Confusing direct and inverse relationships
Q: What's the difference between density and specific gravity?
A: Density is mass per unit volume with units (g/cm³, kg/m³, etc.), while specific gravity is a unitless ratio comparing the density of a substance to the density of water.
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \quad \text{(with units)} \]
\[ \text{Specific Gravity} = \frac{\text{Density of Substance}}{\text{Density of Water}} \quad \text{(unitless)} \]
For example, if a liquid has density 0.85 g/cm³, its specific gravity is 0.85/1.0 = 0.85. Specific gravity is convenient because it eliminates units and allows direct comparison to water (SG = 1.0).
Q: How does temperature affect density measurements?
A: Temperature significantly affects density through thermal expansion. As temperature increases, most materials expand, increasing volume while keeping mass constant, thus decreasing density.
For solids: \( \rho = \frac{\rho_0}{1 + \beta(T - T_0)} \) where β is the volume expansion coefficient.
For liquids and gases, the effect is more pronounced. Water is unusual - it's densest at 4°C (0.9998 g/cm³), becoming less dense as it freezes to ice (0.917 g/cm³), which is why ice floats.
Standard density measurements are typically taken at 20°C or 25°C. Always specify temperature when reporting density values.