Problem Statement
Calculate the specific weight, density, and specific gravity of two litres of a liquid which weighs 15 N.
Given Data
- Volume, \(V = 2.0 \, \text{litres}\)
- Weight, \(W = 15 \, \text{N}\)
Solution
1. Convert Volume to SI Units (m³)
The standard unit for volume in these calculations is cubic metres (m³).
2. Calculate Specific Weight (\(\gamma\))
Specific weight is defined as weight per unit volume.
3. Calculate Density (\(\rho\))
Density is related to specific weight by the acceleration due to gravity, \(g \approx 9.81 \, \text{m/s}^2\).
4. Calculate Specific Gravity (\(S.G.\))
Specific gravity is the ratio of the density of the liquid to the density of water (\(\rho_{\text{water}} \approx 1000 \, \text{kg/m}^3\)).
Specific Weight: \( \gamma = 7500 \, \text{N/m}^3 \)
Density: \( \rho \approx 764.5 \, \text{kg/m}^3 \)
Specific Gravity: \( S.G. \approx 0.765 \)
Explanation of Terms
Specific Weight (\(\gamma\)): This is the weight of a substance per unit of its volume. It's a measure of how heavy a certain volume of the liquid is and depends on gravity.
Density (\(\rho\)): This is the mass of a substance per unit of its volume. It's an intrinsic property and does not depend on gravity.
Specific Gravity (\(S.G.\)): This is a dimensionless value that compares the density of a liquid to the density of water. It provides a quick way to know if the liquid will float on or sink in water.
Physical Meaning
The results provide a complete profile of the liquid's basic physical properties. The specific gravity of approximately 0.765 is less than 1, which immediately tells us that this liquid is less dense than water and will float on top of it. This is characteristic of many oils and fuels.
The density of 764.5 kg/m³ gives a precise measure of its mass, which is crucial for engineering calculations involving mass flow rates, storage tank capacity, and transportation logistics.
