Calculating Specific Gravity of a Liquid from Weight and Volume

Let's break down how to find the specific gravity of a liquid when given its volume and weight. Specific gravity is a measure of how dense a substance is compared to the density of a reference substance, which for liquids and solids is typically water.

Understanding Key Concepts

• Weight (W): The force of gravity acting on an object's mass. It is calculated as \(W = m \times g\), where \(m\) is mass and \(g\) is acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
• Mass (m): The amount of matter in an object.
• Volume (V): The amount of space an object occupies.
• Density (\(\rho\)): Mass per unit volume. It is calculated as \(\rho = \frac{m}{V}\).
• Specific Gravity: The ratio of the density of a substance to the density of a reference substance (usually water at a specific temperature, typically $4^\circ\text{C}$ where its density is maximum, approx. $1000 \, \text{kg/m}^3$). Specific Gravity = \(\frac{\text{Density of Substance}}{\text{Density of Reference Substance}}\).

Given Information

• Volume of the liquid (V) = 1 litre
• Weight of the liquid (W) = 5.5 N

Reference Values

• Density of water (\(\rho_{water}\)) \(\approx 1000 \, \text{kg/m}^3\)
• Acceleration due to gravity (g) \(\approx 9.8 \, \text{m/s}^2\)

Step-by-Step Calculation of Liquid Specific Gravity

To find the specific gravity, we first need the density of the liquid. To find the density, we need the mass and volume of the liquid. We are given the weight and volume.

Step 1: Convert Volume to Standard Units (\(m^3\))

The standard unit for volume in SI is cubic meters (\(m^3\)). 1 litre is equal to 1 cubic decimeter ($1 \, \text{dm}^3$), and $1 \, \text{dm}^3 = (10^{-1} \, \text{m})^3 = 10^{-3} \, \text{m}^3$.

So, \(V = 1 \, \text{litre} = 1 \times 10^{-3} \, \text{m}^3\).

Step 2: Calculate the Mass of the Liquid

We know that Weight = Mass \(\times\) acceleration due to gravity (\(W = m \times g\)). We can rearrange this formula to find the mass:

\(m = \frac{W}{g}\)

Substitute the given values:

\(m = \frac{5.5 \, \text{N}}{9.8 \, \text{m/s}^2}\)

\(m \approx 0.5612 \, \text{kg}\)

Step 3: Calculate the Density of the Liquid

Density is mass per unit volume (\(\rho = \frac{m}{V}\)).

Substitute the calculated mass and the converted volume:

\(\rho_{liquid} = \frac{0.5612 \, \text{kg}}{1 \times 10^{-3} \, \text{m}^3}\)

\(\rho_{liquid} = 561.2 \, \text{kg/m}^3\)

Step 4: Calculate the Specific Gravity of the Liquid

Specific gravity is the ratio of the liquid's density to the density of water.

Specific Gravity = \(\frac{\rho_{liquid}}{\rho_{water}}\)

Using \(\rho_{water} = 1000 \, \text{kg/m}^3\):

Specific Gravity = \(\frac{561.2 \, \text{kg/m}^3}{1000 \, \text{kg/m}^3}\)

Specific Gravity = \(0.5612\)

The specific gravity of the liquid is approximately 0.5612. Looking at the options, the closest value is 0.56.

Final Answer on Specific Gravity

Based on the calculations, the specific gravity of the liquid is approximately 0.56.

Additional Information on Liquid Properties

Specific gravity is a useful dimensionless quantity because it tells you how much denser or less dense a substance is compared to water. A specific gravity less than 1 means the substance is less dense than water and will float (like oil), while a specific gravity greater than 1 means it is denser and will sink (like honey).

The value of specific gravity depends slightly on the temperature and pressure, as these factors affect the density of both the substance and the reference substance (water). However, for most introductory problems, constant values for water density and gravity are used.

Remember that mass is an intrinsic property of an object (it stays the same regardless of location), while weight is a force that depends on the local gravitational acceleration.