Lesson 2: Density and Specific Gravity

Density

The density of a material is its mass per unit volume. That is, $Density = \frac{Mass}{Volume}$
The density of fresh water (FW) = 1000 kg/m3 = 1 tonne/ m³
The density of salt water (SW) = 1025 kg/m3 = 1.025 tonne/ m³

Specific Gravity

Specific gravity or relative density is the density of a substance relative to the density of another substance. For all of our calculations, the 'other substance' will be fresh water. We will calculate the specific gravity of a substance by determining the ratio of the density of that substance to the density of fresh water. $Specific gravity (or Relative Density) of a substance = \frac{Density of substance}{Density of Fresh Water}$ Or, $S G (o r R D) = \frac{D ({in kg/m}^{3})}{1000 {kg/m}^{3}} = \frac{D ({in t/m}^{3})}{1 {t/m}^{3}}$ Specific Gravity is a unitless quantity.

Example 1: A rectangular block of steel has a mass of 0.5 tonnes. If the dimensions of the block are $0.5 m \times 0.4 m \times 0.7 m$ , find the density of the steel.

Solution:

$Volume of a rectangular box = Length \times Width \times Height = 0.5 m \times 0 .4 m \times 0 .7 m = 0 .14 m^{3}$ Also, mass is given to be 0.5 tonnes. Therefore, $Density = \frac{Mass}{Volume} = \frac{0.5 t}{0.14 m^{3}} = 3.57 {t/m}^{3}$

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Example 2: A ship's double bottom tank measures $20 m \times 10 m \times 0.8 m$ . How much fuel oil of density 0.897 t/m³ can be carried in the tank?

Solution:

Volume of tank = $20 m \times 10 m \times 0.8 m = 160 m^{3}$
Amount of oil = $Mass = Volume \times Density = 160 m^{3} \times 0.897 t / m^{3} = 143.52 t$
Hence, the tank can carry 143.52 t of fuel oil.

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Example 3: Find the density of a fuel oil whose specific gravity is 0.85. Express the answer in kg/m³

Solution:

We know that, $Density = R .D . \times Density of Fresh Water$ Also, density of fresh water (FW) = 1000 kg/m³
Therefore, $Density of fuel oil = R .D . of fuel oil \times Density of fresh water = 0.85 \times 1000 k g / m^{3} = 850 k g / m^{3}$

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Example 4: An inverted cone (vertex downwards), has a depth of 100 cm and a radius of 20 cm. Find the mass of oil of density 0.98 t/m³ that can go into the cone.

Solution:

First of let us find the volume of cone.

Volume of cone = $\frac{1}{3} \times π \times r^{2} \times h = \frac{1}{3} \times π \times {(0.2 m)}^{2} \times (1 m) = 0.04189 m^{3}$
Mass of oil = $D e n s i t y \times V o l u m e = 0.98 {t/m}^{3} \times 0.04189 m^{3} = 0.0411 t$

Hence, the cone can hold 0.0411 tonnes of oil.

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Example 5: A rectangular ballast tank is 12 m long, 6 m wide and 8 m deep. Calculate the mass of sea water required to fill the tank to a level of 5 m from the bottom of the tank.

Solution:

We know that: $Density= \frac{Mass}{Volume} or Mass=Density × Volume$ Also, $Density of sea water = 1.025 {tonnes/m}^{3}$ We need to calculate the volume of sea water required so that we can determine its mass.

The dimensions of tank are given to be 12 m × 6 m × 8 m but the water is to be filled up to 5 m from the bottom of the tank. Therefore, we will calculate the volume using 5 m as the value for height.

\begin{array}{l} So, \\ Volume of  sea water = Length × Width × Height of water from bottom \\ = 12 m \times 6 m \times 5 m = 360 m^{3} \\ Therefore, mass of water = 1.025 {tonnes/m}^{3} \times 360 m^{3} = 369 t o n n e s \end{array}

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Lesson 2: Density and Specific Gravity

Density

Specific Gravity

Example 1: A rectangular block of steel has a mass of 0.5 tonnes. If the dimensions of the block are 0.5 m×0.4 m×0.7 m , find the density of the steel.

Example 2: A ship's double bottom tank measures 20 m×10 m×0.8 m . How much fuel oil of density 0.897 t/m3 can be carried in the tank?

Example 3: Find the density of a fuel oil whose specific gravity is 0.85. Express the answer in kg/m3

Example 4: An inverted cone (vertex downwards), has a depth of 100 cm and a radius of 20 cm. Find the mass of oil of density 0.98 t/m3 that can go into the cone.

Example 5: A rectangular ballast tank is 12 m long, 6 m wide and 8 m deep. Calculate the mass of sea water required to fill the tank to a level of 5 m from the bottom of the tank.

Example 1: A rectangular block of steel has a mass of 0.5 tonnes. If the dimensions of the block are $0.5 m \times 0.4 m \times 0.7 m$ , find the density of the steel.

Example 2: A ship's double bottom tank measures $20 m \times 10 m \times 0.8 m$ . How much fuel oil of density 0.897 t/m³ can be carried in the tank?

Example 3: Find the density of a fuel oil whose specific gravity is 0.85. Express the answer in kg/m³

Example 4: An inverted cone (vertex downwards), has a depth of 100 cm and a radius of 20 cm. Find the mass of oil of density 0.98 t/m³ that can go into the cone.