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Specific gravity is a measure of how dense a substance is compared to water. It does not have a unit, as it is a ratio. Since water has a density of approximately \(1000 \, \text{kg/m}^3\), specific gravity can be calculated using the formula: \[ \text{sp gr} = \frac{\text{Density of substance}}{\text{Density of water}} \]

• If the specific gravity is greater than 1, the substance is denser than water.
• If it's less than 1, the substance is less dense than water.

For cast iron, which has a specific gravity of \(7.20\), it's much denser than water. Thus, the density of cast iron can be calculated by multiplying its specific gravity by the density of water: \[ 7200 \, \text{kg/m}^3 = (7.20) \times (1000 \, \text{kg/m}^3) \] Understanding specific gravity is crucial when comparing materials or choosing them for specific applications, since it provides insight into how heavy or light they will be in relation to water.

Mass calculation ties together the concepts of density and volume using the formula \( \rho = \frac{m}{V} \), where \( m \) is mass, \( V \) is volume, and \( \rho \) is density. Rearranging this formula gives us a method to find mass: \( m = \rho \times V \).For cast iron, with a density of \(7200 \, \text{kg/m}^3\), we can calculate the mass of a given volume of this material. For example, to find the mass of \(60.0 \, \text{cm}^3\) of cast iron:

• First, convert the volume to cubic meters (since density is in \(\text{kg/m}^3\)).
• The mass is then calculated as \(m = (7200 \, \text{kg/m}^3) \times (60.0 \times 10^{-6} \, \text{m}^3)\).

Calculating gives the mass of \(0.432 \, \text{kg}\). This process is useful for determining how much material you have when only volume and density are known.

Volume conversion is essential when working with different units of measurement. Often, density is provided in terms of \( \text{kg/m}^3 \), which means volume must be in cubic meters to match the units. This requires converting from more commonly used units like cubic centimeters (\( \text{cm}^3 \)).The basic conversion to keep in mind is:

• \(1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3\)

For example, when you have a volume of \(60.0 \, \text{cm}^3\), you convert it to \(\text{m}^3\) for calculations: \[ 60.0 \, \text{cm}^3 = 60.0 \times 10^{-6} \, \text{m}^3 \]This conversion is necessary so that the unit dimensions are consistent when calculating properties like mass using density. Properly converting volume ensures accuracy in any subsequent calculations involving density.