20,345
edits
Changes
→Finding Density from Mass and Volume
: If an [[object]] is less '''dense''' than [[water]] it will rise through [[water]] and float on the surface.
===EquationFormula===
: Density = Mass/volume
*The [[density]] of a substance can change with [[temperature]] and [[pressure]].
*'''Density''' is used in calculating the [[mass]] of an [[object]] given its [[Volume (Space)|volume]], and vice versa.
*For an [[alloy]] the '''density''' of that [[alloy]] is found by [[sum]]ing the [[mass]] of each [[alloy]] divided by the total [[Volume (Space)|volume]]. However, where the [[Volume (Space)|volume]] of each component is known, not the [[mass]] then the [[mass]] is replaced with the [[Product (Maths)|product]] of the [[density]] and [[Volume (Space)|volume]] of each [[alloy]].
===Formula===
====Density of a single material===='''Density''' of a [[material]] is given by the formula: *<math>\rho = \frac{m}{V}</math> Where: ρ = The [[density]] of the [[object]]. m = The [[mass]] of the [[object]]. V = The [[Volume (Space)|volume]] taken up by the [[object]]. ====Density of a composite or alloy material==== *<math>\rho = \frac{\rho_1V_1+\rho_2V_2}{V}</math> *<math>\rho = \frac{\rho_1V_1}{V}+\frac{\rho_2V_2}{V}</math> *<math>\rho = \frac{\rho_1V_1+\rho_2V_2}{V_1+V_2}</math> ρ = The [[density ]] of water the [[alloy]] ρ<sub>1</sub> = The [[density]] of one of the [[alloy]] composites ρ<sub>2</sub> = The [[density]] of the other [[alloy]] composite V<sub>1</sub> = The [[Volume (Space)|volume]] of one of the [[alloy]] composites V<sub>2</sub> = The [[Volume (Space)|volume]] of the other [[alloy]] composite V = The total [[Volume (Space)|volume]] taken up by the [[alloy]] ===Example Calculations=======Finding Density of an Alloy given the density and volume of its composites===={| class="wikitable"|-| style="height:20px; width:300px; text-align:center;" |'''A [[Volume (Space)|volume]] of 6.35x10<sup>-3</sup>m<sup>3</sup> [[Aluminium]] and 4.10x10<sup>-3</sup>m<sup>3</sup> [[Magnesium]] are [[alloy]]ed together. Given the density of [[Aluminium]] is 2700kgm<sup>-3</sup> and the density of [[Magnesium]] is approximately 1000 kg1700kgm<sup>-3</m³sup> then calculate the density of the [[alloy]]'''.The | style="height:20px; width:300px; text-align:center;" |'''A [[Volume (Space)|volume]] of 4.9x10<sup>-3</sup>m<sup>3</sup> [[Copper]] and 1.1x10<sup>-3</sup>m<sup>3</sup> [[Zinc]] are [[alloy]]ed together. Given the density of [[Copper]] is 8900kgm<sup>-3</sup> and the density of gold [[Zinc]] is much higher than 7100kgm<sup>-3</sup> then calculate the density of the [[alloy]]'''.|-| style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s''' [[Aluminium]] [[Volume (Space)|volume]] = 6.35x10<sup>-3</sup>m<sup>3</sup> [[Aluminium]] [[Density]] = 2700kgm<sup>-3</sup> [[Magnesium]] [[Volume (Space)|volume]] = 4.10x10<sup>-3</sup>m<sup>3</sup> [[Magnesium]] [[Density]] = 1700kgm<sup>-3</sup> | style="height:20px; width:300px; text-align:left;" |'''1. State the known quantities in [[SI Unit]]s''' [[Copper]] [[Volume (Space)|volume]] = 4.9x10<sup>-3</sup>m<sup>3</sup> [[Copper]] [[Density]] = 8900kgm<sup>-3</sup> [[Zinc]] [[Volume (Space)|volume]] = 1.1x10<sup>-3</sup>m<sup>3</sup> [[Zinc]] [[Density]] = 7100kgm<sup>-3</sup> |-| style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' *<math>\rho = \frac{\rho_{Al}V_{Al}+\rho_{Mg}V_{Mg}}{V_{Al}+V_{Mg}}</math> *<math>\rho = \frac{2700\times 6.35 \times 10^{-3} + 1700\times 4.10 \times 10^{-3}}{6.35 \times 10^{-3} + 4.10 \times 10^{-3}}</math> *<math>\rho = 2310kgm^{-3}</math> However, this assumes that the total [[Volume (Space)|volume]] of the [[object]] is simply the [[sum]] of the two [[Volume (Space)|volumes]] which is not necessarily the case due to the way different sized [[particle]]s form a [[aluminiumlattice]]. | style="height:20px; width:300px; text-align:left;" |'''2. [[Substitute (Maths)|Substitute]] the numbers into the [[equation]] and [[Solve (Maths)|solve]].''' *<math>\rho = \frac{\rho_{Cu}V_{Cu}+\rho_{Zn}V_{Zn}}{V_{Cu}+V_{Zn}}</math> *<math>\rho = \frac{8900\times 4.9 \times 10^{-3} + 7100\times 1.1 \times 10^{-3}}{4.9 \times 10^{-3} + 1.1 \times 10^{-3}}</math> *<math>\rho = 8600kgm^{-3}</math> However, making gold heavier for this assumes that the same total [[Volume (Space)|volume]] of the [[object]] is simply the [[sum]] of the two [[Volume (Space)|volumes]] which is not necessarily the case due to the way different sized [[particle]]s form a [[lattice]].|}