Question

can you help me figure out the following:

part a: Calculate the specific gravity for aluminum and for brass.

part b: Calculate the specific gravity for cork.

part c: Calculate the specific gravity of the isopropyl alcohol.

Experiment 9 Specific Gravity (Archimedes Principle) Read the Theory section on pages 9-2 & 9-3. Specific gravity is defined as the density (mass per unit volume) of a substance relative to the density of (liquid) water: s p/po, where pe is the density of liquid water -1.00 gram/cm. Thus the specific gravity will be numerically equal to the density (in grams/cm) with the units suppressed Note that s-(mass of substance) (mass of an equal volume of water). We shall use Archimedes' Principle: The buoyant force (BF) acting on an object immersed in a fluid is equal to the weight of fluid displaced by the immersed object. Let m apparent mass of the substance in airdry weight" and me apparent mass of the substance when fully immersed in water "submerged weight". (Note we are using the terms mass and weight interchangeably here, since when we "weigh" something with pan balance we are really measuring its apparent massie the balance is calibrated to measure in mass units (grams) rather than weight units (newtons.) Then (m-mog - BF - weight of an equal volume of water, or m-mo-mass of an equal volume of water. Thus sm/(m - mo) equation (1) in Lab Manual. Part A Solid More Dense Than Water We use 2 right circular metal cylinders of diameter D and length L. Data: m (grams) me (grams) D (mm) L (mm) Aluminum 11.3 7.2 12.5 31.7 Brass 30.1 12.6 31.7 Calculate the specific gravity for aluminum and for brass. Also, for comparison, calculate the density directly from the mass and volume. Part B Solid Less Dense Than Water We use a piece of cork with a sinker attached to it Let m - apparent mass of cork in air 1.9 grams m; apparent combined mass with only the sinker immersed - 57.5 grams me apparent combined mass with both cork and sinker immersed - 49.6 grams Then sm/(mi-m2) equation (2) in Lab Manual Calculate the specific gravity for cork. Part C Liquid (Isopropyl Alcohol) Let m apparent mass of an object in air = 61.3 grams me apparent mass of the object submerged in water = 55.8 grams ma apparent mass of the object submerged in alcohol- 56.8 grams Then s-( m m ) ( m m ) equation (3) in Lab Manual. Calculate the specific gravity of the isopropyl alcohol.

Answer 1

Part A: Calculate the specific gravity for aluminum and for brass.

Aluminium:

Mass in air, m = 11.3 g;

Mass when fully immersed in water, mo = 7.2 g

Mass of an equal volume of water, m - mo = 11.3 g - 7.2 g = 4.1 g.

Specific gravity for Aluminium:

$s_{Al}=\frac{m}{m-m_{o}}=11.3/4.1=2.756=2.76$

Given, diameter of Aluminium cylinder, D = 12.5 mm = 1.25 cm and length of the aluminium cylinder, L = 31.7 mm = 3.17 cm.

So, the volume of the aluminium cylinder, $V_{Al}=\frac{\pi D^{2}L}{4}=\frac{3.142\times 1.25^{2}\times 3.17}{4}=3.89\:cm^{3}$

Density of Aluminium = $\rho =\frac{m}{V}=\frac{11.3}{3.89}=2.90\: gcm^{-3}$

Brass:

Mass in air, m = 34.1 g;

Mass when fully immersed in water, mo = 30.1 g

Mass of an equal volume of water, m - mo = 34.1 g - 30.1 g = 4.0 g.

Specific gravity for Brass:

SBr = - = 34.1/4.0 = 8.525 = 8.52 m - mo

Given, diameter of Aluminium cylinder, D = 12.6 mm = 1.26 cm and length of the aluminium cylinder, L = 31.7 mm = 3.17 cm.

So, the volume of the Brass cylinder, $V_{Al}=\frac{\pi D^{2}L}{4}=\frac{3.142\times 1.26^{2}\times 3.17}{4}=3.95\:cm^{3}$

Density of Brass = $\rho =\frac{m}{V}=\frac{34.1}{3.95}=8.63\: gcm^{-3}$

Part B: Calculate the specific gravity for cork.

mo = mass of the cork in air = 1.9 g

m1 = apparent combined mass with both the cork and sinker immersed = 57.5 g

m2 = apparent combined mass with only the sinker immersed = 49.6 g

Specific gravity for cork:

$s=\frac{m}{m_{1}-m_{2}}=\frac{1.9}{57.5-49.6}=\frac{1.9}{7.9}=0.24$

Part C: Calculate the specific gravity of the isopropyl alcohol.

m = apparent mass of the object in air = 61.3 g

mo = apparent mass of the object submerged in water = 55.8 g

mA = apparent mass of the object submerged in alcohal = 56.8 g

Specific gravity of isopropyl alcohal:

$s=\frac{m-m_{A}}{m-m_{o}}=\frac{61.3-56.8}{61.3-55.8}=\frac{4.5}{5.5}=0.82$

Regards