Part A: The Density of Water Experimental Data Mass of Empty Cylinder Mass of Cy
ID: 717376 • Letter: P
Question
Part A: The Density of Water Experimental Data Mass of Empty Cylinder Mass of Cylinder + Water Mass of Water only Volume of Water Density of Water Average Density of Water 1* Water Addition 2 Water Addition3 Water Addition llo .90 132. 16P 21558 32.0 /lo. 909 171 .421 154-831 .512, Temperature of Water: Data Analysis 1) Look up the true density of water at the temperature recorded: , - Obtain this value from htlp /jupiter.plymouth edu-jsduncan courses/2012 s Tempadf. Then use this to calculate the percent error in your average density of water. Show your wo Ev-Tv/ Tv Part B: The Density of Aluminum and the Thickness of Foil Experimental Data Table 1 - The Density of Aluminum 5o. 0 75 Mass of Empty Beaker Mass of Beaker and Al pellets Mass of Al pellets Initial volume of water in cylinder Final volume of water and Al pellets Volume of Al pellets 3.3 23Explanation / Answer
Part A - Density of water
1st water addition
Given Mass of Empty Cylinder = W1= 110.909 gms
Mass of Empty Cylinder + Water = W2 = 132.467 gms
Mass of Water = W = W2-W1 =132.467-110.909 = 21.588 gms
Volume of Water (V)= 22 ml
So Density = Mass Of Water/Volume of Water
= 21.588 gm /22 ml
= 0.9813 gm/ml
2nd water addition
Given Mass of Empty Cylinder = W1= 110.909 gms
Mass of Empty Cylinder + Water = W2 = 154.831 gms
Mass of Water = W = W2-W1 =154.831-110.909 =43.922 gms
Volume of Water (V)= 44.8 ml
So Density = Mass Of Water/Volume of Water
= 43.922 gm /44.8 ml
= 0.9804 gm/ml
3rd water addition
Given Mass of Empty Cylinder = W1= 110.909 gms
Mass of Empty Cylinder + Water = W2 = 171.421 gms
Mass of Water = W = W2-W1 =171.421-110.909 = 60.512 gms
Volume of Water (V)= 61.5 ml
So Density = Mass Of Water/Volume of Water
= 60.512 gm /61.5 ml
= 0.9839 gm/ml
Average Density of water Experimental Value (EV) = (1st+2nd+3rd)/3
= (0.9813+0.9804+0.9839)/3 gm/ml
= 0.9819 gm/ml
As True Value (TV) Density of water at temperature 220 C = 0.9978
So Percentage of Error = {( EV-TV)/TV } x100
= {(0.9819-0.9978)/0.9978} X100
= -1.59 %
As error should not be negative it will be 1.59 %
Part B Density Of Aluminum
Mass of Empty Beaker (W1)= 50.075 gms
Mass of Empty Beaker + Al pellets (W2)= 60.266 gms
Mass of Al pellets (W) = W2- W1 = 60.266- 50.075 =10.191 gms
Intial Volume of water in Cylinder (V1) = 32.9 ml
final Volume of water in Cylinder + Al pellets (V2) = 36.2 ml
Volume of Al pellets (V) = 36.2 - 32.9
= 3.3 ml
So Density of Aluminum Expermental Value ( EV)= W/V =10.191 gm/3.3 ml
= 3.088 gm/ cm3 (as ml= cm3 )
As True Value ( TV) Density of Aluminum = 2.702 gm/ cm3
So Percentage of Error = {( EV-TV)/TV } x100
= {(3.088-2.702)/2.702} X100
= 14.29 %
Given Mass of Al Foil (M) = 0.097 gms
Length of Al Foil (L)= 5.5 cm
Width of of Al Foil (W) = 4.5 cm
As true Density of Al = 2.702 gm/ cm3
So Volume V = Mass / Density = 0.097 gm/ 2.702 gm/ cm3 = 0.036 cm3
As Volume = Length X Width X Thickness
So Thickness = VolumeLength X Width = 0.036 cm3 5.5 cm X 4.5 cm
= 0.0014
= 1.4 X 10-3
Part C - Graphical Analysis of mass and Volume of unknown solid
Given X ( volume) Value for Point 1 , Point 2 = 4 ,12 cm3
Y ( mass) Value for Point 1 , Point 2 = 10 ,30 gms
So slope m= (y2-y1)/(x2-x1) = 30-10/ 12-4
= 20/ 8
=2.5
Answer for 4 th part -
Given Length = 1.83 feet = 55.78 cm
True Density = 2.20 gm/cm3
Diameter = 1.5 cm so Radius r = 1.5 /2 = 0.75 cm
So Volume = r2 l
= 3.14 x (0.75)2 55.78
= 98.52 cm3
so mass of cylinder = volume X Density
= 98.52 cm3 x 2.20 gm/cm3
= 216.74 gms