Linear Coefficient Of Expansion Of Metals 10325w

Page |1 I.

Introduction

Most solids expand when heated and this expansion always occurs in three dimensions. However, the thermal expansion of solids is usually too small to be seen, but since the expansion is proportional to the length of the object, the expansion will be considerable if the object is large. This can be attain by using the formula

α= L2- L1L1 (t2- t1) Where:

L2- L1 is change of length t2- t1 is change in temperature L1 is the initial length Linear Expansion: The expansion of metal rod is proportional to its length and the change in temperature. It is increase in length per unit of length at 0℃ per degree change in temperature.

II.

Objective The main objective of the study is to determine the coefficient of linear expansion of certain metals and also to evaluate the length before and after the activity.

III.

Materials i. Burner and Hose ii.Meter Stick iii.Can (or beaker) to catch condensed steam iv.Towel v.Two Metal Rods (different materials) vi.Thermometer vii.Expansion Apparatus viii.Boiler and Hose ix.Boiler Stand

I.

Data / Table

Data

Aluminum

Copper

24℃

24℃

Length of Micrometer screw at Room Temperature

52 cm

52.8 cm

Setting of Micrometer screw at Room Temperature

0.2 mm

0.1 mm

Setting of Micrometer at 100℃

1.2 mm

0.9 mm

1 mm

0.8 mm

Room Temperature

Change of Length of the Rod

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Coefficient of Linear Expansion Value of the coefficient of Linear Expansion Table %Error

0.000025

0.000020

℃-1

℃-1

0.000024

0.000017

℃-1

℃-1

4%

17.6%

Computation:

For Aluminum (Al)

24℃ to 100℃

24℃ to 100℃ L1=52cm at 24℃, t1 L2=52cm+0.1cm L2=52.1cm at 100℃, t2 ∆L= 1.2mm-0.2mm ∆L=1.0mm or 0.1cm α= L2- L1L1 (t2- t1) α=52.1cm-52cm52cm (100℃-24℃) α=0.1cm52cm (76℃) α=0.000025℃-1

L1=52.8cm at 24℃, t1 L2=52.8cm+0.08cm L2=52.88cm at 100℃, t2 ∆L= 0.9mm-0.1mm ∆L=0.8mmor 0.08cm α= L2- L1L1 (t2- t1) α=52.88cm-52.8cm52.8cm (100℃24℃) α=0.08cm52.8cm (76℃) α=0.0000199℃-1

For Copper (Cu)

Percent Error: Al

% error= Experimented Value-Accepted Value Accepted Value ×100% % error= 0.000025℃-1- 0.000024℃-1 0.000024℃-1 ×100% % error=4% Cu

% error= Experimented Value-Accepted Value Accepted Value ×100% % error= 0.000020℃-1- 0.000017℃-1 0.000017℃-1 ×100% % error=17.6%

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II.

Question and Answer (Analysis) 1) To which measurement or reading do you attribute most of your error? Explain. The group find it more difficult to measure the specific reading of the micrometer caliper in order to determine the more accurate and precise measurement.

2)

If you had measured the original length of the rod in inches, and the micrometer screw had given readings in inches, what value for α would you have obtained? Explain.

L1=520mm 1 in25.4 mm=20.47in L2=521mm 1 in25.4 mm=20.51in α= L2- L1L1 (t2- t1) α=20.51in- 20.47in20.47in (100℃-24℃) α=0.039 in20.47in (76℃) α=0.000025℃-1 The alpha didn’t change because ∆L is proportional to the rod’s initial length.

3)

If lengths had been in degrees Fahrenheit, what value of α per degree Fahrenheit would you have observed?

∆t=76℃=1.876+ 32 ∆t=168.8℉ α=0.1cm52cm (168.8 ℉) α=1.13 ×10-5℉-1 The linear coefficient change because α depend on the ∆t.

4)

What unit is given in the tables for the values of α? Why?

℃ because the Room Temperature in measured on the said unit. Therefore the units being use is ℃ not unless if it is converted to ℉ or K. 5) Which is the most serious, an error of 1mm in measuring the original length or an error of 0.01mm in measuring the amount of expansion? Explain.

Page |4 The most serious is having an error of 0.01mmin measuring the expansion. This is because it can affect the linear coefficient α thus having an error in length affect little to α. 6) Do you think something more precise than a meter stick for measuring the length should be used? Why? Micrometer Caliper since it can measure small objects and has a more accurate and more precise usage in measuring regular and irregular objects. 7) What would happen if the two specimens used in this experiment were clamped together, side by side, and heated? If the two specimens used were clamped together, the whole unit will bend towards the material which has smaller expansion.

8) Use the length of one of the rods at room temperature as L2 and the value of α obtained from the tables to calculate, by Equation [3], the value for L1 at 0℃.

Aluminum (Al) Al = 52cm

L1= ______ ; t1=0℃ L2=52cm ; t2=24℃ ∆t=24-0 = 24 α=0.000025℃-1

L2= L1 (1+ α ∆t) 52cm= L1 1+0.000025℃124℃ L1= 52cm0.000025℃-124℃ L1=51.97cm

9)

Substitute the value of the length at 0℃, found from Question 9, for L0 in Equation [1] and solve for the coefficient of linear expansion, using other data the same as before do you think that substituting L1 for L0 makes a serious error in Equation [2]? Why?

α= L2- L1L0 (t2- t1) α= 0.1cm51.97 (76℃) α= 0.000025℃-1 It does not make any serious error since the length of the metal does not affect too much in the linear coefficient.

III.

Conclusion

The construction of common thermometers is based upon the general tendency of matter to expand when heated and to when cooled. The coefficient of linear expansion of a substance is its change in length per unit per degree rise in temperature. A linear expansion coefficient is usually employed in describing the expansion of a solid. The

Page |5 coefficient of linear expansion is constant of proportionally whose value depends on the type of material.