Essay
unit 22 LAA classification of materials assignment
- Module
- Unit 22 - Materials Science
- Institution
- PEARSON (PEARSON)
- distinction level - essay based/ assignment
[Show more]
Content preview
Pearson BTEC Level 3 National Diploma in Applied ScienceQualification
Pearson BTEC Level 3 National Extended Diploma in Applied Science
Unit number and title Unit 22: Materials Science
Learning aim A: Understand the classification and properties of different materials.
Assignment title Classification of materials
Assessor
Issue date
Hand in deadline
You are an apprentice materials science technician working for the
global company 3M. Each year the company organises a
competition for all of the apprentice scientists that it trains across
the world. You are representing the UK division of 3M in the
European heat and the successful winner will go on to a world final.
Vocational Scenario or Context The ultimate winner will be offered a sponsorship to cover the costs
of studying for a degree in materials science. Your first task in the
competition is to classify some unknown materials by testing their
macro and microscopic properties. You will have to provide a report
that uses your experimental data (primary and secondary) to
justify what group you would classify each material into.
Using your lesson notes and research you have carried out into
materials and their properties, produce a report for the judging
panel that includes:
1. The classification of FOUR unknown materials on the basis of
their macroscopic properties.
Examine the macroscopic properties of each material (ie
Task melting point, boiling point, strength, toughness,
malleability, electrical and thermal conductivity, density
and biodegradability) and from this determine the
classification into one of the following groups and name
the material:
o metals and alloys
o glass
o polymers
o ceramics
o carbon allotropes
o composites
Present your findings i.e primary data(collected) and
secondary data(from research) in a coherent results table
and explain clearly your rationale behind your classification
of each material into an appropriate group
2. The analysis of the microscopic and macroscopic structure of
TWO unknown materials
You will be given X-ray diffractions, electron microscopy images,
physical data and spectra of the bonding and atomic structure for
two unknown materials.
Use these to predict each material’s macroscopic properties
, (ie melting point, boiling point, strength, toughness,
malleability, electrical and thermal conductivity, density and
biodegradability.
Based on primary data, compare your predictions to the
actual macroscopic properties of the materials.
Classify the materials into groups according to their
properties.
3. The comparison of the microscopic structural features of:
o Graphite compared to vulcanised rubber OR
o Copper compared to polytetrafluoroethylene
(TEFLON)
o
For your selected pair:
- Classify the 2 materials into their groups
- Compare their microscopic structural features (ie atomic
structure, bonding, additives, reinforcing fibres/particles)
- explain how the microscopic structure determine their
properties at the macroscopic level; which allows them to
be classified into their different groups.
- explain how good their macroscopic properties as a
lubricant and as a conductor of electricity are
- explain how their microscopic properties determine how good
they are at conducting electricity
Checklist of evidence required A report which:
Determines of the classification of FOUR unknown materials
on the basis of their macroscopic properties;
The analysis of the microscopic and macroscopic structure of
TWO unknown materials to classify them into groups
according to their properties
Compares the microscopic properties of TWO materials to
determine how good they are as a lubricant and at conducting
electricity (examples of macroscopic properties and
microscopic properties);
Sources of information must be appropriately referenced.
Correct scientific terminology must be used.
Criteria covered by this task:
Unit/Criteria reference To achieve the criteria, you must show that you are able to:
Analyse the microscopic and macroscopic structure of different given materials
A. D1
and classify them into groups according to their properties.
Compare microscopic structures and classifications of different given materials
A.M1
to determine their macroscopic properties.
,A. P1 Determine the classification of given materials by considering their properties.
Sources of information to http://www.rsc.org/learn-chemistry/resource/res00001755/
support you with this modelling-alloys-with-plasticine?cmpid=CMP00005265
Assignment
http://essentialchemicalindustry.org/polymers/polymers-an-
overview.html
http://pslc.ws/macrog/ptfe.htm
http://nptel.ac.in/courses/112107085/module2/lecture1/lecture1.pdf
https://www.khanacademy.org/test-prep/mcat/physical-sciences-
practice/physical-sciences-practice-tut/e/elasticity-and-kinetics-of-
vulcanized-rubber
http://ceramics.org/learn-about-ceramics/structure-and-properties-
of-ceramics
http://www.nanoscienceworks.org/nanopedia/carbon-allotropes
http://www.explainthatstuff.com/composites.html
Other assessment materials
attached to this Assignment
Brief
Assignment: classification of materials
, Introduction: I am apprentice materials science technicians working for the global company. I will be classifying
unknown materials by predicting their macroscopic properties and then comparing the actual macroscopic properties.
And from this I will be classifying the groups the materials should be in. in addition I would be classifying to unknown
materials based on the images of material A and material B. Microscopic properties are the properties we see on a
microscopic level, such as: atomic structure, SEM, TEM. And the macroscopic properties are the properties that are
made from the microscopic properties, this could be the boiling point, melting point and more.
SECTION 1-The classification of FOUR unknown materials on the basis of their macroscopic
properties.
description Melting Boiling Electrical Thermal Mass Volume Density
point (oC) o
point ( C) conductivity conductivity (g) (cm3) g/cm3
Tin 231.9oC 2602oC Yes, good Yes, good 1.77g 0.540cm3 3.27g/cm3
For the first material, I have identified the unknown material as a metal called tin. I was able to identify the metal as tin as I
examined the macroscopic properties that the table included already. In this table, everything was given, whereas I calculated the
density of the material by using the equation p=m/V (density= mass/ volume), which allowed me to get the answer 3.27 g/cm 3-
which helps me to classify the material even more accurately . I firstly searched if there was any material with a melting point of
231.9 degrees, which gave me an answer of Tin according to Owlaction (https://owlcation.com/stem/Freezing-Melting-and-
Boiling-Points-of-Solids-Liquids-and-Gases-in-general-use-today) and I searched what material has a boiling point of 2602
degrees, giving me the answer- tin according to a site called LiveScience (https://www.livescience.com/37355-tin.html ) . I
searched to see if the other calculations were matching, however since the mass and volume depends on how much tin you have,
we were only able to search and see the density of the material also came up as Tin, which did not match with my answer, since
the value can be judged by me, since I have a different volume and mass volume compared to the sites of information. So far, we
can see that the Tin’s calculation value is all accurate since it was the same as the information from the sites. I classified the
material as a great electrical conductor as it was said according to Quora (https://www.quora.com/Is-tin-can-a-conductor-of-
electricity ), as it can be used as a “critical components in electronics”. In addition, tin was also said to be a good thermal
conductor according to Quora, this is because tin has free moving electrons that can transfer thermal energy quickly and easily-
this can also be implied also to why Tin is a good electrical conductor. With all the information from the secondary information
and primary, I was able to accurately classify the material as tin. Going in more in depth for tin, I was able to find that tin is
ductile and malleable to work with- which would be very useful when it comes to making products or making a mechanism with
the material. This would explain their low boiling point for a metal. This is due to the fact that if force is exerted onto the metal,
there is a Nuclei shift where the bonds do not break, allowing the tin to be malleable. In addition, tin is not biodegradable and can
take up to 100 to 500 years to degenerate, since if they do break-down they could potentially release chemicals or any other
harmful products onto the ground. We also classified that tin is a non-ferrous metal as tin is malleable- which is one advantage of
knowing if a material is a non-ferrous metal.
descriptio Melting Boiling Electrical Thermal Mass (g) Volume Density
n point (oC) point (oC) conductivity conductivit (cm3) g/cm3
y
Glass 1400-1600oC 2230 oC poor poor 2.09g 0.882cm3 2.37g/cm3
For the second material, I identified the unknown material as Glass, I was able to classify this by searching what material matches
the primary data above and calculating the density of the material by using the mass and volume data into the formula p=m/V
(density= mass/ volume). This allowed me to get the answer 2.37g/cm 3- this allowed me to have a general idea where the
material’s density would be when i research the material in order to classify it. When it came to searching the density, I got
2.889g/cm3 according to Libretexts (https://chem.libretexts.org/Ancillary_Materials/Exemplars_and_Case_Studies/Exemplars/
Forensics/Glass_Density_Evidence ), this number is somewhat close to the density that I calculated, implying that glass could
potentially be the unknown material, however I would need more evidence on why I think the unknown material is glass.
According to the primary data, the unknown material’s melting point is 1400-1600 degrees, and with this information I searched
to see what material has a melting point of 1400-600 degrees, which also gave me the answer of Glass.
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller ridings1. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for £11.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
53068 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy revision notes and other study material for 14 years now