Matter and Energy - Chemistry Coursescourses.chem.psu.edu/chem11/pdf's/Lectures/11Lect3_C3_Matter_an… · Matter and Energy Chapter 3. 2 ... Chemistry studies The properties of different

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Matter and EnergyMatter and Energy

Chapter 3Chapter 3

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Universe ClassifiedUniverse Classified

Matter is the part of the universe that has mass and Matter is the part of the universe that has mass and volumevolume

Energy is the part of the universe that has the Energy is the part of the universe that has the ability to do workability to do work

Chemistry studiesChemistry studiesThe properties of different types of matterThe properties of different types of matterThe way matter behaves when influenced by other The way matter behaves when influenced by other

matter and/or energymatter and/or energy

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Changes in Matter

Physical Changes do not change the Physical Changes do not change the fundamental components that make the fundamental components that make the substancesubstanceState Changes State Changes –– boiling, melting, condensingboiling, melting, condensing

Chemical Changes involve a change in the Chemical Changes involve a change in the fundamental components of the substancefundamental components of the substanceProduce a new substanceProduce a new substanceChemical reactionChemical reactionReactants Reactants →→ ProductsProducts

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Classify Each of the following as Classify Each of the following as Physical or Chemical Properties Physical or Chemical Properties

The boiling point of ethyl alcohol is The boiling point of ethyl alcohol is 7878°°C.C.

Diamond is very hard.Diamond is very hard.

Sugar ferments to form ethyl alcohol.Sugar ferments to form ethyl alcohol.

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Classify Each of the following as Physical or Chemical Properties

The boiling point of ethyl alcohol is 78The boiling point of ethyl alcohol is 78°°C.C.Physical property Physical property –– describes inherent characteristic of describes inherent characteristic of

alcohol alcohol –– boiling pointboiling pointDiamond is very hard.Diamond is very hard.Physical property Physical property –– describes inherent characteristic of describes inherent characteristic of

diamond diamond –– hardnesshardnessSugar ferments to form ethyl alcohol.Sugar ferments to form ethyl alcohol.Chemical property Chemical property –– describes behavior of sugar describes behavior of sugar ––

forming a new substance (ethyl alcohol)forming a new substance (ethyl alcohol)

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solid, liquid, gassolid, liquid, gas

States of MatterStates of Matter

State Shape Volume Compress Flow

Solid KeepsShape

KeepsVolume

No No

Liquid TakesShape ofContainer

KeepsVolume

No Yes

Gas TakesShape ofContainer

TakesVolume ofContainer

Yes Yes

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Classify Each of the following as Classify Each of the following as Physical or Chemical Changes Physical or Chemical Changes

Iron metal is melted.

Iron combines with oxygen to form rust.

Sugar ferments to form ethyl alcohol.

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Classify Each of the following as Classify Each of the following as Physical or Chemical Changes Physical or Chemical Changes

Iron is melted.Physical change – describes a state change, but the

material is still iron

Iron combines with oxygen to form rust..Chemical change – describes how iron and oxygen

react to make a new substance, rust

Sugar ferments to form ethyl alcohol.Chemical change – describes how sugar forms a new

substance (ethyl alcohol)

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Elements and CompoundsElements and CompoundsElements: Substances which can not be broken down into simpler substances by chemical reactions

Compounds: Substances that are chemical combinations of elements.

Compounds can be broken down into elements:Properties of the compound not related to the properties of the elements that compose it

Same chemical composition at all times

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Classification of MatterClassification of Matter

Homogeneous = uniform throughout, appears to be one thing

pure substancessolutions (homogeneous mixtures)

Heterogeneous = non-uniform, contains regions with different properties than other regions

Pure SubstanceConstant Composition

Homogeneous

MixtureVariable Composition

Matter

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Pure Substances vs. MixturesPure Substances vs. MixturesPure SubstancesHave the same physical and chemical properties

Have the same compositionHomogeneousSeparate into components based on chemical properties

MixturesDifferent samples may show different propertiesVariable compositionHomogeneous or HeterogeneousSeparate into components based on physical properties

All mixtures are made of pure substances

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Identify the following as a Identify the following as a pure substance, pure substance, homogeneous mixture or homogeneous mixture or heterogeneous mixtureheterogeneous mixture

GasolineGasoline

A stream with gravel on the bottomA stream with gravel on the bottom

Copper metalCopper metal

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Gasolinea homogenous mixture

A stream with gravel on the bottoma heterogeneous mixture

Copper metalA pure substance (all elements are pure substances)

Identify the following as a pure substance, homogeneous mixture or heterogeneous mixture

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Separation of MixturesOne can separate mixtures based on different

physical properties of the components

EvaporationVolatility

ChromatographyAdherence to a Surface

FiltrationState of Matter (solid/liquid/gas)

DistillationBoiling Point

TechniqueDifferent Physical Property

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Energy and Energy ChangesEnergy and Energy Changes

Capacity to do workchemical, mechanical, thermal, electrical, radiant,

sound, nuclear

Energy may affect mattere.g. raise its temperature, eventually causing a

state change

All physical changes and chemical changes involve energy changes

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HeatHeatHeat: a flow of energy due to a temperature difference

1. Exothermic = Process results in the evolution of heat.Example: when a match is struck, it is an

exothermic process because energy is produced as heat.

2. Endothermic = A process that absorbs energy.Example: melting ice to form liquid water is an

endothermic process.

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Units of EnergyUnits of Energy

One calorie is the amount of energy needed to raise the temperature of one gram of water by 1°Ckcal = energy needed to raise the temperature of

1000 g of water 1°C

joule 4.184 J = 1 cal

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Example - Converting Calories to Joules

251Jcal1

J 4.184 60.1cal

joules4.184cal 1

=×

=

Convert 60.1 cal to joules

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Energy and the Temperature of Matter

The increase in the temperature of an object depends on:the amount of heat added (Q)

the object’s mass

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Specific Heat

Specific heat is the amount of energy required to raise the temperature of 1 gram of a substance by 1 Celsius degree

C gJ 4.184 is water ofheat specific the,definitionBy °

Amount of Heat = Specific Heat x Mass x Temperature ChangeQ = s x m x ΔT

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Example – Calculate the amount of heat energy (in joules) needed to raise temperature of 7.40 g of water from 29.0°C to 46.0°C

Mass = 7.40 g

Temperature Change = 46.0°C – 29.0°C = 17.0°C

J 526 C17.07.40gC g

J 4.184 Heat =°××°

=

Specific Heat of Water = 4.184 C-g

J°C g

J°

Q = s x m x ΔT

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Example – A 1.6 g sample of metal appears to be gold and requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold?

C gJ0.20

C18x g 1.6J 5.8 s

C18 C23 - C41 TT m

Q s

TmsQ

°=

°=

°=°°=ΔΔ×

=

Δ××=

Table 3.2 lists the specific heat of gold as 0.13Therefore the metal cannot be pure gold.

C gJ°