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Chapter 2

Energy and Heat

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CHAPTER OUTLINE���

§  Energy & Heat §  Temperature Scales §  Specific Heat §  Energy and Nutrition §  Classification of Matter §  Physical & Chemical Properties §  Physical & Chemical Changes §  Change of State §  Heating & Cooling Curve

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ENERGY & HEAT���

q  Energy is defined as the capacity of matter to do work.

q  There are two types of energy:

Potential (stored)

Kinetic (moving)

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ENERGY & HEAT���

q  Energy possesses many forms (chemical, electrical, thermal, etc.), and can be converted from one form into another.

q  In chemistry, energy is commonly expressed as heat.

PE is converted to

KE

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HEAT vs. ���TEMPERATURE���

q  Heat is measured in SI units of joule or the common unit of calorie.

1 cal = 4.184 J

Although the same amount of heat is added to both containers, the temperature increases more in the container with the smaller amount of water.

Heat & temperature are NOT the same

thing!

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HEAT vs. ���TEMPERATURE���

The difference between Heat and Temperature

A form of energy associated with

particles of matter

A measure of the intensity of heat or how hot or cold a

substance is

Heat is the total energy of all

particles of matter

Temperature is the average kinetic

energy of particles of matter

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TEMPERATURE���

q  Temperature is a measure of how hot or cold a substance is.

q  Thermometer is an instrument used for measuring temperature, and is based on thermometric properties of matter (i.e. expansion of solids or liquids).

q  Three scales are used for measuring temperature.

TEMPERATURE��� SCALES���

32 - 212

Fahrenheit

Celsius

0 - 100

Kelvin

273 - 373

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TEMPERATURE���CONVERSIONS���

q  To convert from one scale to another, the following relationships can be used:

K = °C + 273 °C = K - 273

°F = (1.8 x°C) + 32 °C = (°F-32) 1.8

°F = [(°C+40) x 1.8] - 40 °C = [(°F+40)1.8] - 40

Example 1:���The melting point of silver is 960.8 °C. What is this temperature in Kelvin?

K = °C + 273

K = 960.8 + 273 = 1233.8 K = 1234 K

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Example 2:���Pure iron melts at 1800 K. What is this temperature in Celsius?

°C = K - 273

°C = 1800 - 273 = 1527 °C

Example 3:���On a winter day, the temperature is 5 °F. What is this temperature on the Celsius scale?

°C = [(5 +40) ÷ 1.8]- 40 = -15 °C

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Example 4:���To make ice cream, rock salt is added to crushed ice to reach temperature of -11 °C. What is this temperature in Fahrenheit?

°F = [(-11 + 40) x 1.8]- 40 = 12 °F

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SPECIFIC���HEAT���

q  Different materials have different capacities for storing heat.

q  The specific heat of a substance is the amount of heat required to change the temperature of 1 g of that substance by 1°C.

q  Units of specific heat are:

s = J / g ºC

s = cal / g ºC

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SPECIFIC���HEAT���

Substance (cal/g°C) (J/g°C) Aluminum 0.214 0.897

Copper 0.0920 0.385

Iron 0.0308 0.129

Ammonia 0.488 2.04

Ethanol 0.588 2.46

Water 1.00 4.184

Specific Heat of Some Substances

Most substances have substantially

lower specific heats compared

to water

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SPECIFIC���HEAT���

q  When heated, substances with low specific heat get hot faster, while substances with high specific heat get hot at a slower rate.

q  When cooled, substances with low specific heat cool faster, while substances with high specific heat cool at a slower rate.

Specific heat and change in temperature are inversely proportional

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CALCULATING���HEAT���

q  The amount of heat lost or gained by a substance is related to three quantities:

Mass of substance

Specific heat of substance

Change in its temperature

Heat = x x

Q = m x s x ΔT

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Example 1:���How much heat is needed to raise the temperature of 200. g of water by 10.0 °C. (Specific heat of water is 4.184 J/g°C)

Q = m x s x ΔT

Q = (200. g)(4.184 J/gºC)(10.0 ºC)

Q = 8370 J or 8.37 kJ

m = 200. g

s = 4.184 J/g°C

ΔT = 10.0 °C

Q = ???

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Example 2:���Calculate the specific heat of a solid if 1638 J of heat raise the temperature of 125 g of the solid from 25.0 to 52.6 °C.

s = 0.475 J/g ºC

Qs =m x TD o

1638 J=(125 g)(27.6 C)

Q = 1638 J

m = 125 g

ΔT = 27.6 °C

s = ???

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Example 3:���Ethanol has a specific heat of 2.46 J/g°C. When 655 J are added to a sample of ethanol, its temperature rises from 18.2 °C to 32.8 °C. What is the mass in grams of the ethanol sample?

m = 18.2 g

Qm = =s x TD o o

655 J(2.46 J/g C)(14.6 C)

Q = 655 J

s = 2.46 J/g°C

ΔT = 14.6 °C

m = ???

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ENERGY & NUTRITION���

q  The foods we eat provide energy for our bodies. Vitamins and minerals are necessary for health but have little energy value.

q  Carbohydrates are the main source of fuel for the body, but when their reserves are exhausted, fats and then proteins can be used for energy.

q  In the field of nutrition, the energy from food is measured in nuts of Calories (Cal). One Calorie is equal to 1000 calories or 1 kilocalorie (kcal).

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ENERGY & NUTRITION���

q  In the laboratory, foods are burned in a calorimeter to determine their energy. A sample of food is burned in the calorimeter, and the energy released is absorbed by water surrounding the calorimeter.

q  The energy of the food can be calculated from the mass of the food and the temperature increase of the water.

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Example 1:���A 2-oz serving of pasta provides 200 Cal. What is the energy value of pasta in Cal/g? Step 1: Given: 200 Cal/2 oz Need: Cal/g

Step 2: English-English factor

oz lb

Step 3: 16 oz 1 lb

and 454 g 1 lb

Step 4: 200 Cal 16 oz x

2 oz 1 lb

g Metric-English factor

1 lbx 454 g

= 4 Cal/g

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Example 2:���A 2.3-g sample of butter is placed in a calorimeter containing 1900 g of water at a temperature of 17 °C. After the complete combustion of the butter, the water has a temperature of 28 °C. What is the energy value of butter in Cal/g?

1. Calculate heat absorbed by water

Heat absorbed by water

Heat released by butter =

2. Calculate energy value of butter

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Example 2:���

Q = m x s x ΔT

Q = (1900 g)(1.00 cal/gºC)(11 ºC)

Q = 21000 cal = 21 Cal

m = 1900 g

s = 1.00 cal/g°C

ΔT = 11 °C

Q = ???

1. Calculate heat absorbed by water

2. Calculate energy value of butter 21 Cal 2.3 g = 9.1 Cal/g

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ENERGY CONTENT���OF FOODS���

Food Type (kJ/g) (kcal/g) Carbohydrate 17 4

Fat 38 9

Protein 17 4

The energy value of foods are the kilocalories or kilojoules obtained from the complete combustion of 1 g of a carbohydrate, fat or protein.

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Example 1:���What is the energy value (in Cal) for a piece of chocolate cake that contains 34 g of carbohydrate, 10 g of fat and 5 g of protein? (Calculate answer to 2 sig figs.)

Carbohydrate

Fat

Protein

34 g x 4 kcal

1 g = 140 kcal

10 g x 9 kcal

1 g = 90 kcal

5 g x 4 kcal

1 g = 20 kcal

Total energy content = 140 kcal + 90 kcal + 20 kcal = 250 kcal = 250 Cal

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Example 2:���A 1-oz (28 g) serving of oat-bran hot cereal with half a cup of whole milk contains 22 g of carbohydrate, 7 g of fat, and 10 g of protein. If you eat two servings of the oat bran for breakfast, how many kilocalories will you obtain? (Calculate answer to 2 sig figs)

Carbohydrate

Fat

Protein

44 g x 4 kcal

1 g = 176 kcal

14 g x 9 kcal

1 g = 126 kcal

20 g x 4 kcal

1 g = 80 kcal

Total energy content = 176 kcal + 126 kcal + 80 kcal = 382 kcal = 380 kcal

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CLASSIFICATION���OF MATTER���

q  Matter is anything that has mass, and occupies space.

q  Matter can be classified by its physical state as solid, liquid or gas.

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SOLIDS���

q  Solid particles have strong forces of attraction towards each other.

q  Solids are not very compressible.

q  Solids are densely packed particles with definite shape and volume.

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LIQUIDS���

q  Liquid particles have moderate forces of attraction towards each other and are mobile.

q  Liquids are slightly compressible.

q  Liquids are loosely packed particles with definite volume but indefinite shape.

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GASES���

q  Gas particles have little or no forces of attraction towards each other.

q  Gases are very compressible.

q  Gases are very loosely packed particles with indefinite shape or volume.

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SUMMARY OF���PROPERTIES OF MATTER���

State Shape Volume Particles Compress-ibility

Solid Definite Definite Densely packed

Very slight

Liquid Indefinite Definite Mobile Slight

Gas Indefinite Indefinite Far apart High

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CLASSIFICATION���OF MATTER���

MATTER Anything that has mass

PURE SUBSTANCE

Fixed composition & properties

MIXTURE

Variable composition & properties

Mixtures can be converted into pure substances by simple physical processes (e.g. filtration, evaporation)

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MIXTURES���

HOMOGENEOUS

Uniform composition & properties

MIXTURE

Variable composition & properties

HETEROGENEOUS

Non-uniform composition & properties

Tea, Coke Ink

Salad dressing Cement

Also called solutions

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PURE SUBSTANCES���

PURE SUBSTANCE

Fixed composition & properties

COMPOUNDS

2 or more elements chemically combined

ELEMENTS

Composed of one type of atom

Compounds can be converted into elements by chemical processes or reactions (e.g. electrolysis) hydrogen,

copper, gold water, salt

aspirin

Properties are unique compared to their

components

Smallest particle is a molecule

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CONCEPT���CHECK���

Classify each substance below as element, compound or mixture.

Element: only one type of atom Element: only one type of atom Compound: composition is fixed Mixture: made of two or more types of substances

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MIXTURES���

q  Mixtures are 2 or more substances physically combined together.

q  Mixtures possess properties similar to those of their components.

q  Mixtures can be separated easily by a physical process.

q  Two types of mixtures are possible:

homogeneous heterogeneous

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HETEROGENEOUS���MIXTURES���

q  Heterogeneous mixtures are non-uniform in their composition.

q  Examples include vegetable soup, cement and salad dressing.

Heterogeneous

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HOMOGENEOUS���MIXTURES���

q  Homogeneous mixtures are uniform in their composition.

q  Examples include gasoline, soda pop and salt solution.

Homogeneous

q  Homogeneous mixtures are called solutions.

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MIXTURES vs.���COMPOUNDS���

List 3 differences between compounds & mixtures.

Composition

Compounds have fixed composition while mixtures have varied composition

Properties

Compounds have unique properties while mixtures have blended properties

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MIXTURES vs.���COMPOUNDS���

List 3 differences between compounds & mixtures.

Make-up

Compounds are chemically combined (cannot be easily separated) while mixtures are physically combined (easily separated)

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PHYSICAL & CHEMICAL���PROPERTIES���

q  The characteristics of a substance are called its properties.

q  Physical properties are those that describe the matter without changing its composition.

Examples are density, color, melting and boiling points, and electrical conductivity.

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PHYSICAL & CHEMICAL���PROPERTIES���

q  The characteristics of a substance are called its properties.

q  Chemical properties are those that describe how matter behave in combination with other matter, and involve change in its composition. Examples are flammability, corrosion, and reactivity with acids.

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Examples:���

Identify each of the following properties as physical or chemical: 1.  Oxygen is a gas 2.  Helium is un-reactive 3.  Water has high specific heat 4.  Gasoline is flammable 5.  Sodium is soft & shiny

Chemical

Physical

Physical

Chemical

Physical

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PHYSICAL���CHANGES���

q  Changes in physical properties of matter that do not involve change in its composition are called physical changes.

Examples are melting, evaporation and other phase changes.

q  Physical changes are easily reversible.

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CHEMICAL���CHANGES���

q  A change that alters the chemical composition of matter, and forms new substance is called a chemical change.

Examples are burning, rusting, and reaction with acids.

q  Chemical changes are not easily reversible, and are commonly called chemical reactions.

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Examples:���

Identify each of the following changes as physical or chemical: 1.  Cooking food 2.  Mixing sugar in tea 3.  Carving wood 4.  Burning gas 5.  Food molding

Chemical

Physical

Physical

Chemical

Chemical

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CHANGE���OF STATE���

q  When matter releases or absorbs energy without a change in temperature, phase change occurs (e.g. melting, evaporation).

Solid Liquid Fusion or melting

Freezing

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CHANGE���OF STATE���

Liquid Gas Evaporation

Condensation

Solid Gas Sublimation

Deposition

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ENERGY & ���CHANGE OF STATE���

Less energetic

More energetic

Warming process

Cooling process

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HEATING & COOLING���CURVES���

q  When heat is added to ice, it absorbs the heat without a change in temperature, causing a phase change.

q  Similarly, when heat is added to hot water, a phase change occurs without an increase in temperature.

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HEAT OF ���FUSION���

q  The quantity of heat required to melt 1 g of solid is called the heat of fusion (Hf).

Hf for ice is 80 cal/g

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HEAT OF ���VAPORIZATION���

q  The quantity of heat required to evaporate 1 g of liquid is called the heat of vaporization (Hv).

Hv for water is 540 cal/g

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HEAT OF FUSION &���VAPORIZATION���

q  For any substance the heat of vaporization is greater than the heat of fusion.

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CALCULATION HEAT DURING A PHASE CHANGE���

q  The amount of heat, released or absorbed during phase change, depends on the amount of substance and the heat of vaporization or heat of fusion.

Q = mass x heat of fusion = m x Hf

Q = mass x heat of vaporization = m x Hv

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Example 1:���

m = 50g

Hf = 80 cal/g

Q = ???

Q = 4000 cal

How much heat is required to melt 50 g of ice at 0°C?

Q = m x Hf

H = 50 g x 80 cal/g

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Example 2:���

m = 50g

Hv= 540 cal/g

Q = ???

Q = 27000 cal

How much heat is required to vaporize 50 g of water at 100°C?

H = m x Hv

H = 50 g x 540 cal/g

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Example 3:���Calculate the amount of heat required to change 25 g of ice at 0°C to water at 30.0°C.

melting ice

Qwarming water

QTotalQ = +

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Example 3:���

= m x Hf

= 25 g x 80 cal/g = 2000 cal

= m x c x ∆T

= 25 g x 1.0 cal/gºC x 30.0 ºC = 750 cal

melting ice

Q

warming water

Q

Qtotal = 2000 cal + 750 cal = 2750 cal

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THE END