Causes of ChangeEnergy and Change

Changes and EnergyBoth chemical and physical changes are accompanied

by either an increase or decrease in energy.

Energy is the ability to do work or produce heat.

There are two major types of energy: kinetic and potential.

Kinetic energy is the energy of motion and depends on both the mass and the velocity of the object.

Potential energy is stored energy.

Matter possesses both types of energy.

Heat and temperature are used to describe the energy content of matter.

What is temperature?Temperature is a measure of the average kinetic

energy (energy of motion) of particles in matter.

As the kinetic energy of the particles increases, the temperature increases.

As the kinetic energy of the particles decreases, the temperature decreases.

The lowest possible temperature (the temperature at which the particles stop moving) is called absolute zero.

Temperature is an intensive property and does not depend on the amount of matter present.

How is temperature measured?

Temperature is measured with a thermometer.

Almost all substances expand with an increase in temperature (exception is water).

Thermometers are designed so that the substances they contain (mercury, etc.) expand and contract more than the volume of the glass tube that contains them so that the column height of the substance changes.

Temperature Scales

Several temperature scales have been devised:

- Fahrenheit (oF): weather is measured using this scale.

- Celsius (oC): the scale used in the metric system.

-Kelvin (K): based on absolute zero; used in the International System (SI).

Comparison of Temperature Scales

•Celsius degrees are larger in size than Fahrenheit degrees.

•Celsius degrees are the same size as kelvins.

•The Kelvin scale is the same as the Celsius scale (only 273o higher).

•The only temperature that is the same on the Celsius and Fahrenheit scale is -40o.

Converting Between Scales

Celsius to Fahrenheit

oF = 9/5oC + 32

Fahrenheit to Celsius

oC = (oF – 32)5/9

Celsius to Kelvin

K = oC + 273

Kelvin to Celsius

oC = K - 273

Practice Problems65o C = _______KAnswer: 65oC + 273 = 338 K25oC = _______KAnswer: 25oC + 273 = 298 K-50oC = ______KAnswer: -50oC + 273 = 223 K300 K = _______oCAnswer: 300 K -273 = 27oC273 K = _______oCAnswer: 273 K – 273 = 0oC150 K = _______oCAnswer: 150 K – 273 = -123oC

How Are Heat and Temperature Related?Heat (q) is the total amount of energy that transfers

from one object to another.

Heat is an extensive property and changes based on the amount of matter present.

Temperature measures the intensity of the energy and heat measures the quantity of energy.

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The temperature of an object determines the direction of heat transfer.

When two objects of different temperatures are in contact, heat moves from the object at the higher temperature to the object at the lower temperature until they are both at the same temperature.

Methods of Heat Transfer

Conduction

Convection

Radiation

Conduction

Conduction occurs when heat is transferred by particles that are in direct contact with each other.

Good conductors transfer heat easily. Metals are good conductors.

Insulators do not transfer heat easily. Glass, rubber, wood, and plastic are good insulators.

Examples of conduction: Food cooking on a stove, burning your hand on a hot iron.

Walking on hot coals

Convection

Convection occurs when heat is transferred through the movement of gas or liquid particles.

Warm air/water is less dense and rises.

Cool air/water is more dense and sinks.

These differences in density create a circulating current.

Examples: the heating of your home, deep ocean water is cooler than at the surface.

Lava Lamps and Convection

Radiation

Radiation of heat occurs when heat is transferred in matter or space by means of electromagnetic waves.

This type of heat travels outward from its source in all directions.

Examples: a fireplace heating a room, the sun heating the earth.

A radiator exemplifies all three methods of heat transfer.

1. Heat radiates outward from the radiator to heat the room through radiation.

2. The man is warming his hands over the radiator through the process of convection.

3. If the man accidentally touches the surface of the radiator, he will burn his hands through the process of conduction.

Practice QuestionsIdentify each of the following as conduction, convection,

and radiation.1.A hot air balloon.Answer: Convection2. Warming yourself in front of a fireplace.Answer: Radiation3.Cooking food in a microwave oven.Answer: Radiation4. Frying bacon.Answer: Conduction5. Burning your hand on a curling iron.Answer: Conduction6. It is warmer in the attic than on the first floor of your

house.Answer: Convection

Transfer of Energy in Chemical and Physical Changes

All chemical and physical changes involve the release or absorption of heat.

The system includes the substances involved in the change.

The surroundings include everything else in the universe.

The system and the surroundings make up the universe.

Law of Conservation of Energy states that energy cannot be created or destroyed-instead it is transferred or changed from one form to another.

Endothermic vs. Exothermic

Endothermic processes are those in which heat flows from the surroundings into a system (heat is absorbed).

The system gains heat and the surroundings cool down.

Exothermic processes are those in which heat flows from the system to the surroundings (heat is released).

The system loses heat and the surroundings heat up.

Energy of Phases and Phase Changes Relative energy of particles making up the states (phases) of

matter.

Solids < Liquids < Gases << Plasma

Endothermic Phase Changes (Absorption of Energy)

Melting (SolidLiquid)

Vaporization (Liquid Gas)

Sublimation (Solid Gas)

Exothermic Phase Changes (Release of Energy)

Freezing (Liquid Solid)

Condensation (Gas Liquid)

Deposition (Gas Solid)

Phase Diagrams

•Triple point-the temperature and pressure at which all three phases of matter coexist.

•Critical point-the highest temperature at which a liquid can exist. (The substance exists only as a vapor above this temperature no matter what the pressure).

•Points on the line separating phases of matter represent the melting or boiling points.

•If the slope of the line separating the solid and liquid phase is positive, the solid is more dense than the liquid.

Phase Diagram of WaterNote: The line separating the solid and liquid phases has a negative slope. Therefore, the solid phase is less dense than the liquid phase and the solid floats in the liquid.

Practice Interpreting Phase Diagrams

For additional interactive practice, visit the following website.

http://www.sciencegeek.net/Chemistry/taters/phasediagram.htm

Measuring Heat

Heat is measured with a calorimeter.

The unit used when measuring heat is the calorie (cal) or joule (J).

A calorie is the amount of heat needed to raise the temperature of 1 gram of pure water 1oC.

1 calorie = 4.18 joules.

1000 calories (1 kcal) = 1 dietary calorie (Cal)

Heat CapacityThe amount of heat needed to raise the

temperature of an object 1oC is called the heat capacity of the object.

The heat capacity depends on the mass and the chemical make-up of the substance.

Heat capacity is an extensive property.Specific heat capacity is the amount of heat it

takes to raise the temperature of 1 g of a substance 1oC.

Specific heat capacity is an intensive property.Water has a high specific heat capacity compared

to other substances-therefore, it requires more heat to change the temperature of a sample of water.

Specific Heats of Various Substances

Substance Specific Heat (J/goC)

Water 4.18

Ethanol (grain alcohol) 2.44

Ice 2.03

Steam 2.01

Aluminum 0.897

Concrete 0.84

Iron 0.449

Silver 0.235

Lead 0.129

Gold 0.129

Calculating Heat and Specific HeatThe equation for calculating heat:

heat = mass x change in temp x specific heat

or q = m x ∆T x C

m is the mass and is measured in grams

∆T = final temp – initial temp (oC)

C is the specific heat

**when working with water, C = 1.00 cal/goC or 4.18 J/goC.

q is the heat measured in joules or calories . If q < 0, heat is being released (exothermic) and if q>0, heat is being absorbed (endothermic).

Practice Problem #1

If the temperature of 34.4 g of ethanol increases from 25.5oC to 78.8oC, how much heat has been absorbed by the ethanol?

Answer to Practice Problem #1

q = ?

m= 34.4 g

∆ T = 78.8-25.5oC

C=2.44 J/g oC

q = mx∆TxC

q = 34.4 (78.8-25.5)2.44

q= 34.4(53.3)2.44

q=4,470 J or 4.47 kJ

Practice Problem #2

A 155 g sample of an unknown substance was heated from 25.0oC to 40.0oC. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance?

Answer to Practice Problem 2q= 5696 J

m= 155 g

∆T = 40.0 – 25.0oC

C = ?

q = mx∆TxC

5696 = 155 (40.0-25.0) x

5696 = 155 (15.0) x

5696 = 2325 x

x = 5696/2325 = 2.45 J/g oC

Practice Problem #3

A 4.50 g nugget of pure gold absorbed 276 J of heat. The initial temperature was 25.0oC. What was the final temperature?

Answer to Practice Problem 3

q = 276 Jm = 4.50 g∆T = Tf – 25.0 oCC = 0.129 J/g oC

q = m x ∆T x C

276 = 4.50 (x – 25.0) 0.129276 = .5805 (x – 25)276/.5805 = .5805/.5805 ( x – 25)475 = x – 25X = 500oC

Practice Problem #4

When a 58.8 g piece of hot alloy is placed in 125 g of cold water in a calorimeter, the temperature of the alloy decreases by 106.1oC, while the temperature of the water increases by 10.5oC. What is the specific heat of the alloy?

Answer to Practice Problem #4

In order to solve this problem, it is important to recognize that the alloy lost heat (temperature decreased) and the water gained heat (temperature increased).

Due to the law of conservation of energy, heat lost = heat gainedHeat gained by water: q = m x ∆T x C

q = 125 x 10.5 x 4.18 = 5486 JHeat lost by alloy: q = m x ∆T x C

5486 = 58.8 x (106.1) x CC = 0.879 J/goC

Latent Heat and Changes in StateLatent heat is the energy involved in

matter undergoing changes in state.

Heat of fusion is the energy required to change 1 g of a substance from a solid to a liquid.

The heat of fusion is unique to the substance.

q = Hfusion x mass

Practice Problem

How much energy is required to melt 10 g of ice? (Heat of fusion of ice is 334 J/g)

q = Hfusion x mass

q = 10 x 334

q = 3340 J

Latent Heat and Changes in State

Heat of vaporization is the energy needed to change 1 g of a substance from a liquid to a gas.

The heat of vaporization is also unique to the substance.

q = Hvaporization x mass

Practice Problem

How much energy is required to boil 10 g of water? (Hvap for water is 2260 J/g)

q = Hvap x m

q = 2260 x 10

q = 22600 J

Transfer of Energy in Physical Changes

Wherever there is a change in temperature, there is also a change in kinetic energy of the particles making up the substance. (q=mx∆TxC)

Areas on the graph where there is no change in temperature represent a change in potential energy (or change in state) of the particles making up the substance.(q = Hfus x m or q = Hvap x m)

Practice Problem (Honors Chemistry only)

Calculate the energy required to change 10 g of ice at -10oC to steam at 110oC.

Hint: This problem must be solved in five separate steps.

Answer to practice problemStep 1: Convert 10 g ice at -10oC to ice at 0oC.Answer: q = mx∆TxC

q = 10 x (0 - -10oC) x 2.03 = 203 JStep 2: Melt the 10 g of ice.Answer: q = mxHfus

q = 10 x 334 = 3340 JStep 3: Raise the temp of the water from 0oC to 100oC.Answer: q = mx∆TxC q = 10 x (100-0) x 4.18 = 4180 JStep 4: Boil the 10 g of water.Answer: q= mxHvap

q = 10 x 2260 = 22600JStep 5: Raise the temp of the steam from 100oC to 110oC.Answer: q = mx∆TxC q = 10 x (110 – 100) x 2.01 = 201J Step 6: Calculate the total.Answer: 203 + 3340 + 4180 + 22600 + 201 = 30524 = 31000 J

Practice Problem #2

Calculate the energy required to change 1 mole of ice at 0oC to steam at 100oC.

Answer to Practice Problem #2.

Step 1: Melt the ice.Answer: q = mxHfus

q = 18 g x 334 = 6012 JStep 2: Raise the temp of the water from 0oC to 100oC. Answer: q = m x ∆T x C q = 18 x (100-0) x 4.18 = 7524 JStep 3: Convert the liquid water to steamAnswer: q = mxHvap

q = 18 x 2260 = 40680 J Step 4: Calculate the total.Answer: 6012 + 7524 + 40680 = 54216 J = 54000J

Practice Interpreting Energy Graphs Illustrating Phase Changes

For additional interactive practice, visit the following website.

http://www.sciencegeek.net/Chemistry/taters/phasediagrams2.htm

Thermochemical Equations

An equation that includes the heat change involved in a chemical reaction is called a thermochemical equation.

Heat of reaction (∆H) is the heat change for a particular equation.

∆H = Hproducts - Hreactants

Potential Energy Diagrams

All reactions begin with the breaking of the bonds of the reactants (endothermic step).

In order for those bonds to break, collisions between the reacting molecules must occur and the reacting molecules must possess a minimum amount of energy (activation energy).

During the collision, an unstable intermediate molecule forms (activated complex).

The molecules then rearrange and form new bonds resulting in the formation of products (exothermic step).

•If the initial endothermic step is greater than the final exothermic step, the overall reaction is endothermic.•∆H = energy of products – energy of reactants•If the reaction is endothermic, ∆H > 0.

•If the initial endothermic step is smaller than the final exothermic step the reaction is exothermic.•∆H = energy of products – energy of reactants•If the reaction is exothermic, ∆H < 0.

Effect of a Catalyst• A catalyst is a

substance that speeds up a reaction while remaining unchanged in the process.

• One way in which catalysts accomplish this is by lowering the activation energy.

Practice Interpreting Potential Energy GraphsFor additional interactive practice, visit the

following website.

http://www.sciencegeek.net/Chemistry/taters/energydiagram.htm

Writing Thermochemical Equations

When 2 mol of solid magnesium combines with 1 mole of oxygen gas, 2 mol of solid magnesium oxide is formed and 1204 kJ of heat is released. Write the thermochemical equation for this reaction.

2Mg(s) + O2(g) 2MgO(s) + 1204 kJ

Note: The energy is written on the reactant side of the equation for endothermic reactions.

Practice ProblemUse the following equation to answer the questions

below: 129 kJ + 2 NaHCO3 Na2CO3 + H2O + CO2

1)Is the reaction exothermic or endothermic?Answer: endothermic (heat appears on the left side of

the equation)2) Sketch a potential energy curve that could represent

this reaction.Answer:

3) How much heat is needed to decompose 2.24 moles of NaHCO3?

Answer: x kJ = 2.24 moles NaHCO3

129 kJ 2 moles NaHCO3

x = 144 kJ

Review

For additional interactive practice, visit the following website.

http://www.sciencegeek.net/Chemistry/taters/Unit7Thermochemistry.htm