matter and energy chapter 12.4 chapter 15.1 - 15.3 1

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Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

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Vocabulary – Ch. 3.1 – 3.2 (SIA Review) Physical property Physical property Extensive property Extensive property Intensive property Intensive property Chemical property Chemical property States of matter States of matter Solid Solid Liquid Liquid Gas Gas Vapor Vapor Physical Change Physical Change Chemical Change Chemical Change Law of Conservation of Mass Law of Conservation of Mass Phase Change Phase Change 3

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Page 1: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Matter and EnergyChapter 12.4

Chapter 15.1 - 15.3

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Page 2: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Chapter Objectives Identify observable characteristics of

a chemical reaction Define Energy Show how energy applies to chemical

reactions and physical processes Interpret phase diagrams Interpret heating (cooling) curves

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Page 3: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Vocabulary – Ch. 3.1 – 3.2(SIA Review)

Physical property Extensive property Intensive property Chemical property States of matter Solid Liquid Gas Vapor

Physical Change Chemical Change Law of

Conservation of Mass

Phase Change

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Page 4: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Vocabulary – Chapter 15.1-15.2

Energy Heat Joule Specific Heat Specific Heat Equation (q = mC ∆T)

Heat of Vaporization (∆Hvap) Heat of Fusion (∆Hfus) Heating Curve

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Page 5: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Vocabulary – Ch. 12.4 Pressure Barometer  Atmosphere Melting Point Vaporization Evaporation Vapor Pressure Boiling Point

 Sublimation Freezing point  Condensation  Deposition Phase diagram  Triple Point

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Page 6: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Properties of Matter Chemical Properties: the ability to

combine or change into other substances.– Examples: flammability, oxidation,

rotting

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Page 7: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

States of Matter State of Matter: Its physical form. There are three physical states:

Solid: – Definite shape – Definite volume– Closely packed particles

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Page 8: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

States of MatterLiquid:–particles move past each other

(flow)– definite volume – takes the shape of its container

(indefinite)

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Page 9: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

States of MatterGas: – flows– takes the shape of

its container (indefinite shape)–Fills the container completely. (indefinite volume)

Note: A vapor refers to a gaseous state of a substance that is a solid or liquid at room temperature.

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Page 10: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Ch. 12.4 - Changes in Matter

Physical changes are those which alter the substance without altering its composition. – Change of phase one physical state to

anotherMelting of ice - composition

unchanged, i.e. ice is water in solid form (H2O)

They generally require energy, the ability to absorb or release heat (or work).

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Page 11: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Phase Changes What are the phase changes of water?1. Melting – changing of a solid to a

liquid(heat of fusion = ∆Hf)

2. Vaporization – changing from a liquid to a gas (heat of vaporization = ∆Hvap)

3. Sublimation – Changing from a solid to a gas (heat of sublimation = ∆Hsub)

What do these processes have in common?

Answer:

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Page 12: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Phase Changes Phase changes in the opposite

direction have names too.1. liquid to a solid: 2. gas to a liquid: 3. gas to a solid: What do these have in common?Answer:

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Page 13: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

UNITS OF ENERGY1 calorie = heat required to

raise temp. of 1.00 g of H2O by 1.0 oC.

1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food “calorie”)

But we use the unit called the JOULE

1 cal = 4.184 Joules (exactly)

James Joule1818-1889

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Page 14: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Heats of Fusion & Vaporization

Heat of Fusion (∆Hfus) – The amount of heat (in Joules) needed to melt 1 g of substance.

For ice: 334 J/g q (heat) = ∆Hfus*m (m= mass of

ice/water) Heat of Vaporization (∆Hvap) – The

amount of heat (in Joules) needed to vaporize 1 g of substance

For water: 2260 J/g q (heat) = ∆Hvap*m (m= mass of water/steam)

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Page 15: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Example Problems How much heat does it take to melt

20.5 g of ice at 0⁰C? q = 334 J/g * 20.5 = 6850 J (6.85 kJ)

How much heat is released when 50.0 g of steam at 100 ⁰C condenses to water at 100 ⁰C?

q = - 2260 J/g * 50.0 g = -113,000 J (-113 kJ)

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Page 16: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Specific Heat Capacity Specific Heat Capacity –

amount of heat (q) required to raise the temperature of one gram of a substance by 1 degree. C = J (energy gained or lost) mass (g) * Temp Change(⁰C)

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Page 17: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Heat Capacity ValuesSubstanceSpec. Heat (J/g•⁰C)Water 4.184Ethylene glycol 2.39Al uminum 0.897glass 0.84

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Page 18: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Calculating Heat Gained or lost

The heat, q, gained or lost by a substance can be calculated by knowing the mass of the object, the temperature change, and the heat capacity.

q = mC∆T

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Page 19: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Calculations involving Heat Example 1: A 5.00 g piece of

aluminum is heated from 25.0⁰C to 99.5⁰C. How many joules of heat did it absorb?

q = m * C * ∆T = 5.00 g * 0.897 J/g*⁰C * 74.5⁰C = 334 J

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Page 20: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Calculations involving Heat Example 2: 10.2 g of cooking oil at

25.0 ⁰C is placed in a pan and 3.34 kJ of heat is required to raise the temperature to 196.4 ⁰C. What is the specific heat of the oil?

q = m*C*∆T C = q/(m ∆T) C = 3340 J/(10.2 g * (196.4-25.0) ⁰C) C = 1.91 J/g* ⁰C

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Page 21: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Calculations involving Heat Important Points!

q (heat) is a positive quantity. The sign (+ or -) refers to whether the system you’re looking gained it (+) or lost it (-).

From the previous example, the oil would lose 3340 J of heat upon cooling back to 25.0 ⁰C. (-3340 J heat lost)

Specific heat capacity is like a bucket. It is a measure of how much energy an object absorbs before the temperature changes. 21

Page 22: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Heating Curve for WaterNote that T is constant as ice melts

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Page 23: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Heating/Cooling Curve for Water

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Page 24: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Heat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/g

What quantity of heat is required to melt 500. g of ice (at 0oC) and heat the water to steam at 100oC?

Heat & Changes of State

+333 J/g +2260 J/g

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Page 25: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

What quantity of energy as heat is required to melt 500. g of ice (at 0⁰C) and heat the water to steam at 100 oC?

1. To melt ice at 0⁰C q = (500. g)(333 J/g) = 1.67 x 105 J2.To raise water from 0 oC to 100oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x

105 J3.To vaporize water at 100oC q = (500. g)(2260 J/g) = 1.13 x 106 J4. Total energy = 1.51 x 106 J = 1510 kJ

Heat & Changes of State

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Page 26: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Practice problemIf we add 6050 J of heat to 54.2g of ice at -10.0⁰C, what will it be at the end? What temperature will it be? The specific heat of ice is 2.03 J/g*⁰C.

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Page 27: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

• Pressure is the force acting on an object per unit area:

• Gravity exerts a force on the earth’s atmosphere• A column of air 1 m2 in cross section exerts a force of

about 105 N (101,300 N/m2).• 1 Pascal (Pa) = 1 N/m2 . So, 101,300 N/m2 = 101,300 Pa

or 101.3 kPa.• Since we are at the surface of the earth, we ‘feel’ 1

atmosphere of pressure.

FPA

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Pressure

Page 28: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

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Page 29: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Barometer A barometer

measures atmospheric pressure

The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg.

1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

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Page 30: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Units of pressure 1 atmosphere (atm) = 760 mm Hg = 760

torr 1 atm = 101,300 Pascals = 101.3 kPa Can make conversion factors from these. What is 724 mm Hg in atm ?

What is 724 mm Hg in kPakPa

atmkPax

HgmmatmxHgmm 5.96

13.101

7601724

atmHgmm

atmxHgmm 953.0760

1724

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Page 31: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Phase ChangesVapor pressure is the pressure exerted by a vapor over a liquid.The vapor pressure increases with increasing temperature.This is why water evaporates even though it’s not 212˚F.

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Page 32: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Phase Changes

However, when the vapor pressure of the water is the same as the atmospheric pressure the water is … boiling.

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Page 33: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Phase Diagram (Ch. 12.4)

A phase diagram is a graph of pressure vs temperature that shows in which phase a substance exists under different conditions of T & P.

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Page 34: Matter and Energy Chapter 12.4 Chapter 15.1 - 15.3 1

Temperature

Solid Liquid

Gas

1 Atm

Pres

sure

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Phase Diagram for water

Sublimation Depositio

n

Melting Freezing

Boiling

Condensation