matter and energy chapter 12.4 chapter 15.1 - 15.3 1

Matter and EnergyChapter 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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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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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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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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Properties of Matter Chemical Properties: the ability to

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

rotting

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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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States of MatterLiquid:–particles move past each other

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

(indefinite)

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

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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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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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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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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

Heating Curve for WaterNote that T is constant as ice melts

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Heating/Cooling Curve for Water

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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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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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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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• 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

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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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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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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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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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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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Temperature

Solid Liquid

Gas

1 Atm

Pres

sure

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Phase Diagram for water

Sublimation Depositio

n

Melting Freezing

Boiling

Condensation