1 Chapter 2 Energy and Matter 2.1 Energy. 2 Energy makes objects move makes things stop is needed to “do work”

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Chapter 2 Energy and Matter

2.1Energy

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Energy

Energy• makes objects move• makes things stop• is needed to “do work”

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Work

Work is done when • you climb• you lift a bag of groceries• you ride a bicycle• you breathe• your heart pumps blood• water goes over a dam

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Potential Energy

Potential energy is energystored for use at a later

time.

Examples:• water behind a dam• a compressed spring• chemical bonds in

gasoline, coal, or food

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Kinetic Energy

Kinetic energy is the energy of matter in motion.

Examples:• swimming• water flowing over a dam• working out• burning gasoline

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Chapter 2 Energy and Matter

2.2Temperature

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Temperature

Temperature • is a measure of how hot or cold

an object is compared to another object

• indicates that heat flows from the object with a higher temperature to the object with a lower temperature

• is measured using a thermometer

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Temperature Scales

Temperature scales • are Fahrenheit,

Celsius, and Kelvin

• have reference points for the boiling and freezing points of water

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Units for Measuring Energy or Heat

Heat is measured in joules or calories.• 4.184 joules (J) = 1 calorie (cal) (exact) • 1 kJ = 1000 J• 1 kilocalorie (kcal) = 1000 calories (cal)

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• On the Fahrenheit scale, there are 180 °F between the freezing and boiling points, and on the Celsius scale, there are 100 °C.

180 °F = 9 °F = 1.8 °F 100 °C 5 °C 1 °C

• In the formula for the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0 °C to 32 °F.

TF = 9 (TC) + 32 ° 5 or TF = 1.8(TC) + 32 °

Fahrenheit Formula

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• TC is obtained by rearranging the equation for TF.

TF = 1.8(TC) + 32 °• Subtract 32 from both sides.

TF – 32 ° = 1.8(TC) + (32 ° – 32 °)

TF – 32 ° = 1.8(TC)

• Divide by 1.8. TF – 32 ° = 1.8 TC

1.8 1.8

TF – 32 ° = TC 1.8

Celsius Formula

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Solving a Temperature Problem

A person with hypothermia has abody temperature of 34.8 °C. WhatIs that temperature in °F?

TF = 1.8(TC) + 32 °

TF = (1.8)(34.8 °C) + 32 ° exact tenth’s exact

= 62.6 ° + 32 ° = 94.6 °F tenth’s

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The Kelvin temperature• is obtained by adding 273 to the Celsius temperature

TK = TC + 273

In the Kelvin temperature scale: • There are 100 units between the freezing and boiling

points of water.100 K = 100 °C or 1 K = 1 °C

• 0 K (absolute zero) is the lowest possible temperature. 0 K = –273 °C

Kelvin Temperature Scale

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Chapter 2 Energy and Matter

2.3Specific Heat

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Specific Heat

Specific heat • is different for different substances• is the amount of heat that raises the

temperature of 1 g of a substance by 1 °C• in the SI system has units of J/g C • in the metric system has units of cal/g C

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Example of Calculating Specific Heat

What is the specific heat of a metal if 24.8 g absorbs65.7 cal of energy and the temperature rises from20.2 C to 24.5 C? STEP 1 Given: 24.8 g, 65.7 cal, ΔT = 20.2 C to 24.5 C

Need: SHmetal = cal/g C

STEP 2 Plan: ΔT = 24.5 C – 20.2 C = 4.3 C SH = Heat(cal) g C

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Example of Calculating Specific Heat (continued)

STEP 3 Set up to calculate SH: 65.7 cal = 0.62 cal/g C (24.8 g)(4.3 C)

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Heat Equation

Rearranging the specific heat expression gives theheat equation.

Heat = g x T x cal (or J) = cal (or J) g °C

The amount of heat lost or gained by a substance iscalculated from the• Mass of substance (g)• Temperature change ( T)• Specific heat of the substance (cal/g °C) or (J/g °C)

Guide to Calculations Using Specific Heat

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A hot-water bottle contains 750 g of water at 65 °C. If thewater cools to body temperature (37 °C), how manycalories of heat could be transferred to sore muscles?

STEP 1 Given: 750 g of water cools from 65 °C to 37 °C SHwater = 1.00 cal/g °C

Need: calories of heat transferred

STEP 2 Calculate the temperature change T: 65 °C – 37 °C = 28 °C

Sample Calculation for Heat

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STEP 3 Write the heat equation: Heat (cal) = mass(g) x T x SH

STEP 4 Substitute given values and solve for heat:

750 g x 28 °C x 1.00 cal g °C

= 21 000 cal

Sample Calculation for Heat

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Learning Check

How many calories are obtained from a pat of butter

if it provides 150 J of energy when metabolized?

1) 36 cal2) 150 cal 3) 630 cal

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Solution

STEP 1 Given: 150 J Need: caloriesSTEP 2 Plan: J calSTEP 3 Write equality and conversion factors: 1 cal = 4.184 J 1 cal and 4.184 J 4.184 J 1 calSTEP 4 Set up problem: 150 J x 1 cal = 36 cal (1)

4.184 J

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Learning Check

How many kilojoules are needed to raise thetemperature of 325 g of water from 15.0 °C to

77.0 °C?1) 20.2 kJ2) 84.3 kJ3) 105 kJ

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Solution

2) 84.3 kJ STEP 1 Given: 325 g of water warms from 15.0 °C

to 77.0 °C SHwater = 4.184 J/g °C

1 kJ = 1000 J Need: kilojoules of heat neededSTEP 2 Calculate the temperature change T:

77.0 °C – 15.0 °C = 62.0 °CSTEP 3 Write the heat equation:

Heat (joules) = mass (g) x T x SH

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Solution (continued)

STEP 4 Substitute given values and solve for heat: 325 g x 62.0 °C x 4.184 J x 1 kJ

g °C 1000 J = 84.3 kJ (2)

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Chapter 2 Energy and Matter

2.4Energy and Nutrition

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CalorimetersA calorimeter • is used to measure heat

transfer• contains a reaction

chamber and thermometer in water

• indicates the heat lost by a sample

• indicates the heat gained by water

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Chapter 2 Energy and Matter

2.5Classification of Matter

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MatterMatter • is the material that makes up a substance• makes up the things we see everyday, such as water, wood, cooking pans,

clothes, and shoes

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A pure substance is classified as • matter with a specific composition• an element when composed of one type of atom• a compound when composed of two or more

elements combined in a definite ratio

Pure Substances

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Mixtures

A mixture is matter that consists of• two or more substances that are physically

mixed, not chemically combined• two or more substances in different

proportions• substances that can be separated by

physical methods

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Homogeneous Mixtures

In a homogeneous mixture,

• the composition is uniform throughout

• the different parts of the mixture are not visible

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Heterogeneous Mixtures

In a heterogeneous mixture,

• the composition is not uniform; it varies from one part of the mixture to another

• the different parts of the mixture are visible

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Chapter 2 Energy and Matter

2.6States and Properties

of Matter

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Solids

Solids have • a definite shape• a definite volume• particles that are close

together in a fixed arrangement

• particles that move very slowly

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Liquids

Liquids have• an indefinite shape

but a definite volume• the same shape as

their container• particles that are close

together but mobile• particles that move

slowly

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Gases

Gases have • an indefinite shape• an indefinite volume• the same shape and

volume as their container• particles that are far apart• particles that move very

fast

Physical Properties

Physical properties• are observed or measured without

changing the identity of a substance• include shape and color• include melting point and boiling point

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Physical Change

In a physical change,• the identity and

composition of the substance do not change

• the state can change or the material can be torn into smaller pieces

Chemical PropertiesChemical properties• describe the ability of a substance to change into a new substance

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During a chemical change, reacting substances form new

substances with different compositions and properties

a chemical reaction takes place

IronFe

Iron (III) oxideFe2O3

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Chapter 2 Energy and Matter

2.7Changes of State

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Melting and Freezing

A substance • is melting when it changes from a solid to a

liquid• is freezing when it changes from a liquid to a

solid• such as water has a freezing (melting) point of

0 °C

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Heat of Fusion

The heat of fusion • is the amount of heat released when 1 gram

of liquid freezes (at its freezing point) • is the amount of heat needed to melt 1 gram

of a solid (at its melting point)• for water (at 0 °C) is

334 J or 80 cal 1 g of water

Guide to Calculations Using Heat of Fusion (or Vaporization)

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Calculations Using Heat of Fusion

How much heat in calories is needed to melt 15.0 g ofice at 0 °C ?STEP 1 Given: 15.0 g of water(s)

change of state: melting at 0 °CSTEP 2 Plan: g of water(s) g of water(l)

STEP 3 Write conversion factors:1 g of water = 80 cal

1 g of water and 80 cal 80 cal 1 g of water

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Calculations Using Heat of Fusion (continued)

STEP 4 Set up the problem to calculate calories:

15.0 g water x 80 cal = 1200 cal 1 g water

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SublimationSublimation• occurs when a solid changes directly to a gas• is typical of dry ice, which sublimes at 78 C• takes place in frost-free refrigerators• is used to prepare freeze-dried foods for long-

term storage

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Evaporation and CondensationWater• evaporates when molecules

on the surface gain sufficient energy to form a gas

• condenses when gas molecules lose energy and form a liquid

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Heat of Vaporization

The heat of vaporization is the amount of heat• absorbed to vaporize 1 g of a liquid to gas at the

boiling point• released when 1 g of a gas condenses to liquid at

the boiling point

Boiling Point of Water = 100 °C

Heat of Vaporization (water) = 2260 J or 540 cal 1 g of water

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Classify each of the following as a 1) physical change or 2) chemical change.A. ____ burning a candleB. ____ ice melting on the streetC. ____ toasting a marshmallowD. ____ cutting a pizzaE. ____ polishing a silver bowl

Learning Check

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Classify each of the following as a 1) physical change or 2) chemical change.A. 2 burning a candleB. 1 ice melting on the streetC. 2 toasting a marshmallowD. 1 cutting a pizzaE. 2 polishing a silver bowl

Solution

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Learning Check

How many joules are released when 25.0 g of water

at 0 °C freezes? 1) 334 J

2) 2000 J 3) 8350 J

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Solution

3) 8350 J

STEP 1 Given: 25.0 g of water(l) change of state: freezing at 0 °C

STEP 2 Plan: g of water g of water

STEP 3 Write conversion factors:1 g of water = 334 J 1 g of water and __334 J___ 334 J 1 g of water

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Solution (continued)

STEP 4 Set up the problem to calculate joules:

25.0 g water x 334 J = 8350 J (3) 1 g water

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Learning Check

How many kilocalories (kcal) are released when 50.0 g

of water(g) as steam from a volcano condenses at

100 °C?1) 27 kcal2) 540 kcal 3) 2700 kcal

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Solution

1) 27 kcalSTEP 1 50.0 g of water(g) = 50.0 g of water(l) Change of state: water condensing at 100 °CSTEP 2 Plan: g of water(g) g of water(l)STEP 3 Write conversion factors:

1 g of water = 540 cal 1 g of water and 540 cal

540 cal 1 g of water 1 kcal = 1000 cal

1 kcal and 1000 cal 1000 cal 1 kcal