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Chapter 3: Matter and Energy Classification of Matter Properties of Matter Temperature Energy Specific Heat.

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Presentation on theme: "Chapter 3: Matter and Energy Classification of Matter Properties of Matter Temperature Energy Specific Heat."— Presentation transcript:

1 Chapter 3: Matter and Energy Classification of Matter Properties of Matter Temperature Energy Specific Heat

2 Chapter 3 - Slide 2 of 65 Matter Is the material that makes up all things. Has mass and occupies space. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

3 Chapter 3 - Slide 3 of 65 Matter Anything that has mass and takes up space. Matter Pure Substance Mixture

4 Chapter 3 - Slide 4 of 65 Classification of Matter Pure Substances: A form of matter that always has a definite and constant composition. Properties always the same under a given set of conditions (temperature & pressure)

5 Chapter 3 - Slide 5 of 65 A pure substance is classified as An element when composed of one type of atom. A compound when composed of two or more different elements combined in a definite ratio. Pure Substances

6 Chapter 3 - Slide 6 of 65 Pure Substances: Elements Element –Pure substance that can not be broken down into simpler substances by chemical means. Copper (Cu) Lead (Pb) Aluminum (Al) –The most basic form of matter. –Each element can be found on the periodic table.

7 Chapter 3 - Slide 7 of 65 Pure Substances: Compounds Compounds –A chemical combination of 2 or more different elements. –A pure substance that can be broken down into simpler substances by chemical means. Salt (NaCl) Table sugar (C 12 H 22 O 11 ) Water (H 2 O) Carbon monoxide (CO) NOTE: CO is different from the element Co

8 Chapter 3 - Slide 8 of 65 Elements in a Compound “Table salt” is a compound that contains the elements sodium and chlorine. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

9 Chapter 3 - Slide 9 of 65 Pure Substances Elements and Compounds Elements and compounds have definite compositions, and each has a set of properties that are unique. Which pairs of symbols / formulas represent elements and compounds respectively? (A) (B) (C) (D) (E) P-1

10 Chapter 3 - Slide 10 of 65 Mixtures A mixture is a type of matter that consists of Two or more substances that are physically mixed, not chemically combined. A physical combination of two or more pure substances in which each substance retains its own chemical identity. Two or more substances in different (variable) proportions. Substances that can be separated by physical methods.

11 Chapter 3 - Slide 11 of 65 Mixtures Example of a mixture: Pasta and water can be separated by using a strainer. Uses a physical method to separate the components. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

12 Chapter 3 - Slide 12 of 65 Homogeneous Mixtures In a homogeneous mixture, The composition is uniform throughout. The different components of the mixture are not visible or discernable, one from the other. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

13 Chapter 3 - Slide 13 of 65 Heterogeneous Mixtures In a heterogeneous mixture, The composition of substances is not uniform. The composition varies from one part of the mixture to another. The different parts of the mixture are visible. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

14 Chapter 3 - Slide 14 of 65 Classification of Matter Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

15 Chapter 3 - Slide 15 of 65 Elements v. Compounds v. Mixtures Identify the following as an Q-1 to Q-5

16 Chapter 3 - Slide 16 of 65 Matter Has characteristics called physical and chemical properties.

17 Chapter 3 - Slide 17 of 65 Physical Properties Physical properties are: Characteristics observed or measured without changing the identity of a substance. Shape, physical state, odor, boiling and freezing points (Changes of state), density, and color of that substance.

18 Chapter 3 - Slide 18 of 65 Physical Properties of Copper Copper has the following physical properties: Reddish-orange Very shiny Excellent conductor of heat and electricity Solid at 25  C Melting point 1083  C Boiling point 2567  C Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

19 Chapter 3 - Slide 19 of 65 A physical change occurs in a substance if there is A change in the state. A change in the physical shape. No change in the identity and composition of the substance. Physical Change

20 Chapter 3 - Slide 20 of 65 States of Matter The states of matter are Solid Definite volume and shape Liquid Definite volume, but take the shape of its container Gas No definite volume or shape Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

21 Chapter 3 - Slide 21 of 65 Examples of States of Matter Solids Rocks, shells, baseballs, tennis racquets, books Liquids Lakes, rain, melted gold, mercury in a thermometer Gases Air, helium in a balloon, neon in a neon tube

22 Chapter 3 - Slide 22 of 65 Examples of Changes of State Some changes of state for water: Solid water (ice) melts and forms liquid water. Liquid water boils and forms gaseous water (steam). Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

23 Chapter 3 - Slide 23 of 65 Examples of Physical Changes Examples of physical changes: Paper torn into little pieces (change of size) Copper hammered into thin sheets (change of shape) Water poured into a glass (change of shape ) Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

24 Chapter 3 - Slide 24 of 65 Chemical Properties Chemical properties describe the ability of a substance To interact with other substances To change into a new substance Example: Iron has the ability to form rust when exposed to oxygen. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

25 Chapter 3 - Slide 25 of 65 Chemical Change In a chemical change, a new substance forms that has A new composition New chemical properties New physical properties Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

26 Chapter 3 - Slide 26 of 65 Some Chemical Changes Silver tarnishes Shiny metal reacts to form black, grainy coating. Wood burns A piece of wood burns with a bright flame to form ash, carbon dioxide, water vapor, and heat. Iron rusts A shiny nail combines with oxygen to form orange-red rust.

27 Chapter 3 - Slide 27 of 65 Classify each of the following changes as physical or chemical A.Burning a candle. B. Ice melting on the street. C. Toasting a marshmallow. D. Cutting a pizza. E. Iron rusting in an old car. Learning Check

28 Chapter 3 - Slide 28 of 65 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. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

29 Chapter 3 - Slide 29 of 65 Temperature Scales Temperature scales Are Fahrenheit, Celsius, and Kelvin. Have reference points for the boiling and freezing points of water. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

30 Chapter 3 - Slide 30 of 65 Learning Check P-2

31 Chapter 3 - Slide 31 of 65 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 calculating the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0°C to 32°F. T F = 9/5 T C + 32  orT F = 1.8 T C + 32  Fahrenheit Formula

32 Chapter 3 - Slide 32 of 65 Solving for °F Temperature A person with hypothermia has a body temperature of 34.8°C. What is that temperature in °F? T F = 1.8 T C + 32  T F = 1.8 (34.8°C) + 32° exact tenth's exact = 62.6 + 32° = 94.6°F tenth’s Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

33 Chapter 3 - Slide 33 of 65 T C is obtained by rearranging the equation for T F. T F = 1.8T C + 32 Subtract 32 from both sides. T F - 32 = 1.8T C ( +32 - 32) T F - 32 = 1.8T C Divide by 1.8 =°F - 32 = 1.8 T C 1.8 1.8 T F - 32 = T C 1.8 Celsius Formula

34 Chapter 3 - Slide 34 of 65 The normal temperature of a chickadee is 105.8°F. What is that temperature on the Celsius scale? 1) 73.8 °C 2) 58.8 °C 3) 41.0 °C Learning Check

35 Chapter 3 - Slide 35 of 65 A pepperoni pizza is baked at 455°F. What temperature is needed on the Celsius scale? 1) 423°C 2) 235°C 3) 221°C Learning Check

36 Chapter 3 - Slide 36 of 65 On a cold winter day, the temperature is –15°C. What is that temperature in °F? 1) 19 °F 2) 59°F 3) 5°F Learning Check

37 Chapter 3 - Slide 37 of 65 The kelvin temperature Has 100 units between freezing and boiling points. 100 K = 100°Cor 1 K = 1 °C Adds 273 to the Celsius temperature. T K = T C + 273 0 K (absolute zero) is the lowest possible temperature. 0 K = –273 °C Kelvin Temperature Scale

38 Chapter 3 - Slide 38 of 65 Temperatures Table 3.6 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

39 Chapter 3 - Slide 39 of 65 Learning Check P-8

40 Chapter 3 - Slide 40 of 65 Energy Makes objects move. Makes things stop. Is needed to “do work ”. Energy

41 Chapter 3 - Slide 41 of 65 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. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

42 Chapter 3 - Slide 42 of 65 Potential Energy Potential energy is energy stored for use at a later time. Examples are Water behind a dam. A compressed spring. Chemical bonds in gasoline, coal, or food. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

43 Chapter 3 - Slide 43 of 65 Kinetic Energy Kinetic energy is the energy of matter in motion. Examples are Swimming. Water flowing over a dam. Working out. Burning gasoline. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

44 Chapter 3 - Slide 44 of 65 Learning Check Identify the energy as potential or kinetic. 1. Roller blading. 2. A peanut butter and jelly sandwich. 3. Mowing the lawn. 4. Gasoline in the gas tank.

45 Chapter 3 - Slide 45 of 65 Forms of Energy Energy can take many forms. Heat (thermal) Mechanical (movement) Light Electrical Chemical Nuclear P-3

46 Chapter 3 - Slide 46 of 65 Changes in Forms of Energy Energy can change from one form to another. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

47 Chapter 3 - Slide 47 of 65 Heat is measured in joules or calories. 4.184 Joules (J) = 1 calorie (cal) Exact by definition 1 kJ = 1000 J 1 kilocalorie (kcal) = 1000 calories (cal) Units for Measuring Energy or Heat

48 Chapter 3 - Slide 48 of 65 Examples of Energy In Joules Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

49 Chapter 3 - Slide 49 of 65 Learning Check How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized? Solution →

50 Chapter 3 - Slide 50 of 65 Solution How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized? Given: 150 JNeed: calories Plan: J cal Equality: 1 cal = 4.184 J Set Up:150 J x 1 cal = 36 cal 4.184 J

51 Chapter 3 - Slide 51 of 65 Specific heat Is different for different substances. Is the amount of heat (q) 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. Specific Heat

52 Chapter 3 - Slide 52 of 65 Examples of Specific Heats Table 3.7 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Note: Small values indicate good conductor of heat.

53 Chapter 3 - Slide 53 of 65 Solution What is the specific heat of a metal if 24.8 g absorbs 275 J of energy and the temperature rises from 20.2  C to 24.5  C? Given: 24.8 g, 275 J, 20.2  C to 24.5  C Need: J/g  C Plan: SH = Heat/gΔ  C where Δ o C = (T f – T i ) ΔT = 24.5  C – 20.2  C = 4.3  C SH Equation: SH = heat (q) (mass)(  T) Set Up: 275 J = 2.6 J/g  C (24.8 g)(4.3  C)

54 Chapter 3 - Slide 54 of 65 Rearranging the specific heat expression gives the heat equation. Heat(q) = g x Δ°C x J = J g°C The amount of heat lost or gained by a substance is calculated from the Mass of substance (g). Temperature change (  T or Δ°C ). Specific heat of the substance (J/g°C). Heat Equation

55 Chapter 3 - Slide 55 of 65 A layer of copper on a pan has a mass of 135 g. How much heat in joules will raise the temperature of the copper from 26°C to 328°C if the specific heat of copper is 0.385 J/g°C? The temperature change is 328°C - 26°C = 302°C. heat (J) = g x  T x SH(Cu) 135 g x 302°C x 0.385 J g °C = 15 700 J or 1.57 x 10 4 J Using Specific Heat

56 Chapter 3 - Slide 56 of 65 How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C? 77.5°C – 15.5°C = 62.0°C heat = g x  T x SH 325 g x 62.0°C x 4.184 J x 1 kJ g °C 1000 J = 84.3 kJ Solution

57 Chapter 3 - Slide 57 of 65 Calculating Mass Aluminum is used to make kitchen utensils. What is the mass of an aluminum spatula if 3.25 kJ of heat raise its temperature from 20.0°C to 45.0°C. SH Al = 0. 897 J/g°C? Given: 3.25 kJ (3250 J), 20.0°C to 45.0°C ΔT = 25.0°C Plan: Solve heat equation for mass in grams g = heat ΔT x SH Set Up: 3250 J 25.0°C x 0.897 J/g o C = 145 g Al

58 Chapter 3 - Slide 58 of 65 Transferring Heat Energy Heat energy Flows from a warmer object to a colder object. Provides kinetic energy for the colder object. Energy lost by the warmer object is equal to the heat energy gained by the colder object.

59 Chapter 3 - Slide 59 of 65 Calorimeters and Heat Transfer A calorimeter Is used to measure heat transfer. Can be made with a coffee cup, water, and a thermometer. Indicates the heat lost by a sample and gained by water. Heat lost (-q) = Heat (q) gained Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

60 Chapter 3 - Slide 60 of 65 Measuring Heat Changes A 50.0-g sample of tin is heated to 99.8°C and dropped into 50.0 g water at 15.6°C. If the final temperature is 19.8°C, what is the specific heat of tin? Heat gain (q) by water = 50.0 g x 4.2°C x 4.184 J/g °C = 880 J Heat loss (-q) by tin = -880 J SH tin = -880 J = 0.22 J/g°C (50.0 g)(-80.0°C)

61 Chapter 3 - Slide 61 of 65 Energy and Nutrition On food labels, energy is shown as the nutritional Calorie, written with a capital C. In countries other than the U.S., energy is shown in kilojoules (kJ). 1 Cal =1000 cal 1 Cal = 1 kcal 1 Cal = 4184 J 1 Cal = 4.184 kJ

62 Chapter 3 - Slide 62 of 65 Caloric Food Values The caloric or energy values for 1 g of a food is given in kJ or kcal (Cal) Table 3.8 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

63 Chapter 3 - Slide 63 of 65 Energy Values for Some Foods Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 3.9

64 Chapter 3 - Slide 64 of 65 Energy Requirements The amount of energy needed each day depends on Age Gender Physical activity Table 3.11 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

65 Chapter 3 - Slide 65 of 65 A cup of whole milk contains 12 g carbohydrate, 9.0 g fat, and 9.0 g protein. How many kcal (Cal) does a cup of milk contain? 1) 50 kcal (50 Cal) 2) 80 kcal (80 Cal) 3) 170 kcal (170 Cal) Learning Check Solution →


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