Presentation on theme: "1 Energy and States of Matter Energy. 2 Makes objects move. Makes things stop. Is needed to do work. Energy."— Presentation transcript:
1 Energy and States of Matter Energy
2 Makes objects move. Makes things stop. Is needed to do work. Energy
3 Work is done when You go up stairs. You play soccer. You lift a bag of groceries. You ride a bicycle. You breathe. Your heart pumps blood. Water goes over a dam. Work
4 Potential energy is energy that is stored for use at a later time. Examples are: Water behind a dam A compressed spring Chemical bonds in gasoline, coal, or food Potential Energy
5 Kinetic energy is the energy of motion. Examples are: Hammering a nail Water flowing over a dam Working out Burning gasoline Kinetic Energy
6 Learning Check Identify the energy as 1) potential or 2) kinetic A. Roller blading. B. A peanut butter and jelly sandwich. C. Mowing the lawn. D. Gasoline in the gas tank.
7 Solution Identify the energy as 1) potential or 2) kinetic A. Roller blading. (2 kinetic) B. A peanut butter and jelly sandwich. (1 potential) C. Mowing the lawn. (2 kinetic) D. Gasoline in the gas tank. (1 potential)
8 Energy has many forms: Mechanical Electrical Thermal (heat) Chemical Solar (light) Nuclear Forms of Energy
9 Heat energy flows from a warmer object to a colder object. The colder object gains kinetic energy when it is heated. During heat flow, the loss of heat by a warmer object is equal to the heat gained by the colder object. Heat
10 Heat is measured in calories or joules. 1 kilocalorie (kcal) = 1000 calories (cal) 1 calorie = 4.18 Joules (J) 1 kJ = 1000 J Some Equalities for Heat
11 Energy in Chemical Reactions
12 Reaction Conditions A chemical reaction occurs when the reacting molecules collide. Collisions between molecules must have sufficient energy to break the bonds in the reactants. Once the bonds between atoms of the reactants are broken, new bonds can form to give the product.
13 Chemical Reactions In the reaction H 2 + I 2 2 HI, the bonds of H 2 and I 2 must break, and bonds for HI must form. H 2 + I 2 collision bonds break 2HI
14 Activation Energy The activation energy is the minimum energy needed for a reaction to take place. When a collision has the energy that is equal to or greater than the activation energy, reaction can occur.
15 Exothermic Reactions The heat of reaction is the difference in the energy of the reactants and the products. An exothermic reaction releases heat because the energy of the products is less that the reactants.
16 Endothermic Reactions In an endothermic reaction, heat is absorbed because the energy of the products is greater that that of the reactants.
17 Learning Check Identify each reaction as 1) exothermic or 2) endothermic A. N 2 + 3H 2 2NH kcal B. CaCO kcalCaO + CO 2 C. 2SO 2 + O 2 2SO 3 + heat
18 Solution Identify each reaction as 1) exothermic or 2) endothermic 1 A. N 2 + 3H 2 2NH kcal 2 B. CaCO kcal CaO + CO 2 1 C. 2SO 2 + O 2 2SO 3 + heat
19 Specific heat is the amount of heat (calories or Joules) that raises the temperature of 1 g of a substance by 1°C. Specific Heat
20 A. A substance with a large specific heat 1) heats up quickly2) heats up slowly B. When ocean water cools, the surrounding air 1) cools 2) warms3) stays the same C. Sand in the desert is hot in the day and cool at night. Sand must have a 1) high specific heat 2) low specific heat Learning Check
21 A. A substance with a large specific heat 2) heats up slowly B. When ocean water cools, the surrounding air 2) warms C. Sand in the desert is hot in the day and cool at night. Sand must have a 2) low specific heat Solution
22 When 200 g of water are heated, the water temperature rises from 10°C to 18°C. If 400 g of water at 10°C are heated with the same amount of heat, the final temperature would be 1) 10 °C2) 14°C 3) 18°C 200 g 400 g Learning Check
23 When 200 g of water are heated, the water temperature rises from 10°C to 18°C. If 400 g of water at 10°C are heated with the same amount of heat, the final temperature would be 2) 14°C 200 g 400 g Solution
24 To calculate the amount of heat lost or gained by a substance, we need the grams of substance, temperature change T, and the specific heat of the substance. Heat = g x °C x cal (or J) = cal ( or J) g °C Calculation with Specific Heat
25 A hot-water bottle contains 750 g of water at 65°C. If the water cools to body temperature (37°C), how many calories of heat could be transferred to sore muscles? The temperature change is 65°C - 37°C = 28°C. heat (cal) = g x T x Sp. Ht. (H 2 O) 750 g x 28°C x 1.00 cal g°C = cal Sample Calculation for Heat
26 How many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C? 1) 1.8 kcal 2) 7.2 kcal 3) 9.0 kcal Learning Check
27 How many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C? 2) 7.2 kcal 75°C - 15°C = 60 °C 120 g x (60°C) x 1.00 cal x 1 kcal g °C 1000 cal Solution
28 On nutrition and food labels, the nutritional Calorie, written with a capital C, is used. 1 Cal is actually 1000 calories. 1 Calorie = 1 kcal 1 Cal = 1000 cal Energy and Nutrition
29 The caloric values for foods indicate the number of kcal provided by 1 g of each type of food. Caloric Food Values
30 Calories in Some Foods
31 Energy Requirements The amount of energy needed each day depends on age, sex, and physical activity.
32 Loss and Gain of Weight If food intake exceeds energy use, a person gains weight. If food intake is less than energy use, a person loses weight.
33 A cup of whole milk contains 12 g of carbohydrates, 9.0 g of fat, and 9.0 g of protein. How many kcal (Cal) does a cup of milk contain? 1) 48 kcal 2) 81 kcal 3) 165 kcal Learning Check
34 3) 165 kcal 12 g carb x 4 kcal/g = 48 kcal 9.0 g fat x 9 kcal/g=81 kcal 9.0 g protein x 4 kcal/g=36 kcal Total kcal= 165 kcal Solution
35 States of Matter
36 Solids have A definite shape. A definite volume. Particles that are close together in a fixed arrangement. Particles that move very slowly. Solids
37 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. Liquids
38 Gases have An indefinite shape. An indefinite volume. The same shape and volume as their container. Particles that are far apart. Particles that move fast. Gases
39 Summary of the States of Matter
40 Identify each as: 1) solid 2) liquid or 3) gas. ___ A. It has a definite volume, but takes the shape of the container. __ B. Its particles are moving rapidly. __C. It fills the volume of a container. __D. It has particles in a fixed arrangement. __E. It has particles close together that are mobile. Learning Check
41 Identify each as: 1) solid 2) liquid or 3) gas. 2A. It has a definite volume, but takes the shape of the container. 3B. Its particles are moving rapidly. 3C. It fills the volume of a container. 1D. It has particles in a fixed arrangement. 2E. It has particles close together that are mobile. Solution
42 Attractive Forces between Particles In ionic compounds, ionic bonds are strong attractive forces that hold positive and negative ions together.
43 Attractive Forces between Particles In covalent compounds, polar molecules exert attractive forces called dipole-dipole attractions. Hydrogen bonds are strong dipole attractions between hydrogen atoms and atoms of F, O, or N, which are very electronegative.
44 Attractive Forces between Particles Nonpolar molecules form liquids or solids through weak attractions called dispersion forces. Dispersion forces are caused by temporary dipoles that develop when electrons are not distributed equally.
45 Melting Points and Attractive Forces Ionic compounds require large amounts of energy to break apart ionic bonds. Thus, they have high melting points. Hydrogen bonds are the strongest type of dipole-dipole attractions. They require more energy to break than other dipole attractions. Dispersion forces are weak interactions and very little energy is needed to change state.
46 Melting Points and Attractive Forces of Some Typical Substances
47 Learning Check Identify the type of attractive forces for each: 1) ionic 2) dipole-dipole 3) hydrogen bonds 4) dispersion A. NCl 3 B. H 2 O C. Br-Br D. KCl E. NH 3
48 Solution Identify the type of attractive forces for each: 1) ionic 2) dipole-dipole 3) hydrogen bonds4) dispersion 2 A. NCl 3 3 B. H 2 O 4 C. Br-Br 1 D. KCl 3 E. NH 3
49 Changes of State
50 A substance is melting while it changes from a solid to a liquid. A substance is freezing while it changes from a liquid to a solid. The freezing (melting) point of water is 0°C. Melting and Freezing
51 The heat of fusion is the amount of heat released when 1 gram of liquid freezes at its freezing point. The heat of fusion is the amount of heat needed to melt 1 gram of a solid at its melting point. For water the heat of fusion (at 0°C) is 80. cal 1 g water Calculations Using Heat of Fusion
52 The heat involved in the freezing (or melting) a specific mass of water (or ice) is calculated using the heat of fusion. Heat = g water x 80. cal g water Problem: How much heat in calories is needed to melt 15.0 g of water? 15.0 g water x 80. cal = 1200 cal 1 g water Calculation Using Heat of Fusion
53 A. How many calories are needed to melt 5.0 g of ice of 0°C? 1) 80. cal2) 400 cal 3) 0 cal B. How many calories are released when 25 g of water at 0°C freezes? 1) 80. cal2) 0 cal 3) 2000 cal Learning Check
54 A. How many calories are needed to melt 5.0 g of ice of 0°C? 2) 400 cal5.0 g x 80. cal 1 g B. How many calories are released when 25 g of water at 0°C freezes? 3) 2000 cal 25 g x 80. cal 1 g Solution
55 Boiling & Condensation Water evaporates when molecules on the surface gain enough energy to form a gas. At boiling, all the water molecules acquire enough energy to form a gas.
56 The heat of vaporization Is the amount of heat needed to change 1 g of liquid to gas at the boiling point. Is the amount of heat released when 1 g of a gas changes to liquid at the boiling point. Boiling (Condensing) Point of Water = 100°C Heat of Vaporization (water) = 540 cal 1 g water Heat of Vaporization
57 Learning Check How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C? 1) 27 kcal 2) 540 kcal 3) kcal
58 Solution How many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C? 1) 27 kcal 50.0 g steam x 540 cal x 1 kcal = 27 kcal 1 g steam1000 cal
59 Heating and Cooling Curves
60 Heating Curve A heating curve illustrates the changes of state as a solid is heated. Sloped lines indicate an increase in temperature. Plateaus (flat lines) indicate a change of state.
61 A. A flat line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change 2) a constant temperature 3) a change of state Learning Check
62 A. A flat line on a heating curve represents 2) a constant temperature 3) a change of state B. A sloped line on a heating curve represents 1) a temperature change Solution
63 Cooling Curve A cooling curve illustrates the changes of state as a gas is cooled. Sloped lines indicate a decrease in temperature. This cooling curve for water begins at 140°C and ends at -30°C.
64 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 1) 0°C2) 50°C3) 100°C B. At a temperature of 0°C, water 1) freezes2) melts3) changes to a gas C. At 40 °C, water is a 1) solid 2) liquid3) gas D. When water freezes, heat is 1) removed2) added Learning Check
65 Use the cooling curve for water to answer each. A. Water condenses at a temperature of 3) 100°C B. At a temperature of 0°C, water 1) freezes C. At 40 °C, water is a 2) liquid D. When water freezes, heat is 1) removed Solution
66 To reduce a fever, an infant is packed in 250 g of ice. If the ice at 0°C melts and warms to body temperature (37.0°C), how many calories are removed from the body? Step 1: Diagram the changes. 37°C T = 37.0°C - 0°C = 37.0°C 0°C S L Combined Heat Calculations
67 Combined Heat Calculations (continued) Step 2: Calculate the heat to melt ice (fusion) 250 g ice x 80. cal= cal 1 g ice Step 3: Calculate the heat to warm the water from 0°C to 37.0°C 250 g x 37.0°C x 1.00 cal = 9250 cal g °C Total: Step 2 + Step 3 = cal Final answer= cal
68 Learning Check 150 g of steam at 100°C is released from a boiler. How many kilocalories are lost when the steam condenses and cools to 15°C? 1) 81 kcal 2) 13 kcal 3) 94 kcal
69 Solution 3) 94 kcal Condense: 150 g x 540 cal x 1 kcal = 81 kcal 1 g 1000 cal Cool: 150 g x 85°C x 1 cal x 1 kcal = 13 kcal g °C 1000 cal Total: 81 kcal + 13 kcal = 94 kcal