2 Energy Energy Makes objects move. Makes things stop. Is needed to “do work.”
3 Work 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.
4 Potential EnergyPotential energy is energy that is stored for use at a later time. Examples are:Water behind a damA compressed springChemical bonds ingasoline, coal, or food
5 Kinetic Energy Kinetic energy is the energy of motion. Examples are: Hammering a nailWater flowing over a damWorking outBurning gasoline
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 Forms of Energy Energy has many forms: Mechanical Electrical Thermal (heat)ChemicalSolar (light)Nuclear
9 Heat 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.
10 Some Equalities for Heat Heat is measured in calories or joules.1 kilocalorie (kcal) = 1000 calories (cal)1 calorie = Joules (J)1 kJ = 1000 J
12 Reaction ConditionsA 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 ReactionsIn the reaction H2 + I HI, the bonds of H2 and I2 must break, and bonds for HI must form.H2 + I collision bonds break HI
14 Activation EnergyThe 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 ReactionsThe 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) endothermicA. N2 + 3H NH kcalB. CaCO kcal CaO + CO2C. 2SO2 + O2 2SO3 + heat
18 Solution Identify each reaction as 1) exothermic or 2) endothermic 1 A. N2 + 3H NH kcal2 B. CaCO kcal CaO + CO21 C. 2SO2 + O SO3 + heat
19 Specific HeatSpecific heat is the amount of heat (calories or Joules) that raises the temperature of 1 g of a substance by 1°C.
20 Learning Check A. A substance with a large specific heat 1) heats up quickly 2) heats up slowlyB. When ocean water cools, the surrounding air1) cools 2) warms 3) stays the sameC. Sand in the desert is hot in the day and coolat night. Sand must have a1) high specific heat ) low specific heat
21 Solution A. A substance with a large specific heat 2) heats up slowly B. When ocean water cools, the surrounding air2) warmsC. Sand in the desert is hot in the day and coolat night. Sand must have a2) low specific heat
22 Learning CheckWhen 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 be1) 10 °C 2) 14°C 3) 18°C400 g200 g
23 SolutionWhen 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 be2) 14°C400 g200 g
24 Calculation with Specific Heat To calculate the amount of heat lost or gained by a substance, we need thegrams of substance,temperature change T, and thespecific heat of the substance.Heat = g x °C x cal (or J) = cal ( or J)g °C
25 Sample Calculation for Heat 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. (H2O)750 g x 28°C x cal g°C= cal
26 Learning CheckHow many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C?1) 1.8 kcal2) 7.2 kcal3) 9.0 kcal
27 SolutionHow many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C?2) 7.2 kcal75°C - 15°C = 60 °C120 g x (60°C) x cal x 1 kcalg °C cal
28 Energy and NutritionOn nutrition and food labels, the nutritional Calorie, written with a capital C, is used.1 Cal is actually 1000 calories.1 Calorie = 1 kcal1 Cal = cal
29 Caloric Food ValuesThe caloric values for foods indicate the number of kcal provided by 1 g of each type of food.
40 Learning Check Identify each as: 1) solid 2) liquid or 3) gas. ___ A. It has a definite volume, but takes theshape 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 aremobile.
41 Solution Identify each as: 1) solid 2) liquid or 3) gas. 2 A. It has a definite volume, but takes theshape of the container.3 B. Its particles are moving rapidly.3 C. It fills the volume of a container.1 D. It has particles in a fixed arrangement.2 E. It has particles close together that aremobile.
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 ) dipole-dipole3) hydrogen bonds 4) dispersionA. NCl3B. H2OC. Br-BrD. KClE. NH3
48 Solution Identify the type of attractive forces for each: 1) ionic 2) dipole-dipole3) hydrogen bonds 4) dispersion2 A. NCl33 B. H2O4 C. Br-Br1 D. KClE. NH3
50 Melting and FreezingA 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.
51 Calculations Using Heat of Fusion 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) is80. cal1 g water
52 Calculation Using Heat of Fusion 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. calg waterProblem: How much heat in calories is needed to melt 15.0 g of water?15.0 g water x cal = cal1 g water
53 Learning CheckA. How many calories are needed to melt 5.0 g of ice of 0°C?1) 80. cal 2) 400 cal 3) 0 calB. How many calories are released when 25 g of water at 0°C freezes?1) 80. cal 2) 0 cal 3) 2000 cal
54 Solution A. How many calories are needed to melt 5.0 g of ice of 0°C? 2) 400 cal 5.0 g x 80. cal1 gB. How many calories are released when 25 g of water at 0°C freezes?3) 2000 cal 25 g x 80. cal
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 Heat of Vaporization 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°CHeat of Vaporization (water) = cal1 g water
57 Learning CheckHow many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C?1) 27 kcal2) 540 kcal3) kcal
58 SolutionHow many kilocalories (kcal) are released when 50.0 g of steam in a volcano condenses at 100°C?1) 27 kcal50.0 g steam x 540 cal x 1 kcal = 27 kcal1 g steam 1000 cal
60 Heating CurveA 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 Learning Check A. A flat line on a heating curve represents 1) a temperature change2) a constant temperature3) a change of stateB. A sloped line on a heating curve represents
62 Solution A. A flat line on a heating curve represents 2) a constant temperature3) a change of stateB. A sloped line on a heating curve represents1) a temperature change
63 Cooling CurveA 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 Learning Check Use the cooling curve for water to answer each. A. Water condenses at a temperature of1) 0°C 2) 50°C 3) 100°CB. At a temperature of 0°C, water1) freezes 2) melts 3) changes to a gasC. At 40 °C, water is a1) solid 2) liquid 3) gasD. When water freezes, heat is1) removed 2) added
65 Solution Use the cooling curve for water to answer each. A. Water condenses at a temperature of3) 100°CB. At a temperature of 0°C, water1) freezesC. At 40 °C, water is a2) liquidD. When water freezes, heat is1) removed
66 Combined Heat Calculations 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°CT = 37.0°C - 0°C = °C0°C S L
67 Combined Heat Calculations (continued) Step 2: Calculate the heat to melt ice (fusion)250 g ice x cal = cal1 g iceStep 3: Calculate the heat to warm the water from0°C to 37.0°C250 g x 37.0°C x cal = calg °CTotal: Step 2 + Step 3 = calFinal answer = cal
68 Learning Check150 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 kcal2) 13 kcal3) 94 kcal
69 Solution 3) 94 kcal Condense: 150 g x 540 cal x 1 kcal = 81 kcal 1 g calCool: 150 g x 85°C x 1 cal x 1 kcal = 13 kcalg °C calTotal: kcal + 13 kcal = 94 kcal
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