# Energy and States of Matter

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Energy and States of Matter

Energy Energy Makes objects move. Makes things stop.
Is needed to “do work.”

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.

Potential Energy 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

Kinetic Energy Kinetic energy is the energy of motion. Examples are:
Hammering a nail Water flowing over a dam Working out Burning gasoline

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.

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)

Forms of Energy Energy has many forms: Mechanical Electrical
Thermal (heat) Chemical Solar (light) Nuclear

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.

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

Energy in Chemical Reactions

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.

Chemical Reactions In 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

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.

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.

Endothermic Reactions
In an endothermic reaction, heat is absorbed because the energy of the products is greater that that of the reactants.

Learning Check Identify each reaction as
1) exothermic or 2) endothermic A. N2 + 3H NH kcal B. CaCO kcal CaO + CO2 C. 2SO2 + O2 2SO3 + heat

Solution Identify each reaction as 1) exothermic or 2) endothermic
1 A. N2 + 3H NH kcal 2 B. CaCO kcal CaO + CO2 1 C. 2SO2 + O SO3 + heat

Specific Heat Specific heat is the amount of heat (calories or Joules) that raises the temperature of 1 g of a substance by 1°C.

Learning Check A. A substance with a large specific heat
1) heats up quickly 2) heats up slowly B. When ocean water cools, the surrounding air 1) cools 2) warms 3) 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 ) low specific heat

Solution 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

Learning Check 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 °C 2) 14°C 3) 18°C 400 g 200 g

Solution 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 400 g 200 g

Calculation with Specific Heat
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

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

Learning Check 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

Solution 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 cal x 1 kcal g °C cal

Energy and Nutrition 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 = cal

Caloric Food Values The caloric values for foods indicate the number of kcal provided by 1 g of each type of food.

Calories in Some Foods

Energy Requirements The amount of energy needed each day depends on age, sex, and physical activity.

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.

Learning Check 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

Solution 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 = kcal

States of Matter

Solids Solids have A definite shape. A definite volume.
Particles that are close together in a fixed arrangement. Particles that move very slowly.

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.

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 fast.

Summary of the States of Matter

Learning Check 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.

Solution Identify each as: 1) solid 2) liquid or 3) gas.
2 A. It has a definite volume, but takes the shape 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 are mobile.

Attractive Forces between Particles
In ionic compounds, ionic bonds are strong attractive forces that hold positive and negative ions together.

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.

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.

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.

Melting Points and Attractive Forces of Some Typical Substances

Learning Check Identify the type of attractive forces for each:
1) ionic ) dipole-dipole 3) hydrogen bonds 4) dispersion A. NCl3 B. H2O C. Br-Br D. KCl E. NH3

Solution Identify the type of attractive forces for each:
1) ionic 2) dipole-dipole 3) hydrogen bonds 4) dispersion 2 A. NCl3 3 B. H2O 4 C. Br-Br 1 D. KCl E. NH3

Changes of State

Melting and Freezing 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.

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) is 80. cal 1 g water

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. cal g water Problem: How much heat in calories is needed to melt 15.0 g of water? 15.0 g water x cal = cal 1 g water

Learning Check A. How many calories are needed to melt 5.0 g of ice of 0°C? 1) 80. cal 2) 400 cal 3) 0 cal B. How many calories are released when 25 g of water at 0°C freezes? 1) 80. cal 2) 0 cal 3) 2000 cal

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. 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

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.

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°C Heat of Vaporization (water) = cal 1 g water

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

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 steam 1000 cal

Heating and Cooling Curves

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.

Learning Check 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

Solution 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

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.

Learning Check Use the cooling curve for water to answer each.
A. Water condenses at a temperature of 1) 0°C 2) 50°C 3) 100°C B. At a temperature of 0°C, water 1) freezes 2) melts 3) changes to a gas C. At 40 °C, water is a 1) solid 2) liquid 3) gas D. When water freezes, heat is 1) removed 2) added

Solution 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

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°C T = 37.0°C - 0°C = °C 0°C S L

Combined Heat Calculations (continued)
Step 2: Calculate the heat to melt ice (fusion) 250 g ice x 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 cal = cal g °C Total: Step 2 + Step 3 = cal Final answer = cal

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

Solution 3) 94 kcal Condense: 150 g x 540 cal x 1 kcal = 81 kcal
1 g cal Cool: 150 g x 85°C x 1 cal x 1 kcal = 13 kcal g °C cal Total: kcal + 13 kcal = 94 kcal

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