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Energy and States of Matter

Energy

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Energy

Makes objects move.

Makes things stop.

Is needed to “do work.”

Energy

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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

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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

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Kinetic energy is the energy of motion. Examples are:

Hammering a nail Water flowing over a dam Working out Burning gasoline

Kinetic Energy

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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.

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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)

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

Forms of Energy

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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

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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

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Energy in Chemical Reactions

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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.

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Chemical Reactions In the reaction H2 + I2 2 HI, the bonds of H2

and I2 must break, and bonds for HI must form.

H2 + I2 collision bonds break 2HI

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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.

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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.

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Endothermic Reactions

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

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Learning Check

Identify each reaction as

1) exothermic or 2) endothermic

A. N2 + 3H2 2NH3 + 22 kcal

B. CaCO3 + 133 kcal CaO + CO2

C. 2SO2 + O2 2SO3 + heat

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Solution

Identify each reaction as

1) exothermic or 2) endothermic

1 A. N2 + 3H2 2NH3 + 22 kcal

2 B. CaCO3 + 133 kcal CaO + CO2

1 C. 2SO2 + O2 2SO3 + heat

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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

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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 2) low specific heat

Learning Check

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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

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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 be1) 10 °C 2) 14°C 3) 18°C

200 g400 g

Learning Check

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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 be2) 14°C

200 g400 g

Solution

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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

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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 1.00 cal g°C

= 21 000 cal

Sample Calculation for Heat

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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

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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

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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

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

Caloric Food Values

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Calories in Some Foods

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Energy Requirements The amount of energy needed each day

depends on age, sex, and physical activity.

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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.

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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

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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

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States of Matter

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Solids have

A definite shape.

A definite volume. Particles that are close together in a fixed

arrangement.

Particles that move very slowly.

Solids

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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

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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

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Summary of the States of Matter

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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

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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.

Solution

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Attractive Forces between Particles

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

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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.

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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.

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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.

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Melting Points and Attractive Forces of Some Typical Substances

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Learning Check

Identify the type of attractive forces for each:1) ionic 2) dipole-dipole 3) hydrogen bonds 4) dispersion A. NCl3

B. H2O

C. Br-BrD. KClE. NH3

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Solution

Identify the type of attractive forces for each:1) ionic 2) dipole-dipole3) hydrogen bonds 4) dispersion 2 A. NCl3

3 B. H2O

4 C. Br-Br 1 D. KCl 3 E. NH3

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Changes of State

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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

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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

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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

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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

Learning Check

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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 1 g

Solution

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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.

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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) = 540 cal

1 g water

Heat of Vaporization

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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) 27 000 kcal

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Solution

How 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 kcal

1 g steam 1000 cal

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Heating and Cooling Curves

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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.

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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

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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

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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.

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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

Learning Check

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Use the cooling curve for water to answer each.A. Water condenses at a temperature of

3) 100°CB. At a temperature of 0°C, water

1) freezesC. At 40 °C, water is a

2) liquidD. When water freezes, heat is

1) removed

Solution

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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°C0°C S L

Combined Heat Calculations

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Combined Heat Calculations (continued)

Step 2: Calculate the heat to melt ice (fusion)

250 g ice x 80. cal = 20 000 cal 1 g iceStep 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 = 29 250 cal

Final answer = 29 000 cal

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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

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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