2 Energy AllSlides

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Chapter 2 Energy and Matter

2.1Energy

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Energy

• makes objects move.

• makes things stop.

• is needed to “do work.”

Energy

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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 © 2009 by Pearson Education, Inc.

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Potential Energy

Potential energy is• stored energy.

Examples are

• water behind a dam.• a compressed spring.• chemical bonds in

gasoline, coal, or food. Copyright © 2009 by Pearson Education, Inc.

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Kinetic Energy

Kinetic energy is the • energy of motion.

Examples are

• swimming.• water flowing over a dam.• working out.• burning gasoline. Copyright © 2009 by Pearson Education, Inc.

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Learning Check

Identify the energy as potential or kinetic.

A. RollerbladingB. a peanut butter and jelly sandwichC. mowing the lawnD. gasoline in the gas tank

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Solution

Identify the energy as potential or kinetic.

A. Rollerblading (kinetic)B. a peanut butter and jelly sandwich (potential)C. mowing the lawn (kinetic)D. gasoline in the gas tank (potential)

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Heat is measured in joules or calories.

4.184 Joules (J) = 1 calorie (cal)

1 kJ = 1000 J

1 kilocalorie (kcal) = 1000 calories (cal)

Units for Measuring Energy or Heat

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Examples of Energy In Joules

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Learning Check

How many cal are obtained from a pat of butterif it provides 150 J of energy when metabolized?

1) 0.36 cal

2) 36 cal

3) 630 cal

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Solution

How many cal are obtained from a pat of butterif it provides 150 J of energy when metabolized?

Answer 2)Solution:Given: 150 JNeed: caloriesPlan: J -> calEquality: 1 cal = 4.184 JSetup: 150 J x 1 cal = 36 cal

4.184 J

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2.2 Energy and

Nutrition


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Calorimeters

A calorimeter

• is used to measure heat transfer.

• can be made with a coffee cup and a thermometer.

• indicates the heat lost by a sample

• indicates the heat gained by water


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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 calories1 Cal = 1 kcal1 Cal = 1000 cal1 Cal = 4184 J

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Caloric Food Values

The caloric or energy values for foods indicate the

number of kcal (Cal) provided by 1 g of each type of food.

Carbohydrate: 4 kcal 1 g

Fat (lipid): 9 kcal 1 g

Protein: 4 kcal 1 g

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Energy Values for 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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A cup of whole milk contains 12 g of carbohydrate, 9 g of fat, and 5 g of protein. How many kcal (Cal) does a cup of milk contain (round answer to the tens place)?

1) 50 kcal (or Cal)2) 80 kcal (or Cal)3) 150 kcal (or Cal)

Learning Check

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A cup of whole milk contains 12 g of carbohydrate, 9 g of fat, and 5 g of protein. How many kcal (Cal) does a cup of milk contain?

3) 150 kcal (or Cal)

12 g carbohydrates x 4 kcal/g = 50 kcal9 g fat x 9 kcal/g = 80 kcal5 g protein x 4 kcal/g = 20 kcal

150 kcal

Solution

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Chapter 2 Energy and Matter2.3

Temperature Conversions


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Temperature

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.


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Temperature Scales

Temperature Scales

• are Fahrenheit, Celsius, and Kelvin.

• have reference points for the boiling and freezing points of water.


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A. What is the temperature of freezing water? 1) 0 °F 2) 0 °C 3) 0 K

B. What is the temperature of boiling water? 1) 100 °F 2) 32 °F 3) 373 K

C. How many Celsius units are between the boiling and freezing points of water?1) 100 2) 180 3) 273

Learning Check

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A. What is the temperature of freezing water? 2) 0 °C

B. What is the temperature of boiling water? 3) 373 K

C. How many Celsius units are between the boiling and freezing points of water?1) 100

Solution

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• On the Fahrenheit scale, there are 180 °F between the freezing and boiling points; 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 the Fahrenheit temperature, adding32 ° adjusts the zero point of water from 0 °C to 32 °F.

TF = 9/5 TC + 32 °or

TF = 1.8 TC + 32 °

Fahrenheit Formula

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• TC is obtained by rearranging the equation for TF.TF = 1.8TC + 32 °

• Subtract 32 ° from both sides.TF - 32 ° = 1.8 TC ( + 32 ° – 32 °)TF - 32 ° = 1.8 TC

• Divide by 1.8 = °F - 32 ° = 1.8 TC1.8 1.8

TF - 32 ° = TC1.8

Celsius Formula

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Solving A Temperature Problem

A person with hypothermia has abody temperature of 34.8 °C. What is that temperature in °F?

TF = 1.8 TC + 32 °

TF = 1.8 (34.8 °C) + 32 °exact 3 SFs exact

= 62.6 + 32 ° (addition)= 94.6 °F

tenth’s


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The normal body temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale?

1) 73.8 °C 2) 58.8 °C3) 41.0 °C

Learning Check

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3) 41.0 °C

TC = (TF – 32 °)1.8

= (105.8 ° – 32 ° exact)1.8

= 73.8 °F (3 SFs) = 41.0 °C1.8 (exact) (division, 3SFs)

Solution

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A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale?

1) 423 °C2) 235 °C3) 221 °C

Learning Check

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A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale?

2) 235 °C

TF - 32 ° = TC1.8

(455 °– 32 °) = 235 °C1.8

Solution

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On a cold winter day, the temperature is –15 °C.What is that temperature in °F?

1) 19 °F2) 59 °F3) 5 °F

Learning Check

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Answer 3) 5 °FSolution:

TF = 1.8TC + 32 °TF = 1.8 (–15 °C) + 32 °

= – 27 °F + 32 °= 5 °F

Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign. 1.8 x 15 +/ – = –27

Solution

Study Tip: Temperature Math

The temperature equation involves the exact numbers 1.8 and 32. Only the temperature is measured. To convert °C to °F, a multiplication rule is followed by anaddition rule.multiplication step

1.8 (–15 °C) = – 27 °F (2 SF)addition step

– 27 °F (ones place)+ 32 ° (exact)

= 5 °F (ones place)

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The Kelvin temperature scale• has 100 units between the freezing and boiling points

of water. 100 K = 100 °C or 1 K = 1 °C

• is obtained by adding 273 to the Celsius temperature.TK = TC + 273

• contains the lowest possible temperature, absolute zero (0 K).

0 K = –273 °C

Kelvin Temperature Scale

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Temperatures

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What is normal body temperature of 37 °C in kelvins?

1) 236 K2) 310 K 3) 342 K

Learning Check

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What is normal body temperature of 37 °C in kelvins?

2) 310 K

TK = TC + 273= 37 °C + 273= 310. K (to ones place)

Solution

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2.4 Specific Heat


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Specific heat

• is different for different substances.• is the amount of heat 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

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Examples of Specific Heats

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A. When ocean water cools, the surrounding air 1) cools. 2) warms. 3) stays the same.

B. Sand in the desert is hot in the day, and coolat night. Sand must have a1) high specific heat. 2) low specific heat.

Learning Check

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A. When ocean water cools, the surrounding air 2) warms.

B. Sand in the desert is hot in the day, and coolat night. Sand must have a2) low specific heat.

Solution

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Learning Check

What is the specific heat if 24.8 g of a metal

absorbs 275 J of energy and the temperature rises

from 20.2 °C to 24.5 °C?

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Solution

Given: 24.8 g metal, 275 J of energy, 20.2 °C to 24.5 °C

Need: Specific heat J/g °C

Plan: specific heat (SH) = Heat (J)g °C

∆T = 24.5 �C – 20.2 °C = 4.3 °C

Setup: 275 J = 2.6 J/g °C(24.8 g)(4.3 °C)

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Heat Equation

The amount of heat lost or gained by a substance iscalculated from the• mass of substance (g).• temperature change (°T).• specific heat of the substance (J/g °C).

This is expressed as the heat equation.

Heat = g x °C x J = Jg °C

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How many kJ are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C?

1) 20.4 kJ2) 77.7 kJ3) 84.3 kJ

Learning Check

Study Tip: Using Specific Heat

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Solution

Answer: 3) 84.3 kJ

77.0 °C – 15.0 °C = 62.0 °C325 g x 62.0 °C x 4.184 J x 1 kJ

g °C 1000 J= 84.3 kJ

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Coral• contains algae that produce sugars (food) and the

bright red and orange pigments of coral. • expels the algae when water temperatures increase

as little as 1 °C.• bleaches as it loses its food supply and color. • dies if the stress of higher temperatures continues. • reefs in Australia and the Indian Ocean have been

badly damaged by increases in ocean temperatures.

Coral Bleaching

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How many kcal are absorbed by ocean water if 3 x 1018 L of water in the Caribbean has an increase of 1 °C. Assume the specific heat of ocean water is the same as water. Assume the density of ocean water is 1.0 g/mL.1) 3 x 1015 kcal2) 3 x 1018 kcal3) 3 x 1021 kcal

Learning Check

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Solution

183 10 L×1000 mL ×

1 L1 g

1 mL×

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= 3 10 g of sea water

3 10 g

×

× 1 °C×1 cal ×g °C

1 kcal 1000 cal

×

18= 3 10 kcal of heat absorbed (Answer 2)×

seawaterseawater

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2.5 States of Matter


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Solids

Solids have • a definite shape.• a definite volume. • particles that are close

together in a fixed arrangement.

• particles that move very slowly.

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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.

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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 very fast.

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Three States of Matter for Water


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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 ofthe 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 ofthe 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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