Energy. Work. Potential Energy. Kinetic Energy. Learning Check 2.1. Energy. Energy. makes objects move. makes things stop. is needed to do work.

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1 Chapter 2 Energy and Matter Energy 2.1 Energy Energy makes objects move. makes things stop. is needed to do work. 1 2 Work Potential Energy 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. Potential energy is stored energy. Examples are water behind a dam. a compressed spring. chemical bonds in gasoline, coal, or food. 3 4 Kinetic Energy Kinetic energy is the energy of motion. Examples are swimming. water flowing over a dam. working out. burning gasoline. Identify the energy as potential or kinetic. A. Rollerblading B. a peanut butter and jelly sandwich C. mowing the lawn D. gasoline in the gas tank 5 6

2 Units for Measuring Energy or Heat 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) Heat is measured in joules or calories Joules (J) = 1 calorie (cal) 1 kj = 1000 J 1 kilocalorie (kcal) = 1000 calories (cal) 7 8 Examples of Energy In Joules How many cal are obtained from a pat of butter if it provides 150 J of energy when metabolized? 1) 0.36 cal 2) 36 cal 3) 630 cal 9 10 How many cal are obtained from a pat of butter if it provides 150 J of energy when metabolized? Answer 2) : Given: 150 J Need: calories Plan: J -> cal Equality: 1 cal = J Setup: 150 J x 1 cal = 36 cal J Chapter 2 Energy and Matter 2.2 Energy and Nutrition 11 12

3 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. 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 calories 1 Cal = 1 kcal 1 Cal = 1000 cal 1 Cal = 4184 J 1 Cal = kj Caloric Food Values Energy Values for Some Foods The caloric or energy values for foods indicate the number of kcal (Cal) provided by 1 g of each type of food. Carbohydrate: Fat (lipid): Protein: 4 kcal 1 g 9 kcal 1 g 4 kcal 1 g Energy Requirements The amount of energy needed each day depends on age, sex, and physical activity. 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) 17 18

4 Chapter 2 Energy and Matter 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) 2.3 Temperature Conversions 12 g carbohydrates x 4 kcal/g = 50 kcal 9 g fat x 9 kcal/g = 80 kcal 5 g protein x 4 kcal/g = 20 kcal 150 kcal Temperature Temperature Scales 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. Temperature Scales are Fahrenheit, Celsius, and Kelvin. have reference points for the boiling and freezing points of water. Copyright 2009 by Pearson Education, Inc A. What is the temperature of freezing water? 1) 0 F 2) 0 C 3) 0 K A. What is the temperature of freezing water? 2) 0 C B. What is the temperature of boiling water? 1) 100 F 2) 32 F 3) 373 K 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 2) 180 3) 273 C. How many Celsius units are between the boiling and freezing points of water? 1)

5 Fahrenheit Formula Celsius Formula 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, adding 32 adjusts the zero point of water from 0 C to 32 F. T F = 9/5 T C + 32 or T F = 1.8 T C + 32 T C is obtained by rearranging the equation for T F. T F = 1.8T C + 32 Subtract 32 from both sides. T F - 32 = 1.8 T C ( ) T F - 32 = 1.8 T C Divide by 1.8 = F - 32 = 1.8 T C T F - 32 = T C Solving A Temperature Problem A person with hypothermia has a body temperature of 34.8 C. What is that temperature in F? T F = 1.8 T C + 32 T F = 1.8 (34.8 C) + 32 exact 3 SFs exact The normal body temperature of a chickadee is F. What is that temperature on the Celsius scale? 1) 73.8 C 2) 58.8 C 3) 41.0 C = (addition) = 94.6 F tenth s ) 41.0 C T C = (T F 32 ) 1.8 = ( exact) 1.8 = 73.8 F (3 SFs) = 41.0 C 1.8 (exact) (division, 3SFs) A pepperoni pizza is baked at 455 F. What temperature is needed on the Celsius scale? 1) 423 C 2) 235 C 3) 221 C 29 30

6 A pepperoni pizza is baked at 455 F. What temperature is needed on the Celsius scale? 2) 235 C T F - 32 = T C 1.8 On a cold winter day, the temperature is 15 C. What is that temperature in F? 1) 19 F 2) 59 F 3) 5 F ( ) = 235 C Study Tip: Temperature Math Answer 3) 5 F : T F = 1.8T C + 32 T F = 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 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 an addition rule. multiplication step 1.8 ( 15 C) = 27 F (2 SF) addition step 27 F (ones place) + 32 (exact) = 5 F (ones place) Kelvin Temperature Scale Temperatures 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. T K = T C contains the lowest possible temperature, absolute zero (0 K). 0 K = 273 C 35 36

7 What is normal body temperature of 37 C in kelvins? 1) 236 K 2) 310 K 3) 342 K What is normal body temperature of 37 C in kelvins? 2) 310 K T K = T C = 37 C = 310. K (to ones place) Chapter 2 Energy and Matter 2.4 Specific Heat Specific Heat 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 Examples of Specific Heats 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 cool at night. Sand must have a 1) high specific heat. 2) low specific heat

8 A. When ocean water cools, the surrounding air 2) warms. B. Sand in the desert is hot in the day, and cool at night. Sand must have a 2) low specific heat. 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? Heat Equation 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 ΔT = 24.5 C 20.2 C = 4.3 C Setup: 275 J = 2.6 J/g C (24.8 g)(4.3 C) The amount of heat lost or gained by a substance is calculated 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 = J g C Study Tip: Using Specific Heat 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 kj 2) 77.7 kj 3) 84.3 kj 47 48

9 Coral Bleaching Answer: 3) 84.3 kj 77.0 C 15.0 C = 62.0 C 325 g x 62.0 C x J x 1 kj g C 1000 J = 84.3 kj 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 How many kcal are absorbed by ocean water if 3 x 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 kcal 2) 3 x kcal 3) 3 x kcal ml 3 10 L 1 L 21 = 3 10 g of sea seawater water 1 C g seawater 1 g 1 ml 1 cal 1 kcal g C 1000 cal 18 = 3 10 kcal of heat absorbed (Answer 2) Chapter 2 Energy and Matter 2.5 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

10 Liquids Gases 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 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 Three States of Matter for Water Summary of the States of Matter 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. 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

11 Chapter 2 Energy and Matter Melting and Freezing 2.6 Changes of State A substance is melting while it changes from a solid to a liquid. is freezing while it changes from a liquid to a solid. such as water has a freezing (melting) point of 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). is the amount of heat needed to melt 1 gram of a solid (at its melting point). for water (at 0 C) is 80. cal 1 g water Calculation Using Heat of Fusion The heat needed to freeze (or melt) a specific mass of water (or ice) is calculated using the heat of fusion. Heat = g water (ice) x 80. cal 1 g water (ice) Example: How much heat in cal is needed to melt 15. g of ice? 15. g ice x 80. cal = 1200 cal 1 g ice F Study Tip: Using Heat of Fusion A. How many calories are needed to melt 5.0 g of ice at 0 C? 1) 80. cal 2) 4.0 x 10 2 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) 2.0 x 10 3 cal 65 66

12 Sublimation A. How many calories are needed to melt 5.00 g of ice at 0 C? 2) 4.0 x 10 2 cal 5.0 g x 80. cal 1 g B. How many calories are released when 25.0 g of water at 0 C freezes? 3) 2.0 x 10 3 cal 25 g x 80. cal 1 g Sublimation occurs when particles change directly from solid to a gas. is typical of dry ice, which sublimes at -78 C. takes place in frost-free refrigerators. is used to prepare freezedried foods for long-term storage Evaporation and Condensation Boiling Water evaporates when molecules on the surface gain sufficient energy to form a gas. condenses when gas molecules lose energy and form a liquid. At boiling, all the water molecules acquire enough energy to form a gas. bubbles appear throughout the liquid Heat of Vaporization The heat of vaporization is the amount of heat absorbed to vaporize 1 g of a liquid to gas at the boiling point. released when 1 g of a gas condenses to liquid at the boiling point. Boiling Point of Water = 100 C How many kilocalories (kcal) are released when 50.0 g of steam from a volcano condenses at 100 C? 1) 27 kcal 2) 540 kcal 3) 2700 kcal Heat of Vaporization (water) = 540 cal 1 g water 71 72

13 Summary of Changes of State 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 Curve A heating curve illustrates the changes of state as a solid is heated. uses sloped lines to show an increase in temperature. uses plateaus (flat lines) to indicate a change of state. 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 Cooling Curve 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. A cooling curve illustrates the changes of state as a gas is cooled. uses sloped lines to indicate a decrease in temperature. uses plateaus (flat lines) to indicate a change of state

14 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, liquid 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. 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, liquid 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 Δ T = 37.0 C - 0 C = 37.0 C 0 C solid liquid melting temperature increase 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 = cal g C Total: Step 2 + Step 3 = cal (rounded to 2 SF) 81 82

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