CH104 Chapter 2: Energy & Matter Heating & Cooling Curves

Name: CH104 Chapter 2: Energy & Matter Heating & Cooling Curves
Uploaded: 2017-07-24T21:58:26+00:00
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Description: CH104 Chapter 2: Energy & Matter Heating & Cooling Curves

CH104 Chapter 2: Energy & Matter Heating & Cooling Curves
Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves

Chapter 2 Energy and Matter
2.2 Temperature

Units of Measurement Metric SI Common Conversions Length Volume Mass
Temperature meter (m) m = 1.09 yd liter (L) L = 1.06 qt gram (g) kg = 2.2 lb Celsius (oC) C = (F-32)/1.8 Kelvin (K) K = C + 273

Temperature 180 o 100 o

Learning Check 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

Solution 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

Trivial Pursuit Question
At what temperature does oF = oC? - 40 oF = - 40 oC - 40 o

Temperature conversion
Common scales used Fahrenheit, Celsius and Kelvin. oF = 1.8 oC + 32 oC = (oF - 32) 1.8 K = oC + 273 SI unit

Solving a Temperature Problem
A person with hypothermia has a body temperature of 34.8 °C. What Is that temperature in °F? F = 1.8(C) ° F = (1.8)(34.8 °C) ° exact tenth’s exact = 62.6 ° ° = 94.6 °F tenth’s

Practice: Temp Conversion
What is 75.0 º F in ºC? ºC = (75.0 º F -32 º) = 23.9 ºC 1.8 What is -12 º F in ºC? º F = 1.8 (-12) + 32 º F = 10 º C

Learning Check The normal 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

Solution The normal temperature of a chickadee is °F. What is that temperature on the Celsius scale? 1) °C 2) °C 3) °C TC = TF – 32 ° 1.8 = (105.8 – 32 °) = 73.8 °F = 41.0 °C 1.8 ° tenth’s place

Learning Check 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

Solution 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 TF – 32 ° = TC 1.8 (455 – 32 °) = 235 °C 1.8 one’s place

Learning Check 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

Solution 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 TF = 1.8TC ° TF = 1.8(–15 °C) + 32 ° = – ° = 5 °F one’s place Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign. 1.8 x 15 +/ – = –27

Temperatures

Learning Check What is normal body temperature of 37 °C in kelvins?
1) 236 K 2) 310 K 3) 342 K

Solution What is normal body temperature of 37 °C in kelvins?
1) 236 K 2) 310 K 3) 342 K TK = TC = 37 °C = 310. K one’s place

Energy Energy = The capacity to cause change Heat Light Wind

Potential Energy = stored Energy
(Has potential for motion) X Kinetic Energy = Energy in motion (Fulfilling its potential)

Learning Check Identify each of the following as potential energy or kinetic energy. A. roller blading B. a peanut butter and jelly sandwich C. mowing the lawn D. gasoline in the gas tank

Solution Identify each of the following as potential energy or kinetic energy. A. roller blading kinetic B. a peanut butter and jelly sandwich potential C. mowing the lawn kinetic D. gasoline in the gas tank potential

Kinetic Energy KE = 1 mv2 2 Which has more E? Truck moving at 5 mph
Bicycle moving at 5 mph

Units of Measurement Metric SI Common Conversions Length Volume Mass
Temperature Energy meter (m) m = 1.09 yd liter (L) L = 1.06 qt gram (g) kg = 2.2 lb Celsius (oC) C = (F-32)/1.8 Kelvin (K) K = C + 273 calorie (cal) 1Kcal = 1000 cal = 1Cal Joule (J) cal = 4.18 J

Examples of Energy Values in Joules

Learning Check How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized?

Solution How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized? Given Need 150 J 1 cal 4.184 J = cal 36 Conversion Factors: 1 cal and J 4.184 J cal

2.3 Specific Heat

Energy Units of Energy = calorie cal kilocalorie kcal
1000 cal = 1 kcal Calorie Cal Cal = kcal joule J 4.18 J = 1 cal British Thermal Unit BTU

Energy calorie: E to raise 1 g H20 by 1 oC Specific Heat H2O 1 cal
1g 1oC 1oC

Examples of Specific Heats

Specific Heat E to raise Temp of 1g substance by 1 oC H2O 1.00 cal
g oC H2O 1.00 Fe 0.11 Cu 0.093 Ag 0.057 Au 0.031 Al 0.22 Sand 0.19 Add 1 cal 0oC = start

Specific Heat E to raise Temp of 1g substance by 1 oC H2O 1.00 30o Au
0.031 cal g oC Ag 0.057 20o Cu 0.093 10o Fe 0.11 Sand 0.19 Al 0.22 1o H2O 1.00 Add 1 cal 0oC = start

Specific Heat H2O 1.00 30o Au 0.031 Low SpHt; Heats quickly Ag 0.057
Cu 0.093 10o Fe 0.11 Sand 0.19 Al 0.22 1o H2O 1.00 High SpHt; Resists change

Resists change in body temp
Specific Heat Dehydrated person Body temp rises quickly Hydrated person Resists change in body temp Heats quickly Gets hot Sand 0.19 H2O 1.00 Resists change Stays cold

Learning Check A. For the same amount of heat added, a substance with a large specific heat 1) has a smaller increase in temperature 2) has a greater increase in temperature 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. For the same amount of heat added, a substance with a large specific heat 1) has a smaller increase in temperature 2) has a greater increase in temperature 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

Specific Heat 24.8g Sample Problem:
What is the specific heat of a metal if 24.8 g absorbs 65.7 cal of energy and the temp rises from 20.2 C to 24.5 C? 65.7 0.62 cal g oC = cal 24.8g 4.3oC = cal 1g 1oC SpHt = 24.5oC DT = 4.3 Co m = 24.8g 20.2oC

Specific Heat E = m DT SpHt
Sample Problem: How much energy does is take to heat 50 g’s of water from 75oC to 87oC? 87oC DT = 12 Co m = 50g H2O 75oC 1 cal 1g 1oC SpHt =

Specific Heat E = m DT SpHt
Sample Problem: How much energy does is take to heat 50 g’s of water from 75oC to 87oC? m DT SpHt 12 Co 1 cal 1g 1oC 50g H2O = cal to heat water 600

Specific Heat E = m DT SpHt 750g H2O Sample Problem:
A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles? 65oC DT = 28 Co m = 750g H2O 37oC 1 cal 1g 1oC SpHt =

Specific Heat E = m DT SpHt 750g H2O = cal 21000 Sample Problem:
A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles? m DT SpHt 28 Co 1 cal 1g 1oC 750g H2O = cal from cool water 21000

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

Solution E = m DT SpHt How many kilojoules are needed to raise the
temperature of 325 g of water from 15.0 °C to 77.0 °C? m DT SpHt = KJ 84.3 62 Co 4.184 J 1g 1oC 1 KJ 1000 J 325g H2O 77oC DT = 62 Co 15oC

2.4 Energy and Nutrition 1 Cal = 1000 calories 1 Cal = 1 kcal 1 Cal = J 1 Cal = kJ

Energy Units of Energy = 1 Cal 1000 cal = 1 kcal 1 cal = 4.18 J 1 Cal
4184 J = kJ

Calorimeters A calorimeter
A reaction chamber & thermometer in H2O used to measure heat transfer indicates the heat lost by a sample indicates the heat gained by water

Energy from Food 4 kcal g 17 kJ g Carbohydrate 9 kcal g 38 kcal g Fat
Protein 4 kcal g 17 kcal g

Calories in Some Foods Energy Requirements

Energy from Food Sample Problem:
How much energy in kcal (= Cal) would be obtained from 2 Tbl peanut butter containing 6 g carb, 16 g fat, & 7 g protein? 6 g carb 4 kcal 1 g carb = 24 kcal = 196 kcal = 196 Cal 16 g fat 9 kcal 1 g fat = 144 kcal = 200 Cal 7 g protein 4 kcal 1 g protein = 28 kcal

Learning Check A cup of whole milk contains 12 g of carbohydrate, 9.0 g of fat, and 9.0 g of protein. How many kcal (Cal) does a cup of milk contain? (Round to the nearest 10 kcal.)

Solution A cup of whole milk contains 12 g of carbohydrate, 9.0 g of fat, and 5.0 g of protein. How many kcal (Cal) does a cup of milk contain? (Round to the nearest 10 kcal.) 12 g carb 4 kcal 1 g carb = 48 kcal = 149 kcal = 149 Cal 9 g fat 9 kcal 1 g fat = 81 kcal = 150 Cal 5 g protein 4 kcal 1 g protein = 20 kcal

2.5 Classification of Matter

Matter Weight on earth. The stuff things are made of.
Has Mass and takes up space. (Air, water, rocks, etc..) The amount of stuff (in g’s) (Bowling Ball > Balloon) Matter: The stuff things are made of. Has Mass and takes up space. Mass: The amount of stuff. Usually measured in grams. Bowling ball has more mass than Weight on earth. Weight on earth. Pull of Gravity on matter.

Classification of matter

Classification of matter
Pure Substance Mixture Element Compound Fe FeS Fe + S Mg MgO Mg + O2

Non-uniform composition
Mixtures Mixture Homogeneous (Solution) Heterogeneous Non-uniform composition Uniform composition Pizza Gasoline Tea w/ice Air Fe + S Urine Sand

Physical Separation of A Mixture
Mixtures can be separated involves only physical changes Like Filtering & distilling Like when pasta and water are separated with a strainer

Classification of Matter

Learning Check Identify each of the following as a pure substance or a mixture. A. pasta and tomato sauce B. aluminum foil C. helium D. Air

Solution Identify each of the following as a pure substance or a mixture. A. pasta and tomato sauce mixture B. aluminum foil pure substance C. helium pure substance D. Air mixture

Learning Check Identify each of the following as a homogeneous or heterogeneous mixture. A. hot fudge sundae B. shampoo C. sugar water D. peach pie

Solution Identify each of the following as a homogeneous or heterogeneous mixture. A. hot fudge sundae heterogeneous B. Shampoo homogeneous C. sugar water homogeneous D. peach pie heterogeneous

2.6 States and Properties of Matter

Elemental states at 25oC Solid Liquid Gas H He Li Be B C N O F Ne Na
Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba Ls Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 6 - 2 2

Properties of matter Solids, liquids and gases can easily be recognized by their different properties. Density The mass of matter divided by it’s volume. Shape Is it fixed or take the shape of the container? Compressibility If we apply pressure, does the volume decrease? Thermal expansion How much does the volume change when heated? 3

Properties of matter High High Large Physical State Solid Liquid Gas
Property High like solids High Density Shape Compressibility Thermal expansion Low expands to fill container Fixed Shape of container Large Small Small Very small Moderate Small 4

Summary of the States of Matter

Three States of Water

Hydrogen Bonding of Water
Frozen H2O: Slow moving molecules H-Bond in patterns

Learning Check Identify each as a S) solid, L) liquid, or G) gas.
__ A. It has a definite volume but takes the shape of the container. __ B. Its particles are moving very rapidly. __ C. It fills the volume of a container. __ D. It has particles in a fixed arrangement. __ E. It has particles that are close together and are mobile.

Solution Identify each as a S) solid, L) liquid, or G) gas.
L A. It has a definite volume but takes the shape of the container. G B. Its particles are moving very rapidly. G C. It fills the volume of a container. S D. It has particles in a fixed arrangement. L E. It has particles that are close together and are mobile.

Physical Properties of Copper
are observed or measured without changing the identity of a substance include shape and color include melting point and boiling point Physical Properties of Copper

Physical Change In a physical change,
the identity and composition of the substance do not change the state can change or the material can be torn into smaller pieces

Examples of Physical Change

Chemical Properties Chemical properties
describe the ability of a substance to change into a new substance

Chemical Change During a chemical change,
reacting substances form new substances with different compositions and properties a chemical reaction takes place Iron Fe Iron (III) oxide Fe2O3

Examples of Chemical Change

Summary

Learning Check Classify each of the following as a 1) physical change or 2) chemical change. A. ____ burning a candle B. ____ ice melting on the street C. ____ toasting a marshmallow D. ____ cutting a pizza E. ____ polishing a silver bowl

Solution Classify each of the following as a
1) physical change or 2) chemical change. A. ____ burning a candle 2) chemical B. ____ ice melting on the street 1) physical C. ____ toasting a marshmallow 2) chemical D. ____ cutting a pizza 1) physical E. ____ polishing a silver bowl 2) chemical

2.7 Changes of State

Condense Vaporize Melt Freeze
Changes of State Fast, far apart, Random Vapor Condense Vaporize Boiling Pt Moderate, close, Random arrangement Liquid Melting Pt = Freezing Pt Melt Freeze Slow, close, Fixed arrangement Solid

Changes of State Vapor Deposit Frost Liquid Sublime Freeze Dry Solid

Heating Curve Specific Heat of Steam Heat of Vaporization
0.48 cal to heat vapor g o C Heat of Vaporization 540 cal to vaporize water g Specific Heat of H2O 1.00 cal to heat water g o C Heat of Fusion 80 cal to melt ice g Specific Heat of Ice 0.50 cal to heat ice g o C

Heating Curve 0.48 cal to heat vapor g o C 540 cal to vaporize water g
1.00 cal to heat water g o C 80 cal to melt ice g 0.50 cal to heat ice g o C

Example: Calculate the total amount of heat needed to change 500
Example: Calculate the total amount of heat needed to change 500. g of ice at –10 oC into 500. g of steam at 120oC. 500g 20 o C cal g o C = 4800 cal to heat vapor 500g cal g = 270,000 cal to vaporize water 500g 100 o C cal g o C = 50,000 cal to heat water 500g cal g = 40,000 cal to melt ice 500g 10 o C cal g o C = 2500 cal to heat ice 367,300 cal = 3.67 x 105 cal

Learning Check A. A plateau (horizontal 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 plateau (horizontal 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

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, 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

Solution 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

Learning Check To reduce a fever, an infant is packed in 250 g of ice. If the ice (at 0 °C) melts and warms to body temp (37.0 °C), how many calories are removed?

Solution To reduce a fever, an infant is packed in 250 g of ice. If the ice (at 0 °C) melts and warms to body temp (37.0 °C), how many calories are removed? 37.0 °C 250g o C cal g o C = 9,250 cal to heat water STEP 2 250g cal g = 20,000 cal to melt ice 0.0 °C STEP 1 29,000 cal = 2.9 x 104 cal

CH104 Chapter 2: Energy & Matter Heating & Cooling Curves

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CH104 Chapter 2: Energy & Matter Heating & Cooling Curves

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