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![Page 1: CH104 CH104 Chapter 2: Energy & Matter Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves.](https://reader037.fdocuments.in/reader037/viewer/2022102713/56649d085503460f949d9eb4/html5/thumbnails/1.jpg)
CH104
CH104 CH104 Chapter 2: Energy & MatterChapter 2: Energy & Matter
TemperatureTemperature
Energy in ReactionsEnergy in Reactions
Specific HeatSpecific Heat
Energy from FoodEnergy from Food
States of MatterStates of Matter
Heating & Cooling CurvesHeating & Cooling Curves
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CH104
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.2Temperature
2
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CH104
Metric
SI
Common
ConversionsLengthVolumeMassTemperature
Units of MeasurementUnits of Measurement
meter (m) 1 m = 1.09 yd
liter (L) 1 L = 1.06 qt
gram (g) 1 kg = 2.2 lb
Celsius (oC) C = (F-32)/1.8
Kelvin (K) K = C + 273
![Page 4: CH104 CH104 Chapter 2: Energy & Matter Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves.](https://reader037.fdocuments.in/reader037/viewer/2022102713/56649d085503460f949d9eb4/html5/thumbnails/4.jpg)
CH104
180180 oo100100 oo
TemperatureTemperature
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CH104
Learning CheckLearning 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
5
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CH104
SolutionSolution
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
6
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CH104
• At what temperature does At what temperature does ooF = F = ooC?C?
Trivial Pursuit QuestionTrivial Pursuit Question
- 40 - 40 ooF = - 40 F = - 40 ooCC
- 40 - 40 oo
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CH104
Temperature conversionTemperature conversion
• Common scales used• Fahrenheit,Fahrenheit, CelsiusCelsius andand KelvinKelvin..
ooFF = 1.8 = 1.8 ooC + 32 C + 32
ooCC = = ((ooF - 32)F - 32)1.81.8
KK = = ooC + 273C + 273 SI unitSI unit
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CH104
Solving a Temperature ProblemSolving a Temperature ProblemA person with hypothermia has abody temperature of 34.8 °C. WhatIs that temperature in °F?
F = 1.8(C) + 32 °
F = (1.8)(34.8 °C) + 32 ° exact tenth’s exact
= 62.6 ° + 32 ° = 94.6 °F tenth’s
9
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CH104
Practice: Temp ConversionPractice: 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
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CH104
Learning CheckLearning Check The normal temperature of a chickadee is 105.8
°F. What is that temperature on the Celsius scale?
1) 73.8 °C 2) 58.8 °C 3) 41.0 °C
11
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CH104
SolutionSolution
TC = TF – 32 °
1.8
= (105.8 – 32 °)
1.8
= 73.8 °F = 41.0 °C
1.8 ° tenth’s place
12
The normal temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale?
1) 73.8 °C 2) 58.8 °C 3) 41.0 °C
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CH104
Learning Check 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
13
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CH104
Solution Solution
TF – 32 ° = TC
1.8
(455 – 32 °) = 235 °C
1.8 one’s place
14
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
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CH104
Learning Check 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
15
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CH104
Solution Solution
TF = 1.8TC + 32 °
TF = 1.8(–15 °C) + 32 °
= – 27 + 32 °
= 5 °F one’s place
16
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
Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign.
1.8 x 15 +/ – = –27
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CH104
TemperaturesTemperatures
17
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CH104
Learning CheckLearning Check
What is normal body temperature of 37 °C in kelvins?
1) 236 K 2) 310 K 3) 342 K
18
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CH104
Solution Solution
TK = TC + 273
= 37 °C + 273
= 310. K
one’s place
19
What is normal body temperature of 37 °C in kelvins?
1) 236 K 2) 310 K 3) 342 K
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CH104 20
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.1Energy
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CH104
EnergyEnergy
EnergyEnergy = The capacity to cause change
HeatLightWind
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CH104
Potential Energy Potential Energy = stored Energystored Energy (Has potential for motion)
Kinetic Energy Kinetic Energy = Energy in motionEnergy in motion (Fulfilling its potential)
X
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CH104 23
Learning CheckLearning 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
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CH104 24
SolutionSolution
Identify each of the following as potential energy or kinetic energy.
A. roller blading kinetic kinetic
B. a peanut butter and jelly sandwich potential potential
C. mowing the lawn kinetic kinetic
D. gasoline in the gas tank potential potential
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CH104
Kinetic EnergyKinetic EnergyKEKE = 1 mv2
2 Which has more E?
•Truck moving at 5 mph
•Bicycle moving at 5 mph
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CH104
MetricMetric
SISI
CommonCommon
ConversionsConversions
LengthLength
VolumeVolume
MassMass
TemperatureTemperature
EnergyEnergy
Units of MeasurementUnits of Measurement
meter (m) 1 m = 1.09 ydmeter (m) 1 m = 1.09 yd
liter (L) 1 L = 1.06 qtliter (L) 1 L = 1.06 qt
gram (g) 1 kg = 2.2 lbgram (g) 1 kg = 2.2 lb
calorie (cal) 1Kcal = 1000 cal = 1Calcalorie (cal) 1Kcal = 1000 cal = 1CalJoule (J) 1 cal = 4.18 JJoule (J) 1 cal = 4.18 J
Celsius (oC) C = (F-32)/1.8
Kelvin (K) K = C + 273
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CH104
Examples of Energy Values in JoulesExamples of Energy Values in Joules
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CH104 28
Learning CheckLearning CheckHow many calories are obtained from a pat
of butter if it provides 150 J of energy when metabolized?
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CH104
SolutionSolutionHow many calories are obtained from a pat
of butter if it provides 150 J of energy when metabolized?
Given
150 J
Need
= cal
1 cal and 4.184 J4.184 J 1 cal
1 cal4.184 J
36
Conversion Factors:
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CH104 30
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.3Specific Heat
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CH104
EnergyEnergyUnits of Energy =Units of Energy =
caloriecalorie calcal
kilocaloriekilocalorie kcalkcal 1000 cal = 1 kcal1000 cal = 1 kcal
CalorieCalorie CalCal Cal = kcalCal = kcal
joulejoule JJ 4.18 J = 1 cal4.18 J = 1 cal
British British
Thermal UnitThermal UnitBTUBTU
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CH104
EnergyEnergycalorie: calorie:
•E E toto raise 1 g H raise 1 g H220 0 byby 1 1 ooCC
H2O 11 cal
1g 1oC
11 cal
1g 1oC
Specific HeatSpecific Heat
1oC
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CH104
Specific HeatSpecific HeatE to raise Temp of 1g substance by 1 E to raise Temp of 1g substance by 1 ooCC
Fe Fe
0.110.11
Cu
0.093H2O
1.00
Ag
0.057
Au
0.031Sand
0.19
Al
0.22
0oC = start Add 1 cal
cal
g oC
cal
g oC
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CH104
Specific HeatSpecific Heat
Fe Fe
0.110.11
Cu
0.093
Ag
0.057
Au
0.031
Sand
0.19
Al
0.22
0oC = start
H2O
1.00
E to raise Temp of 1g substance by 1 E to raise Temp of 1g substance by 1 ooCC
cal
g oC
cal
g oC
1o
10o
30o
20o
Add 1 cal
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CH104
High SpHt; Resists changeHigh SpHt; Resists change
Specific HeatSpecific Heat
Fe Fe
0.110.11
Cu
0.093
Ag
0.057
Au
0.031
Sand
0.19
Al
0.22H2O
1.00
1o
10o
30o
20o
Low SpHt; Heats quickly
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CH104
Specific HeatSpecific Heat
Sand
0.19H2O
1.00
Resists change
Stays cold
Heats quickly
Gets hot
Hydrated person
Resists change in body temp
Dehydrated person
Body temp rises quickly
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CH104
Learning CheckLearning CheckA. 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 2) low specific heat
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CH104
SolutionSolutionA. 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 2) low specific heat
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CH104
What is the specific heat of a metal if 24.8 g absorbs 65.7 cal 65.7 cal of energy and the temp rises from 20.2 C to 24.5 C?
20.2oC
Specific HeatSpecific HeatSample Problem:Sample Problem:
24.5oCT = 4.3 CoT = 4.3 Co
cal
g oC
cal
g oC
m =m =
SpHt =SpHt =SpHt =SpHt =
24.8g
= cal
24.8g 4.34.3oC
= cal
24.8g 4.34.3oC
65.765.7 = cal
1g 1oC
= cal
1g 1oC
0.620.62
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CH104
75oC
Specific HeatSpecific HeatSample Problem:Sample Problem:
How much energy does is take to heat 50 g’s of water from 75oC to 87oC?How much energy does is take to heat 50 g’s of water from 75oC to 87oC?
87oCT = 12 CoT = 12 Co
E = m E = m T SpHtT SpHt
11 cal
1g 1oC
11 cal
1g 1oC
m =m =
SpHt =SpHt =SpHt =SpHt =
50g H2O
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CH104
50g H2O
Specific HeatSpecific HeatSample Problem:Sample Problem:
How much energy does is take to heat 50 g’s of water from 75oC to 87oC?How much energy does is take to heat 50 g’s of water from 75oC to 87oC?
12 Co 12 Co
E = m E = m T SpHtT SpHt
11 cal
1g 1oC
11 cal
1g 1oC
mm SpHtSpHtTT
= = cal calto heat waterto heat water
600600
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CH104
37oC
Specific HeatSpecific HeatSample Problem: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?
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?
65oCT = 28 CoT = 28 Co
E = m E = m T SpHtT SpHt
11 cal
1g 1oC
11 cal
1g 1oC
m =m =
SpHt =SpHt =SpHt =SpHt =
750g H2O
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CH104
750g H2O
Specific HeatSpecific HeatSample Problem:Sample Problem:
28 Co 28 Co
E = m E = m T SpHtT SpHt
11 cal
1g 1oC
11 cal
1g 1oC
mm SpHtSpHtTT
= = cal calfrom cool waterfrom cool water
2100021000
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?
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?
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CH104 45
Learning CheckLearning CheckHow many kilojoules are needed to raise thetemperature of 325 g of water from 15.0 °C to
77.0 °C?
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CH104 46
SolutionSolutionHow many kilojoules are needed to raise thetemperature of 325 g of water from 15.0 °C to
77.0 °C?
15oC
77oC
E = m E = m T SpHtT SpHt
T = 62 CoT = 62 Co
325g H2O 62 Co 62 Co 4.1844.184 J
1g 1oC
4.1844.184 J
1g 1oC
mm SpHtSpHtTT
== KJ KJ84.384.311 KJ
1000 J
11 KJ
1000 J
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CH104 47
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.4Energy and Nutrition
1 Cal = 1000 calories
1 Cal = 1 kcal
1 Cal = 4184 J
1 Cal = 4.184 kJ
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CH104
EnergyEnergyUnits of Energy =Units of Energy =
1000 cal = 1 kcal1000 cal = 1 kcal
1 Cal1 Cal
1 Cal1 Cal
4184 J = 4.184 kJ4184 J = 4.184 kJ
1 cal = 4.18 J 1 cal = 4.18 J
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CH104
CalorimetersCalorimetersA 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
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CH104
Energy from FoodEnergy from Food
CarbohydrateCarbohydrate
FatFat
ProteinProtein
4 kcal
g
4 kcal
g
9 kcal
g
9 kcal
g
4 kcal
g
4 kcal
g
17 kJ
g
17 kJ
g
38 kcal
g
38 kcal
g
17 kcal
g
17 kcal
g
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CH104
= 196 Cal= 196 Cal
Energy from FoodEnergy from FoodSample Problem: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?
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 carb6 g carb
16 g fat16 g fat
4 kcal
1 g carb
4 kcal
1 g carb
9 kcal
1 g fat
9 kcal
1 g fat
4 kcal
1 g protein
4 kcal
1 g protein7 g protein7 g protein
= 24 kcal= 24 kcal
= 144 kcal= 144 kcal
= 28 kcal= 28 kcal
= 196 kcal= 196 kcal
= 200 Cal= 200 Cal
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CH104 53
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.)
Learning CheckLearning Check
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CH104 54
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.)
SolutionSolution
= 149 Cal= 149 Cal
12 g carb12 g carb
9 g fat9 g fat
4 kcal
1 g carb
4 kcal
1 g carb
9 kcal
1 g fat
9 kcal
1 g fat
4 kcal
1 g protein
4 kcal
1 g protein5 g protein5 g protein
= 48 kcal= 48 kcal
= 81 kcal= 81 kcal
= 20 kcal= 20 kcal
= 149 kcal= 149 kcal
= 150 Cal= 150 Cal
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CH104 55
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.5Classification of Matter
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CH104
The stuffstuff things are made of.
Has MassMass and takes up space.
(Air, water, rocks, etc..)
• MatterMatter
The amountamount of stuff (in g’s) (Bowling Ball > Balloon)
WeightWeight on earth.
Pull of Gravity on matter.
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CH104
Classification of matterClassification of matter
MatterMatter
Pure Substance Mixture
Element Compound
FeFe FeSFeS
MgMg MgOMgO Mg + OMg + O22
Fe + SFe + S
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CH104
Pizza
MixtureMixture
HomogeneousHomogeneous
(Solution)(Solution)HeterogeneousHeterogeneous
Fe + SFe + S
MixturesMixtures
NNoonn--uunniiffoorrmm ccoommppoossiittiioonnUniform compositionUniform composition
Air
UrineGasoline
Sand
Tea w/ice
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CH104
Physical Separation of A MixturePhysical Separation of A MixtureMixtures can be separated• involves only physical changes– Like Filtering & distilling
Like when pasta and water are separated with a strainer
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CH104
Identify each of the following as a pure substance or a mixture.
A. pasta and tomato sauce
B. aluminum foil
C. helium
D. Air
Learning CheckLearning Check
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CH104
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
SolutionSolution
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CH104 64
Identify each of the following as a homogeneous or heterogeneous mixture.
A. hot fudge sundae
B. shampoo
C. sugar water
D. peach pie
Learning CheckLearning Check
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CH104 65
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
SolutionSolution
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CH104
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.6States and Properties
of Matter
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CH104
Elemental states at 25Elemental states at 25ooCC
He
Rn
XeI
KrBrSe
ArClS
NeFO
P
NC
H
Li
Na
Cs
Rb
K
TlHgAuHfLsBa
Fr
PtIrOsReWTa PoBiPb
Be
Mg
Sr
Ca
CdAgZrY PdRhRuTcMoNb
AcRa
ZnCuTiSc NiCoFeMnCrV
In SbSn
Ga Ge
Al
Gd
Cm
Tb
Bk
Sm
Pu
Eu
Am
Nd
U
Pm
Np
Ce
Th
Pr
Pa
Yb
No
Lu
Lr
Er
Fm
Tm
Md
Dy
Cf
Ho
Es
At
Te
As
Si
B
6 - 2
SolidLiquid
Gas
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CH104
DensityDensity
ShapeShape
CompressibilityCompressibility
ThermalThermalexpansionexpansion
Properties of matterProperties of matter
PropertyPhysical StatePhysical State
SolidSolid LiquidLiquid Gas Gas
High Highlike solids
Low
Fixed Shape ofcontainer
expandsto fillcontainer
Small Small Large
Verysmall
Small Moderate
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CH104 71
Three States of WaterThree States of Water
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CH104
•
Frozen HFrozen H22O:O: Slow moving moleculesSlow moving moleculesH-BondH-Bond in patterns in patterns
Hydrogen Bonding of Water Hydrogen Bonding of Water
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CH104
Learning CheckLearning CheckIdentify 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.
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CH104
SolutionSolutionIdentify each as a S) solid, L) liquid, or GG) gas.L A. It has a definite volume but takes the
shape of the container.GG B. Its particles are moving very rapidly.G 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.
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CH104
Physical PropertiesPhysical PropertiesPhysical properties• are observed or measured without changing the
identity of a substance• include shape and color• include melting point and boiling point
Physical Properties of CopperPhysical Properties of Copper
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CH104 76
Physical ChangePhysical ChangeIn 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
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CH104
Examples of Physical ChangeExamples of Physical Change
77
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CH104
Chemical PropertiesChemical PropertiesChemical properties
• describe the ability of a substance to change into a new substance
78
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CH104
Chemical ChangeChemical ChangeDuring a chemical
change, • reacting substances
form new substances with different compositions and properties
• a chemical reaction takes place
Iron
Fe
Iron (III) oxide
Fe2O3
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CH104
Examples of Chemical ChangeExamples of Chemical Change
80
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CH104 81
SummarySummary
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CH104 82
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
Learning CheckLearning Check
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CH104 83
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
SolutionSolution
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CH104 84
Chapter 2 Energy and MatterChapter 2 Energy and Matter
2.7Changes of State
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CH104
Changes of StateChanges of State
Melting Pt = Melting Pt = Freezing PtFreezing Pt
Boiling PtBoiling Pt
SolidSolid
LiquidLiquid
VaporVapor
CondenseCondenseCondenseCondense
FreezeFreezeFreezeFreezeMeltMeltMeltMelt
VaporizeVaporizeVaporizeVaporize
Slow, close,Slow, close,Fixed Fixed
arrangementarrangement
Moderate, close,Moderate, close,Random Random
arrangementarrangement
Fast, far apart,Fast, far apart,RandomRandom
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CH104
SolidSolid
LiquidLiquid
VaporVapor
Changes of StateChanges of State
FrostFrostDepositDepositDepositDeposit
SublimeSublimeSublimeSublimeFreeze DryFreeze Dry
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CH104
Heating CurveHeating Curve
Specific Heat of IceSpecific Heat of Ice0.50 cal0.50 cal to heat ice to heat ice g g o o CC
Specific Heat of IceSpecific Heat of Ice0.50 cal0.50 cal to heat ice to heat ice g g o o CC
Heat of FusionHeat of Fusion80 cal80 cal to melt ice to melt ice gg
Heat of FusionHeat of Fusion80 cal80 cal to melt ice to melt ice gg
Specific Heat of HSpecific Heat of H22OO
1.00 cal1.00 cal to heat water to heat water g g o o CC
Specific Heat of HSpecific Heat of H22OO
1.00 cal1.00 cal to heat water to heat water g g o o CC
Heat of VaporizationHeat of Vaporization540 cal540 cal to vaporize water to vaporize water gg
Heat of VaporizationHeat of Vaporization540 cal540 cal to vaporize water to vaporize water gg
Specific Heat of SteamSpecific Heat of Steam0.48 cal0.48 cal to heat vapor to heat vapor g g o o CC
Specific Heat of SteamSpecific Heat of Steam0.48 cal0.48 cal to heat vapor to heat vapor g g o o CC
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CH104
Heating CurveHeating Curve
0.50 cal0.50 cal to heat ice to heat ice g g o o CC
80 cal80 cal to melt ice to melt ice gg
1.00 cal1.00 cal to heat water to heat water g g o o CC
540 cal540 cal to vaporize water to vaporize water gg
0.48 cal0.48 cal to heat vapor to heat vapor g g o o CC
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CH104
Example:Example: Calculate the total amount of heat needed to change 500. g of ice at –10 oC into 500. g of steam at 120oC.
500g500g 10 10 o o CC 0.50 cal 0.50 cal g g o o CC
500g 500g 80 cal80 cal gg
500g500g 100 100 o o CC 1.00 cal 1.00 cal g g o o CC
500g500g 540 cal540 cal gg
500g500g 20 20 o o CC 0.48 cal 0.48 cal g g o o CC
== 2500 cal 2500 cal to heat iceto heat ice
== 40,000 cal 40,000 cal to melt iceto melt ice
= = 50,000 cal50,000 calto heat waterto heat water
= = 270,000 cal270,000 calto vaporize waterto vaporize water
= = 4800 cal4800 calto heat vaporto heat vapor
367,300 cal = 3.67 x 10367,300 cal = 3.67 x 1055 cal cal
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CH104 90
Learning CheckLearning CheckA. 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 1) a temperature change 2) a constant temperature 3) a change of state
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CH104
SolutionSolutionA. 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 1) a temperature change 2) a constant temperature 3) a change of state
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CH104 92
Learning CheckLearning CheckUse 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
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CH104 93
SolutionSolutionUse 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
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CH104 94
Learning CheckLearning CheckTo 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?
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CH104
SolutionSolutionTo 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?
STEP 1
37.0 °C
0.0 °C
STEP 2250g 250g 80 cal80 cal gg
250g250g 37 37 o o CC 1.00 cal 1.00 cal g g o o CC
== 20,000 cal 20,000 cal to melt iceto melt ice
= = 9,250 cal9,250 calto heat waterto heat water
29,000 cal = 2.9 x 1029,000 cal = 2.9 x 1044 cal cal