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CH104 Chapter 2: Energy & Matter. 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. meter (m) 1 m = 1.09 yd. liter (L) 1 L = 1.06 qt.

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CH104 Chapter 2: Energy & Matter

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Ch104 chapter 2 energy matter

CH104 Chapter 2: Energy & Matter

Temperature

Energy in Reactions

Specific Heat

Energy from Food

States of Matter

Heating & Cooling Curves


Chapter 2 energy and matter

Chapter 2 Energy and Matter

2.2Temperature


Ch104 chapter 2 energy matter

Units 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


Ch104 chapter 2 energy matter

Temperature

180o

100o


Learning check

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) 1002) 1803) 273


Solution

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) 1002) 1803) 273


Trivial pursuit question

Trivial Pursuit Question

  • At what temperature does

    oF = oC?

- 40 oF = - 40 oC

- 40 o


Temperature conversion

Temperature conversion

  • Common scales used

    • Fahrenheit,CelsiusandKelvin.

oF= 1.8 oC + 32

oC = (oF - 32)

1.8

K = oC + 273

SI unit


Solving a temperature problem

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) + 32 °

F = (1.8)(34.8 °C) + 32 °

exact tenth’s exact

= 62.6 ° + 32 °

= 94.6 °F

tenth’s


Practice temp conversion

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 check1

    Learning 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 °C3) 41.0 °C


    Solution1

    Solution

    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 °C3) 41.0 °C

    TC = TF – 32 °

    1.8

    =(105.8 – 32 °)

    1.8

    =73.8 °F = 41.0 °C

    1.8 °tenth’s place


    Learning check2

    Learning Check

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

    1) 423 °C2) 235 °C 3) 221 °C


    Solution2

    Solution

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

    1) 423 °C2) 235 °C 3) 221 °C

    TF – 32 °= TC

    1.8

    (455 – 32 °) = 235 °C

    1.8one’s place


    Learning check3

    Learning Check

    On a cold winter day, the temperature is –15 °C.

    What is that temperature in °F?

    1) 19 °F2) 59 °F3) 5 °F


    Solution3

    Solution

    On a cold winter day, the temperature is –15 °C.

    What is that temperature in °F?

    1) 19 °F2) 59 °F3) 5 °F

    TF = 1.8TC + 32 °

    TF = 1.8(–15 °C) + 32 °

    = – 27 + 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

    Temperatures


    Learning check4

    Learning Check

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

    1) 236 K2) 310 K 3) 342 K


    Solution4

    Solution

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

    1) 236 K2) 310 K 3) 342 K

    TK = TC + 273

    =37 °C + 273

    =310. K

    one’s place


    Chapter 2 energy and matter1

    Chapter 2 Energy and Matter

    2.1Energy


    Energy

    Energy

    Energy = The capacity to cause change

    Heat

    Light

    Wind


    Ch104 chapter 2 energy matter

    Potential Energy = stored Energy

    (Has potential for motion)

    X

    Kinetic Energy = Energy in motion

    (Fulfilling its potential)


    Learning check5

    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


    Solution5

    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

    Kinetic Energy

    KE = 1mv2

    2

    Which has more E?

    • Truck moving at 5 mph

    • Bicycle moving at 5 mph


    Ch104 chapter 2 energy matter

    Units 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

    calorie (cal) 1Kcal = 1000 cal = 1Cal

    Joule (J) 1 cal = 4.18 J


    Examples of energy values in joules

    Examples of Energy Values in Joules


    Learning check6

    Learning Check

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


    Solution6

    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 4.184 J

    4.184 J 1 cal


    Chapter 2 energy and matter2

    Chapter 2 Energy and Matter

    2.3Specific Heat


    Energy1

    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


    Ch104 chapter 2 energy matter

    H2O

    Energy

    • calorie:

      • E to raise 1 g H20 by 1 oC

    Specific Heat

    1 cal

    1g 1oC

    1oC


    Ch104 chapter 2 energy matter

    Au

    0.031

    Ag

    0.057

    Cu

    0.093

    Fe

    0.11

    Sand

    0.19

    Al

    0.22

    H2O

    1.00

    Add 1 cal

    Specific Heat

    E to raise Temp of 1g substance by 1 oC

    cal

    g oC

    0oC = start


    Specific heat

    Au

    0.031

    Ag

    0.057

    Cu

    0.093

    Fe

    0.11

    Sand

    0.19

    Al

    0.22

    H2O

    1.00

    Specific Heat

    E to raise Temp of 1g substance by 1 oC

    30o

    cal

    g oC

    20o

    10o

    1o

    Add 1 cal

    0oC = start


    Ch104 chapter 2 energy matter

    Au

    0.031

    Ag

    0.057

    Cu

    0.093

    Fe

    0.11

    Sand

    0.19

    Al

    0.22

    H2O

    1.00

    Specific Heat

    30o

    Low SpHt; Heats quickly

    20o

    10o

    1o

    High SpHt; Resists change


    Ch104 chapter 2 energy matter

    Sand

    0.19

    H2O

    1.00

    Specific Heat

    Dehydrated person

    Body temp rises quickly

    Hydrated person

    Resists change in body temp

    Heats quickly

    Gets hot

    Resists change

    Stays cold


    Learning check7

    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) warms3) 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


    Solution7

    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) warms3) 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


    Ch104 chapter 2 energy matter

    24.8g

    Specific Heat

    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 =

    20.2oC


    Ch104 chapter 2 energy matter

    50g H2O

    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 =

    75oC

    1 cal

    1g 1oC

    SpHt =


    Ch104 chapter 2 energy matter

    50g H2O

    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

    = cal

    to heat water

    600


    Ch104 chapter 2 energy matter

    750g H2O

    Specific Heat

    E = m DT SpHt

    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 =

    37oC

    1 cal

    1g 1oC

    SpHt =


    Ch104 chapter 2 energy matter

    750g H2O

    Specific Heat

    E = m DT SpHt

    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

    = cal

    from cool water

    21000


    Learning check8

    Learning Check

    How many kilojoules are needed to raise the

    temperature of 325 g of water from 15.0 °C to 77.0 °C?


    Solution8

    325g H2O

    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

    77oC

    DT = 62 Co

    15oC


    Chapter 2 energy and matter3

    Chapter 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


    Energy2

    Energy

    Units of Energy =

    1 Cal

    1000 cal = 1 kcal

    1 cal = 4.18 J

    1 Cal

    4184 J = 4.184 kJ


    Calorimeters

    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

    Energy from Food

    4 kcal

    g

    17 kJ

    g

    Carbohydrate

    9 kcal

    g

    38 kcal

    g

    Fat

    Protein

    4 kcal

    g

    17 kcal

    g


    Ch104 chapter 2 energy matter

    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 check9

    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.)


    Solution9

    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


    Chapter 2 energy and matter4

    Chapter 2 Energy and Matter

    2.5

    Classification of Matter


    Ch104 chapter 2 energy matter

    • Matter

    • 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)

    Weight on earth.

    Pull of Gravity on matter.


    Classification of matter

    Matter

    Mixture

    Pure Substance

    Element

    Compound

    Fe + S

    FeS

    Fe

    Mg + O2

    MgO

    Mg

    Classification of matter


    Ch104 chapter 2 energy matter

    Mixture

    Homogeneous

    (Solution)

    Heterogeneous

    Pizza

    Fe + S

    Gasoline

    Tea w/ice

    Sand

    Mixtures

    Non-uniformcomposition

    Uniform composition

    Air

    Urine


    Physical separation of a mixture

    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


    Learning check10

    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


    Solution10

    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 check11

    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


    Solution11

    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


    Chapter 2 energy and matter5

    Chapter 2 Energy and Matter

    2.6States and Propertiesof Matter


    Elemental states at 25 o c

    Elemental states at 25oC

    H

    He

    Solid

    Liquid

    Li

    Be

    B

    C

    N

    O

    F

    Ne

    Gas

    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


    Properties of matter

    Properties of matter

    Physical State

    SolidLiquidGas

    Property

    High

    like solids

    High

    Low

    Density

    Shape

    Compressibility

    Thermal

    expansion

    expands

    to fill

    container

    Fixed

    Shape of

    container

    Large

    Small

    Small

    Moderate

    Very

    small

    Small


    Three states of water

    Three States of Water


    Ch104 chapter 2 energy matter

    Hydrogen Bonding of Water

    Frozen H2O:

    Slow moving molecules

    H-Bond in patterns


    Learning check12

    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.


    Solution12

    Solution

    Identify each as a S) solid, L) liquid, or G) gas.

    LA. It has a definite volume but takes the shape of the container.

    GB. Its particles are moving very rapidly.

    G C. It fills the volume of a container.

    SD. It has particles in a fixed arrangement.

    LE. It has particles that are close together and are mobile.


    Physical properties

    Physical Properties

    Physical 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 Copper


    Physical change

    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

    Examples of Physical Change


    Chemical properties

    Chemical Properties

    Chemical properties

    • describe the ability of a substance to change into a new substance


    Chemical change

    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

    Examples of Chemical Change


    Summary

    Summary


    Learning check13

    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


    Solution13

    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


    Chapter 2 energy and matter6

    Chapter 2 Energy and Matter

    2.7Changes of State


    Ch104 chapter 2 energy matter

    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


    Ch104 chapter 2 energy matter

    Changes of State

    Vapor

    Deposit

    Frost

    Liquid

    Sublime

    Freeze Dry

    Solid


    Ch104 chapter 2 energy matter

    Heating Curve

    Specific Heat of Steam

    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


    Ch104 chapter 2 energy matter

    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


    Ch104 chapter 2 energy matter

    Example: Calculate the total amount of heat needed to change 500. g of ice at –10 oC into 500. g of steam at 120oC.

    500g20 o C 0.48 cal

    g o C

    = 4800 cal

    to heat vapor

    500g540 cal

    g

    = 270,000 cal

    to vaporize water

    500g100 o C 1.00 cal

    g o C

    = 50,000 cal

    to heat water

    500g 80 cal

    g

    = 40,000 cal

    to melt ice

    500g 10 o C 0.50 cal

    g o C

    = 2500 cal

    to heat ice

    367,300 cal = 3.67 x 105 cal


    Learning check14

    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

    1) a temperature change

    2) a constant temperature

    3) a change of state


    Solution14

    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

    1) a temperature change

    2) a constant temperature

    3) a change of state


    Learning check15

    Learning Check

    Use the cooling curve for water to answer each.

    A. Water condenses at a temperature of

    1) 0 °C2) 50 °C3) 100 °C

    B. At a temperature of 0 °C, liquid water

    1) freezes2) melts3) changes to a gas

    C. At 40 °C, water is a

    1) solid 2) liquid3) gas

    D. When water freezes, heat is

    1) removed2) added


    Solution15

    Solution

    Use the cooling curve for water to answer each.

    A. Water condenses at a temperature of

    1) 0 °C2) 50 °C3) 100 °C

    B. At a temperature of 0 °C, liquid water

    1) freezes2) melts3) changes to a gas

    C. At 40 °C, water is a

    1) solid 2) liquid3) gas

    D. When water freezes, heat is

    1) removed2) added


    Learning check16

    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?


    Solution16

    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

    250g37 o C 1.00 cal

    g o C

    = 9,250 cal

    to heat water

    STEP 2

    250g 80 cal

    g

    = 20,000 cal

    to melt ice

    0.0 °C

    STEP 1

    29,000 cal = 2.9 x 104 cal


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