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# CH104 Chapter 2: Energy & Matter PowerPoint PPT Presentation

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.

CH104 Chapter 2: Energy & Matter

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#### Presentation Transcript

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

2.2Temperature

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

Temperature

180o

100o

### 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

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

• At what temperature does

oF = oC?

- 40 oF = - 40 oC

- 40 o

### 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

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

• 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 105.8 °F. What is that temperature on the Celsius scale?

1) 73.8 °C 2) 58.8 °C3) 41.0 °C

### 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 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

### 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 Check

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

What is that temperature in °F?

1) 19 °F2) 59 °F3) 5 °F

### 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

### Learning Check

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

1) 236 K2) 310 K 3) 342 K

### 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

2.1Energy

### 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.

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.

B. a peanut butter and jelly sandwich potential

C. mowing the lawn kinetic

D. gasoline in the gas tank potential

### Kinetic Energy

KE = 1mv2

2

Which has more E?

• Truck moving at 5 mph

• Bicycle moving at 5 mph

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

### 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 4.184 J

4.184 J 1 cal

### Chapter 2 Energy and Matter

2.3Specific 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

H2O

Energy

• calorie:

• E to raise 1 g H20 by 1 oC

Specific Heat

1 cal

1g 1oC

1oC

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

cal

g oC

0oC = start

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

0oC = start

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

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 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

### 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

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

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 =

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

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 =

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 Check

How many kilojoules are needed to raise the

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

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 Matter

2.4Energy and Nutrition

1 Cal =1000 calories

1 Cal = 1 kcal

1 Cal = 4184 J

1 Cal = 4.184 kJ

### Energy

Units of Energy =

1 Cal

1000 cal = 1 kcal

1 cal = 4.18 J

1 Cal

4184 J = 4.184 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

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

### Chapter 2 Energy and Matter

2.5

Classification of 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.

Matter

Mixture

Pure Substance

Element

Compound

Fe + S

FeS

Fe

Mg + O2

MgO

Mg

### Classification of 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

Mixtures can be separated

• involves only physical changes

• Like Filtering & distilling

Like when pasta and water are separated with a strainer

### 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

### Chapter 2 Energy and Matter

2.6States and Propertiesof Matter

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

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

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.

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

• 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

### 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

### 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

### Chapter 2 Energy and Matter

2.7Changes of State

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

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. 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 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

### 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 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

### 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

### 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

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