friday, april 11, 2014 grab a lab from my desk: –read through the front page. today: –system vs....

Friday, April 11, 2014

• Grab a Lab from my Desk:– Read through the front page.

• Today:– System vs. Surroundings– Endothermic vs. Exothermic– Chemical vs. Physical– Perform Lab

• Water means tap water

Unit 11Hot/Cold Packs

7.1 Endothermic and Exothermic

7.2 Calorimetry and Heat Capacity

7.3 Changes in State

Introductory Activity

• How many things can you think of in everyday life that either give off heat or absorb heat?

• Which of these things are physical processes?• Which are chemical processes?

Chemical vs. Physical

• Physical and chemical processes both can both produce heat (which will make it feel hot) or absorb heat (which will make it feel cold)

• Physical – The actual chemicals are not changed, and can often be re-cooped.– Examples: scratching, cutting, dissolving, breaking.

• Chemical – The actual chemicals react to form new different chemicals.

Hot/Cold Packs

Transfer of energyTransfer of energy

Use

System & Surroundings


between

Materials ability to absorb energy

without noticeable temperature change

Materials ability to absorb energy

without noticeable temperature change

Effect on temperature depends on

Physical change

Physical change

Is determined with

CalorimetryCalorimetry

Can be done in

Chemical change

Chemical change

System vs. Surroundings

• System:– The chemicals that are involved in the chemical/physical

process

• Surroundings:– The things that are surrounding those chemicals:

• Water, beaker, air, lab . . . . the universe

Endothermic & Exothermic

• Endothermic:– When the system absorbs energy from the surroundings.– Tends to feel cold.

• Exothermic:– When the system releases energy to the surroundings.– Tends to feel hot.

SECTION 7.1 -ENDOTHERMIC AND EXOTHERMIC


• The system is only made of the molecules undergoing the change

• The surroundings are everything else– The water molecules & the container– Your hand, the air, the thermometer

• Note that water is made up of water molecules—not a solid chunk of water…but for this picture, it’s best to represent water as one thing since it’s the surroundings and focus on the molecules reacting as the system.

Exothermic & You

• You touch the beaker and it feels hot• Energy is being transferred TO YOU• You are the surroundings• When energy moves from system to

surroundings, it’s exothermic

Exothermic & the Thermometer

• The temperature (measured by the thermometer) is related to the average kinetic energy of the molecules in the container

• The majority of the molecules in a solution are water

• If the temperature is increasing, the energy of the water molecules is increasing

• Since water is the surrounding (it’s not actually reacting), energy is being transferred to the surroundings

• Exothermic shows an increase in the temperature within the container

Endothermic

• The opposite is also true• If the container feels cold to you, energy is

being transferred FROM YOU (the surroundings) into the system—endothermic

• If the thermometer goes down, energy is being transferred FROM the water molecules (surroundings) into the system--endothermic

Let’s Practice

Example:Identify the system and surroundings when you hold an ice cube while it

melts. Is this endo- or exothermic?

System: Water molecules in the form of iceSurroundings: You and the air

It feels cold to you…so energy is leaving you (surroundings)

When energy goes from surroundings to system it’s endothermic

Please sit in your new seat

• Remember:– Your row is your lab group– No more than 4 people per row– No empty seats in front of row

Monday, April 14, 2014

• Today:– Notes/Demonstrations of:

• Energy Flow (Change in Energy)• Exothermic and endothermic.• Units of Energy

• Homework:– Energy Conversions

Enthalpy

http://commons.wikimedia.org/wiki/Image:Hot_Horseshoe_%28stevefe%29.jpg

• Enthalpy (H) - total energy of system• ’s in enthalpy are usually measured:

– As ’s occur H can be measured– H is measured in kJ

Fe(s, 300K) Fe(s, 1100K) H=20.1 kJ– indicates that as it is heated by 800 K, its

enthalpy increases by 20,100 J for each mole of iron

Types of Processes (Reactions)

Exothermic• Exo – “out of”• Energy flows out of the

reaction into surroundings.– Surroundings get warmer– Heat, light, vibrations,

explosions

• H is negative

Endothermic• Endo – “into”, “inside of”• Energy flows into the

reaction from the surroundings– Surroundings get colder

• H is positive

Energy Diagram 1

Exothermic!

Energy Diagram 2

Endothermic!

Demonstrations

Demo 1

• 20 g Ba(OH)2

• 10 g ammonium thiocyanate

• Place in beaker on a wet wooden board.

• Mix thoroughly.

• * Danger! Energy Change!

Demo 2• 5.0 g KMnO4• 1 mL glycerin• Evaporating dish• Place KMnO4 in a pile in

the evaporating dish• Make a well w/ scoopula• Add glycerin & stand

back

• * Danger! Energy Change!

Temperature

• Temperature – proportional to the average kinetic energy of the molecules

• Kinetic Energy - Energy due to motion (Related to how fast the molecules are moving)

As temperature increases

Molecules move faster

Heat & Enthalpy

• Heat (q)– The flow of energy from higher temperature particles to lower temperature particles

• Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume

• Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”

Energy Units

• The most common energy units are Joules (J) and calories (cal)

• These equivalents can be used in dimensional analysis to convert units

4.184 J 1.00 cal

1000 J

1000 cal

1 kJ

1 Cal (food calorie)

=

=

=

Energy Equivalents

Tuesday, April 15, 2015

• Pickup Papers on Red Lab Table• Today:

– Go Over Test– Go Over Energy Conversions/Endo Exothermic– Tutoring/Extra Help– ThermoStoich

• Homework:– Complete ThermoStoich Worksheet– Check @lths.org email – sent on Friday

Wednesday, April 16, 2014

–Turn in ThermoStoich on my Desk–Grab a Heat Capacity Worksheet

• Today:–Notes: Heat Capacity–Homework:

• Heat Capacity Worksheet• Bring Books/Notes/Calc. tomorrow

Heat Capacity

• Specific Heat Capacity (c) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C

• c for liquid water – 4.184 J/g°C

Heat Capacity

High Heat Capacity• Takes a large amount of

energy to noticeably change temp

• Heats up slowly

• Cools down slowly

• Maintains temp better with small condition changes

Low Heat Capacity• Small amount of energy

can noticeably change temperature

• Heats up quickly

• Cools down quickly

• Quickly readjusts to new conditions

• A pool takes a long time to warm up and remains fairly warm over night.

• Lake Michigan . . .• Pizza . . . • The air warms quickly

on a sunny day, but cools quickly at night

• A cast-iron pan stays hot for a long time after removing from oven.

• Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly

Substance Heat Capacity

J/g°CH2O(l) 4.184

Aluminum 0.890

Iron 0.450

Mercury 0.140

Carbon 0.710

Silver 0.240

What things affect temperature change?

• Heat Capacity of substance– The higher the heat capacity, the slower

the temperature change

• Mass of sample– The larger the mass, the more molecules

there are to absorb energy, so the slower the temperature change

TCmq

Energy added or removed

Mass of sample

Specific heat capacity of substance

Change in temperature

Example

TCmq

q = change in energym = massC = heat capacity DT = change in temperature (T2 - T1)

TCg

JgJ

184.445285

Example:If 285 J is added to 45 g of water at 25°C, what is the

final temperature? C water = 4.184 J/g°C

T

CgJg

J

184.445

285

12 TTT

CT 51.1

12 TTT

CCCT 51.262551.12

Let’s Practice #1

TCmq

q = change in energym = massCp = heat capacity DT = change in temperature (T2 - T1)

q = - 4707.00J

CCg

Jgq 0.750.3018.40.25

Example:How many joules must be

removed from 25.0 g of water at 75.0°C to drop the

temperature to 30.0°? Cp water = 4.184 J/g°C

Let’s Practice #2

TCmq

q = change in energym = massC = heat capacity DT = change in temperature (T2 - T1)

TCg

JgJ

900.00.30437

Example:If the specific heat capacity of

aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g

sample at 15.0°C

T

CgJg

J

900.00.30

437

12 TTT

CT 19.16

12 TTT

CCCT 19.310.1519.162

Thursday, April 17, 2014

• Grab a worksheet • Today:

– Pick up a handout.– Begin Calorimetry

• Homework:– Calorimeter Problems

• Tuesday:– Meet in the Maroon Room. Cool Demos.

CALORIMETRY

Intro to Calorimetry Problem: How much water was in the cup?

25 g Fe

T of Electrified device = 780 °C

Initial T of water = 25 °CFinal T of water = 42 °C

Remember:• q=mc∆T is an equation for

1 substance, not 2!• Energy is always

conserved.• Ciron=0.46 J/g°C

Conservation of Energy

• 1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes– This is also referred to as the Law of


• If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

Calorimetry

• Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system

• The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.

qsurroundings = - qsystem

(m×C×DT)surroundings = - (m×C×DT)system

• Don’t forget the “-” sign on one side• Make sure to keep all information about

surroundings together and all information about system together.

Two objects at different temperatures . . .

• Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature

• So you know that T2 for the system is the same as T2 for the surroundings!

Metal:m = 23.8 gT1 = 100.0°CT2 = 32.5°CC = ? Water:m = 50.0 gT1 = 24°CT2 = 32.5°CC = 4.184 J/g°C

Cp = 1.04 J/g°C

watermetal qq

watermetal TCmTCm

CC

CgJgCCCg

5.245.32184.40.500.1005.328.23

CCg

CCCg

JgC p

0.1005.328.23

5.245.32184.40.50

An example of CalorimetryExample:

A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of

the metal?

Example:A 10.0 g of aluminum (specific heat capacity

is 0.900 J/g°C) at 95.0°C is placed in a container of 100.0 g of water (specific heat capacity is 4.184 J/g°C) at 25.0°C. What’s

the final temperature?

Metal:m = 10.0 gT1 = 95.0°CT2 = ?C = 0.900 J/g°C Water:m = 100.0 gT1 = 25.0°CT2 = ?C = 4.184 J/g°C T2 = 26.47 °C

watermetal qq

watermetal TCmTCm

CT

CgJgCTCg

Jg

0.25184.40.1000.95900.00.10 22

Let’s Practice #2

CTCT 0.254.418950.9 22

104604.4188550.9 22 TT

113150.427 2 T

4.427

113152 T


• Today:– Nova Video on Making “Stuff”

• Next week:– Calorimetry Lab– Heating/Cooling Curves– Review ThermoChemistry Unit– Test on ThermoChemistry


• Today:– Nova Video on Making “Stuff” #2

• This Week:– Calorimetry Lab– Heating/Cooling Curves– Review ThermoChemistry Unit– Test on ThermoChemistry


• Grab a lab from my desk• Today:

– Spend the first 3 min starting the prelab– Go over the prelab– Work in PAIRS on the lab.

• Tomorrow:– Heating/Cooling Curves

• Reminder: Test on Friday

Thursday, April 25, 2013• Have out your Calorimetry Worksheet from last

week.• Pick up a Virtual Calorimetry Lab.• Today:

– Go Over Homework– Work on Lab

• Work in partners• Each is responsible for writing• Staple together and hand in before you leave• Finish INDIVIDUALLY for homework if you don’t

Calorimetry Worksheet

1. 0.451 J/g○C

2. 0.401 J/g○C

3. 40.0 g

4. 0.836 J/g○C

5. a. 627,600 J

b. 627,600 J

c. 150,000 J

d. 150 Cal


• Turn in your lab if you didn’t yesterday.• Pick up papers from the Red Lab Table• Today:

– Questions from papers that were passed back?– Thermo-stoichiometry– Work on Thermo-stoich worksheet – turn in before you leave– Begin working on a review

• Monday:– Review

• Tuesday:– Test

Enthalpy & Rx Type

• Enthalpy is abbreviated H

• We can measure changes in enthalpy . . .

• If H = + value– Reaction requires energy– endothermic

• If H = - value– Reaction looses energy– exothermic

What does this mean?H2 + Br2 2HBr + 36.4 kJ

• If 1 moles of H2 are reacted with excess Br2, how much energy is produced?– 36.4 kJ

• If 2 moles of H2 are reacted with excess Br2, how much energy is produced?– 72.8 kJ

• We have a mole/energy ratio!

What does this mean Continued?H2 + Br2 2HBr + 36.4 kJ

• If 1 moles of Br2 are reacted with excess H2, how much energy is produced?

• If 2 moles of Br2 are reacted with excess H2, how much energy is produced?– 36.4 kJ

• We have a mole/energy ratio!

An example of Thermo w/ Stoichiometry

• How much heat will be released when 6.44 g of sulfur reacts with excess O2 according to the following equation?

2S + 3O2 2SO3 ∆H=-791.4 kJ

6.44 g S 1 mol S

32.07 g S

kJ

mol S79.5 kJ

791.4

2


• Pickup Heating/Cooling Curve Worksheet• Today:

– Go Over Thermo-Stoichiometry– Notes on Heating/Cooling Curves– Work on Worksheet for Heating/Cooling Curves

• Tomorrow:– Review

• Wednesday:– Test on ThermoStoichiometry

SECTION 7.3 -CHANGES IN STATE

What’s happening when a frozen ice pack melts?

The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)

Change in State

To melt or boil, intermolecular forces must be broken

Breaking intermolecular forces requires energy

A sample with solid & liquid will not rise above the melting point until all the solid is gone.

The same is true for a sample of liquid & gas

Melting

• When going from a solid to a liquid, some of the intermolecular forces are broken

• All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy).

Vaporization

• When going from a liquid to a gas, all of the rest of the intermolecular forces are broken

• All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy).

Solid

Liquid

Gas

Melting

Vaporizing or Evaporating

Condensing

Freezing

Incr

easi

ng m

olec

ular

mot

ion

(tem

pera

ture

)

Changes in State go in Both Directions

Going the other way

• The energy needed to melt 1 gram is the same as the energy released when 1 gram freezes.– If it takes adding 547 J to melt a sample,

then removing 547 J would be released when the sample freezes.

• The energy needed to boil 1 gram is the same as the energy released when 1 gram is condensed.– If it takes 2798 J to boil a sample, then

2798 J will be released when a sample is condensed.

Example

vapHmH

H = enthalpy (energy)m = mass of sampleHfus = enthalpy of fusion

gcalgH 2.547)5.157(

DH = - 8.6×104 cal

Example:How much energy is

released with 157.5 g of water is condensed?

Hvap water = 547.2 cal/g

Since we’re condensing, we need to “release” energy…DH will be negative!

Heating curve of water

-50

0

50

100

150

Energy input

Te

mp

era

ture

Heating Curves

Melting & Freezing

Point

Boiling & Condensing

Point

Heating curves show how the temperature changes as energy is added to the sample


-50

0

50

100

150

Energy input

Te

mp

era

ture

Going Up & Down

+DH

-DH

Moving up the curve requires energy, while moving down releases energy


-50

0

50

100

150

Energy input

Te

mp

era

ture

States of Matter on the Curve

Gas OnlyEnergy added

increases temp

Liquid & gasEnergy added breaks remaining IMF’s

Liquid OnlyEnergy added

increases temp

Solid & LiquidEnergy added breaks IMF’s

Solid OnlyEnergy added

increases temp


-50

0

50

100

150

Energy input

Te

mp

era

ture

Different Heat Capacities

Gas OnlyC = 2.01 J/g°C

Liquid OnlyC = 4.184 J/g°C

Solid OnlyC = 2.13 J/g°C

The solid, liquid and gas states absorb water differently—use the correct C!


• Pick up work on Red Lab Table• Have out your homework• Today:

– Go Over Answers to Homework– Work on Review

• Tomorrow:– Test Unit 11

1. 1,3,5

2. 2,4

3. 2,4

4. 1,3,5

5. 1,3,5

6. 2,4

7. 2 min

8. 6.5 min

9. 22 min

10.endothermic

11. Scratch out

12. 1.8 min

13.13.5 min

14.1

15.5

16.Enthalpy of vaporization is greater than Enthalpy of fusion.

1. 5 C

2. 15 C

3. 5 C

4. A

5. C

6. E

7. B

8. D

9. B, D

10.A, C, E

11.D

12.B


• BoHA:– Turn in your March Madness if you have it complete. Do not turn it

in later.– Pick up a packet, and scantron.

• Today:– New MoleBucks– Makeups for Unit 11 Test?– ACT Practice

• Mark answers both in your booklet and your ScanTron!

• Homework:– Try to work out a logical solution the answer to the problem given.

If you don’t have an answer, at least come up with some logic to help you along tomorrow.


25 g Fe





concerved.• Ciron=0.46 J/g°C


• BoHA:– If you weren’t here yesterday, pickup ACT,

ScanTron and a handout.

• Today:– Pass Back ACT (I need ScanTron back)– New Seats– Introductory Calorimetry Problem (solve in

groups)– Notes on Calorimetry

• Homework:– Calorimetry Problems Worksheet

CALORIMETRY


25 g Fe





concerved.• Ciron=0.46 J/g°C


• 1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes– This is also referred to as the Law of


• If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

Calorimetry

• Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system

• The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.

qsurroundings = - qsystem

(m×C×DT)surroundings = - (m×C×DT)system

• Don’t forget the “-” sign on one side• Make sure to keep all information about

surroundings together and all information about system together.

Two objects at different temperatures

• Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature

• So you know that T2 for the system is the same as T2 for the surroundings!

Metal:m = 23.8 gT1 = 100.0°CT2 = 32.5°CC = ? Water:m = 50.0 gT1 = 24°CT2 = 32.5°CC = 4.184 J/g°C

Cp = 1.04 J/g°C

watermetal qq

watermetal TCmTCm

CC

CgJgCCCg

5.245.32184.40.500.1005.328.23

CCg

CCCg

JgC p

0.1005.328.23

5.245.32184.40.50

An example of CalorimetryExample:

A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of

the metal?

Example:A 10.0 g of aluminum (specific heat capacity

is 0.900 J/g°C) at 95.0°C is placed in a container of 100.0 g of water (specific heat capacity is 4.184 J/g°C) at 25.0°C. What’s

the final temperature?

Metal:m = 10.0 gT1 = 95.0°CT2 = ?C = 0.900 J/g°C Water:m = 100.0 gT1 = 25.0°CT2 = ?C = 4.184 J/g°C T2 = 26.47 °C

watermetal qq

watermetal TCmTCm

CT

CgJgCTCg

Jg

0.25184.40.1000.95900.00.10 22

Let’s Practice #3

CTCT 0.254.418950.9 22

104604.4188550.9 22 TT

113150.427 2 T

4.427

113152 T

Wednesday, April 6, 2011

• BoHA:–Grab a RSG for Unit 12

• Today:–Go Over Calorimetry Problems–Questions from Homework–ACT Questions?–Find specific heat of a metal.–Start RSG for Next Unit

Answers to Homework

1. 0.451 J/g°C

2. 0.401 J/g°C

3. 40.0 g

4. 0.836 J/g°C

5. multipart answers:a) 1,230,000 J are absorbed

b) 1,230,000 J are released

c) 294,000 calories

d) 294 Calories

friday, april 11, 2014 grab a lab from my desk: –read through the front page. today: –system vs....

Documents

surroundings system

surroundings exothermic

water slide

surroundings endothermic

water molecules surroundings

exothermic slide

system endothermic

endothermic slide