1. Friday, April 11, 2014 Grab a Lab from my Desk: Today:
Read through the front page.
Today:
System vs. Surroundings
Endothermic vs. Exothermic
Chemical vs. Physical
Perform Lab
Water means tap water

2. Unit 11 Hot/Cold Packs 7.1 Endothermic and Exothermic
7.2 Calorimetry and Heat Capacity
7.3 Changes in State

3. 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?

4. 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.

5. Hot/Cold Packs Physical change Transfer of energy Chemical change
Use
Physical change
Can be done in
Transfer of energy
Chemical change
Effect on temperature depends on
between
System & Surroundings
Materials ability to absorb energy without noticeable temperature change
Is determined with
Calorimetry

6. System vs. Surroundings
The chemicals that are involved in the chemical/physical process
Surroundings:
The things that are surrounding those chemicals:
Water, beaker, air, lab the universe

7. Endothermic & Exothermic
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.

8. Section 7.1 - Endothermic and Exothermic

9. System & Surroundings
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.

10. 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

11. 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

12. 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

13. 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 ice
Surroundings: 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

15. 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

17. Monday, April 14, 2014 Today: Homework: Notes/Demonstrations of:
Energy Flow (Change in Energy)
Exothermic and endothermic.
Units of Energy
Homework:
Energy Conversions

18. Enthalpy 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

19. Types of Processes (Reactions)
Exothermic
Endothermic
Exo – “out of”
Energy flows out of the reaction into surroundings.
Surroundings get warmer
Heat, light, vibrations, explosions
H is negative
Endo – “into”, “inside of”
Energy flows into the reaction from the surroundings
Surroundings get colder
H is positive

20. Energy Diagram 1
Exothermic!

21. Energy Diagram 2
Endothermic!

22. Demonstrations * Danger! Energy Change! Demo 1 Demo 2 20 g Ba(OH)2
10 g ammonium thiocyanate
Place in beaker on a wet wooden board.
Mix thoroughly.
* Danger! Energy Change!
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!

23. 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

24. 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”

25. Energy Units
The most common energy units are Joules (J) and calories (cal)
These equivalents can be used in dimensional analysis to convert units
Energy Equivalents =
4.184 J
1.00 cal
1000 J =
1 kJ =
1000 cal
1 Cal (food calorie)

27. 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
– sent on Friday

29. Today: Homework: 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

30. 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

31. Heat Capacity High Heat Capacity Low 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
Small amount of energy can noticeably change temperature
Heats up quickly
Cools down quickly
Quickly readjusts to new conditions

33. Substance Heat Capacity J/g°C
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°C
H2O(l)
4.184
Aluminum
0.890
Iron
0.450
Mercury
0.140
Carbon
0.710
Silver
0.240

34. 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
Energy added or removed
Specific heat capacity of substance
Change in temperature
Mass of sample

35. Example
Example:
If 285 J is added to 45 g of water at 25°C, what is the final temperature? C water = J/g°C
q = change in energy
m = mass
C = heat capacity
DT = change in temperature (T2 - T1)

36. Let’s Practice #1 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 = J/g°C
q = change in energy
m = mass
Cp = heat capacity
DT = change in temperature (T2 - T1)
q = J

37. Let’s Practice #2 Example:
If the specific heat capacity of aluminum is J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15.0°C
q = change in energy
m = mass
C = heat capacity
DT = change in temperature (T2 - T1)

39. Thursday, April 17, 2014 Grab a worksheet Today: Homework: Tuesday:
Pick up a handout.
Begin Calorimetry
Homework:
Calorimeter Problems
Tuesday:
Meet in the Maroon Room. Cool Demos.

40. Calorimetry

41. Intro to Calorimetry Problem:  How much water was in the cup?
T of Electrified device = 780 °C
Remember:
q=mc∆T is an equation for 1 substance, not 2!
Energy is always conserved.
Ciron=0.46 J/g°C
25 g Fe
Initial T of water = 25 °C
Final T of water = 42 °C

42. 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 Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

43. 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.

44. 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!

45. An example of Calorimetry
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?
Metal:
m = 23.8 g
T1 = 100.0°C
T2 = 32.5°C
C = ?
Water:
m = 50.0 g
T1 = 24°C
C = J/g°C
Cp = 1.04 J/g°C

46. Let’s Practice #2 Example:
A 10.0 g of aluminum (specific heat capacity is J/g°C) at 95.0°C is placed in a container of g of water (specific heat capacity is J/g°C) at 25.0°C. What’s the final temperature?
Metal:
m = 10.0 g
T1 = 95.0°C
T2 = ?
C = J/g°C
Water:
m = g
T1 = 25.0°C
C = J/g°C
T2 = °C

48. Friday, April 25, 2014 Today: Next week: Nova Video on Making “Stuff”
Calorimetry Lab
Heating/Cooling Curves
Review ThermoChemistry Unit
Test on ThermoChemistry

49. Monday, April 28, 2014 Today: This Week:
Nova Video on Making “Stuff” #2
This Week:
Calorimetry Lab
Heating/Cooling Curves
Review ThermoChemistry Unit
Test on ThermoChemistry

50. Tuesday, April 29, 2014 Grab a lab from my desk Today: Tomorrow:
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

55. 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

56. Calorimetry Worksheet
0.451 J/g○C
0.401 J/g○C
40.0 g
0.836 J/g○C
627,600 J
150,000 J
150 Cal

61. Friday, April 26, 2013 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

62. 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

63. 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!

64. 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!

65. 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= kJ
6.44 g S
1 mol S
791.4 kJ
79.5 kJ
32.07 g S
mol S 2

69. Monday, April 29, 2013 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

70. Section 7.3 - Changes in State
What’s happening when a frozen ice pack melts?
Section Changes in State

71. Change in State
The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)
Breaking intermolecular forces requires energy
To melt or boil, intermolecular forces must be broken
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

72. 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).

73. 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).

74. Changes in State go in Both Directions
Increasing molecular motion (temperature)
Changes in State go in Both Directions
Gas
Vaporizing or
Evaporating
Liquid
Melting
Condensing
Solid
Freezing

75. 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.

76. Example
Example:
How much energy is released with g of water is condensed? Hvap water = cal/g
H = enthalpy (energy)
m = mass of sample
Hfus = enthalpy of fusion
Since we’re condensing, we need to “release” energy…DH will be negative!
DH = - 8.6×104 cal

77. Heating Curves
Heating curves show how the temperature changes as energy is added to the sample
Boiling & Condensing
Point
Melting & Freezing
Point

78. Going Up & Down
Moving up the curve requires energy, while moving down releases energy
+DH
-DH

79. States of Matter on the Curve
Liquid & gas
Energy added breaks remaining IMF’s
Liquid Only
Energy added increases temp
Gas Only Energy added increases temp
Solid Only Energy added increases temp
Solid & Liquid
Energy added breaks IMF’s

80. Different Heat Capacities
The solid, liquid and gas states absorb water differently—use the correct C!
Liquid Only
C = J/g°C
Gas Only C = 2.01 J/g°C
Solid Only C = 2.13 J/g°C

82. Pick up work on Red Lab Table Have out your homework Today:
Tuesday, April 30, 2013
Pick up work on Red Lab Table
Have out your homework
Today:
Go Over Answers to Homework
Work on Review
Tomorrow:
Test Unit 11

83. 11. Scratch out
min
13.5 min 1 5
Enthalpy of vaporization is greater than Enthalpy of fusion.
5 C
15 C A C E B D
B, D
A, C, E
1,3,5
2,4
2 min
6.5 min
22 min
endothermic

85. Monday, April 4, 2011 BoHA: Today: Homework:
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.

86. Intro to Calorimetry Problem:  How much water was in the cup?
T of Electrified device = 780 °C
Remember:
q=mc∆T is an equation for 1 substance, not 2!
Energy is always concerved.
Ciron=0.46 J/g°C
25 g Fe
Initial T of water = 25 °C
Final T of water = 42 °C

88. Tuesday, April 5, 2011 BoHA: Today: Homework:
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

89. Calorimetry

90. Intro to Calorimetry Problem:  How much water was in the cup?
T of Electrified device = 780 °C
Remember:
q=mc∆T is an equation for 1 substance, not 2!
Energy is always concerved.
Ciron=0.46 J/g°C
25 g Fe
Initial T of water = 25 °C
Final T of water = 42 °C

91. 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 Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

92. 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.

93. 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!

94. An example of Calorimetry
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?
Metal:
m = 23.8 g
T1 = 100.0°C
T2 = 32.5°C
C = ?
Water:
m = 50.0 g
T1 = 24°C
C = J/g°C
Cp = 1.04 J/g°C

95. Let’s Practice #3 Example:
A 10.0 g of aluminum (specific heat capacity is J/g°C) at 95.0°C is placed in a container of g of water (specific heat capacity is J/g°C) at 25.0°C. What’s the final temperature?
Metal:
m = 10.0 g
T1 = 95.0°C
T2 = ?
C = J/g°C
Water:
m = g
T1 = 25.0°C
C = J/g°C
T2 = °C

97. BoHA: Today: Wednesday, April 6, 2011 Grab a RSG for Unit 12
Go Over Calorimetry Problems
Questions from Homework
ACT Questions?
Find specific heat of a metal.
Start RSG for Next Unit

98. Answers to Homework 0.451 J/g°C 0.401 J/g°C 40.0 g 0.836 J/g°C
multipart answers:
1,230,000 J are absorbed
1,230,000 J are released
294,000 calories
294 Calories