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

Unit 11 Hot/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 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

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

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.

Section 7.1 - Endothermic and Exothermic

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.

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 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

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: Homework: Notes/Demonstrations of: Energy Flow (Change in Energy) Exothermic and endothermic. Units of Energy Homework: Energy Conversions

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

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

Energy Diagram 1 Exothermic!

Energy Diagram 2 Endothermic!

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!

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 Energy Equivalents = 4.184 J 1.00 cal 1000 J = 1 kJ = 1000 cal 1 Cal (food calorie)

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

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

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 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

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

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

Example 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 q = change in energy m = mass C = heat capacity DT = change in temperature (T2 - T1)

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 = 4.184 J/g°C q = change in energy m = mass Cp = heat capacity DT = change in temperature (T2 - T1) q = - 4707.00J

Let’s Practice #2 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 q = change in energy m = mass C = heat capacity DT = change in temperature (T2 - T1)

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.

Calorimetry

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

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

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!

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 = 4.184 J/g°C Cp = 1.04 J/g°C

Let’s Practice #2 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 g T1 = 95.0°C T2 = ? C = 0.900 J/g°C Water: m = 100.0 g T1 = 25.0°C C = 4.184 J/g°C T2 = 26.47 °C

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

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

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

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 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

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

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 791.4 kJ 79.5 kJ 32.07 g S mol S 2

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

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

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

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).

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

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 Example: How much energy is released with 157.5 g of water is condensed? Hvap water = 547.2 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

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

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

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

Different Heat Capacities The solid, liquid and gas states absorb water differently—use the correct C! Liquid Only C = 4.184 J/g°C Gas Only C = 2.01 J/g°C Solid Only C = 2.13 J/g°C

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

11. Scratch out 12. 1.8 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

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.

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

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

Calorimetry

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

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

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!

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 = 4.184 J/g°C Cp = 1.04 J/g°C

Let’s Practice #3 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 g T1 = 95.0°C T2 = ? C = 0.900 J/g°C Water: m = 100.0 g T1 = 25.0°C C = 4.184 J/g°C T2 = 26.47 °C

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

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