1. PHALL not integrated into presentation
Chapter
Thermochemistry

2. THERMODYNAMICS
Courtesy of lab-initio.com

3. 6.2 Enthalpy and Calorimetry 6.3 Hess’s Law
Chapter 6
Table of Contents
6.1 The Nature of Energy
6.2 Enthalpy and Calorimetry
6.3 Hess’s Law
6.4 Standard Enthalpies of Formation
6.5 Present Sources of Energy
6.6 New Energy Sources
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4. Finishing ch. 5 test - on computer
Chapter 6
Table of Contents
Finishing ch. 5 test - on computer
Notes with computer for section including problems (note to me: assigned as HW last year - incorporated into notes this year)
HW: Pre-lab Calorimetry
Turn in part 2 ch. 5 take home Test
Turn in previous lab reports
Turn in ch. 5 Interactive Notes packet for grade.
Turn in any missing or late assignments - check Pinnacle for grades
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5. Capacity to do work or to produce heat.
Section 6.1
The Nature of Energy
Energy
Capacity to do work or to produce heat.
That which is needed to oppose natural attractions.
Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed.
The First Law of Thermodynamics
The total energy content of the universe is constant.
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6. Thermodynamics is the study of energy and its interconversions.
Section 6.1
The Nature of Energy
Thermodynamics
Thermodynamics is the study of energy and its interconversions.
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7. Definitions #1 Energy: The capacity to do work or produce heat
Potential Energy: Energy due to position or composition
Kinetic Energy: Energy due to the motion of the object

8. Potential energy – energy due to position or composition.
Section 6.1
The Nature of Energy
Energy
Potential energy – energy due to position or composition.
Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.
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9. Section 6.1
The Nature of Energy
Initial Position
In the initial position, ball A has a higher potential energy than ball B.
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10. The Nature of Energy Final Position
Section 6.1
The Nature of Energy
Final Position
After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B.
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11. Energy Transformations
Section 10.1
The Nature of Energy
Energy Transformations
Types of Energy
1. Kinetic energy
2. Potential energy
3. Chemical energy
4. Heat energy
5. Electric energy
6. Radiant energy
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12. Section 6.1
The Nature of Energy
What are 2 ways energy is transferred? 1) By doing work 2) by Heat transfer - thermal energy is transferred
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13. Energy Transfer by Doing work
Section 6.1
The Nature of Energy
Energy can be transferred from one object to another by doing work. When work is done on an object, it results in a change in the object's motion (more specifically, a change in the object's kinetic energy).
Energy is often defined as the ability to do work. Work equals force multiplied by distance.
An illustration of how doing work is an example of energy transfer
Suppose that a person exerts a force on the wheelbarrow that is initially at rest, causing it to move over a certain distance. Recall that the work done on the wheelbarrow by the person is equal to the product of the person's force multiplied by the distance traveled by the wheelbarrow. Notice that when the force is exerted on the wheelbarrow, there's a change in its motion. Its kinetic energy increases. But where did the wheelbarrow get its kinetic energy? It came from the person exerting the force, who used chemical energy stored in the food they ate to move the wheelbarrow. In other words, when the person did work on the wheelbarrow, they transferred a certain amount of chemical energy stored in the person was transferred to the wheelbarrow, causing its kinetic energy to increase. As a result, the person's store of chemical energy decreases and the wheelbarrow's kinetic energy increases.
Wherever you look, you can see examples of energy transfers. When you turn on a light, you see result of energy being transferred from the sun to the plants to the coal to electricity and finally to light you see. During each of these transfers, energy changes form. There are two main forms of energy, kinetic energy (motion) and potential energy (position). To further classify energy, these forms are sometimes further described as thermal (heat), elastic, electromagnetic (light, electrical, magnetic), gravitational, chemical (food), and nuclear energy.
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14. Work – force acting over a distance.
Section 6.1
The Nature of Energy
Energy
Heat involves the transfer of energy between two objects due to a temperature difference.
Work – force acting over a distance.
Energy is a state function; work and heat are not:
State Function – property that does not depend in any way on the system’s past or future (only depends on present state).
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15. State Functions depend ONLY on the present state of the system
ENERGY IS A STATE FUNCTION
A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane 
WORK IS NOT A STATE FUNCTION
WHY NOT???

16. System – part of the universe on which we wish to focus attention.
Section 6.1
The Nature of Energy
Chemical Energy
System – part of the universe on which we wish to focus attention.
Surroundings – include everything else in the universe.
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17. Energy flows out of the system.
Section 6.1
The Nature of Energy
Chemical Energy
Exothermic Reaction:
Energy flows out of the system.
Energy gained by the surroundings must be equal to the energy lost by the system.
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18. Endothermic Reaction: Heat flow is into a system.
Section 6.1
The Nature of Energy
Chemical Energy
Endothermic Reaction:
Heat flow is into a system.
Absorb energy from the surroundings.
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19. Section 6.1
The Nature of Energy
Concept Check
Is the freezing of water an endothermic or exothermic process? Explain.
Exothermic process because you must remove energy in order to slow the molecules down to form a solid.
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20. Section 6.1
The Nature of Energy
Concept Check
Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.
Your hand gets cold when you touch ice.
The ice gets warmer when you touch it.
Water boils in a kettle being heated on a stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
Exothermic (heat energy leaves your hand and moves to the ice)
Endothermic (heat energy flows into the ice)
Endothermic (heat energy flows into the water to boil it)
Exothermic (heat energy leaves to condense the water from a gas to a liquid)
Endothermic (heat energy flows into the ice cream to melt it)
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21. Section 6.1
The Nature of Energy
Concept Check
Classify each process as exothermic or endothermic. Explain. The system is underlined in each example.
Your hand gets cold when you touch ice.
The ice gets warmer when you touch it.
Water boils in a kettle being heated on a stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
Exo
Endo
Exothermic (heat energy leaves your hand and moves to the ice)
Endothermic (heat energy flows into the ice)
Endothermic (heat energy flows into the water to boil it)
Exothermic (heat energy leaves to condense the water from a gas to a liquid)
Endothermic (heat energy flows into the ice cream to melt it)
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22. Section 6.1
The Nature of Energy
Concept Check
For each of the following, define a system and its surroundings and give the direction of energy transfer.
Methane is burning in a Bunsen burner in a laboratory.
Water drops, sitting on your skin after swimming, evaporate.
System – methane and oxygen to produce carbon dioxide and water; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the system to the surroundings (exothermic)
System – water drops; Surroundings – everything else around it; Direction of energy transfer – energy transfers from the surroundings (your body) to the system (water drops) (endothermic)
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23. Hydrogen gas and oxygen gas react violently to form water. Explain.
Section 6.1
The Nature of Energy
Concept Check
Hydrogen gas and oxygen gas react violently to form water. Explain.
Which is lower in energy: a mixture of hydrogen and oxygen gases, or water?
Water is lower in energy because a lot of energy was released in the process when hydrogen and oxygen gases reacted.
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24. To change the internal energy of a system: ΔE = q + w
Section 6.1
The Nature of Energy
Internal Energy
Law of conservation of energy is often called the first law of thermodynamics.
Internal energy E of a system is the sum of the kinetic and potential energies of all the “particles” in the system.
To change the internal energy of a system: ΔE = q + w
q represents heat
w represents work
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25. Work vs. Energy Flow Click here to watch visualization.
Copyright © Houghton Mifflin Company. All rights reserved. 6–

26. Sign reflects the system’s point of view. Endothermic Process:
Section 6.1
The Nature of Energy
Internal Energy
Sign reflects the system’s point of view.
Endothermic Process:
q is positive (heat enters the system)
Exothermic Process:
q is negative (heat exits the system)
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27. Sign reflects the system’s point of view.
Section 6.1
The Nature of Energy
Internal Energy
Sign reflects the system’s point of view.
Surroundings do work on the system:
w is positive
System does work on surroundings:
w is negative
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28. Work = F x d (from 9th grade) - applying to a gas
Section 6.1
The Nature of Energy
Work = F x d (from 9th grade) - applying to a gas
Work = P × A × Δh = PΔV
P is pressure.
A is area.
Δh is the piston moving a distance.
ΔV is the change in volume.
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29. To convert between L·atm and Joules, use 1 L·atm = 101.3 J.
Section 6.1
The Nature of Energy
Work
For an expanding gas, ΔV is a positive quantity because the volume is increasing. Thus ΔV and w must have opposite signs:
w = –PΔV
To convert between L·atm and Joules, use 1 L·atm = J.
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30. Work, Pressure, and Volume
Expansion
Compression
+ΔV (increase)
-ΔV (decrease)
-w results
+w results
Esystem decreases
Esystem increases
Work has been done by the system on the surroundings
Work has been done on the system by the surroundings

31. Figure 6.4 The Piston, Moving a Distance Against a Pressure P, Does Work On the Surroundings
Copyright © Houghton Mifflin Company. All rights reserved. 6–

32. Which of the following performs more work?
Section 6.1
The Nature of Energy
Exercise
w = –PΔV
Which of the following performs more work?
a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.
A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.
They both perform the same amount of work. w = -PΔV
a) w = -(2 atm)( ) = -6 L·atm
b) w = -(3 atm)( ) = -6 L·atm
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33. Which of the following performs more work?
Section 6.1
The Nature of Energy
Exercise
w = –PΔV
Which of the following performs more work?
a) A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.
A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.
They perform the same amount of work.
(-6 L*atm)
They both perform the same amount of work. w = -PΔV
a) w = -(2 atm)( ) = -6 L·atm
b) w = -(3 atm)( ) = -6 L·atm
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34. ΔE = q + w ΔE = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving to the surroundings
+q if energy is entering from the surroundings
w = work done by, or on, the system
-w if work is done by the system on the
surroundings
+w if work is done on the system by the
surroundings

35. Section 6.1
The Nature of Energy
Concept Check
ΔE = q + w
w = –PΔV
Determine the sign of ΔE for each of the following with the listed conditions:
a) An endothermic process that performs work on surrounding.
|work| > |heat|
|work| < |heat|
b) Work is done on a gas and the process is exothermic.
Δ E = negative wk (-) heat +
Δ E = positive
Δ E = positive wk + and heat -
Δ E = negative
a) q is positive for endothermic processes and w is negative when system does work on surroundings; first condition – ΔE is negative; second condition – ΔE is positive
b) q is negative for exothermic processes and w is positive when surroundings does work on system; first condition – ΔE is positive; second condition – ΔE is negative
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36. Thermodynamic Quantities
Section 6.1
The Nature of Energy
Thermodynamic Quantities
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37. Energy Change in Chemical Processes
Endothermic:
Reactions in which energy flows into the
system as the reaction proceeds.
+ qsystem qsurroundings

38. Energy Change in Chemical Processes
Exothermic:
Reactions in which energy flows out of the system as the reaction proceeds.
- qsystem qsurroundings

39. Exercise 1 Internal Energy
Section 6.1
The Nature of Energy
Exercise 1
Exercise 1 Internal Energy
Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.
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40. Exercise 1 Internal Energy
Section 6.1
The Nature of Energy
Exercise 1
Exercise 1 Internal Energy
Calculate ΔE for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.
ANSWER: kJ
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41. The Nature of Energy Exercise 2 Exercise 2 PV Work
Section 6.1
The Nature of Energy
Exercise 2
Exercise 2 PV Work
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.
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42. The Nature of Energy Exercise 2 Exercise 2 PV Work
Section 6.1
The Nature of Energy
Exercise 2
Exercise 2 PV Work
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.
ANSWER: ‒270 L•atm
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43. Exercise 3 Internal Energy, Heat, and Work
Section 6.1
The Nature of Energy
Exercise 3 Internal Energy, Heat, and Work
Exercise 3 A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the addition of 1.3 • 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅ atm = J.)
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44. Exercise 3 Internal Energy, Heat, and Work
Section 6.1
The Nature of Energy
Exercise 3 Internal Energy, Heat, and Work
Exercise 3 A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the addition of 1.3 • 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅ atm = J.)
ANSWER: 8.0 • 107 J
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45. Stopped here 2-5-13 - need to make ch.3-4 test shorter.

46. Click on the following link and
Section 6.2
Temperature and Heat
Temperature - Check out this simulation regarding the effect temperature on molecular motion.
Click on the following link and
basics
Click on Run Now. Open with Java.
Adjust the temperatures, elements, and states of matter to see what happens from the solid, liquid, gas tab.
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47. A measure of the random motions of the components of a substance.
Section 6.2
Temperature and Heat
Temperature
A measure of the random motions of the components of a substance.
Temperature (T)—is proportional to the average kinetic energy of the molecules, KEave .
“Heat ‘em up and speed ‘em up”.
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48. Temperature and Heat Heat
Section 6.2
Temperature and Heat
Heat
A flow of energy between two objects due to a temperature difference between the objects.
Heat is the way in which thermal energy is transferred from a hot object to a colder object.
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49. Section 6.2
Temperature and Heat
Heat
Two systems with different temperatures that are in thermal contact will exchange thermal energy, the quantity of which is call heat. This transfer of energy in a process (flows from a warmer object to a cooler one, transfers heat because of temperature difference but, remember, temperature is not a measure of energy—it just reflects the motion of particles)
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50. Enthalpy and Calorimetry
Section 6.2
Enthalpy and Calorimetry
Enthalpy (H)– flow of energy (heat exchange) at constant pressure when two systems are in contact.
 Enthalpy of reaction (ΔHrxn) – amount of heat released (negative values) or absorbed (positive values) by a chemical reaction at constant pressure in kJ/molrxn
 Enthalpy of combustion (ΔHcomb)—heat absorbed or released by burning (usually with O2) in kJ/molrxn; note that combustion reactions yield oxides of that which is combusted
 Enthalpy of formation (ΔHf) – heat absorbed or released when ONE mole of compound is formed from elements in their standard states in kJ/molrxn
 Enthalpy of fusion (ΔHfus)—heat absorbed to melt (overcome IMFs) 1 mole of solid to MP expressed in kJ/molrxn
 Enthalpy of vaporization (ΔHvap)—heat absorbed to vaporize or boil (overcome IMFs) 1 mole liquid to in kJ/molrxn
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51. Change in Enthalpy (ΔH )
Section 6.2
Enthalpy and Calorimetry
Change in Enthalpy (ΔH )
State function
ΔH = q at constant pressure (ex. atmospheric pressure)
ΔH = Hproducts – Hreactants
Measure only the change in enthalpy, ΔH ( the difference between the potential energies of the products and the reactants)
A system that releases heat to the surroundings, an exothermic reaction, creates a negative ΔH because the enthalpy of the products is lower than the enthalpy of the reactants of the system (Figure 1). 
A system that absorbs heat from the surroundings, an endothermic reaction, creates a positive ΔH, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.
A positive ΔH is endothermic rxn
A negative ΔH is exothermic rxn
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52. Enthalpy Calculations
Section 6.2
Enthalpy and Calorimetry
Enthalpy Calculations
Enthalpy can be calculated from several
sources including:
 Stoichiometry
 Calorimetry
 From tables of standard values
 Hess’s Law
 Bond energies
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53. Enthalpy and Calorimetry
Stoichiometrically:
Section 6.2
Enthalpy and Calorimetry
Exercise 4
Upon adding solid potassium hydroxide pellets to water the following reaction takes place:
KOH(s) → KOH(aq) + 43 kJ/mol
Answer the following questions regarding the addition of 14.0 g of KOH to water:
a) Does the beaker get warmer or colder? How do you know?
b) Is the reaction endothermic or exothermic?
c) What is the enthalpy change for the dissolution of the
14.0 grams of KOH?
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54. Enthalpy and Calorimetry
Stoichiometrically:
Section 6.2
Enthalpy and Calorimetry
Exercise 4
Upon adding solid potassium hydroxide pellets to water the following reaction takes place:
KOH(s) → KOH(aq) + 43 kJ/mol
Answer the following questions regarding the addition of 14.0 g of KOH to water:
a) Does the beaker get warmer or colder?
b) Is the reaction endothermic or exothermic?
c) What is the enthalpy change for the dissolution of the
14.0 grams of KOH?
ANSWER: (a) warmer (the +43 kJ/mole is heat on the product side) (b) exothermic (c) −10.7 kJ/molrxn (take 14 g divide by the molar mass of KOH - to get # moles - 1 to 1 ratios and thus if 1 mole produces 43 kJ, then approx. .25 mol would be (.25 mol x 43 kJ)
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55. Consider the combustion of propane:
Section 6.2
Enthalpy and Calorimetry
Exercise 5
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = – 2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which g of propane is burned in excess oxygen at constant pressure.
(5.00 g C3H8)(1 mol / g C3H8)(-2221 kJ / mol C3H8)
ΔH = -252 kJ
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56. Consider the combustion of propane:
Section 6.2
Enthalpy and Calorimetry
Exercise 5
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = – 2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which g of propane is burned in excess oxygen at constant pressure.
5.00 g C3H8 x 1 mol C3H8 /44 g x kJ/1 mol C3H8 = –252 kJ
(5.00 g C3H8)(1 mol / g C3H8)(-2221 kJ / mol C3H8)
ΔH = -252 kJ
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57. Science of measuring heat
Section 6.2
Calorimetry
Enthalpy and Calorimetry
Science of measuring heat
Heat capacity – energy required to raise temp. by 1 degree (Joules/ °C)
Specific heat capacity: (s)
The energy required to raise the temperature of one gram of a substance by one degree Celsius.
Molar heat capacity: (Cp)
The energy required to raise the temperature of one mole of substance by one degree Celsius.
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58. An endothermic reaction cools the solution.
Section 6.2
Enthalpy and Calorimetry
Calorimetry
If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.
An endothermic reaction cools the solution.
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59. Calorimetry
The amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimeter.

60. Enthalpy and Calorimetry
Section 6.2
Enthalpy and Calorimetry
A Coffee–Cup Calorimeter Made of Two Styrofoam Cups Coffee-cup calorimetry – in the lab this is how we experiment to find energy of a particular system. We use a Styrofoam® cup, reactants that begin at the same temperature and look for change in temperature. After all data is collected (mass or volume; initial and final temperatures) we can use the specific formula to find the energy released or absorbed. We refer to this process as constant pressure calorimetry. ** q = these conditions**
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61. q = Energy released (heat) (or gained)
Section 6.2
Enthalpy and Calorimetry
Calorimetry
q = s × m × ΔT
q = Energy released (heat) (or gained)
s = specific heat capacity (J/°C·g)
m = mass (g)
ΔT = change in temperature (°C)
(Q = Cp x m × ΔT) - some use Cp as specific heat variable.
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62. Units for Measuring Heat
The Joule is the SI system unit for measuring heat:
The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree
1 BTU (British Thermal Unit) is the heat required to raise the temperature of 1 pound of water by 1 °F

63. The common energy units for heat are the calorie and the joule.
Section 6.2
Measuring Energy Changes
The common energy units for heat are the calorie and the joule.
calorie – the amount of energy (heat) required to raise the temperature of one gram of water 1oC.
Joule – 1 calorie = joules
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64. Measuring Energy Changes
Section 6.2
Measuring Energy Changes
Exercise 6
Convert 60.1 cal to joules and kJ
1000 J = 1 kiloJoule (kJ)
251 J = kJ
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65. Measuring Energy Changes
Section 6.2
Measuring Energy Changes
 Energy (q) released or gained at constant pressure:
q = m x s x ΔT
q = quantity of heat (Joules or calories)
m = mass in grams
ΔT = Tf −Ti (final – initial)
s = specific heat capacity ( J/g°C) (or Cp with some)
 Specific heat of water (liquid state) = J/g°C (or cal/g°C)
*Water has one of the highest specific heats known! This property makes life on earth possible and regulates earth’s temperature year round!
 Heat lost by substance = heat gained by water
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66. Measuring Energy Changes
Section 6.2
Measuring Energy Changes
Exercise 7
Calculate the amount of heat energy (in joules) needed to raise the temperature of g of water from 21.0°C to 39.0°C.
Where are we going?
We want to determine the amount of energy needed to increase the temperature of 6.25 g of water from 21.0°C to 39.0°C.
What do we know?
The mass of water and the temperature increase.
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67. q = s × m × ΔT Measuring Energy Changes Example 7
Section 6.2
Measuring Energy Changes
q = s × m × ΔT
Example 7
Calculate the amount of heat energy (in joules) needed to raise the temperature of g of water from 21.0°C to 39.0°C.
What information do we need?
We need the specific heat capacity of water.
4.184 J/g°C
How do we get there?
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68. q = s × m × ΔT Measuring Energy Changes Exercise 8 Section 6.2
A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.)
a) g
b) 4.6 g
c) 19 g
d) 590 g
The correct answer is c.
Q = s×m×ΔT
m = Q/ s×ΔT
m = {(142 cal)(4.184 J/1 cal)} / {(0.45 J/g°C)(92°C–23°C)}
m = 19 g
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69. q = s × m × ΔT Measuring Energy Changes Exercise 8 Section 6.2
A sample of pure iron requires 142 cal of energy to raise its temperature from 23ºC to 92ºC. What is the mass of the sample? (The specific heat capacity of iron is 0.45 J/gºC.)
a) g
b) 4.6 g
c) 19 g
d) 590 g
The correct answer is c.
Q = s×m×ΔT
m = Q/ s×ΔT
m = {(142 cal)(4.184 J/1 cal)} / {(0.45 J/g°C)(92°C–23°C)}
m = 19 g
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70. The final temperature of the water is: a) Between 50°C and 90°C
q = s × m × ΔT
Section 6.2
Enthalpy and Calorimetry
Concept Check
Example 9
A g sample of water at 90°C is added to a g sample of water at 10°C.
The final temperature of the water is:
a) Between 50°C and 90°C
b) 50°C
c) Between 10°C and 50°C
The correct answer is b).
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71. The final temperature of the water is: a) Between 50°C and 90°C
q = s × m × ΔT
Section 6.2
Enthalpy and Calorimetry
Concept Check
Example 9
A g sample of water at 90°C is added to a g sample of water at 10°C.
The final temperature of the water is:
a) Between 50°C and 90°C
b) 50°C - work on next screens
c) Between 10°C and 50°C
The correct answer is b).
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72. Enthalpy and Calorimetry
Section 6.2
Enthalpy and Calorimetry
Problem
We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Δt, but the FINAL temperature. This is what we are solving for.
The warmer water goes down from to 90 to x, so this means its Δt equals 90 minus x. The colder water goes up in temperature, so its Δt equals x minus 10.
That last paragraph may be a bit confusing, so let's compare it to a number line: x
To compute the absolute distance, it's the larger value minus the smaller value, so 90 to x is 90 minus x and the distance from x to 10 is x - 10.
These two distances on the number line represent our two Δt values:
a) the Δt of the warmer water is 90 minus x
b) the Δt of the cooler water is x minus 10
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73. Enthalpy and Calorimetry
Problem
Section 6.2
Enthalpy and Calorimetry
q lost = q gain
q = (mass) (Δt) (s)
So:
(mass) (Δt) (s) = (mass) (Δt) (s)
With qlost on the left side and qgain on the right side. Since these are both liquid water the specific heat capacity (s) are equal and can be eliminated from the equation. Also, since the amount of water is the same, it too could be eliminated. Then we let x be final temperature
Substituting values into the above, we then have
90 - x = x Then solve for x (the final temperature)
100 = 2x so x = 50
The final temperature is at 50 degrees C.
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74. q = s × m × ΔT Concept Check Enthalpy and Calorimetry Exercise 10
Section 6.2
Enthalpy and Calorimetry
Exercise 10
A g sample of water at 90.°C is added
to a g sample of water at 10.°C.
The final temperature of the water is:
a) Between 50°C and 90°C
b) 50°C
c) Between 10°C and 50°C
Calculate the final temperature of the water.
100 x 4.19 x (90-t) = 500 x 4.19 x (t - 10) =
t = 500t so = 600t
t = = 23°C
q = s × m × ΔT
The correct answer is c). The final temperature of the water is 23°C.
- (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf – 10.)
Tf = 23°C
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75. q = s × m × ΔT Enthalpy and Calorimetry Concept Check Exercise 11
Section 6.2
Enthalpy and Calorimetry
Concept Check
q = s × m × ΔT
Exercise 11
You have a Styrofoam cup with 50.0 g of water at 10.°C. You add a 50.0 g iron ball at 90.°C to the water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g)
The final temperature of the water is:
a) Between 50°C and 90°C
b) 50°C
c) Between 10°C and 50°C
Calculate the final temperature of the water.
(4.18) (t-10) = (.45) (90-t) t = = 18°C
The correct answer is c). The final temperature of the water is 18°C.
- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.)
Tf = 18°C
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76. Table 6.1 The Specific Heat Capacities of Some Common Substances
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77. Enthalpy and Calorimetry Type 2 Problem Exercise 12
Section 6.2
Enthalpy and Calorimetry
Type 2 Problem Exercise 12
A g sample of Cu (specific heat capacity = .20 J/°C·g is heated to 82.4 °C and then placed in a container of water at 22.3°C. The final temperature of the water and the copper is 24.9°C. What was the mass of the water in the original container, assuming that all of the heat lost by the copper is gained by the water?
The heat lost by copper is equal to:
110.0g Cu x 0.20 J/°C·g x (82.4°C °C)= 1265 J
The heat lost (-value) by Cu is equal to the heat gained by water.
mH2O = q/(s x Δt) so 1265 J / (4.184 J x ( )) = g water.
The correct answer is c). The final temperature of the water is 18°C.
- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf – 10.)
Tf = 18°C
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78. The Heat Curve for Water, going from -20 to 120 oC,78

79. The liquid temperature is rising from 0 to 100 oC (q = s x m x ΔT)
120
The liquid is boiling at 100o C; no temperature change (use q = mol x ΔHvaporization)
The liquid is boiling at 100o C; no temperature change (use q = mol x ΔHvap.)
The gas temperature is rising from 100 to 120 oC (use (use q = s x m x ΔT)
The liquid temperature is rising from 0 to 100 oC (q = s x m x ΔT)
The solid is melting at 0o C; no temperature change (use q = mol x ΔHfusion)
The solid temperature is rising from -20 to 0 oC (use q = s x m x ΔT) 79

80. Enthalpy and Calorimetry
Exercise 13
Section 6.2
Enthalpy and Calorimetry
20.0 mL of pure water at 285 K is mixed with 48.0 mL of water at 315 K. What is the final temperature of the mix? Requires 5 steps. Assume all heat from warm water transferred to cold water. q gain = q lost (Density pure water is 1.0 g/mL so 1 gm/L = m/20 mL
a) m × ΔT x s = m × ΔT x s
b) (20 g) (tf-285) (s) = (48 g) (315 - tf) (s)
c) 20tf = tf
d) 68tf = 20820
e) tf = = 306 K
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81. Enthalpy and Calorimetry
Exercise 14
Section 6.2
Enthalpy and Calorimetry
20.0 g of water at 28.5 °C is mixed with 32.5 g of water at °C. What is the final temperature after complete mixing? (Assume no energy is lost to the surroundings.)
q cold = q warm
a) m × ΔT x s = m × ΔT x s
b) (20 g) (tf-28.5) (s) = (32.5 g) (77 - tf) (s) (liquids - s cancels)
c) 20tf = tf
d) tf =
e) tf = 58.5 °C
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82. Enthalpy and Calorimetry
Exercise 15
Section 6.2
Enthalpy and Calorimetry
Exercise 15 In a coffee cup calorimeter, mL of 1.0 M NaOH and mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.
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83. Enthalpy and Calorimetry
Exercise 15
Section 6.2
Enthalpy and Calorimetry
Exercise 15 In a coffee cup calorimeter, mL of 1.0 M NaOH and mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6°C. After the reaction, the final temperature is 31.3°C. Assuming that all solutions have a density of 1.0 g/cm3 and a specific heat capacity of J/g°C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the calorimeter.
ANSWER: ‒5.6 kJ/mol (m = mass = 200 g (because we have 100 mL NaOH plus 100 mL HCl each with a density of 1 g / mL) Then using the equation q = m * s * ΔT = (200 g) (4.184 J/g*C) * 6.7 C) = J = 5.6 kJ
Since this is an exothermic reaction, q is negative.
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84. Exercise 16
Section 6.2
Enthalpy and Calorimetry
When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8 gram sample of methane is burned at constant pressure.
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85. Exercise 16
Section 6.2
Enthalpy and Calorimetry
When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ/mol of energy is released as heat. Calculate ΔH for a process in which a 5.8 gram sample of methane is burned at constant pressure.
ANSWER: ΔH = heat flow = ‒320 kJ/mol
5.8 g / 16.0 g = moles so set up proportion
1 mole/890 kJ = 0.36 mol/x
so x = (890)(.36)/(1) = = 320 kJ/mol and is -320 kJ/mol since exothermic reaction
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86. Enthalpy and Calorimetry
Exercise 17
Section 6.2
Enthalpy and Calorimetry
Exercise 17 Constant-Pressure Calorimetry
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/°C ⋅ g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.
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87. Enthalpy and Calorimetry
Exercise 17
Section 6.2
Enthalpy and Calorimetry
Exercise 17 Constant-Pressure Calorimetry
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, and that the specific heat capacity of the solution is 4.18 J/°C ⋅ g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.
ANSWER = ‒26 kJ/mol
2 L total volume = 2000 mL x (1.0 g/mL) = 2000 grams of sol.
ΔH = q at constant pressure = q = s × m × ΔT
= (4.18 J/C*g) (2000 g) ( C) = 25,916 Joules or 25.9 = 26 kJ per mole
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88. Specific Heat
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·C)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)

89. Enthalpy and Calorimetry
Section 6.2
Enthalpy and Calorimetry
END OF NOTES
HW: Read Section by next Tuesday.
HW: Pre-lab Calorimetry due on Friday for lab
Turn in assignments including take home test ch. 5 if not done already no later than Friday if not turned in today.
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