energy and changes in matter todays objectives differentiate between heat and temperature. interpret...
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Energy and Changes in Matter
Today’s Objectives
• Differentiate between heat and temperature.
• Interpret the heating and cooling curves of various substances.
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Heat versus Temperature
Heat Temperature
Definition
Energy Transfer between
Substances, Related to the Total Kinetic
Energy
A Measure of the Average
Kinetic Energy of a Substance
Units of Measurement
Joules (J)Calories (cal)
Celsius (oC)Kelvin (K)
How to Measure?
Indirectly Directly
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Match versus Ice Sculpture
Which Has a Higher Temperature?A Match
Which Has More Heat Energy?The Ice Sculpture
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Heat
Heat always flows from high temperatures to low temperatures.
Heat flows from the fire
to the marshmallo
w.
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Units of Measurement for Heat
1 cal = 4. 18 J1000 cal = 1 Cal
How many calories are in 54.0J?
calJ
calJ9.12
18.4
10.54
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Heat and Physical Changes
Add or Remove Heat
Change in Kinetic Energy of the
Substance
Change in Potential Energy of the Substance
Change in Temperature
Change in Phase
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Heating Curve
Boiling Point
MeltingPoint
Solid
Liqu
id
Gas
Vaporization
Melting
PE Potential Energy Changing
with Phase Changes
KEKinetic Energy Changing
with Temperature Changes
KE
KE
KE
PE
PE
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Energy and Changes in Matter
Today’s Objective
Perform calculations related to heat and changes in temperature or phase.
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Calculating Heat
To calculate the amount of heat energy when temperature changes,
Q = m Cp DT
Q = heatm = mass
Cp = specific heat
DT = Change in Temperature = Tfinal - Tinitial
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What is Specific Heat (Cp)?
Specific heat is the amount of energy required to change the temperature of a substance by 1oC.
A substance with a high specific heat requires more energy to change its temperature than a substance with a low specific heat.
Low Specific Heat = Good Conductors
High Specific Heat = Good Insulators
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ExamplesUn-Popable Balloon• Which balloon will pop first?– One with Only Air – One with Air and Water
Which substance is the best conductor?– Copper– Gold– Air
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Example Problem #1How much energy is required to raise a 34.0g sample of copper metal from 20.0oC to 45.0oC? The specific heat of copper is0.385 .
Q = m Cp DT
Q = (34.0g)(0.385 ) (45.0oC -20.0oC)
= (34.0g)(0.385 ) (25.0oC )
= 327 J
Jg oC
Jg oC
“to” – “from”
Jg oC
Q > 0 = heat is being absorbed.
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Example Problem #2How much heat is released when a 3.20g sample of water is cooled from 83.0oC to 54.0oC? The specific heat of water is4.18 .
Q = m Cp DT
Q = (3.20g)(4.18 ) (54.0oC -83.0oC)
= (3.20g)(4.18 ) (-29.0oC )
= -388 J
Jg oC
Jg oC
“to” – “from”
Jg oC
Q < 0 = heat is being released.
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Calculating Heat
To calculate the amount of heat energy when phase changes,
Q = mHQ = heat m = mass
Phase Change H Value for H
MeltingHeat of Fusion (Hf)
+Hf
Freezing -Hf
Vaporization Heat of Vaporization (Hv)
+Hv
Condensation -Hv
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Example Problem #3
How much heat is needed to melt 56.2g of ice at 0oC?
Q = mHf
Q = (56.2g)(334 )
= 18,800 J
J g
Q > 0 = endothermic phase changes
Use heat of fusion!
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Example Problem #4
How much heat is released when 120g of steam condenses?
Q = mHV
Q = (120g) (-2260 )
= -270,000 J
J g
Q < 0 = exothermic phase changes
Use a negative heat of vaporization!
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Important Reminders
When the Temperature Changes
Q = mCpDT
When the Phase Changes Q = mH
Phase Change H Value for H
MeltingHeat of Fusion (Hf)
+Hf
Freezing -Hf
Vaporization Heat of Vaporization (Hv)
+Hv
Condensation -Hv
DT = Temperature You Are Going To – Temperature You are Coming From
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Energy and Changes in Matter
Today’s Objective
Determine experimentally the heat of fusion of water.
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Important Reminders about Your Experiment
• Make sure that you always have ice in your calorimeter at ALL times!
• Your constant temperature should be between -4oC and 4oC.
• If you spill the water, you will have to re-do the lab.– This is especially important at the end when you
are measuring the volume. Your volume will exceed 100mL!
• For your calculations, only use the accepted value for the heat of fusion when you are calculating percent error.
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Energy and Changes in Matter
Today’s Objectives
• Solve problems related to heat and physical changes.
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Example #1
Determine the identity of a substance that requires 89.1J to raise 53.8g sample from 18.0oC to 22.3oC.
Q = m Cp DT
Cp = =
Cp = = 0.385
mDTmDT
(53.8g)(22.3oC-18.0oC)
89.1J
mDT
Q
(53.8g)(4.3oC)
89.1J Jg oC
Copper
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Example #2Determine the amount of heat energy required to change 4.50g of ice from -15.0oC to water at 56.0oC.
This problem is different because we are changing phase AND
temperature.
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Heating Curve of Water
Step 1: Ice at -15oC Ice at 0oC Q=mCpDT
Step 2: Ice at 0oC Water at 0oC Q=mHf
Step 3: Water at 0oC Water at 56oC Q=mCpDT
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Example #2 - Continued
Q = m Cp DT
Q = (4.5g)(2.05 )(0oC - - 15oC)
Q = (4.5g)(2.02 )(15oC)
Q = 138 J
Step 1: Ice at -15oC Ice at 0oC Q=mCpDT
Jg oC
Jg oC
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Example #2 - Continued
Q = m Hf
Q = (4.5g)(334 )
Q = 101700 J
Step 2: Ice at 0oC Water at 0oC Q=mHf
J g
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Example #2 - Continued
Q = m Cp DT
Q = (4.5g)(4.18 )(56oC - 0oC)
Q = (4.5g)(4.18 )(56oC)
Q = 1050 J
Step 3: Water at 0oC Water at 56oC Q=mCpDT
Jg oC
Jg oC
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Example #2 - Continued
Determine the amount of heat energy required to change 4.50g of ice from -15.0oC to water at 56.0oC.Calculate the total amount of heat.
Step 1 Q = 138 JStep 2 Q = 101700 JStep 3 Q = 1050J
103, 000J
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Example #3
How much energy is released when 4.00g of steam at 110.0oC is cooled and condensed to form water at 90.0oC?
Step 1: Steam at 110oC to 100oCStep 2: Steam Condenses to Water
Step 3: Water at 100oC to 90oC
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Example #3 - ContinuedStep 1: Steam at 110oC to 100oC
Q = mCpDT
Q = (4.00g)(2.02 )(100oC - 110oC)
Q = -80.8 J
Step 2: Steam Condenses to WaterQ = mHv
Q = (4.00g)(-2260 )
Q = -9040 J
J
g oC
J
g
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Example #3 - Continued
Step 3: Water at 100oC to 90oCQ = mCpDT
Q = (4.00g)(4.18 )(90oC - 100oC)
Q = -167 J
Total Energy ReleasedQ = -80.8 J + - 9040 J + -167 J
Q = -9290 J
Jg oC
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Thermochemistry
the study of the heat changes during chemical reactions, and the effects on
chemical and physical processes.
Today’s Objectives
• Differentiate between exothermic and endothermic reactions
• Interpret the potential energy diagrams for chemical reactions.
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Heat and Chemical Reactions
In chemical reactions two things occur,
1) The original chemical bonds are broken. This requires energy.
2) New chemical bonds are formed. This releases energy.
Because the energy it takes to break the bonds and the energy that is released when the new bonds are formed are not always equal, heat is either absorbed or released in chemical reactions.
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Heat and Chemical Reactions
Endothermic Reactions
• Absorb Energy
• Feel cold to the touch
• Heat is treated as a reactant.
Exothermic Reactions
• Release Energy
• Feel warm to the touch
• Heat is treated as a product.
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Potential Energy Diagram for an Endothermic Reaction
PE ofReactants
PE of Products
Heat of Reaction
(DH)
DH = Products – Reactants = 350kJ – 200kJ = 150 kJ
(kJ)
PEReactants < PEProducts
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Potential Energy Diagram for an Exothermic Reaction
PE ofReactants
PE of Product
s
Heat of Reaction
(DH)
DH = Products – Reactants = 15kJ – 40kJ = -25kJ
(kJ)
PEReactants > PEProducts
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Law of Conservation of Energy
Any chemical or physical process does not create or destroy
energy.
Energy Released or Absorbed by the System
Energy Absorbed or Released by
the Surroundings
=
System = What You are Studying
Surroundings = Everything Else/Generally Water
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CalorimetryHeat must be calculated indirectly. In other words, scientists use experimental means to calculate the heat energy associated
with a process. These experimental means are generally referred to as calorimetry.
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Tips for Solving Calorimetry Problems Successfully
• Calculate how much heat is gained or lost by the water.
• Assume that the amount of heat gained or lost by the water is equal to the amount of heat lost or gained by the other substance or reaction.
• Solve for the missing value for the other substance or reaction.