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DO NOW: Hand in specific heat lab & answer the following question on “Do Now” sheet:. 56 grams of hot copper are added to 140 grams of water in a 92 g aluminum calorimeter causing the temperature of the water to rise from 22.0 o C to 29.5 o C. What was the initial temperature of the hot copper? - PowerPoint PPT Presentation
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DO NOW: Hand in specific heat lab &answer the following question on “Do Now” sheet:
56 grams of hot copper are added to 140 grams of water in a 92 g aluminum calorimeter causing the temperature of the water to rise from 22.0oC to 29.5oC. What was the initial temperature of the hot copper?
Cwater=4180 J/(kg K) CAl=897 J/(kg K) CCu=385 J/(kg K)
Ti= ?
Q1=(mcDT)water
Q2=(mcDT)Al
Q3=(mcDT)Cu
Q1+Q2+Q3=0
22.0o
C
29.5o
C
140 g water
92 g Al
56 g Cu
Agenda 3.6.2014Describe states of matter
Explore what happens when a substance changes from one state of matter to another.
Label and explain “heating curves”
Use heat of vaporization and heat of fusion to calculate energy absorbed or released when a substance undergoes a change of state.AnnouncementsQuiz tomorrow on Chapter 12.1
HW – 12.1 Review Worksheet
List and Photo of Materials – Extended to Monday (3/10)
State of matter with fixed volume and shape. Molecules are in motion but are tightly packed and maintain their locations relative to neighboring molecules.1. Solid2. Liquid3. Gas4. Plasma
Measure of the average kinetic energy per randomly wiggling particle
1. Heat2. Thermal Energy3. Temperature4. Specific Heat
Measure of the total amount of kinetic energy associated with the random motion of particles plus the total potential energy associated with the intermolecular bonds
1. Heat2. Thermal Energy3. Temperature4. Specific Heat
State of matter that takes on the shape of its container but has a constant volume. Although molecules are tightly packed bonds are continuously changing as they change their locations relative to neighboring molecules.
1. Solid2. Liquid3. Gas4. Plasma
In this state of matter molecules do not form lasting bonds with neighbors and spread out to occupy all the space available. With plenty of extra room between molecules this state of matter can be compressed into a smaller volume.1. Solid2. Liquid3. Gas4. Plasma
At high enough temperatures particles in a gas have enough kinetic energy to knock electrons free from neutral atoms resulting in this “hot soup” of charged ions.1. Solid2. Liquid3. Gas4. Plasma
Heat is added at a uniform rate to a pot of water that starts off at room
temperature. The temperature of the water is shown for the first few
minutes. Sketch what you think the rest of the graph look like if the pot is
left on the stove for 20 minutes
0 2 4 6 8 10 12 14 16 18 200
20
40
60
80
100
120
Time (minutes)
Tem
pera
ture
(oC
)
Take 3 minutes to discuss in small groups what is going on here.
Would the formula Q=mcDT work for the entire 20 minutes? Why or
why not?
0 5 10 15 200
20
40
60
80
100
120
Time (minutes)
Tem
pera
ture
(oC
)
Why does boiling water stay at 100oC
when you continue to add heat?In liquid state molecules are “stuck”
to neighbors by temporary bondsAddition of heat at the boiling point “un-sticks” molecules from neighboring particles WITHOUT increasing kinetic energy.
BIG IDEA – At the boiling point adding heat results in a change of state. As H2O changes from liquid to gas thermal energy increases but temperature stays the same!
Changes of State Phase
Diagram - Water
Melting Point
Boiling Point
DO NOW Take out your homework and answer the following question (3 min).
How much heat is required to raise the temperaure of 1 kg liquid water (C = 4180 J/kg °C) from 0 °C to 100 °C ?
Q = mCwaterΔT
Heating curves
http://www.kentchemistry.com/links/Matter/HeatingCurve.htmHEAT
Heat of fusion (Hf)Amount of heat required to melt 1 kg
at the melting point Heat required to melt something:
Q=mHf
Heat required to freeze something: Q=m(-Hf)
Example: Adding 668 kJ to a block of ice at 0oC causes 2.0 kg to melt. What is the heat of fusion of ice?
Q= 6.68x105 Jm=2.0 kg
Hf= Q/m =3.34x105 J/kg
Heat of Vaporization (Hv)Amount of heat required to change 1 kg of
liquid to a gas at the boiling point. Heat required to vaporize something:
Q=mHv
Heat required to condense something: Q=m(-Hv)
Example: The boiling point of ethanol is 78oC. How much heat must be added to 750 g of ethanol at 78oC to turn it into a gas if Hv=846 kJ/kg ?
m=0.75 kg Hv=846 kJ/kg
Q=mHv=635 kJ
Definition1. Change from solid to
liquid
2. Energy required to convert 1 kg from a solid to liquid at the melting point
3. Liquid phase and gaseous phase can exist in equilibrium
4. Change from liquid to gas
5. Energy required to convert 1 kg from a liquid to a gas at the boiling point.
6. Solid Phase and liquid phase can exist in equililbrium.
TermA. Melting Point
B. Boiling Point
C. Heat of Vaporization
D. Vaporization
E. Melting
F. Heat of Fusion
DO NOW – Have your photo and list of materials ready for me to check and match the definitions with the terms
E
F
B
C
A
D
-40
0
100
130
Tem
pera
ture
of H
20 (°
C)
Heat added Q = mCiceΔT
Q = mCwaterΔT
Q = mCsteamΔT
Q = mHf
Q = mHv
Heating Curve for Water
-400
100
130
Tem
pera
ture
of H
20 (°
C)
Heat added Q = mCiceΔT
Q = mCwaterΔT
Q = mCsteamΔT
Q = mHf
Q = mHv
Heating Curve for Water
Q = mCiceΔT
Q = mHf
Q = mCwaterΔT
Q = mHv
Q = mCsteamΔT
Drawing Heat Diagrams
-40°C
0°C
100°C
130°C
What is the total amount of heat required to raise the temperature 1 kg of water from -40 °C to 20 °C?Cice= 2060 J/kg°C Cwater = 4180 J/kg°CHf = 3.35x105 J/kgQ = mCiceΔT
Q = mHf
Q = mCwaterΔT
-40°C
0°C
20°C
Qtotal = mCiceΔT + mHf + mCwaterΔT
Qtotal = 82400 J + 3.35x105 J + 83600 J
Qtotal = 5.01x105 J
HomeworkChange of State Problems – p. 337 #61, 62, 64