a 50.0 g ball is dropped from an altitude of 2.0 km. calculate: u i, k max, & w done through the...
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A 50.0 g ball is dropped from an altitude of 2.0 km.
Calculate: Ui, Kmax, & W done through the fall
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Chapter 12Thermal Energy
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Thermodynamics
•The movement of heat
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Kinetic Theory1)All matter is made up of
tiny particles2)All particles are in
constant motion3)All collisions are elastic
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Temperature•A measure of average kinetic
energy
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Temperature•A measure of heat intensity
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Thermal Equilibrium• When the average kinetic
energy of two or more substances become equal; thus their particles have the same exchange rate
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•Because it is a measure of average kinetic
energy, temperature is related to the motion of
particles (atoms, molecules, ions, etc)
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Thermometer•A device, calibrated to some temp scale, that is
allowed to come to thermal equilibrium with
something else
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Temperature Scales• Celcius (oC)
–Based on MP & BP of water
• Kelvin (K)–Based of absolute temperature
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Temperature Scales
•K = oC + 273
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Convert Temperatures
100 K = ___ oC
100 oC = ___ K
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Heat•A form of energy
that flows due to temperature differences
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Heat (Q)•Because particle at higher temp. move
faster than particles at a lower temp., the net
flow of heat is H C
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Heat (Q)•Heat will continue to have net flow from H C as long as there is
a temperature difference
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Heat (Q)•When there is no
temperature differences, the system
has reached thermal equilibrium
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Work•The movement of
energy by means other than temperature
difference
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1st Law of Thermo.•The increase in thermal energy =
sum of heat added & work done to a
system
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1st Law of Thermo.
E = Q + W
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In Most Engines•Heat is added by some
high energy source (gas)
•Work is done by the engine
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In Most Engines
E = Q + WBut W < 0
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Entropy
•A measure of the disorder in a
system
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2nd Law of Thermo.
•In natural processes,
entropy increases
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Entropy
•When fuel is burned, entropy is
increased
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Specific Heat (C)•The thermal energy
required to raise 1 unit mass of matter
1 degree
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Specific Heat (C)•The thermal energy
required to raise 1 kg of matter 1
degree K
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Heat (Q or H)•Heat transfer = mass
x specific heat x the temperature change
Q = mCT
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Calculate the heat required to raise 50.0
g of water from 25.0oC to 65.0oC.
Cwater = 4180 J/kgK
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Calculate the heat required to raise
250.0 g of lead from -25.0oC to 175.0oC. Clead = 130 J/kgK
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28 kJ of heat was required to raise the
temperature of 100.0 g of a substance from
-125oC to 575oC. Calculate: C
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3.6 kJ of heat was required to raise the
temperature of 10.0 g of a substance from -22oC to 578oC.
Calculate: C
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Conservation of Heat
•The total energy of an isolated system
is constant
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Conservation of HeatBecause the total amount of heat is
constant
q orHsystem = 0
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Conservation of Heatq orHsystem = 0
Hsys = H1 + H2 + ..
qsys = q1 + q2 + ..= 0
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Conservation of Heatqsys = q1 + q2 = 0
mCT1 + mCT2 = 0
mCT1 = - mCT2
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Conservation of Heatqsys = qgained + qlost
qgained = - qlost
mCTgain = - mCTlost
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A 50.0 g slug of metal at 77.0 oC is added to
500. g water at 25.0oC.Teq= 27.0oC.
Calculate: Cmetal
Cwater = 4180 J/kgK
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A 200.0 g slug of metal at 77.5 oC is added to
400. g water at 25.0oC.Teq= 27.5oC.
Calculate: Cmetal
Cwater = 4180 J/kgK
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Solving Mixture Temperatures
qsystem = 0
qsystem = qhot + qcold
mCThot = -mCTcold
T = Tf – Ti
mC(Tf – Ti)hot = -mC(Tf – Ti)cold
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Conservation of Heat
mChTf - mChTh
+mCcTf - mCcTc
= 0
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Conservation of Heat
mChTf - mChTh
=
-mCcTf + mCcTc
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20.0 g of water at 25.0oC is added to
30.0 g water at 75.0oC. Calculate: Teq
Cwater = 4180 J/kgK
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500. g of water at 75.0oC is added to 300. g water in a 200. g calorimeter all at 25.0oC. Calculate: Teq
Cwater = 4180 J/kgKCcal = 1000 J/kgK
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A 500.0 g slug of metal at 87.5.oC is added to 4.0 kg water in a 1.0 kg can at
25.0oC. Teq= 27.5oC.Calculate: Cmetal
Cwater = 4180 J/kgKCcan = 1.0 J/gK
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States of Matter•Solid
•Liquid
•Gas
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Solid•Has definite size & definite shape
•Particles vibrate at fixed positions
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Liquid•Has definite size but no definite shape
•Particles vibrate at moving positions
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Gas•Has neither size nor shape
•Particles move at random
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Change of State•When a substance changes from one state of matter to
another
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Change of State•Change of state
involves an energy change
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Changes of State•Melting-Freezing
•Boiling-Condensation
•Sublimation-Deposition
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Melting Point•The temperature at which a solid is at dynamic equilibrium with its liquid.
•Freezing Point (Same)
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Boiling Point•The temperature at which a liquid is at dynamic equilibrium with its gas.
•Condensationing Point (Same)
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Changes of State•During changes of state, the temperature remains constant; all energy is used to change the state
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Heat of Fusion (Hf)•The heat required to melt one unit mass of a substance at its MP
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Heat of Fusion (Hf)•Hf water = 3.34 x 105 J/kg
•Hf water = 334 J/g
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Heat of Vaporization (HV)
•The heat required to vaporize one unit mass of a substance at its BP
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Heat of Vaporization (HV)
•Hv water = 2.26 x 106 J/kg
•Hv water = 2260 J/g
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Temperature vs Heat Plot
160
210
260
310
360
410
460
510
0 50 100 150 200 250 300 350 400
Heat (kJ)
Tem
pera
ture
(K)
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Change of State
q = mH
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Changes of State
qf = mHf
qv = mHv
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Calculate the heat required to change
250 g ice to water at its MP:
Hf = 3.34 x 105 J/kg
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Calculate the heat required to boil 400 g of water at its BP:HV = 2.26 x 106 J/kg
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Calculate the heat change when the
temperature of 2.0 kg H2O is changed from
50oC to 150oC:
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Calculate the heat change when the
temperature of 4.0 kg H2O is changed from-25.0oC to 125.0oC:
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Constants for Water• Hf = 3.34 x 105 J/kg
• Hv = 2.26 x 106 J/kg
• Cice = 2060 J/kgK
• Cwater = 4180 J/kgK
• Csteam = 2020 J/kgK
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1st Law of ThermoTotal E equal work done
plus heat added to it
E = Q + W
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Heat Engine•Any engine that converts heat energy to mechanical energy (Steam, internal combustion, etc.)
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Heat Pumps & Refrigerators
•Use pressure changes & the heat of vaporization to transfer heat from cold to hot
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2nd Law of Thermo
The total entropy of an isolated system always increases
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20.0 g of lead at 75.0oC is added to 100.0 g
water at 25.0oC. Calculate: Teq
Cwater = 4180 J/kgKClead = 130. J/kgK
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50.0 g of milk at 5.00oC is added to 500.0 g coffee
in a 400.0 g cup at 75.0oC. Calculate: Teq
Ccoffee = 4.00 J/gKCcup = 1.50 J/gKCmilk = 3.50 J/gK
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Ti = 25.0oC Tf = 200.0oC BP = 100.0oC MP = 0.0oCMass of H2O = 5.00 kgCalculate: Qtotal
Cice= 2.06 J/gK, Hv = 2260 J/gCwater= 4.18 J/gK, Hf = 334 J/gCsteam= 2.02 J/gK
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Ti = -50.0oC Tf = 300.0oC BP = 100.0oC MP = 0.0oCMass of H2O = 5.00 kgCalculate: Qtotal
Cice= 2.06 J/gK, Hv = 2260 J/gCwater= 4.18 J/gK, Hf = 334 J/gCsteam= 2.02 J/gK
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20.0 g of lead at 75.0oC is added to 100.0 g
water at 25.0oC. Calculate: Teq
Cwater = 4180 J/kgKClead = 130. J/kgK
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A 500.0 g slug of metal at 86.5.oC is added to 4.0 kg water in a 2.0 kg can at
24.0oC. Teq= 26.5oC.Calculate: Cmetal
Cwater = 4180 J/kgKCcan = 1.0 J/gK
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A 50.0 g of ice at -20.0 oC is added to 2.0 kg water in
a 1.0 kg can at 25.0oC. Calculate: Teq
Cw = 4180 J/kgK Cc = 1.0 J/gK
Cice = 2.06 J/gK Hf = 340 J/g
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A 50.0 g of steam at 120.0 oC is added to 2.0 kg water in a 1.0 kg can at 20.0oC.
Calculate: Teq
Cw = 4180 J/kgKCc = 1.0 J/gKHV = 2260 J/g
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A 400.0 g of steam at 125.0 oC is added to 2.0 kg ice in a 1.0 kg can at -20.0oC. Calculate: Teq
Constants will be on the board