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