Chapter 12 Thermal Energy
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Transcript of Chapter 12 Thermal Energy
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Chapter 12 Thermal Energy
Thermodynamics - The study of heat
Kinetic Theory - All matter consists of minute
particles which are in constant motion
Thermal Energy - The overall energy of motion of the particles that make up an object.
Temperature Scale
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Temperature - The average kinetic energy of the particles.
Heat - The transfer of thermal energy because of a difference in temperature.
Thermal Energy Transfer• Conduction - The transfer of energy when particles collide.• Convection - The transfer of heat by means of motion in a fluid.• Radiation - The transfer of energy by electromagnetic waves.
Chapter 12 Thermal Energy
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Chapter 12Thermal Energy
Temperature Scales•Kelvin•Fahrenheit•Celsius
Absolute Zero - The temperature at which all the thermal energy has been removed from the gas.
Absolute Zero -273.15º C or 0 K
Tk = Tc + 273
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Temperatures Fahrenheit ºF Celsius ºC KelvinH20 Boils 212 100 373H20 Freezes 32 0 273CO2 Freezes -189 -123 150Nitrogen Boils -320 -196 77Absolute Zero -459 -273 0
Chapter 12 Thermal Energy
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•Rumford’s Experiment•Disproved Caloric Theory•Showed a relationship between heat and work
1st Law of Thermodynamics – When energy is converted to heat, all energy is conserved.
•Joules’ Experiment
•Related heat to energy
•N(mgh) converts to heat
Chapter 12 Thermal Energy
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2nd Law of Thermodynamics – Heat flows from hot to cold.
Law of Entropy The universe is continuously going from a state of order to disorder.
Chapter 12 Thermal Energy
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Chapter 12 Thermal Energy
ΔQ The transfer of thermal energy, measured in joules 1 calorie = 4.18 joules
• Specific Heat ( C ) The amount of energy needed to to raise the temperature of a unit mass one temperature unit.
• Specific Heat is measured in J/kg·K or J/kg·°C• Table Pg 279
Heat Transfer Q = mCΔT = mC(Tfinal-Tinitial)
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Chapter 12 Thermal Energy
How much heat is needed to raise 20 grams of water from 40ºCto 70 ºC?
m = 20 gC = 4.18 J/g C ºTi = 40 ºCTf = 70 ºC
ΔQ = mCΔT
ΔQ = 2500 J
ΔQ = (20g)(4.18 J/g C)(70 ºC - 40 ºC )
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Chapter 12 Thermal Energy
Find the specific heat of tungsten if it takes 100 joules of energyto raise 20 grams of the material from 20 ºC to 57 ºC.
m = 20 gC =Tf = 57 ºCTi = 20 ºCQ = 100 J
ΔQ = mCΔT
C = .135 J/g ºC
C = 100 J/(20 g*37 ºC)
C = Q/(mΔT)
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Chapter 12 Thermal Energy
149,400 J of heat are added to a 5 kg mass of a substance that raisesthe temperature from -25ºC to 20º C. What is the material?
m = 5000 gC =Tf = 20 ºCTi = -25 ºCQ = 149,400 J
ΔQ = mCΔT
C = .644 J/g ºC
C = 149,400 J/(5000 g*45 ºC)
C = Q/(mΔT)
The material is glass
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Chapter 12 Thermal Energy
Method of mixtures Heat gained plus heat loss in a closed systemis zero.
ΔQgained+ΔQlost = 0
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Chapter 12 Thermal Energy
What is the final temperature of a mixture where 100 grams of ironat 80ºC is added to 53.5 g of water at 20ºC?
ΔQgained+ΔQlost = 0
(53.5 g)(4.18 J/gºC)(Tf - 20ºC ) + (100g)(.45 J/g ºC)(Tf - 80ºC) = 0
(223.63 J/ºC)(Tf – 20ºC )+(45 J/ ºC)(Tf – 80 ºC) = 0
Tf = 30ºC
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Chapter 12 Thermal Energy
What is the final temperature of a mixture where 400 grams of alcohol at 16ºC is added to 400 g of water at 85ºC?
ΔQgained+ΔQlost = 0
(400 g)(2.45 J/g ºC)(Tf - 16 ºC ) + (400g)(4.18 J/g º C)(Tf - 85 ºC) = 0
(980 J/ ºC)(Tf – 16 ºC )+(1672 J/ ºC)(Tf – 85 ºC) = 0
Tf= 59.5 ºC
980 Tf – 15680 + 1672 Tf -142120 = 0 2652 Tf = 157800
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What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0º C to 70.0° C?
Qmct
c Q
mt
c 2500J
(10000g)(70C 10C)
c .0042J
gC
Chapter 12 Thermal Energy
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A 1.0 kg sample of metal with a specific heat of 0.50 J/g°C is heated to 100.0º C and then placed in a 50.0 g sample of water at 20.0°C. What is the final temperature of the metal and the water?
Qgained Qlost 0
mctgained mctlost 0
(50g)(4.18J
gC)(TF 20C) (1000g)(.5
J
gC)(TF 100C) 0
Chapter 12 Thermal Energy
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A 1.0 kg sample of metal with a specific heat of 0.50 J/g°C is heated to 100.0º C and then placed in a 50.0 g sample of water at 20.0°C. What is the final temperature of the metal and the water?
(50g)(4.18J
gC)(TF 20C) (1000g)(.5
J
gC)(TF 100C) 0
209TF 4180 500TF 50000 0
709TF 54180
TF 76C
Chapter 12 Thermal Energy
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Chapter 12Change of State
Solid Liquid Gas
Melting
CondensationFreezing
Sublimation
Supercooled
Vaporization
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Chapter 12Thermal Energy
Heat of Fusion (Hf) The amount of energy needed to melt a unit mass of a substance. Melting Point
Heat of Vaporization (Hv) The amount of heat needed to vaporizea unit mass of a liquid. Boiling Point
There is no temperature change in changing between states.
Q = m Hf Q = m Hv Chart Pg 287
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Chapter 12Thermal Energy
How much energy is needed to melt 20 grams of ice at 0º?
Q = mHf = (20g)(334 J/g) = 6680 J
How much energy is needed to change 30 grams of waterat 100 ºC to steam?
Q = mHv = (30g)(2260 J/g) = 67800 J
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Chapter 12Thermal Energy
How much energy is needed to melt 40 grams of ice at -60ºCto steam at 150ºC?
1. Warm the ice Q = mCΔT = (40 g)(2.06 J/gºC)(60ºC) = 4944 J2. Melt the ice Q = mHf = (40 g)(334 J/ g) = 13360 J 3. Warm the water Q = mCΔT = (40g)(4.18 J/g ºC)(100 ºC) = 16720 J4. Vaporize the water Q = mHv = (40g)(2260 J/g) = 90400 J5. Warm the steam Q = mCΔT = (40g)(2.02 J/ gºC)(50ºC) = 4040 J
Qtotal = 4944 J + 13360 J + 16720 J + 90400 J + 4040 J = 129464 J
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Chapter 12Thermal Energy
How much energy is needed to melt 20 grams of ice at -10ºCto steam at 130ºC?
1. Warm the ice Q = mCΔT = (20 g)(2.06 J/gºC)(10ºC) = 412 J2. Melt the ice Q = mHf = (20 g)(334 J/ g) = 6680 J 3. Warm the water Q = mCΔT = (20g)(4.18 J/g ºC)(100 ºC) = 8360 J4. Vaporize the water Q = mHv = (20g)(2260 J/g) = 45200 J5. Warm the steam Q = mCΔT = (20g)(2.02 J/ gºC)(30ºC) = 1212 J
Qtotal = 412 J + 6680 J + 8360 J + 45200 J + 1212 J = 61864 J