Download - Chapter 12 Thermal Energy
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
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
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
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
•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
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
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)
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 )
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)
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
Chapter 12 Thermal Energy
Method of mixtures Heat gained plus heat loss in a closed systemis zero.
ΔQgained+ΔQlost = 0
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
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
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
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
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
Chapter 12Change of State
Solid Liquid Gas
Melting
CondensationFreezing
Sublimation
Supercooled
Vaporization
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
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
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
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