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Transcript of 12 Thermal Physics
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Physics 1
Thermal Physics
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Assignment
• P&P12:
• In class:1(a),3,4(a,b),
• 6,19,22,23,32,58,59
• 72oF tooC and -10oC to oF
• HW:1(b-d),5,6,8,20,24,37,40,41,61,63,66
• Convert 25oF to oC and -210oC to oF
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What do you think? know?
1. Why does popcorn pop?
2. On a camping trip, your friend
tells you that fluffing up a downsleeping bag before you go to
bed will keep you warmer than
sleeping in the same bag when it
is still crushed. Why?
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3. Why is it difficult to
build a fire with dampwood?
4. Why does steam at
100oC cause more
severe burns than liquid
water does at 100
o
C?
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5. Until refrigerators were invented,many people stored fruits and
vegetables in underground cellars.Why was this more effective thankeeping them in the open air?
6. In the past, when a baby had ahigh fever, the doctor might havesuggested gently sponging off thebaby with rubbing alcohol. Whywould this help?
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7. Why does water expand when it freezes?
8. Why, during the final construction of theSt. Louis arch, was water sprayed on theprevious sections as the last section wasput in place?
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Objectives
• 1. Describe thermal energy and compare it
to potential and kinetic energies.
• 2. Describe changes in temperatures of
two objects reaching thermal equilibrium
• 3. Identify various temperature scales, and
convert between them
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Objectives
• 4. Explain heat as energy transferred
between substances at different
temperatures
• 5. Relate heat and temperature
• 6. Apply principle of energy conservation
to calculate changes in potential, kinetic, &
internal energy
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Objectives
• 7. Perform calculations with specific heat
capacity
• 8. Interpret the various sections of a
heating curve
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Objectives
• 9. Recognize that a system can absorb or release energy as heat in order for work tobe done on or by that system
• 10. Compute work done duringthermodynamic process
• 11. Distinguish between isovolumetric,
isothermal, and adiabatic thermodynamicprocesses
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Objectives
• 12. Illustrate how the first law of
thermodynamics is a statement of
energy conservation
• 13. Calculate heat, work, and change in
internal energy using lst law of T-D
• 14. Apply 1st law of T-D to describe
cyclic processes
• 15. Recognize why 2nd law of T-D
requires 2 bodies at different temps.
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1.Relate temperature to the kinetic
energy of atoms and molecules
• Temperature scales
• In the USA, the Fahrenheit temperature scale is used.Most of the rest of the world uses Celsius, and in scienceit is often most convenient to use the Kelvin scale.
• The Celsius scale is based on the temperatures at whichwater freezes and boils. 0°C is the freezing point of water, and 100° C is the boiling point. Room temperatureis about 20° C, a hot summer day might be 40° C, and acold winter day would be around -20° C.
• To convert between Fahrenheit and Celsius, use theseequations:
•
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• Convert 72 oF to oC
• C = 5/9 (F-32)
• C = 5/9 (72-32) = 22oC
• Convert -10 oC to oF
• F = 9/5 C + 32
• F = 9/5(-10) + 32 = 14oF
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Temperature degree scales comparison
Objective 3:
Temperature Scales
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Celsius to Kelvin: T = Tc + 273.15
Problem:
1. The lowest outdoor temperature everrecorded on Earth is -128.6 o F.,recorded at Vostok Station, Antarctica,in 1983. What is this temperature onthe Celsius and Kelvin scales?
Answers: -89.22oC, 183.93 K
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• Obj. #2 - Hotter temperature means
• – more heat present in a substance
• – the faster the molecules of thesubstance move.
• Obj #5 – Relate heat and temperature
• Heat units: calorie or joule (amount of
heat energy present in a substance)
• Temperature units: degree
(proportional to heat energy present in
a substance)
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2.Describe changes in temperatures oftwo objects reaching thermal equilibrium
• The temperature of the hotter substance willdecrease. The temperature of the coldersubstance will increase. Each change will stopwhen the temperatures are the same – thermal equilibrium. In other words, thermalenergy travels from hot to cold.
• Obj. #4 - Heat energy can be transferred
by– Convection (motion of fluid), conduction(touching), or radiation (electromagneticwaves)
C ti
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ConvectionHeat transfer in fluids generally takes place via convection.
Convection currents are set up in the fluid because thehotter part of the fluid is not as dense as the cooler part,
so there is an upward buoyant force on the hotter fluid,making it rise while the cooler, denser, fluid sinks. Birds andgliders make use of upward convection currents to rise, andwe also rely on convection to remove ground-level pollution.
ConductionWhen heat is transferred via conduction, thesubstance itself does not flow; rather, heat istransferred internally, by vibrations of atoms andmolecules. Electrons can also carry heat, which is thereason metals are generally very good conductors of
heat.
R di i
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RadiationThe third way to transfer heat,in addition to convection and
conduction, is by radiation, inwhich energy is transferred inthe form of electromagneticwaves.
More about electromagneticwaves in a lot more detail in alater chapter; an e-m wave isbasically an oscillating electric
and magnetic field travelingthrough space at the speed oflight.
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Specific heat capacity
The amount of energy that must be added to
raise the temperature of a unit mass of a
substance by one temperature unit.
• Units: J/kg K
• For Water: 4180 J/kg K
• For Aluminum: 897 J/kg K• Which one heats faster?
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This table is on page 318.
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Heat Transfer
Q = mCΔT = mC (T f – T i )• Heat Transfer
• Q = mCΔT = mC (Tf – Ti)
• Q, quantity of heat in joule
• m, mass of substance in kg
• c, specific heat for water in 4186 j/kg K
• t, temperature in Celsius
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See table 12-1 on page 318
Find the amount of heat needed to change thetemperature of 5.0 g of liquid water from 8.0oC to
100oC.
Q = mcDt = .005kg(4186 j/kgoC) (92oC) = 1.9 x 103 j
Again,
specific heat is the amount of
heat necessary to change one kgof a substance 1 degree Celsius orKelvin.
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• 12/3 When you turn on the hot water to
wash dishes, the water pipes have to heat
up. How much heat is absorbed by acopper water pipe with a mass of 2.3 kg
when its temperature is raised from 20.0oC
to 80.0oC?
• Q = mcDt
• Q = (2.3kg)(390J/kgoC)(60.0oC)
• Q = 53820 J or 5.4x104
J
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12/4b
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12/4b
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Specific Heat Capacities
Which
one is
greatestthat you
use
everyday?
OR
J/kgoC
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Law of Heat Exchange
Q lost = Q gained
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• 0.300 kg of coffee, at a temperature of 95 °C, is
poured into a room-temperature (20.oC) steelmug, of mass 0.125 kg. Assuming no energy islost to the surroundings, what does thetemperature of the mug filled with coffee cometo?
• Applying conservation of energy, the totalchange in energy of the system must be zero.So, we can just add up the individual energychanges (the Q's) and set the sum equal to zero.
The subscript c refers to the coffee, and m to themug.
•
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Note that room temperature in Celsius is about20°. Re-arranging the equation to solve for thefinal temperature gives:
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The temperature of the coffee doesn't drop by
much because the specific heat of water (or
coffee) is so much larger than that of steel.This is too hot to drink, but if you wait, heat will
be transferred to the surroundings and the
coffee will cool.
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Latent Heat is energy transferred
during phase changes• Latent Heat• Crystalline materials change phase -- melt and freeze or vaporize
and condense -- at a single, fixed temperature.Energy is required for a change of phase of a substance. The ratioof the energy to the mass of the substance involved is called thelatent heat of the substance. This is much like the specific heat we
have just discussed. It is called latent heat because there is nochange or difference in temperature.
• Latent heat of fusion Hf describes the heat necessary to melt (or freeze) a unit mass of a substance.
Q = m Hf • Latent heat of vaporization Hv describes the heat necessary to
vaporize (or condense) a unit mass of a substance.
• Q = m Hv•
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Temperature vs Heat
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Formulas
• Temperature change use: Q = mc D t
• Melting or freezing: Q = m Hf
• Evaporation or Condensation: Q = mHv
• Hf is latent heat of fusion, 3.33 x 105 J/kg
• Hv is latent heat of vaporization, 2.26 x 106 J/kg
• These values are for water.• Problem #19 on page 325.
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Problem
• A jar of tea is placed in sunlight until it
reaches an equilibrium temperature of
32oC. In an attempt to cool the liquid to
0oC, which has a mass of l80 g, how much
ice at 0oC is needed? Assume the specificheat capacity of the tea to be that of pure
liquid water.
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m tea = 180g m ice = ? which is the mass of the water thathas melted
c tea = c water = 4186 J/kgoC
H f = 3.33 x 105 J/kg and t final = 32oC
Q lost = Q gained tea loses and water gains only meltingthe ice
(mcDt)tea = (mHf )ice
m ice = (mcDt)tea
Hf
ice
m ice = (.180kg)(4186 J/kgoC)(32oC)
3.33x105 J/kg
= 7.2 x 10-2
kg
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1st Law of Thermodynamics
DU = Q - W
The change in thermal energy of
an object is equal to the heat
added to the object minus thework done by the object.
See Fig 12-11 on
Page 326
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A heat engine
Transforms heat at hightemperature into mechanical energyand low-temperature waste heat
A heat pump (refrigerator) absorbsheat from the cold reservoir andgives off heat to the hot reservoir.
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2nd Law of Thermodynamics
• In the 19th Century French engineerSadi Carnot studied the ability ofengines to convert thermal energy into
mechanical energy.• He developed a logical proof that evenan ideal engine would generate some
waste heat.• Carnot’s result is best described by theterm entropy which is the measure ofthe disorder in a system.
Th h i D S i h b h
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The change is entropy, D S, is shown by theequation:
D
S = Q / T Units: J/K
The change in entropy of an object is equal to
the heat added to the object divided by thetemperature of the object.
Natural processes occur in a direction that
increases the entropy of the universe. Allprocesses tend toward disorder unless someaction occurs to keep them ordered. i.e., heatcan flow only from hot to cold naturally.
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Practice Problems
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12/32 answer
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•The End
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