Transcript 05Thermal_PhysicsALT

Temperature
• Temperature (T): Measure of the average kinetic
energy of the particles composing an object or
system due to
–Vibrations
–Rotations
–Random translations
•Does a system’s temperature depend on your
relative velocity to it?
–No: The center of mass (CM) motion of the system
does not contribute to the system’s temperature.
Temperature Scales
• Water based: freezing and boiling points
– Fahrenheit: 32 oF to 212 oF
– Celsius (Centigrade): 0 oC to 100 oC
– Which scale is more precise?
• Kelvin scale
– Directly proportional to the average kinetic energy of
the system.
– 0K is the unreachable absolute zero of temperature
where all particle motion stops.
– Increments are the same as Celsius: 1K = 1 Co
– Conversion between the two: T(K) = T(oC)+273 Co
Internal Energy of a System (U)
• Includes:
– Internal translational K.E. (no CM motion)
– Vibrational and Rotational K.E. of atoms about
their bonds in molecules.
– P.E. associated with bonds within molecules.
– P.E. due to attraction between molecules.
• Processes that change Internal Energy:
– Work (Force applied through a distance)
– Heat (Differences in Temperature)
Heat
• Heat: Process of energy transfer due to
temperature differences.
• Measured in calories or kilocalories
• Mechanical Equivalent of heat: 1 cal = 4.186 J
– Work and heat are both processes of energy
transfer.
– Discovered by James Joules
Joule's Experiment: Falling mass
(work) used to heat water.
Examples
• How many reps of pressing 50 kg must you
do to burn one pound of body fat?
– One of gram of fat is burned by 9 kilocalories
– There are 454 grams in a pound.
– Assume 1 meter long arms.
• A 3.0 gram bullet passes through a tree,
slowing down from 400 m/s to 200 m/s.
How much K.E. is lost to heating the tree in
calories?
Calorimetry
• Deals with change in a system’s energy due to
heating or cooling.
• Specific Heat of a substance (c): measure of the
amount of energy per unit mass needed to cause a
rise of one degree in Temperature. (Q = m c ∆T)
• Latent heat of fusion (Hf): amount of energy per
unit mass needed to melt a solid. (Qf = mHf)
• Latent heat of vaporization (Hv): amount of
energy per unit mass needed to vaporize a liquid.
(Qv = mHv)
• No Temperature changes occur during the phase
transitions.
Calorimetry Example
• We initially have 2 kg of ice at -20°C. How
much energy is needed to turn the ice into
steam at 100°C?
• Given:
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cice = 2100 J/kg°C
cwater = 4186 J/kg°C
Hf = 3.33 ×105 J/kg
Hv = 22.6 ×105 J/kg
Heat Transfer
• Thermodynamics: deals with properties of
different states and amounts of heat or work to go
between them.
• Heat Transfer: deals with the rate at which energy
(as heat) is transferred.
• Three modes of heat transfer:
– Conduction: propagation of vibrations through
adjacent atoms.
– Convection: energy carried by actual transfer of mass.
– Radiation: energy carried by electromagnetic waves.
Conduction
• How fast does heat flow through a wall with the
ends kept at different temperatures?
• Fourier’s Law of Heat Conduction:
(∆Q/∆t) = k A (TH-TC)/l
• ∆Q/∆t = energy transferred per unit time.
• A = area; more atoms available to vibrate.
• TH – TC ; higher K.E. differences
• l = length or thickness of wall.
• k = thermal conductivity; atomic structure.
• Hydraulic and Electric Analogs exist.
Example 1: Heat Loss through a Window
• The window has an area of 3 m2, a thickness of
3.2 mm, and is made of glass with a thermal
conductivity of .84 W/m °C . The temperature
inside is 15 °C and outside it is 14 °C.
• How fast does the window lose heat?
• An equilibrium is reached at which there is a
steady and constant heat flow.
• A temperature profile can be drawn, showing
the steady state temperature as a function of
distance from one end.
Ex2: Composite Material
• Realistic Situation: Walls with insulation
between them.
• Inside Temperature: TH= 25 °C
• Outside Temperature: TC=5 °C
• Insulation (inner wall): l1=1m, k1=.048 W/m °C
• Brick (outer wall):l2=.5m, k2=.84 W/m °C
• Find Temperature at junction and
temperature profile at steady state.
• Key Point: At steady state: ∆Q/∆t = constant
throughout the thickness of the walls.
Other Forms of Heat Transfer
• Convection: (combines with fluid dynamics)
– Free: natural currents due to temperature and density differences.
– Forced: accelerated through fans or turbines.
• Radiation:
– Needs no medium to travel.
– An object emits radiation at a rate dependent on temperature (in
Kelvins): (∆Q/∆t) = AσeT4
• Emissivity (e) = 0 to 1; measure of surface’s ability to emit.
• Stefan-Boltzmann Constant: =5.67×10-8 W/(m2/K4)
• Second Sound:
– Very fast heat transfer in superfluid (cold < 2.7K) helium.
– Travels as waves of varying temperature and specific entropy
– Analogous to sound waves of varying pressure and density.
How Does a Solid Expand When Heated?
• We know solids expand when heated.
Experimentally: ∆L = α L0 ∆T
• ∆L = change in length of object (+ or -)
• α = coefficient of linear expansion
• L0 = original length of the object
• ∆T = change in temperature (+ or -)
• L = L0 + ∆L = (1+ α ∆T) L0 = final length.
• Why should elongation depend on
temperature change?
• Why should elongation depend on original
length of object?
Examples
• Ex1: A plate with a hole cut out. When we heat
the plate how does the hole behave?
• Ex2: A ring of material with α1 stuck around a
tube with α2 and (α2 > α1 ). How can I remove
the ring from the tube?
More Examples
• Uniform rectangular plate of area: A = l×w
• Show ∆A = 2αlw∆T (neglecting small
quantities)
Thermal Stress (ref. ex 10.5 pg 358)
• Ex: A beam bridges two walls. The walls cannot
expand or contract compared to the beam
(αwall << αbeam)
• What will happen if I heat the beam?
– Beam wants to expand by ∆L from L0
– Inability to expand causes a compressive strain: ∆L / L0
– and a resulting compressive stress: F/A = Y (∆L / L0 )
(where Y = Elastic or Young’s Modulus; Hooke’s Law)
– Since: ∆L = α L0 ∆T,
– We have: F/A = Y α ∆T = thermal stress.
– Can be compressive or tensile. Usually enormous.
How do we describe large numbers
of atoms and molecules?
• New unit of mass:
– Atomic mass unit (amu) or (μ)
– 1 μ = 1/12 the mass of a carbon-12 atom (definition)
– Atomic masses of other elements are often given in
amu, expressed relative to the mass of 12C atom.
(example: the periodic table).
• A mole = the number of 12C atoms that has a mass
of 12 grams. (definition)
The Mole
• A “mole” is convenient number, similar to
“dozen” eggs or a “gross” of nails.
• A mole of atoms of an element has a mass in
grams equal to the element’s atomic mass.
• A mole of molecules of a compound has a
mass in grams equal to the compound’s
molecular mass.
• The number to which a mole is equal is:
NA= 6.022×1023 (Avogadro’s Number)
Examples
• Atomic Hydrogen (H):
– What is the mass of an H atom in μ?
– What is the mass of a mole of H atoms in grams?
– What is the mass of one H atom in grams?
• Molecular Hydrogen (H2):
– What is the mass of one H2 molecule in μ?
– What is the mass of a mole of H2 molecules in
grams?
– What is the mass of one H2 molecule in grams?
More Examples
• What is the mass of two moles of H2 O?
• A copper penny has a mass of 3.4 grams.
– How many moles of copper atoms are in
the penny?
– How many atoms of copper are in the
penny?
How is a gas described by its properties?
• How is a gas different from a liquid or solid?
– Fills its container
– The K.E. of its motion is large compared to P.E.
between particles.
• How does a gas fill its container?
– How is volume for a gas defined?
– How is this different from solids and liquids?
– If we put a few atoms in a container, how do they
“fill the container”?
– Gas atoms bounce off walls; resulting impulses on
walls exert pressure.
Summarizing: Properties of a gas system
• Volume (V): The extent of space through which
the gas particles can move. (m3 or L )
• Temperature (T): describes the kinetic energy
associated with molecular and atomic motions. (K)
• Particle Number
– In moles (n) or
– In particles (N)
• Pressure (P): measure of forces involved in hitting
walls and bouncing off. (N/m2 or atm)
Equation of State
• Relates the properties of a system to each other.
• For an “ideal” gas: PV = nRT
– R = ideal gas constant (.0821 L-atm/mol-K)
= (8.31 J/mol-K)
(mol=mole)
– For a set n; fixing two properties sets the other two.
– P = absolute pressure, T must be in kelvins.
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Boyle’s Law: If T is constant: V~ 1/P
Charles’ Law: If P is constant: V ~ T
Gay-Lussac’s Law: If V is constant: P~T
STP: “Standard Temperature and Pressure
T=273K, P= 1 atm (sea level)
Examples
• What is the volume (in liters) of one mole of a
gas at STP?
• Suppose we have a piston in a cylinder and we
introduce at STP:
1 mole of nitrogen gas (N2)
3 moles of hydrogen gas (H2)
This reaction occurs (still at STP):
N2(g) + 3H2(g)  2 NH3(g)
What happens to the volume?
Examples (cont’d)
• An auto tire is inflated to a gauge pressure of
200 kPa at 10oC. After driving a bit, the
temperature rises to 40oC.
What is the new gauge pressure?
Ideal Gas Law
(in particles instead of moles)
• PV = nRT
• Let N = number of atoms or molecules
N = (# moles) (# particles/mole)
= n
NA
• So : n = N/ NA
• PV = (N/NA) = N(R/NA)T = NkbT
• kb= Boltzmann’s constant = 1.38×10-23J/K
What makes an ideal gas “ideal”?
• Why does PV=nRT hold for all ideal
gases?
• Why is R the same for all ideal gases?
• How is an ideal gas characterized?
• The ideal gas law makes no use of the
identity of the particles.
What makes an ideal gas “ideal”?
• There must be a large number of particles in
the system.
• The particles in the gas must be much smaller
than their separations.
• The attractive P.E. between particles must be
much lower than their K.E.
• Collisions of particles with container walls and
with each other must be elastic. (No energy loss
through friction.)
Departures from Ideal Gas Behavior
• Modifications can be made to the Equation of State for
non ideal gases.
• Most common is the “Van der Waals” gas, where a
and b are parameters of the particular gas :
•What is the physical significance of these constants?
• b: decreases effective volume of the container. Why?
• a/V2 : increases effective pressure. Diminished effect
with larger volume. Why?
Thermodynamic Systems
• A system is a part of the universe under consideration.
The rest of the universe is called the “environment” or
the “surroundings”.
• Isolated system: No matter or energy is exchanged with
the environment. (ex: thermos)
• Closed system (or “control mass”): no matter is
exchanged with the environment. (ex: gas in a cylinder
with a piston.)
• Open system (or “control volume”): Allows exchange of
both matter and energy with the environment.
(ex: animal cell surrounded by a membrane)
Properties of a Thermodynamic System
Two types of properties are used to describe systems:
• Extensive Properties:
– Ex: mass (M), volume (V), internal energy (U), heat
capacity (C)
– Depend on system size
• Intensive Properties (also called “point functions”):
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–
–
–
Ex: pressure (P), temperature (T), density (ρ)
Independent of system size
Have a value at every point in the system.
An extensive property can be made intensive by dividing
by the mass of the system (v=specific volume, u=specific
energy, c=specific heat)
Internal Energy of a System (U)
• Includes:
– Internal translational K.E. (no CM motion)
– Vibrational and Rotational K.E. of atoms about
their bonds in molecules.
– P.E. associated with bonds within molecules.
– P.E. due to attraction between molecules.
• Processes that change Internal Energy:
– Work (Force applied through a distance)
– Heat (Differences in Temperature)
Kinetic Theory of an Ideal gas
• How is Temperature related to K.E. and U in an
ideal gas?
• Suppose we have a cubic box:
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–
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Side length (l)
N atoms of mass (m)
Average K.E. of the atoms:
Speeds in all directions are the same on average:
• We see: T~ (K.E.)AV and U depends only on T
– (K.E.)AV = (3/2)kbT
– U = (3/2)NkbT = (3/2)nRT
First Law of Thermodynamics
•
Internal Energy is
–
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–
–
•
Increased by “adding heat”
Decreased by “removing heat”
Increased when positive work is done on the system
Decreased when positive work is done by the system
Two conventions
1. ∆U = Q + W ; W = work done on the system
2. ∆U = Q – W; W = work done by system
In both cases: Q = heat absorbed by system
• An explicit question would ask for the work
done on or by the system.
How is a system affected by processes
of heat and work?
• We characterize a system by its properties.
• Usually, specifying two properties will define the
state of the system.
– All other properties are then determined.
– Ex: PV=nRT; for a system of n moles, specifying P and
V determines T.
• Processes involving heat and/or work
– take us from one state to another
– Result in changes of the energy of the system governed
by the First Law: ∆U = Q + W
How is a system affected by processes
of heat and work? (cont’d)
• We can draw a diagram of P vs. V and represent
– States as points on the diagram.
– Processes as lines or curves between points (states) with an
arrow specifying direction.
• Properties fix a system’s state without regard to the
process taking us there.
– Think of an initial state A as a starting point.
– Think of a final state B as an ending point.
– Different routes on a P-V diagram can be taken from A to B.
• We will study various processes in which heat is
absorbed or expelled from the system and work is done
on or done by the system.
How can we apply processes of heat and work
to take a system to a desired state?
• Types of processes we’ll study:
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Isothermal (constant Temperature: ∆T = 0)
Adiabatic (no heat absorbed or expelled: Q=0)
Isobaric (constant Pressure: P = const)
Isochoric (constant Volume: V=const)
• (also called “isometric” or “isovolumetric”)
• Relationships we’ll use frequently
– PV=nRT, U= (3/2) nRT, ∆U = Q + W
• Questions we’ll be asking:
– What are the properties at a state?
– What work is done on or by the system for a process?
– What heat is absorbed or expelled by the system for a
process?
Isothermal Process
(constant Temperature)
• ∆T = 0, ∆U = 0 (for ideal monatomic gas)
– Heat absorbed = work done by the system
– Isothermal expansion: must add heat
– Isothermal compression: must remove heat.
Adiabatic Process
(Heat is neither gained nor lost)
• Q = 0;
– ∆U = work done on system
– System is thermally insulated or
– Work is done too quickly for heat to transfer
• How does an adiabatic expansion compare
to an isothermal expansion?
Isobaric Process
(Pressure is constant)
• P is constant
– Heat is added (expansion) or removed (compression) to
keep pressure constant.
– Expansion: ∆V > 0; W < 0 (work done on system)
– Compression: ∆V < 0; W > 0 (work done on system)
– Work is easy to calculate:
W = -P∆V = work done on system
• In general:
– Work for a process equals the area under its PV-curve
– The sign depends on the direction of the process
(expansion or compression)
Isochoric Process
(Volume is constant)
• Also called isovolumetric or isometric
• W=0; no work is done on or by the system.
• System neither expands nor contracts; closed
rigid container.
• Change in internal energy equals heat
absorbed: ∆U = Q
Thermodynamic Cycle
• A series of processes that begin and end at the same state:
– (add arrows)
• The area enclosed equals the net work done on or by the cycle.
• Goal of an Engine: to extract useful work done by the system.
Example: Adiabatic Expansion
• In an engine cylinder, .25 moles expands rapidly and
adiabatically against a piston. The temperature drops from
1150K to 400K.
– Does something like this really happen?
– How much work does the gas do?
Example: Change of State
• One liter of water (mass= 1kg) is boiled at
100°C into 1671 liters of steam at 100°C. The
process occurs at 1 atm.
– What is the heat absorbed by the water?
– What is the work done by the steam in expanding?
– What is the change in internal energy of the
system?
Schematic of an Engine
• Goal: To extract useful
work from the engine.
• Price: Energy input as
heat (QH) obtained by
burning fuel.
• Wasted Energy: Energy
lost through heat (QC)
• Efficiency (e):
e = W(OUT)/ QH
= (QH-QC)/ QH
Operation of a Heat Engine
• A “working substance” (recall the control mass or
closed system) is taken through a thermodynamic
cycle.
• Since Initial State = Final State:
– ΔUCYCLE = 0 = QNET - WNET(BY)
– QNET = WNET(BY)
• The net heat absorbed by the system equals the
net work done by the system.
Analyzing a Thermodynamic Cycle
• How do we analyze a cycle for its state properties
and usefulness?
• Example: One mole of an ideal gas goes through a
cycle with the following processes:
– AB: Isobaric expansion at 2P0 from a volume of V0 to
2V0.
– BC: Adiabatic expansion from 2V0 to 3V0 and P0.
– CD: Isobaric compression from 3V0 to V0.
– DA: Isometric compression from P0 to 2P0.
• First Step: Draw the cycle on a PV-diagram.
Analyzing a Thermodynamic Cycle
• For the preceding cycle, find the
following in terms of P0, V0, and R:
a)
b)
c)
d)
Find T at every state.
For each process, find Q, W(BY), ΔU
Find: QH, QC, WNET(BY) , ΔUCYCLE
What is the thermal efficiency?
Heat Capacity of an Ideal Gas
• Heat Capacity of a system (C) is the amount of
energy needed to cause a rise in temperature of
one degree.
• Heat Capacity per unit mass is specific heat.
• For an ideal gas: Heat capacity depends on the
process during which the heat is added.
• CV = (Q/ΔT) V=constant = (3/2)nR
• CP= (Q/ΔT) P=constant = (5/2)nR
• CP= CV +R;
γ= CP / CV = (5/3)
How Efficient can we make an Engine?
• Carnot worked to increase engine efficiency
• Ideal Engine: all processes occur reversibly
• Reversible Process: Carried out in
infinitesimally small steps:
– No friction
– Equilibrium achieved at each step.
Carnot Engine
• Carnot (Ideal) Efficiency achieved with the following cycle:
• The Carnot Efficiency depends only on the temperatures
between which engine is operated:
e (Carnot) = [(QH-QC)/ QH] (Carnot)
= [(TH-TC)/TH] = 1 - TC/TH
How do We Describe Disorder in a System?
• Second Law of Thermodynamics
– Heat flows sponraneously from hot to cold.
(Clausius)
– Heat can’t be converted completely into work.
(Kelvin-Planck)
– A system spontaneously tends towards a state of
higher disorder.
• Examples: Liquids mixing, gases mixing,
gases spontaneously expand, solids melting.
Two Views of a System
• Systems can be described microscopically or
macroscopically.
• Rolling Two Dice:
– 36 possible combos; each is a “microstate”(µ-state),
specified by two numbers (1-6 and 1-6).
– Each microstate is equally probable
– 11 possible “macrostates”
(each specifed by one number: 2-12)
– 7 is most probable macrostate; most disordered.
• Ideal Gas of N particles:
– Specify a microstate by listing positions and momenta
of all particles (6N numbers)
– Specify a macrostate by two numbers (e.g. P,V)
Entropy: Quantitative Measure of Disorder
• Microscopically: S = kb ln Ω (Ω = number of
possible microstates)
• Macroscopically (for an ideal gas):
Consider an isothermal expansion over a very small ∆V:
Q = W(by system) = P∆V =(nRT/V) ∆V
Then : n R (∆V/V) ~ Q/T (both reflect disorder)
(n ∆V/V) = relative increase in V and in n reflect
increase in disorder as process progresses.
Q/T = relative increase in K.E. of system also reflects
increasing disorder.
Change in Entropy
• ∆S = (Q/T)(REVERSIBLE PATH)
• By the Second Law of Thermodynamics
(For processes)
– For an isolated system: ∆S >= 0
– For other systems and their environments:
∆S (TOT) = ∆S (SYSTEM) + ∆S(ENVIRON) >= 0
For Real Processes: ∆S (TOT) > 0