Physics 151: Lecture 38 Today

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Physics 151 Lecture 38 Todays Agenda

• Todays Topics (Chapter 20)
• Internal Energy and Heat
• Heat Capacity
• First Law of Thermodynamics
• Special Processes

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Lecture 37 ACT 2Thermal expansion

• An aluminum plate has a circular hole cut in it.
  A copper ball (solid sphere) has exactly the same
  diameter as the hole when both are at room
  temperature, and hence can just barely be pushed
  through it. If both the plate and the ball are
  now heated up to a few hundred degrees Celsius,
  how will the ball and the hole fit ?

(a) ball wont fit (b) fits more easily
(c) same as before
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Lecture 37 ACT 2Solution
Before
After (higher T)
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Ideal gas / Review

• Equation of state for an ideal gas

PV nRT
R is called the universal gas constant
In SI units, R 8.315 J / molK
kB is called the Boltzmanns constant
kB R/NA 1.38 X 10-23 J/K
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Lecture 37 Problem 3
To

• The mass of a hot-air balloon and its cargo (not
  including the air inside) is 200 kg. The air
  outside is at 10.0C and 101 kPa. The volume of
  the balloon is 400 m3. To what temperature must
  the air in the balloon be heated before the
  balloon will lift off ?
• (Air density at 10.0C is 1.25 kg/m3.)

V, T
m
T 472 K !
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Internal Energy

• Internal energy is all the energy of a system
  that is associated with its microscopic
  components
• These components are its atoms and molecules
• The system is viewed from a reference frame at
  rest with respect to the center of mass of the
  system
• Internal energy does include kinetic energies due
  to
• Random translational motion (not motion through
  space)
• Rotational motion
• Vibrational motion
• Potential energy between molecules

Animation
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Heat

• Heat is defined as the transfer of energy across
  the boundary of a system due to a temperature
  difference between the system and its
  surroundings
• The term heat will also be used to represent the
  amount of energy transferred by this method
• Units of Heat historically-gt the calorie
• One calorie is the amount of energy transfer
  necessary to raise the temperature of 1 g of
  water from 14.5oC to 15.5oC
• In the US Customary system, the unit is a BTU
  (British Thermal Unit)
• One BTU is the amount of energy transfer
  necessary to raise the temperature of 1 lb of
  water from 63oF to 64oF
• The SI uinits are Joules, as we used before !

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Changing Internal Energy

• Both heat and work can change the internal energy
  of a system
• The internal energy can be changed even when no
  energy is transferred by heat, but just by work
• Example, compressing gas with a piston
• Energy is transferred by work

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Mechanical Equivalent of Heat

• James Joule in 1843 established the equivalence
  between mechanical energy and internal energy
• His experimental setup is shown at right
• The loss in potential energy associated with the
  blocks equals the work done by the paddle wheel
  on the water

• The amount of mechanical energy needed to raise
  the temperature of water from 14.5oC to 15.5oC is
  4.186 J

1 cal 4.186 J
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Heat Capacity

• The heat capacity (C) of a particular sample is
  defined as the amount of energy needed to raise
  the temperature of that sample by 1oC
• If energy Q produces a change of temperature of
  DT, then
• Q C DT
• Specific heat (c) is the heat capacity per unit
  mass

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Some Specific Heat Values
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ACT-1

• The Nova laser at Lawrence Livermore National
  Laboratory in California is used in studies of
  initiating controlled nuclear fusion. It can
  deliver a power of 1.60 x 1013 W over a time
  interval of 2.50 ns. Compare its energy output
  in one such time interval to the energy required
  to make a pot of tea by warming 0.800 kg of water
  from 20.0oC to 100oC.
• Which one is larger ?

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Calorimetry

• One technique for measuring specific heat
  involves heating a material, adding it to a
  sample of water, and recording the final
  temperature
• This technique is known as calorimetry
• A calorimeter is a device in which this energy
  transfer takes place
• The system of the sample and the water is
  isolated
• Conservation of energy requires that the amount
  of energy that leaves the sample equals the
  amount of energy that enters the water
• Cons. of Energy Qcold -Qhot

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Phase Changes

• A phase change is when a substance changes from
  one form to another. Two common phase changes are
• Solid to liquid (melting)
• Liquid to gas (boiling)
• During a phase change, there is no change in
  temperature of the substance
• If an amount of energy Q is required to change
  the phase of a sample of mass m, we can specify
  the Latent Heat associated with this transition
  is L Q /m
• The latent heat of fusion is used when the phase
  change is from solid to liquid
• The latent heat of vaporization is used when the
  phase change is from liquid to gas

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Graph of Ice to Steam
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Problem

• An ice cube (m0.070 kg) is taken from a freezer
  ( -10o C) and dropped into a glass of water at 0o
  C. How much of water will freeze ? (C(ice)
  2,000 J/kg K L(water) 334 kJ/kg)

m 4.19 g
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State Variables

• State variables describe the state of a system
• In the macroscopic approach to thermodynamics,
  variables are used to describe the state of the
  system
• Pressure, temperature, volume, internal energy
• These are examples of state variables
• The macroscopic state of an isolated system can
  be specified only if the system is in thermal
  equilibrium internally

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Transfer Variables

• Transfer variables are zero unless a process
  occurs in which energy is transferred across the
  boundary of a system
• Transfer variables are not associated with any
  given state of the system, only with changes in
  the state
• Heat and work are transfer variables
• Example of heat we can only assign a value of
  the heat if energy crosses the boundary by heat

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Work in Thermodynamics

• Work can be done on a deformable system, such as
  a gas
• Consider a cylinder with a moveable piston
• A force is applied to slowly compress the gas
• The compression is slow enough for all the system
  to remain essentially in thermal equilibrium
• This is said to occur quasi-statically

Therefore, the work done on the gas is dW -P dV
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PV Diagrams

• The state of the gas at each step can be plotted
  on a graph called a PV diagram
• This allows us to visualize the process through
  which the gas is progressing

• The work done on a gas in a quasi-static process
  that takes the gas from an initial state to a
  final state is the the area under the curve on
  the PV diagram, evaluated between the initial and
  final states

• This is true whether or not the pressure stays
  constant
• The work done does depend on the path taken

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Work Done By Various Paths
W
W -Pf (Vf Vi)
W -Pi (Vf Vi)

• Each of these processes has the same initial and
  final states
• The work done differs in each process
• The work done depends on the path

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The First Law of Thermodynamics

• The First Law of Thermodynamics is a special case
  of the Law of Conservation of Energy
• It takes into account changes in internal energy
  and energy transfers by heat and work
• Although Q and W each are dependent on the path,
  Q W is independent of the path
• The First Law of Thermodynamics states that
  DEint Q W
• All quantities must have the same units of
  measure of energy
• One consequence gtgt there must exist some
  quantity known as internal energy which is
  determined by the state of the system

Animation
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ACT

• Which statement below regarding the First Law of
  Thermodynamics is most correct ?
• a. A system can do work externally only if its
  internal energy decreases.
• b. The internal energy of a system that interacts
  with its environment must change.
• c. No matter what other interactions take place,
  the internal energy must change if a system
  undergoes a heat transfer.
• d. The only changes that can occur in the
  internal energy of a system are those produced by
  non-mechanical forces.
• e. The internal energy of a system cannot change
  if the heat transferred to the system is equal to
  the work done by the system.

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Adiabatic Process

• An adiabatic process is one during which no
  energy enters or leaves the system by heat
• Q 0
• This is achieved by
• Thermally insulating the walls of the system
• Having the process proceed so quickly that no
  heat can be exchanged
• Since Q 0, DEint W
• If the gas is compressed adiabatically, W is
  positive so DEint is positive and the temperature
  of the gas increases
• If the gas expands adiabatically, the temperature
  of the gas decreases

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Isothermal Process

• An isothermal process is one that occurs at a
  constant temperature
• Since there is no change in temperature, DEint
  0
• Therefore, Q - W
• Any energy that enters the system by heat must
  leave the system by work

Isothermal Expansion for an ideal gas PV
nRT and
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Isobaric Processes

• An isobaric process is one that occurs at a
  constant pressure
• The values of the heat and the work are generally
  both nonzero
• The work done is W P (Vf Vi) where P is the
  constant pressure

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Problem

• Identify processes A-D in the pV diagram below

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ACT

• In an adiabatic free expansion
• a. no heat is transferred between a system and
  its surroundings.
• b. the pressure remains constant.
• c. the temperature remains constant.
• d. the volume remains constant.
• e. the process is reversible.

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Cyclic Processes

• A cyclic process is one that starts and ends in
  the same state
• On a PV diagram, a cyclic process appears as a
  closed curve
• The change in the internal energy must be zero
  since it is a state variable
• If DEint 0, Q -W
• In a cyclic process, the net work done on the
  system per cycle equals the area enclosed by the
  path representing the process on a PV diagram

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ACT-2

• An ideal gas is carried through a thermodynamic
  cycle consisting of two isobaric and two
  isothermal processes as shown in Figure .
• What is the work done in this cycle, in terms of
  p1, p2, V1, V2 ?

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Recap of todays lecture

• Chap. 20
• Internal Energy and Heat
• Heat Capacity
• First Law of Thermodynamics
• Special Processes