Transcript Thermodynamics

Laws of Thermodynamics
Thermal Physics, Lecture
4
Internal Energy, U
Internal Energy is the total energy in a
substance, including thermal, chemical
potential, nuclear, electrical, etc.
Thermal Energy is that portion of the internal
energy that changes when the temperature
changes
Internal Energy
Thermal Energy
Heat, Q
The flow of energy into or out of a substance
due to a difference in temperature.
It results in a loss or gain in the Thermal
Energy
Work, W
The flow of energy into or out of a substance
that is NOT due to a difference in
temperature.
It results in a loss or gain in the Internal
Energy
First Law
U  Q  W
The first law of thermodynamics states that the
internal energy of a system is conserved.
Q is the heat that is added to the system
• If heat is lost, Q is negative.
W is the work done by the system.
• If work is done on the system, W is negative
Example using First Law
2500 J of heat is added to a system, and 1800 J
of work is done on the system. What is the
change in the internal energy of this system?
Signs: Q=+2500 J, W= -1800 J
U  Q  W
  2500 J    1800 J 
 4300 J
Thermodynamic Processes
We will consider a
system where an
ideal gas is contained
in a cylinder fitted
with a movable piston
Isothermal Processes
Consider an isothermal process
iso = same, so isothermal process happens
at the same temperature.
Since PV=nRT, if n and T are constant then
PV = constant
We can plot the
pressure and volume
of this gas on a PV
diagram
The lines of constant
PV are called
isotherm’s
Pressure
Isothermal Processes
Volume
Isothermal Processes
For an ideal gas, the internal energy U depends
only on T, so the internal energy does not change.
3
U  Nk BT for an ideal gas
2
Q must be added to increase the pressure, but the
volume expands and does work on the environment.
Therefore, Q=W
Adiabatic Processes
In an adiabatic process, no heat is allowed to
flow into or out of the system.
Q=0
Examples:
• well-insulated systems are adiabatic
• very rapid processes, like the expansion
of a gas in combustion, don’t allow time
for heat to flow (Heat transfers relatively
slowly)
Adiabatic Processes
U  Q  W
but Q  0 for
adiabatic processes
U  W
Pressure
A
Isothermal
B
Adiabatic
C
Volume
Isobaric Processes
A
B
Pressure
Isobaric processes
happen when the
pressure is
constant
Volume
Isovolumetric Processes
A
Pressure
Isovolumetric
processes
happen when the
volume is
constant
B
Volume
Calculating Work
For our ideal gas undergoing
an isobaric process
W  F d
 PAd
W  P  V
Calculating Work
A
Pressure
What if the pressure is
not constant?
The work is the area
under the curve of the
PV diagram.
B
Work
Volume
Adiabatic Process
Stretch a rubber band
suddenly and use your
lips to gauge the
temperature before and
after.
Summary of Processes
Isothermal: T is constant and Q=W since
U=0
Isobaric: P is constant and W=PV
Isovolumetric: V is constant so W=0 and Q =
U
Adiabatic: Q=0 so U = -W
Example
An ideal gas is slowly compressed at constant
pressure of 2 atm from 10 L to 2 L. Heat is
then added to the gas at constant volume
until the original temperature is reached.
What is the total work done on the gas?
Example
Example (15-5 in textbook)
Work is area under the graph.
Convert pressure to Pa and Volume to cubic
meters.
W  P  V
  2.02 10 Pa  8.0 10 m
5
 1.6 10 J
3
3
3

Example (15-5 in textbook)
How much heat flows into the gas?
U  Q  W
but T f  Ti so U  0
Q W
Q  1.6 10 J
3
Example 2: Boiling Water
1 kg (1 L) of water at 100 C is boiled away at
1 atm of pressure. This results in 1671 L of
steam. Find the change in internal energy.
Example 2: Boiling Water
U  Q  W
Q  mL  1 kg   2.26 106 J/kg 
 2.26 10 J
6
W  PV  1.01105 Pa 1670 10 3 m 3 
 1.69 105 J
U  Q  W
 2.09 106 J
Engines and Refrigerators
HEAT ENGINE
REFRIGERATOR
TH
TH
QH
system
QH
W
QC
TC
W
QC
TC
system taken in closed cycle  Usystem = 0
 therefore, net heat absorbed = work done
QH - QC = W (engine)
QC - QH = -W (refrigerator)

Heat Engine: Efficiency
The objective: turn heat from hot
reservoir into work
The cost: “waste heat”
HEAT ENGINE
TH
QH
1st Law: QH -QC = W
efficiency e  W/QH
W
QC
TC
Heat Engine
1500 J of energy, in the form of heat, goes
into an engine, which is able to do a total of
300 J of work. What is the efficiency of this
engine?
What happens to the rest of the energy?
Heat Engine
Can you get “work” out of a heat engine, if
the hottest thing you have is at room
HEAT ENGINE
temperature?
TH 300K
A) Yes
B) No
QH
W
QC
TC = 77K
Refrigerator
REFRIGERATOR
The objective: remove heat from
cold reservoir
The cost: work
TH
QH
1st Law: QH = W + QC
coeff of performance Kr  QC/W
W
QC
TC
New concept: Entropy (S)
A measure of “disorder”
A property of a system (just like p, V, T, U) related to
number of number of different “states” of system
Examples of increasing entropy:
• ice cube melts
• gases expand into vacuum
Change in entropy:
• S = Q/T
• >0 if heat flows into system (Q>0)
• <0 if heat flows out of system (Q<0)
Second Law of Thermodynamics
The entropy change (Q/T) of the system+environment
is always greater than zero (positive)
• never < 0
• Result: order to disorder
Consequences
• A “disordered” state cannot spontaneously
transform into an “ordered” state
• No engine operating between two reservoirs can
be more efficient than one that produces 0 change
in entropy. This is called a “Carnot engine”
Carnot Cycle
Idealized (Perfect) Heat Engine
• No Friction, so S = Q/T = 0
• Reversible Process
• Isothermal Expansion
• Adiabatic Expansion
• Isothermal Compression
• Adiabatic Compression
Perpetual Motion Machines?
Carnot Efficiency
The absolute best a heat engine can do is
given by the Carnot efficiency:
eideal
TH  TL

TH
Carnot Efficiency
A steam engine operates at a temperature of
500 C in an environment where the
surrounding temperature is 20 C. What is
the maximum (ideal) efficiency of this
engine?
What is the operating temperature were
increased to 800 C ?
Engines and the 2nd Law
The objective: turn heat from
hot reservoir into work
The cost: “waste heat”
HEAT ENGINE
TH
QH
1st Law: QH -QC = W
efficiency e  W/QH =W/QH = 1QC/QH
W
QC
TC
Summary
First Law of thermodynamics: Energy Conservation
• Q = U + W
Heat Engines
• Efficiency = 1-QC/QH
Refrigerators
• Coefficient of Performance =
QC/(QH - QC)
Entropy S = Q/T
2nd Law: Entropy always increases!
Carnot Cycle: Reversible, Maximum Efficiency
e = 1 – Tc/Th