Lecture 38.2ndLawThe..

```nd
2
Law of Thermodynamics
Lecturer:
Professor Stephen T. Thornton
Which of the following is most true?
A) The second law of thermodynamics is
simply a statement of the conservation
of energy.
B) The Carnot Engine is the most efficient
engine possible.
C) Carnot built his engine as a gift to King
Henry VIII.
D) Heat engines normally operate between
three thermal reservoirs.
The Carnot cycle and Carnot engine is it!
Last Time
Heat transfer
Conduction
Convection
Today
Second Law of Thermodynamics
Heat engines
Carnot cycle and Carnot engine
If the first law of thermodynamics is about energy
conservation, then the 2nd law is about the way in
which energy flows.
Examples:
A bowl of water sitting in this room does not
spontaneously freeze.
It is impossible to construct an engine that can
extract thermal energy from a system and turn all
that energy into work.
Thermal systems spontaneously change in only
certain ways.
2nd Law of Thermodynamics
We can discuss this law in a number of ways.
The law basically states the way in which heat
flow occurs.
Heat flow between two objects brought
together in thermal contact always goes from
the hotter object to the colder object.
Lots of ways to say the same thing!
Heat Engines
An engine is a device that converts heat into
mechanical work.
Engines must operate in cycles in order to be
original position. The change in internal
energy is zero.
An engine operates between two thermal
reservoirs.
Schematic Diagram
of Heat Engine
W  Qh  Qc
W , Qh , Qc are positive.
Efficiency e
W Qh  Qc
e

Qh
Qh
Qc
e  1
Qh
Heat Engines
A steam engine is one type of heat engine.
Do demos
• Heat engine
• Steam engine
Our favorite
heat engine.
Q
h
Reversible
processes.
Qc
X Tc
Remember that the
work is equal to the
area under the P - V
curve. Total work
here is work enclosed
in cycle.
Q
h
Qc
X Tc
Carnot Cycle
• Carnot’s cycle represents the most efficient
engine possible.
• It operates between two heat reservoirs.
• All the processes are reversible – two
• We can show Qc = Tc for the Carnot
Qh Th
cycle.
Qc
Tc
e  1
 1
Qh
Th
For the highest efficiency, we need the
maximum difference of temperatures in
thermal reservoirs.
Tc
emax  1 
for Carnot cycle
Th
Because e  W / Qh , we have W  eQh
Wmax
 Tc 
 emax Qh  1   Qh
 Th 
Conceptual Quiz:
A heat engine absorbs 150 J of heat from a
hot reservoir and rejects 90 J of it to a cold
reservoir. What is the efficiency of this
engine?
Qc
Tc
e  1
 1
Qh
Th
A)
20%
B)
40%
C)
60%
D)
67%
E)
90%
Qc
90
e  1
 1
 0.40
Qh
150
Conceptual Quiz:
For the previous heat engine, you are told the
temperature of the hot reservoir is 200 oC
and that of the cold reservoir is 11oC. Your
response is to
Qc
Tc
e  1
 1
Qh
Th
A) believe that this is possible.
B) laugh at the idea.
C) contact a patent lawyer immediately.
Tc
11  273
284
e  1  1
 1
 0.4
Th
200  273
473
Another statement
of 2nd Law of
Thermodynamics
It is not possible to construct an engine
whose sole effect is to transform a given
amount of heat completely into work!
Thermography—the detailed measurement of
radiation from the body—can be used in medical
imaging. Warmer areas may be a sign of tumors
or infection; cooler areas on the skin may be a
sign of poor circulation.
Conceptual Quiz
what feels colder when you
walk on it, which of the
surfaces would have the
highest thermal conductivity?
A) a rug
B) a steel surface
C) a concrete floor
D) has nothing to do with
thermal conductivity
Conceptual Quiz
what feels colder when you
walk on it, which of the
A) a rug
B) a steel surface
surfaces would have the
C) a concrete floor
highest thermal conductivity?
E) has nothing to do with
thermal conductivity
The heat flow rate is k A (T1 − T2)/L. All things being
equal, bigger k leads to bigger heat loss.
From the book: Steel = 40, Concrete = 0.84,
Human tissue = 0.2, Wool = 0.04, in units of J/(s.m.C°).
Heat engine and refrigerator
This figure shows more details
of a typical refrigerator.
We analyze refrigerators differently.
We want to remove as much heat Qc as
possible for the least amount of work.
Coefficient of Performance or COP
Qc
COP =
W
Remember that Qh  Qc  W
This is the amount of heat exhausted
into kitchen.
For an air conditioner, this is the heat
exhausted to the outside.
Air conditioner and heat pump
inside house
Maximize Qh
Maximize Qc
Heat
house
A heat pump can heat a house in the winter:
For an ideal, reversible heat pump (i.e.
Carnot cycle), we have Qc  Tc
Q
h
T
h
 Qc 
W  Qh  Qc  Qh 1  
 Qh 
 Tc 
W  Qh 1  
 Th 
To minimize W we
want temperatures to
be similar.
Conceptual Quiz:
A heat engine exhausts heat QC to a cold
reservoir. The amount of work done by the
W  Qh  Qc
A)
B)
C)
D)
must be QC .
must be greater than QC .
must be less than QC .
could be greater than QC .
W = Qh – Qc > 0
We know that Qh > Qc, but that is about all
we know.
The work could be Qc, but we can’t tell.
The work can be greater or less than Qc, but
we can’t know.
The only reasonable answer is D.
Conceptual Quiz
The heat engine below is:
A) a reversible (Carnot) heat engine
B) an irreversible heat engine
C) a hoax
D) none of the above
Qc
Tc
e  1
 1
Qh
Th
For what??
TC = 310 K
Conceptual Quiz
The heat engine below is:
A) a reversible (Carnot) heat engine
B) an irreversible heat engine
C) a hoax
D) none of the above
Carnot e = 1 − TC/TH = 1 − 270/600 = 0.55.
But by definition e = 1 − QL/QH
= 1 − 4000/8000 = 0.5, smaller
than Carnot e, thus irreversible.
TC = 310 K
Hiker as Heat Engine. Assume that a 65
kg hiker needs 4.3 x 103 kcal of energy to
supply a day’s worth of metabolism.
Estimate the maximum height the person
can climb in one day, using only this
amount of energy. As a rough prediction,
treat the person as an isolated heat engine,
operating between the internal
temperature of 37°C (98.6°F) and the
ambient air temperature of 20°C.
Heat Pump. A heat pump is used to
keep a house warm at 22°C. How
much work is required of the pump to
deliver 3100 J of heat into the house
if the outdoor temperature is (a) 0°C,
(b) -15°C. Assume ideal (Carnot)
behavior.
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