# PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 19

```PHYSICS 231
INTRODUCTORY PHYSICS I
Lecture 19

Last Lecture
•
First Law of Thermodynamics
Q  U  W by the gas
•

Work done by/on a gas
W by the gas  PV  W on the gas
Some Vocabulary
P
• Isobaric
V
• P = constant
• Isovolumetric
• V = constant
• W = 0
• Isothermal
• T = constant
• U = 0 (ideal gas)
P
V
P
V
• Q = 0
P
V
P-V Diagrams
P
Path moves to right:
Wby the gas = Area under curve
P
V
Path moves to left:
Wby the gas = - Area under curve
V
(Won the gas = - Wby the gas)
Work from closed cycles
Clockwise cycle:
WA-&gt;B-&gt;A= Area
Counterclockwise cycle:
WA-&gt;B-&gt;A= -Area
U  0
in closed cycles
(work done by gas)
Example 12.8a
Consider an ideal gas
undergoing the trajectory
through the PV diagram.
In going from A to B to
C, the work done BY the
gas is _______ 0.
C
P
A
B
V
a) &gt;
b) &lt;
c) =
Example 12.8b
In going from A to B to
C, the change of the
internal energy of the
gas is _______ 0.
C
P
A
B
V
a) &gt;
b) &lt;
c) =
Example 12.8c
In going from A to B to
C, the amount of heat
_______ 0.
C
D
P
A
B
V
a) &gt;
b) &lt;
c) =
Example 12.8d
In going from A to B to
C to D to A, the work
done BY the gas is
_______ 0.
C
D
P
A
B
V
a) &gt;
b) &lt;
c) =
Example 12.8e
In going from A to B to
C to D to A, the change
of the internal energy of
the gas is _______ 0.
C
D
P
A
B
V
a) &gt;
b) &lt;
c) =
Example 12.8f
In going from A to B to
C to D to A, the heat
_______ 0.
C
D
P
A
B
V
a) &gt;
b) &lt;
c) =
Consider a monotonic ideal gas.
Example 12.7
a) What work was done by
P (kPa)
the gas from A to B?
A
75
20,000 J
b) What heat was added to
the gas between A and B?
50
20,000 J
c) What work was done by
the gas from B to C?
25
-10,000 J
C
d) What heat was added to
the gas between B and C?
-25,000 J
0.2
e) What work was done by
the gas from C to A?
0
f) What heat was added to
the gas from C to A?
15,000 J
B
V (m3)
0.4
0.6
Example 12.7 (Continued)
g) What was total work done
by gas in cycle?
P (kPa)
|Qin|
A
WAB + WBC + WCA = 10,000 J
h) What was total heat added
to gas in cycle?
C
B
QAB + QBC + QCA = 10,000 J
This does NOT mean
that the engine is 100%
efficient!
|Qin| = QAB + QCA= 35,000 J
|Qout| = |QBC| = 25,000 J
V (m3)
|Qout|
Exhaust!!!
Weng = |Qin|-|Qout|
Heat Engines
• Described by a cycle with:
Qhot= heat that flows into engine
from source at Thot
Qcold= heat exhausted from engine at
lower temperature, Tcold
W= work done by engine
Qhot
engine
W
Qcold
• Efficiency is defined:
engine:
Qhot  Qcold
W
Qcold

e
 1
Qhot
Qhot
Qhot
using
W  Qhot  Qcold
2nd Law of Thermodynamics
(version 1)
No heat engine can be 100% efficient
The most efficient engine is the Carnot Engine
(an idealized engine), for which:
Qcold Tcold

Qhot Thot
 eCarnot
W
Qcold
Tcold

1
1
Qhot
Qhot
Thot
(T in Kelvin)
In practice, we always have e  eCarnot

Carnot Cycle
Example 12.9
An ideal engine (Carnot) is rated at 50% efficiency
when it is able to exhaust heat at a temperature of 20
&ordm;C. If the exhaust temperature is lowered to -30 &ordm;C,
what is the new efficiency.
e = 0.585
Refrigerators
Just a heat engine run in reverse!
• Pull Qcold from fridge
• Exhaust Qhot to outside
Coefficient of Performance:
fridge
Qcold
COP(cooling) 
W
W
Qcold
Most efficient is Carnot refrigerator:
COP(cooling)  COPCarnot
Qhot
Tcold

Thot  Tcold
Note: Highest COP for small T differences
Heat Pumps
Same as refrigerator, except
• Pull Qcold from environment
• Exhaust Qhot to inside of house
Coefficient of Performance:
Qhot
COP(heating) 
W
Qhot
heat
pump
Qcold
Again, most efficient is Carnot:
COP(heat)  COPCarnot
W
Thot

Thot  Tcold
Like Refrigerator: Best performance for small T
Example 12.10
A modern gas furnace can work at practically 100%
efficiency, i.e., 100% of the heat from burning the
gas is converted into heat for the home. Assume
that a heat pump works at 50% of the efficiency of
an ideal heat pump.
If electricity costs 3 times as much per kw-hr as
gas, for what range of outside temperatures is it
advantageous to use a heat pump?
Assume Tinside = 295 &ordm;K.
5
T  295  245.8K  -27 C
6
Entropy
• Measure of Disorder of the system
(randomness, ignorance)
• S = kBlog(N)
N = # of possible arrangements for fixed E and Q
1000
900
800
700
600
500
400
300
200
100
0
(0,12) (1,11) (2,10) (3,9) (4,8) (5,7) (6,6) (7,5) (8,4) (9,3) (10,2) (11,1) (12,0)
Relative probabilities for 12 molecules to
arrange on two halves of container.
2nd Law of Thermodynamics
(version 2)
The Total Entropy of the Universe can never
decrease.
On a macroscopic level, one finds that adding heat
raises entropy:
S  Q / T
Defines temperature in Kelvin!
Why does Q flow from hot to cold?
• Consider two systems, one with TA and one with TB
• Allow Q &gt; 0 to flow from TA to TB
• Entropy changes by:
S = Q/TB - Q/TA
• This can only occur if S &gt; 0, requiring TA &gt; TB.
• System will achieve more randomness by exchanging
heat until TB = TA

Carnot Engine
Carnot cycle is most efficient possible, because
the total entropy change is zero.
It is a “reversible process”.
For real engines:
Qcold Qhot
S  Senvironment 

0 
Tcold Thot
W
Qcold
Tcold
e
1
1
 eCarnot
Qhot
Qhot
Thot
Example 12.11a
An engine does an amount of work W, and exhausts
heat at a temperature of 50 degrees C. The
chemical energy contained in the fuel must be
greater than, and not equal to, W.
a) True
b) False
Example 12.11b
exhausts heat into a large heat exchanger that
stays close to the temperature of the atmosphere.
The engine should be more efficient on a very cold
day than on a warm day.
a) True
b) False
Example 12.11c
An air conditioner uses an amount of electrical
energy U to cool a home. The amount of heat
removed from the home must be less than or equal
to U.
a) True
b) False
Example 12.11d
A heat pump uses an amount of electrical energy U
to heat a home. The amount of heat added to a
home must be less than or equal to U.
a) True
b) False
```