lecture 18 second law of thermodynamics. carnot's cycle
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Lecture 18 second law of thermodynamics. carnot's cycleTRANSCRIPT
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Lecture 18Second law of
thermodynamics. Carnot cycle. Absolute zero.
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ACT: Leaving the fridge open
If you leave the door of your fridge open, you will get a heart-stopping electricity bill, but you will also:
A. Freeze the kitchen
B. Warm up the kitchen
H C C Q Q W Q
Now the fridge and the kitchen are one system. We are taking QC out of this system and dumping QH into it. Overall, heat is added!
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Second law of thermodynamics
It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body (“refrigerator” or Clausius statement)
i.e.,
It is impossible to build a 100%-efficient heat engine (e = 1)
It is impossible for any system to undergo a process in which it absorbs heat form a reservoir at a single temperature and convert the heat completely into mechanical work, with the system ending in the same state as it began (“engine” or Kelvin-Plank statement)
Or
i.e.,
It is impossible to build a workless refrigerator (K ∞)
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Clausius Kelvin-Plank
100%-efficient engineWorkless refrigerator
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Kelvin-Plank Clausius
100%-efficient engine
Workless refrigerator
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The Carnot cycle
Cycle made with reversible isothermal and adiabatic processes.
A
B
D
C
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ACT: Carnot cycle
Where in the cycle is heat absorbed? 1
2
4
3
A. At point 1
B. Between 1 and 2
C. Between 4 and 1
To absorb heat, we need a process between two states. 1 is a state.
No heat transfer between 4 and 1 (adiabatic)
Between 1 and 2, temperature is alwaysTH
1 2
1 2 1 2
0
0
U
Q W
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Heat exchange in Carnot’s cycle
1 2
21 2 1 2 H
1
0
ln 0
U
VQ W nRT
V
1
2
4
3
3 4
43 4 3 4 C
3
0
ln 0
U
VQ W nRT
V
1 1H 1 C 4T V T V 1 1
H 4 3
C 1 2
T V V
T V V
4 3
1 2
V V
V V1 1
H 2 C 3T V T V
2H
H 1 2 1
C 3 4 4C
3
ln
ln
VnRT
Q Q V
Q Q VnRT
V
32
1 4
VV
V V
H
C
T
T H H
C C
Q T
Q T
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Carnot efficiency
H
We
Q C
H
1T
T C H
H
Q Q
Q
C
H
1Q
Q C
CarnotH
1T
eT
Carnot’s cycle is completely reversible. Run backwards, it is a Carnot refrigerator
CC C
H CH C
QQ TK
T TW Q Q
C
CarnotH C
TK
T T
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In-class example: Carnot engine
A Carnot engine can operate between different sets of reservoirs.
1: TH = 80°C, TC = 10°C
2: TH = 10°C, TC = −50°C
3: TH = − 60°C, TC = −100°C
Rank their efficiencies (largest to smallest).
A. 1,2,3
B. 3,2,1
C. 3,1,2
D. 2,1,3
E. 2,3,1
1: TH = 353K, TC = 283K
2: TH = 283K, TC = 223K
3: TH = 213K, TC = 173KC
1
2831 0.198
353e C
CarnotH
1T
eT
2
2231 0.212
283e
3
1731 0.188
213e
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Carnot is the ideal cycle
The Carnot cycle is completely reversible. So let’s use it as a refrigerator, and couple it to a hypothetical superengine with eSE > eCarnot.
Carnot Super engine
QH
QC
QH + Δ
WΔ
QC
TC
TH
Δ
Δ
TC
TH
Superengine produces more work than Carnot for these temperatures
Required to balance energy in SE
This is equivalent to this!
Impossi
ble
No engine can be more efficient than a Carnot engine operating between the same two temperatures.
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Absolute zero
The colder you try to go, the less efficient the refrigerator gets:
Since heat leaks will not disappear as the object is cooled, you need more cooling power the colder it gets.The integral of the power required diverges as T 0.
Therefore you cannot cool a system to absolute zero
The best refrigerator you can get (Carnot) has performance
CCarnot
H C
TK
T T
C
CCarnot 0
H C
0T
TK
T T