Lecture 18 second law of thermodynamics. carnot's cycle

1. Lecture 18
Second law of thermodynamics.
Carnot cycle. Absolute zero.

2. 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
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!
QH = QC +W
> QC

3. Second law of thermodynamics
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)
i.e.,
It is impossible to build a 100%-efficient heat engine (e = 1)
Or
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 workless refrigerator (K →∞)

4. Clausius ⇒ Kelvin-Plank
100%-efficient engine
Workless
refrigerator

5. Kelvin-Plank ⇒ Clausius
Workless
refrigerator
100%-efficient
engine

6. The Carnot cycle
Cycle made with reversible isothermal and adiabatic
processes.
A
B
D
C

7. ACT: Carnot cycle
Where in the cycle is heat absorbed?
1
To absorb heat, we need a
process between two
states. 1 is a state.
2
No heat transfer between
4 and 1 (adiabatic)
4
3
A. At point 1
B. Between 1 and 2
C. Between 4 and 1
Between 1 and 2,
temperature is alwaysTH
∆U1→2 = 0
Q1→2 = W1→2 > 0

8. Heat exchange in Carnot’s cycle
∆U1→2 = 0
Q1→2 = W1→2
∆U3→ 4 = 0
Q3→ 4 = W3→ 4
1
V2
= nRTH ln > 0
V1
V4
= nRTC ln < 0
V3
THV1 γ −1 =TCV4γ −1
THV2
γ −1
=TCV3
γ −1
QH Q1→2
=
QC Q3→ 4
2
TH V4
=
TC  V1

γ −1

÷
÷

4
3
V3
=
V
 2
V
nRTH ln 2
V1
T
=
=− H
V
TC
nRTC ln 4
V3
γ −1

÷
÷

V4 V3
=
V1 V2
V2 V3
=
V1 V4
QH
TH
=−
QC
TC

9. Carnot efficiency
QC + QH
QC
TC
W
e=
=
=1+
=1−
QH
QH
QH
TH
eCarnot
TC
=1−
TH
Carnot’s cycle is completely reversible. Run
backwards, it is a Carnot refrigerator
QC
QC
TC
K =
=
=
W
QH − QC TH −TC
K Carnot
TC
=
TH −TC

10. In-class example: Carnot engine
A Carnot engine can operate between different sets of
reservoirs.
1: TH = 80°C, TC = 10°C
1: TH = 353K, TC = 283K
2: TH = 10°C, TC = −50°C
2: TH = 283K, TC = 223K
3: TH = − 60°C, TC = −100°C
3: TH = 213K, TC = 173KC
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
eCarnot = 1 −
TC
TH
283
= 0.198
353
223
e2 = 1 −
= 0.212
283
e1 = 1 −
173
e3 = 1 −
= 0.188
213

11. 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.
TH
QH + Δ
QH
Carnot
W
QC
Super
engine
QC
Δ
Required to balance
energy in SE
Superengine
produces more work
than Carnot for
these temperatures
TC
TH
Δ
sib
os
p
Im
TC
This is equivalent to
this!
No engine can be more efficient than a Carnot engine
operating between the same two temperatures.
le
Δ

12. Absolute zero
The best refrigerator you can get (Carnot) has performance
K Carnot
TC
=
TH −TC
The colder you try to go, the less efficient the refrigerator
gets:
T
K Carnot =
C
TH −TC
→ 0
T →0
C
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