4Classical thermodynamics
II Statistical Physics
4.3 Carnot cycles
We now construct our “canonical” examples of a heat engine, known as the
Carnot cycle. The important feature of this heat engine is that it is reversible,
so we can run it backwards, and turn it into a fridge.
The system consists of a box of gas that can be expanded or compressed.
The Carnot cycle will involve compressing and expanding this box of gas at
different environments in a clever way in order to extract work out of it. Before
we explain how it works, we can plot the “trajectory” of the system on the
p

V
plane as follows:
V
p
C
B
D
A
T = T
H
T = T
C
–
We start at point
A
, and perform isothermal expansion to get to
B
. In
other words, we slowly expand the gas while in thermal contact with the
heat reservoir at hot temperature
T
H
. As it expands, the gas is doing work,
which we can use to run a TV. Since it remains in constant temperature,
it absorbs some heat Q
H
.
–
When we reach
B
, we perform adiabatic expansion to get to
C
. To do this,
we isolate the system and allow it to expand slowly. This does work, and
no heat is absorbed.
–
At
C
, we perform isothermal compression to get to
D
. We put the system
in thermal contact with the cold reservoir at the cold temperature
T
C
.
Then we compress the gas slowly. This time, we are doing work on the gas,
and as we compress it, it gives out heat Q
C
to the cold reservoir.
–
From
D
to
A
, we perform adiabatic compression. We compress the gas
slowly in an isolated way. We are doing work on the system. Since it is
isothermal, no heat is given out.
Note that we are not just running in a cycle here. We are also transferring heat
from the hot reservoir to the cold reservoir.
Since these combined things form a cycle, we saw that we can write
−
I
¯dW =
I
¯dQ.
The total amount of work done is
W = Q
H
− Q
C
.
If we are building a steam engine, we want to find the most efficient way of
doing this. Given a fixed amount of
Q
H
, we want to minimize the heat
Q
C
we
return to the cold reservoir.
Definition (Efficiency). The efficiency of a heat engine is
η =
W
Q
H
= 1 −
Q
C
Q
H
.
Ideally, we would want to build a system where
η
= 1. This is, of course,
impossible, by Kelvin’s second law. So we can ask — what is the best we can
do? The answer is given by Carnot’s theorem.
Theorem.
Of all engines operating between heat reservoirs, reversible engines
are the most efficient. In particular, all reversible engines have the same efficiency,
and is just a function of the temperatures of the two reservoirs.
Proof.
Consider any other engine, Ivor. We use this to drive Carnot (an arbitrary
reversible engine) backwards:
hot reservoir
cold reservoir
Ivor
reverse
Carnot
Q
0
H
Q
0
C
W
Q
C
Q
H
Now we have heat
Q
0
H
− Q
H
extracted from the hot reservoir, and
Q
0
C
− Q
C
deposited in the cold, and we have
Q
0
H
− Q
H
= Q
0
C
− Q
C
.
Then Clausius’ law says we must have
Q
0
H
− Q
H
≥
0. Then the efficiency of
Igor is
η
Ivor
=
Q
0
H
− Q
0
C
Q
0
H
=
Q
H
− Q
C
Q
0
H
≤
Q
H
− Q
C
Q
H
= η
Carnot
.
Now if Ivor is also reversible, then we can swap the argument around, and deduce
that η
Carnot
≤ η
Ivor
. Combining the two inequalities, they must be equal.
In this case, we know
η
is just a function of
T
H
and
T
C
. We call this
η(T
H
, T
C
). We can use this to define T .
To do so, we need three heat reservoirs,
T
1
> T
2
> T
3
, and run some Carnot
engines.
T
1
T
2
T
3
Carnot
Carnot
Q
1
Q
2
Q
2
Q
3
W
W
0
Then we have
Q
2
= Q
1
(1 − η(T
1
, T
2
))
and thus
Q
3
= Q
2
(1 − η(T
2
, T
3
)) = Q
1
(1 − η(T
1
, T
2
))(1 − η(T
2
, T
3
)).
But since this composite Carnot engine is also reversible, we must have
Q
3
= Q
1
(1 − η(T
1
, T
3
)).
So we deduce that
1 − η(T
1
, T
3
) = (1 − η(T
1
, T
2
))(1 − η(T
2
, T
3
)). (∗)
We now fix a T
∗
. We define
f(T ) = 1 − η(T, T
∗
)
g(T ) = 1 − η(T
∗
, T ).
So we have
1 − η(T, T
0
) = f(T )g(T
0
).
Plugging this into (∗), we find
f(T
1
)g(T
3
) = f(T
1
)g(T
2
)f(T
2
)g(T
3
).
So we find that
g(T
2
)f(T
2
) = 1
for any T
2
.
Thus, we can write
1 − η(T, T
0
) = f(T )g(T
0
) =
f(T )
f(T
0
)
.
The ideal is to use this to define temperature. We now define T such that
f(T ) ∝
1
T
.
In other words, such that a reversible engine operating between
T
H
and
T
C
has
η = 1 −
T
C
T
H
.
Of course, this is defined only up to a constant. We can fix this constant by
saying the “triple point” of water has
T = 273.16 K.
The triple point is a very precisely defined temperature, where water, ice and
water vapour are in equilibrium. We will talk about this later when we discuss
phase transitions.
This is rather remarkable. We started from a rather wordy and vague
statement of the second law, and we came up with a rather canonical way of
defining temperature.