📂Thermal Physics

Carnot's Theorem Proof

Theorem

No engine can be more efficient than a Carnot engine.

Description

Even though it’s impossible to actually implement a Carnot engine, the fact that it represents a theoretical limit is very meaningful.

Proof

Let’s assume that there is an engine $E$ more efficient than a Carnot engine $C$.

$E$ takes in heat $Q_{h} '$ and does work $W$, and $C$ takes in heat $Q_{l}$ and work $W$ to output heat $Q_{h}$.

If we consider both engines as one, $E + C$ takes in heat $Q_{h}’ - Q_{h}$ and outputs heat $Q_{l}’ - Q_{l}$.

First Law of Thermodynamics $$d U = \delta Q + \delta W$$

According to the first law of thermodynamics, since the change in internal energy (work) depends only on the change in heat energy, it follows:

$$W = Q_{h}’ - Q_{l} ' = Q_{h} - Q_{l}$$

By organizing $h$ and $l$ respectively, we get:

$$Q_{h}’ - Q_{h} = Q_{l}’ - Q_{l}$$

Furthermore, the efficiency of $C$ and $E$ is $\eta_{C} < \eta_{E}$, and since $C$ is a Carnot engine, we have:

$$\eta_{C} = 1 - \dfrac{Q_{l}}{Q_{h}}$$

Therefore, the following inequality holds:

$${{W} \over {Q_{h}}} = {{Q_{h} - Q_{l}} \over {Q_{h}}} = \eta_{C} < \eta_{E} = {{W} \over {Q_{h}’ }}$$

Eliminating $W$ from both ends and taking the reciprocal gives:

$$Q_{h}’ < Q_{h}$$

This means $Q_{h}’ - Q_{h} < 0$, but since it was also $Q_{l}’ - Q_{l} = Q_{h}’ - Q_{h}$, we conclude that heat flows from a lower to a higher temperature as shown:

Second Law of Thermodynamics

Clausius: A process that transfers heat from a cold to a hot body by itself does not occur.

This violates Clausius’ second law of thermodynamics, therefore by reductio ad absurdum, we know the assumption was false. Hence, there is no engine more efficient than a Carnot engine like $E$.

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