Chapter 16: The Second Law of Thermodynamics – Heat Engines, Entropy, and the Carnot Cycle

Chapter 16: The Second Law of Thermodynamics

Chapter 16.1: Reversible Processes

In thermodynamics, a reversible process is an idealized process that can be completely undone, returning both the system and its surroundings to their original states. Such processes must occur so slowly that the system remains in thermal (constant temperature, T) and mechanical (constant pressure, p) equilibrium throughout.

• Definition: A reversible process is one that can be reversed by infinitesimal changes in a variable, with no net change in the universe.

• Equilibrium: The system must stay in thermal and mechanical equilibrium at all times.

• Cyclic Process: For a cyclic process, the change in internal energy is zero: $\Delta U = Q - W = 0 \implies Q = W$

Chapter 16.2: Heat Engines

Heat engines are devices that convert heat energy into mechanical work. Their invention was pivotal to the industrial revolution and the development of thermodynamics.

• Operation: A heat engine absorbs heat $Q_H$ from a hot reservoir at temperature $T_H$, uses part of it to do work $W$, and expels the remaining heat $Q_C$ to a cold reservoir at temperature $T_C$.

• Net Heat Absorbed: $Q = Q_H + Q_C$ (with $Q_C$ negative, as it is expelled)

• Thermal Efficiency: The ratio of useful work to heat absorbed: $e = \frac{W}{Q_H}$

• Efficiency Range: $0 \leq e < 1$ (cannot convert all heat to work)

• Efficiency in Terms of Heats: $e = \frac{W}{Q_H} = \frac{Q_H + Q_C}{Q_H} = 1 - \left|\frac{Q_C}{Q_H}\right|$

Example: Truck Engine

• Given: $Q_H = 2500$ J, $W = 500$ J, $L_c = 5 \times 10^4$ J/g

• a) Thermal Efficiency: $e = \frac{500}{2500} = 0.2$

• b) Heat Discarded: $Q_C = Q_H - W = 2500 - 500 = 2000$ J

• c) Gasoline Burned: $m = \frac{Q_H}{L_c} = \frac{2500}{5 \times 10^4} = 0.05$ g

Chapter 16.3: Internal Combustion Engines

Internal combustion engines operate on cycles that convert chemical energy into mechanical work. The two main types are the Otto and Diesel cycles.

• Otto Cycle: Used in gasoline engines; requires a spark to ignite the fuel-air mixture.

• Diesel Cycle: Used in diesel engines; relies on high compression to ignite the mixture without a spark.

• Compression Ratio: Efficiency depends on the compression ratio $R = \frac{V_2}{V_1}$, where $V_2$ is the maximum and $V_1$ the minimum cylinder volume.

Chapter 16.4: Refrigerators

Refrigerators are devices that transfer heat from a cold reservoir to a hot reservoir, essentially operating as heat engines in reverse.

• Performance Coefficient (K): Measures efficiency: $K = \frac{|Q_C|}{W} = \frac{|Q_C|}{|Q_H| - |Q_C|}$

• Range: $K$ is between zero and infinity; higher $K$ means greater efficiency.

Chapter 16.5: The Second Law of Thermodynamics

The second law of thermodynamics states that it is impossible to construct a heat engine that operates in a cycle and converts all absorbed heat into work. Similarly, no process can transfer heat from a cooler to a hotter object without external work.

• Heat Engines: No engine can convert all heat from a single reservoir into work.

• Refrigerators: No process can transfer heat from cold to hot without work.

Chapter 16.6: The Carnot Cycle

The Carnot cycle is an idealized cycle that represents the most efficient possible heat engine. It consists of two isothermal and two adiabatic processes.

• Efficiency: $\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$ $e_{Carnot} = 1 - \left|\frac{Q_C}{Q_H}\right| = 1 - \frac{T_C}{T_H}$

• Kelvin Scale: Always use absolute temperatures (Kelvin).

• Practicality: Carnot engines are not practical due to the requirement for infinitely slow processes.

The Carnot Refrigerator

• Performance Coefficient: $K_{Carnot} = \frac{T_C}{T_H - T_C}$

• Kelvin Scale: Use Kelvin for all temperature calculations.

Chapter 16.7: Entropy

Entropy is a measure of the disorder or randomness of a system. It can also be interpreted as the amount of information missing about a system's microstate.

• High Entropy: Shuffled deck of cards, mixed coffee and creamer.

• Low Entropy: Ordered deck, unmixed coffee and creamer.

Change in Entropy

• Reversible Isothermal Process: $\Delta S = S_2 - S_1 = \frac{Q}{T}$

• Example: Melting 1.00 kg of ice at 0°C to water at 0°C: $\Delta S \approx 1220$ J/K (entropy increases as molecules become less ordered)

Reversible Adiabatic Process

• Adiabatic Process: $Q = 0$, $\Delta S = 0$ (no change in entropy)

• Second Law: For all other processes, net entropy change (system + environment) is positive: $\Delta S_{net} \geq 0$

Consequences of Increasing Entropy

• Creamer disperses in coffee, never the reverse.

• Rooms become dirty over time.

• A house of cards will naturally fall.

• A pressurized gas will spontaneously expand.

• The universe will eventually reach a state of maximum entropy (cold, dark, empty).

Bonus Slide: The Laws of Thermodynamics

• 0. There is a game. (Zeroth law)

• 1. You can't win. (First law: energy conservation)

• 2. You can't break even. (Second law: entropy increases)

• 3. You can't get out of the game. (Third law: absolute zero is unattainable)

Bonus Slide: Ludwig Boltzmann

• Boltzmann's entropy formula: $S = k \log W$ where $S$ is entropy, $k$ is Boltzmann's constant, and $W$ is the number of microstates.

Additional info: The notes provide a concise summary of the second law of thermodynamics, heat engines, entropy, and the Carnot cycle, suitable for college-level physics students. All equations are presented in LaTeX format for clarity.