Entropy and Second Law of Thermodynamics: Complete Physics HRK Notes & Formulas

House of Physics October 01, 2025
Entropy and Second Law of Thermodynamics: Complete Physics Guide | HRK Chapter 26
Mastering Heat Engines, Refrigerators, Carnot Cycle, Thermodynamic Temperature Scale, and Entropy Concepts
Entropy Second Law of Thermodynamics Heat Engine Carnot Cycle Thermodynamic Temperature Scale Reading Time: 25 min

📋 Table of Contents

📜 Historical Background

The development of the second law of thermodynamics and entropy concept spanned the 19th century:

  • Sadi Carnot (1824): Published "Reflections on the Motive Power of Fire" establishing principles of heat engines
  • Rudolf Clausius (1850): Formulated the second law and introduced the concept of entropy
  • Lord Kelvin (1851): Proposed an alternative statement of the second law
  • Ludwig Boltzmann (1870s): Developed statistical interpretation of entropy

These developments fundamentally changed our understanding of energy conversion and the direction of natural processes.

Introduction to Second Law of Thermodynamics

🔬 What is the Second Law of Thermodynamics?

The second law of thermodynamics deals with the direction of natural processes. While the first law concerns energy conservation, the second law determines which processes can occur spontaneously in nature.

Many processes that conserve energy (satisfy the first law) are never observed in nature. For example, a glass of cool water never spontaneously transforms into an ice cube in warmer water. The second law gives a preferred direction to the "arrow of time."

📝 The Arrow of Time

The second law tells us that systems naturally evolve with time in one direction but not the other. It introduces the concept of entropy, which measures the disorder or randomness of a system, just as:

  • The zeroth law led to the concept of temperature
  • The first law led to the concept of internal energy

Heat Engines and Efficiency

🔥 What is a Heat Engine?

A heat engine is a device that converts heat energy into mechanical energy. Every heat engine:

  • Takes heat from a hot body (source)
  • Converts part of it into work
  • Rejects the remaining part to a cold body (sink)

⚙️ Heat Engine Operation

Hot Reservoir at \( T_h \)
↓ \( Q_h \)
ENGINE
\( W_{eng} \)
↓ \( Q_c \)
Cold Reservoir at \( T_c \)

Working Principle: A heat engine carries a working substance through a cyclic process during which:

  1. The working substance absorbs energy by heat from a high-temperature reservoir
  2. Work is done by the engine
  3. Energy is expelled by heat to a lower-temperature reservoir

Example: In a steam engine, water absorbs energy in a boiler, evaporates to steam, does work by expanding against a piston, then condenses and returns to the boiler.

Efficiency of Heat Engine

🧮 Efficiency Calculation

Step 1: Work Done in One Cycle

After completing one complete cycle, the system returns to its initial state, so its internal energy remains constant:

\[ \Delta U = 0 \]
\[ W = Q_h - Q_c \]

Step 2: Efficiency Definition

Efficiency is defined as the ratio of work done to heat absorbed:

\[ e = \frac{W}{Q_h} \]
\[ = \frac{Q_h - Q_c}{Q_h} \]
\[ = 1 - \frac{Q_c}{Q_h} \]

💡 Key Insight

The efficiency of a heat engine is always less than 100%. A perfectly efficient engine (\( e = 1 \)) would require \( Q_c = 0 \), meaning no heat is transferred to the cold body. However, experiments show that no such heat engine can be made.

Kelvin Statement of Second Law of Thermodynamics

📜 Kelvin Statement

"It is impossible to make a heat engine which operating in a cycle can go on doing work by taking heat from a single body, without other body at lower temperature."

This means that two bodies at different temperatures are necessary for the working of a heat engine. The hot body is called the source, while the cold body is called the sink.

Sample Problem: Automobile Engine

Sample Problem 1: Automobile Engine

An automobile engine, whose thermal efficiency e is 22%, operates at 95 cycles per second and does work at the rate of 120 hp.

(a) How much work per cycle is done on the system by the environment?

(b) How much heat enters and leaves the engine in each cycle?

Given Data:
\[ e = 22\% = 0.22 \]
\[ \text{Number of cycles per second} = 95 \]
\[ \text{Rate of Work} = 120 \, \text{hp} = 120 \times 746 = 89520 \, \text{J/s} \]
(a) Work done per cycle:
\[ W = \frac{\text{Rate of Work}}{\text{Number of cycles}} \]
\[ = \frac{89520}{95} \]
\[ = 942 \, \text{J} \]
(b) Heat entering the system:
\[ e = \frac{W}{Q_h} \Rightarrow Q_h = \frac{W}{e} \]
\[ = \frac{942}{0.22} \]
\[ = 4300 \, \text{J} \]
Heat leaving the system:
\[ W = Q_h - Q_c \Rightarrow Q_c = Q_h - W \]
\[ = 4300 - 942 \]
\[ = 3358 \, \text{J} \]

Refrigerators and Coefficient of Performance

❄️ What is a Refrigerator?

A refrigerator is essentially a heat engine run in reverse. It needs work to transfer heat from a cold body to a hot body.

If \( Q_c \) is the amount of heat removed from the cold body at temperature \( T_c \) and \( Q_h \) is the amount of heat given to the hot body at temperature \( T_h \), then the difference \( Q_h - Q_c \) is the work done on the system.

⚙️ Refrigerator Operation

Hot Reservoir at \( T_h \)
↑ \( Q_h \)
REFRIGERATOR
\( W \)
↑ \( Q_c \)
Cold Reservoir at \( T_c \)

Working Principle: A refrigerator operates by:

  1. Absorbing heat from a cold reservoir
  2. Receiving work input from an external source
  3. Rejecting heat to a hot reservoir

Energy Balance: \( Q_h = Q_c + W \)

Example: In a household refrigerator, heat is removed from the interior (cold reservoir) and released to the kitchen (hot reservoir) with the help of electrical work.

Coefficient of Performance

🧮 Coefficient of Performance Calculation

Step 1: Definition

The coefficient of performance (K) of a refrigerator is defined as:

\[ K = \frac{\text{Desired effect}}{\text{Work input}} \]
\[ = \frac{Q_c}{W} \]

Step 2: Alternative Form

\[ W = Q_h - Q_c \]
\[ K = \frac{Q_c}{Q_h - Q_c} \]

Clausius Statement of Second Law of Thermodynamics

📜 Clausius Statement

"It is impossible to make a refrigerator which operating in a cycle can transfer heat from a cold body to a hot body without any external work."

This means that heat cannot flow spontaneously from a cold body to a hot body. Work must be done to achieve this transfer.

Sample Problems

Sample Problem 2: Refrigerator Performance

A refrigerator has a coefficient of performance K = 5.0. If the temperature of the kitchen is 20°C, what is the lowest possible temperature inside the refrigerator?

Given:
\[ K = 5.0 \]
\[ T_h = 20^\circ C = 293 \, K \]
For an ideal refrigerator (Carnot refrigerator):
\[ K = \frac{T_c}{T_h - T_c} \]
\[ 5.0 = \frac{T_c}{293 - T_c} \]
\[ 5(293 - T_c) = T_c \]
\[ 1465 - 5T_c = T_c \]
\[ 1465 = 6T_c \]
\[ T_c = \frac{1465}{6} \]
\[ = 244.17 \, K \]
\[ = -28.83^\circ C \]
Sample Problem 3: Heat Pump

A heat pump is used to maintain the inside temperature of a house at 20°C when the outside temperature is 0°C. The heat losses from the house are 100,000 kJ/h. What is the minimum power required to operate the heat pump?

Given:
\[ T_h = 20^\circ C = 293 \, K \]
\[ T_c = 0^\circ C = 273 \, K \]
\[ Q_h = 100,000 \, \text{kJ/h} \]
For an ideal heat pump (Carnot heat pump):
\[ K = \frac{T_h}{T_h - T_c} \]
\[ = \frac{293}{293 - 273} \]
\[ = \frac{293}{20} \]
\[ = 14.65 \]
Coefficient of performance for heat pump:
\[ K = \frac{Q_h}{W} \]
\[ W = \frac{Q_h}{K} \]
\[ = \frac{100,000}{14.65} \]
\[ = 6825.94 \, \text{kJ/h} \]
\[ = \frac{6825.94}{3600} \, \text{kJ/s} \]
\[ = 1.896 \, \text{kW} \]

Equivalence of Clausius and Kelvin Statements

🔄 Statement Equivalence

The Kelvin and Clausius statements of the second law are equivalent. If one is violated, the other is also violated. This can be demonstrated by showing that:

  1. A violation of the Kelvin statement implies a violation of the Clausius statement
  2. A violation of the Clausius statement implies a violation of the Kelvin statement

🧮 Proof of Equivalence

Part 1: Kelvin Violation → Clausius Violation

Assume a Kelvin-violating engine E that extracts heat Q from a cold reservoir and converts it completely to work W = Q.

Now use this work to drive a refrigerator R that extracts heat Q_c from a cold reservoir and rejects heat Q_h = Q_c + W to a hot reservoir.

The net effect: Heat Q_c + Q is transferred from cold to hot reservoir without any external work, violating the Clausius statement.

Part 2: Clausius Violation → Kelvin Violation

Assume a Clausius-violating refrigerator R that transfers heat Q from cold to hot reservoir without work.

Now connect a heat engine E between the same reservoirs, extracting heat Q_h from hot reservoir, doing work W, and rejecting heat Q_c to cold reservoir.

Adjust E so that Q_c = Q. Then the cold reservoir is unaffected, and we have extracted work W = Q_h - Q_c from a single reservoir, violating the Kelvin statement.

Carnot Engine and Carnot Cycle

🔬 What is a Carnot Engine?

A Carnot engine is an ideal reversible heat engine operating between two temperatures. It was proposed by Sadi Carnot in 1824 and has the maximum possible efficiency for given reservoir temperatures.

The Carnot cycle consists of four reversible processes:

  1. Isothermal expansion at temperature \( T_h \)
  2. Adiabatic expansion
  3. Isothermal compression at temperature \( T_c \)
  4. Adiabatic compression

📈 Carnot Cycle on PV Diagram

[Graph: PV diagram showing the four processes of Carnot cycle]

The Carnot cycle forms a closed loop on a PV diagram, with two isotherms and two adiabats. The area enclosed represents the net work done per cycle.

Efficiency of Carnot Engine

🧮 Carnot Efficiency Derivation

Step 1: Working Substance

Consider an ideal gas as the working substance. For isothermal expansion at \( T_h \):

\[ Q_h = nRT_h \ln\left(\frac{V_2}{V_1}\right) \]

Step 2: Isothermal Compression

For isothermal compression at \( T_c \):

\[ Q_c = nRT_c \ln\left(\frac{V_3}{V_4}\right) \]

Step 3: Adiabatic Relations

For adiabatic processes:

\[ T_h V_2^{\gamma-1} = T_c V_3^{\gamma-1} \]
\[ T_h V_1^{\gamma-1} = T_c V_4^{\gamma-1} \]

Step 4: Volume Ratio Relation

\[ \frac{V_2}{V_1} = \frac{V_3}{V_4} \]

Step 5: Efficiency Calculation

\[ e = 1 - \frac{Q_c}{Q_h} \]
\[ = 1 - \frac{nRT_c \ln(V_3/V_4)}{nRT_h \ln(V_2/V_1)} \]
\[ = 1 - \frac{T_c}{T_h} \]

💡 Carnot's Theorem

"No heat engine operating between two given temperatures can be more efficient than a Carnot engine operating between the same two temperatures."

This theorem establishes the Carnot engine as the standard for maximum possible efficiency.

Sample Problems

Sample Problem 4: Carnot Engine Efficiency

A Carnot engine operates between temperatures of 500 K and 300 K. If the engine absorbs 1000 J of heat from the high-temperature reservoir, calculate:

(a) The efficiency of the engine

(b) The work done by the engine

(c) The heat rejected to the low-temperature reservoir

Given:
\[ T_h = 500 \, K \]
\[ T_c = 300 \, K \]
\[ Q_h = 1000 \, J \]
(a) Efficiency:
\[ e = 1 - \frac{T_c}{T_h} \]
\[ = 1 - \frac{300}{500} \]
\[ = 1 - 0.6 \]
\[ = 0.4 = 40\% \]
(b) Work done:
\[ W = e \times Q_h \]
\[ = 0.4 \times 1000 \]
\[ = 400 \, J \]
(c) Heat rejected:
\[ Q_c = Q_h - W \]
\[ = 1000 - 400 \]
\[ = 600 \, J \]

Thermodynamic Temperature Scale

🌡️ Kelvin Temperature Scale

Lord Kelvin proposed a temperature scale based entirely on the properties of heat engines, independent of the properties of any particular substance.

The thermodynamic temperature scale is defined such that:

\[ \frac{Q_h}{Q_c} = \frac{T_h}{T_c} \]

This ratio depends only on the temperatures of the reservoirs, not on the working substance.

🧮 Derivation of Thermodynamic Temperature

Step 1: Carnot Engine Property

For any Carnot engine operating between temperatures \( \theta_1 \) and \( \theta_2 \):

\[ \frac{Q_2}{Q_1} = f(\theta_1, \theta_2) \]

where f is a universal function of the temperatures

Step 2: Three Reservoirs

Consider three reservoirs at temperatures \( \theta_1 > \theta_2 > \theta_3 \). Connect engines between them:

\[ \frac{Q_1}{Q_2} = f(\theta_1, \theta_2) \]
\[ \frac{Q_2}{Q_3} = f(\theta_2, \theta_3) \]
\[ \frac{Q_1}{Q_3} = f(\theta_1, \theta_3) \]

Step 3: Functional Relation

\[ f(\theta_1, \theta_3) = f(\theta_1, \theta_2) \cdot f(\theta_2, \theta_3) \]

This functional equation has the solution:

\[ f(\theta_1, \theta_2) = \frac{\phi(\theta_1)}{\phi(\theta_2)} \]

Step 4: Thermodynamic Temperature

Define thermodynamic temperature T proportional to \( \phi(\theta) \):

\[ \frac{Q_h}{Q_c} = \frac{T_h}{T_c} \]

Absolute Zero and Negative Temperature

❄️ Absolute Zero

Absolute zero is the temperature at which a Carnot engine would have 100% efficiency. From the Carnot efficiency formula:

\[ e = 1 - \frac{T_c}{T_h} \]

If \( T_c = 0 \), then \( e = 1 \), meaning all heat input is converted to work.

⚠️ Negative Temperatures

In the thermodynamic temperature scale, negative absolute temperatures are not possible for normal systems. However, in certain specialized systems with population inversion (like lasers), negative temperatures can be defined, but these are "hotter" than any positive temperature in terms of heat flow.

Entropy Function

🔬 What is Entropy?

Entropy is a state function that measures the disorder or randomness of a system. It was introduced by Rudolf Clausius and is defined for reversible processes as:

\[ dS = \frac{dQ_{rev}}{T} \]

where \( dQ_{rev} \) is the heat transferred reversibly at temperature T.

Entropy in Reversible Processes

🧮 Entropy Change Calculation

Step 1: Definition

The entropy change in a reversible process is:

\[ \Delta S = \int \frac{dQ_{rev}}{T} \]

Step 2: Important Properties

  • Entropy is a state function - its change depends only on initial and final states
  • For a cyclic process, \( \oint dS = 0 \)
  • Entropy is extensive - it scales with the size of the system
Sample Problem 5: Entropy Change in Isothermal Expansion

Calculate the entropy change when 1 mole of an ideal gas expands isothermally and reversibly from volume V₁ to volume V₂.

For isothermal expansion of ideal gas:
\[ \Delta U = 0 \]
\[ dQ_{rev} = dW_{rev} = PdV \]
\[ = \frac{nRT}{V} dV \]
Entropy change:
\[ \Delta S = \int \frac{dQ_{rev}}{T} \]
\[ = \int_{V_1}^{V_2} \frac{nR}{V} dV \]
\[ = nR \ln\left(\frac{V_2}{V_1}\right) \]

Entropy in Irreversible Processes

🔥 Entropy and Irreversibility

For irreversible processes, the entropy change is greater than the integral of dQ/T:

\[ \Delta S \geq \int \frac{dQ}{T} \]

where the equality holds for reversible processes and the inequality for irreversible processes.

Sample Problem 6: Entropy Change in Free Expansion

Calculate the entropy change when 1 mole of an ideal gas undergoes free expansion into vacuum from volume V₁ to volume V₂.

In free expansion:
\[ W = 0, Q = 0, \Delta U = 0 \]
\[ \text{Temperature remains constant} \]
Since entropy is a state function, we can calculate it using a reversible path between the same states:
\[ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) \]
\[ \text{(same as reversible isothermal expansion)} \]
Note: For the actual irreversible process:
\[ \int \frac{dQ}{T} = 0 \]
\[ \Delta S > \int \frac{dQ}{T} \]
\[ \text{confirming the inequality for irreversible processes} \]

Entropy and the Second Law

📜 Entropy Statement of Second Law

"The entropy of an isolated system never decreases. In any natural process, the total entropy of the system and its surroundings either increases or remains constant."

\[ \Delta S_{universe} \geq 0 \]

where the equality holds for reversible processes and the inequality for irreversible processes.

🌍 Universe's Entropy

The entropy of the universe is constantly increasing. This gives a direction to time - the "arrow of time" - explaining why we remember the past but not the future.

🔥 Heat Death

If the universe is an isolated system, its entropy will continue to increase until it reaches a maximum. At this point, no more work can be extracted - this is called the "heat death" of the universe.

🔬 Statistical Interpretation

Boltzmann showed that entropy is related to the number of microstates Ω corresponding to a macrostate:

\[ S = k_B \ln \Omega \]

where \( k_B \) is Boltzmann's constant.

Frequently Asked Questions

❓ Why can't a heat engine be 100% efficient?

A 100% efficient heat engine would violate the second law of thermodynamics. According to the Kelvin statement, it's impossible to convert heat completely into work in a cyclic process without rejecting some heat to a colder reservoir.

Even an ideal Carnot engine, which has the maximum possible efficiency, must reject some heat to the cold reservoir:

\[ e_{Carnot} = 1 - \frac{T_c}{T_h} < 1 \]

This limitation arises because heat is a form of energy associated with random molecular motion, and converting it completely to ordered work would require decreasing the entropy of the universe, which is impossible.

❓ What is the difference between a refrigerator and a heat pump?

Both refrigerators and heat pumps operate on the same principle - they transfer heat from a colder to a hotter reservoir using work input. The difference is in their purpose:

  • Refrigerator: Designed to extract heat from the cold interior (to keep it cold)
  • Heat Pump: Designed to deliver heat to the hot interior (to keep it warm)

Their coefficients of performance are defined differently:

\[ K_{refrigerator} = \frac{Q_c}{W} \]
\[ K_{heat pump} = \frac{Q_h}{W} \]

Note that \( K_{heat pump} = K_{refrigerator} + 1 \)

❓ Why does entropy always increase in irreversible processes?

Entropy increases in irreversible processes because these processes represent a movement from less probable to more probable states. From the statistical perspective:

  • Ordered states have fewer microstates and lower entropy
  • Disordered states have more microstates and higher entropy
  • Systems naturally evolve toward states with higher probability

For example, when two gases mix, the mixed state has many more possible molecular arrangements than the separated state, so the entropy increases. This increase is irreversible because returning to the separated state would require a highly improbable spontaneous rearrangement.

The second law can thus be stated as: "Natural processes tend to move toward states of higher probability."

📚 Master Thermodynamics

Understanding entropy and the second law is fundamental to thermodynamics, statistical mechanics, and many areas of modern physics and engineering. Continue your journey into the fascinating world of energy conversion and the direction of natural processes.

Read More: Physics HRK Notes of Thermodynamics

© House of Physics | HRK Physics Chapter 26: Entropy and Second Law of Thermodynamics

Based on Halliday, Resnick, and Krane's "Physics" with additional insights from university physics curriculum

House of Physics | Contact: aliphy2008@gmail.com

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