The Carnot Cycle: Principles, Derivations, and Solved Problems
Proposed by French engineer Sadi Carnot in 1824, the Carnot Cycle represents the theoretical limit of thermodynamic efficiency. It establishes the foundational boundary for how much useful work can be extracted from heat, setting the ultimate benchmark for all heat engines and refrigerators.
Table of Contents
1. Introduction to the Carnot Engine
A heat engine is a device that continuously converts thermal energy (heat) into mechanical energy (work) by operating in a cyclic process. However, the Second Law of Thermodynamics dictates that no heat engine can convert 100% of the absorbed heat into work; some heat must always be rejected to a colder reservoir.
Sadi Carnot proposed an idealized, theoretical engine that operates on a perfectly reversible cycle. The Carnot Engine consists of a working substance (ideally, an ideal gas) enclosed in a cylinder with a frictionless piston, operating between two thermal reservoirs:
- A Source (Hot Reservoir) at a constant high temperature, $T_H$.
- A Sink (Cold Reservoir) at a constant low temperature, $T_C$.
Because it is completely reversible and lacks dissipative losses (like friction or sudden heat transfers across finite temperature differences), the Carnot engine achieves the maximum possible theoretical efficiency of any engine operating between $T_H$ and $T_C$.
2. The Four Reversible Steps of the Carnot Cycle
The Carnot cycle consists of four distinct, perfectly reversible thermodynamic processes. Let us assume the working substance is $n$ moles of an ideal gas.
Step 1: Reversible Isothermal Expansion (State A to B)
The cylinder is placed in thermal contact with the hot reservoir at $T_H$. The gas is allowed to expand slowly from volume $V_A$ to $V_B$. Because the expansion is isothermal, the internal energy change is zero ($\Delta U = 0$).
- Heat Absorbed: The gas absorbs heat $Q_H$ from the source.
- Work Done: Work $W_1$ is done by the gas on the surroundings.
Step 2: Reversible Adiabatic Expansion (State B to C)
The cylinder is removed from the hot reservoir and perfectly insulated. The gas continues to expand from $V_B$ to $V_C$. Because it is adiabatic, no heat enters or leaves the system ($q = 0$).
- Temperature Drop: The expansion causes the gas to cool down from $T_H$ to the temperature of the sink, $T_C$.
- Work Done: Work $W_2$ is done by the gas at the expense of its internal energy.
Step 3: Reversible Isothermal Compression (State C to D)
The insulation is removed, and the cylinder is placed in contact with the cold reservoir (sink) at $T_C$. The gas is slowly compressed from $V_C$ to $V_D$. Again, $\Delta U = 0$ for this isothermal process.
- Heat Rejected: The gas rejects heat $Q_C$ to the sink.
- Work Done: Work $W_3$ is done on the gas by the surroundings.
Step 4: Reversible Adiabatic Compression (State D to A)
The cylinder is insulated once more. The gas is compressed from $V_D$ back to its initial volume $V_A$ without any heat exchange ($q = 0$).
- Temperature Rise: The compression causes the temperature of the gas to rise from $T_C$ back to $T_H$, completing the cycle.
- Work Done: Work $W_4$ is done on the gas, increasing its internal energy.
3. Visualizing the Cycle: The P-V Diagram
The indicator diagram (Pressure vs. Volume curve) beautifully illustrates the Carnot cycle. The area enclosed by the four curves $ABCD$ represents the net mechanical work done by the engine in one complete cycle.
P-V Diagram of Carnot Cycle
The area enclosed ABCD equals the net work done.
4. Mathematical Derivation of Net Work Done
Let us calculate the work done in each of the four thermodynamic steps using standard conventions (work done by the system is positive, work done on the system is negative).
1. Isothermal Expansion (A $\rightarrow$ B)
Since temperature $T_H$ is constant, the work done $W_1$ and heat absorbed $Q_H$ are equal (First Law: $\Delta U = Q - W$, and $\Delta U = 0$).
2. Adiabatic Expansion (B $\rightarrow$ C)
In an adiabatic process, $Q=0$. Therefore, work is done entirely at the expense of internal energy.
Where $C_v$ is the molar heat capacity at constant volume.
3. Isothermal Compression (C $\rightarrow$ D)
Work is done on the gas, and heat $Q_C$ is rejected to the sink at constant temperature $T_C$.
Since $V_4 < V_3$, $\ln(V_4/V_3)$ is negative, meaning $W_3$ is intrinsically negative (work done on the gas).
4. Adiabatic Compression (D $\rightarrow$ A)
Work is done on the gas, raising its temperature from $T_C$ back to $T_H$.
Total Net Work Done ($W_{net}$)
The total net work done by the engine in one cycle is the sum of the individual works:
Notice that the adiabatic work terms $W_2$ and $W_4$ perfectly cancel each other out! Therefore:
5. Deriving the Carnot Efficiency Formula
Efficiency ($\eta$) of any heat engine is defined as the ratio of useful net work output to the total heat input from the source.
From the First Law applied to the whole cycle ($\Delta U = 0$), we know $W_{net} = Q_H - |Q_C|$.
Connecting Volumes and Temperatures:
To express efficiency purely in terms of temperature, we use the temperature-volume relationship for adiabatic processes ($TV^{\gamma - 1} = \text{constant}$).
- For adiabatic expansion B $\rightarrow$ C: $\quad T_H V_2^{\gamma - 1} = T_C V_3^{\gamma - 1}$
- For adiabatic compression D $\rightarrow$ A: $\quad T_C V_4^{\gamma - 1} = T_H V_1^{\gamma - 1}$
Dividing the first equation by the second gives:
Now, substitute this ratio into our expressions for $Q_H$ and $Q_C$:
Plugging these back into the efficiency formula:
(Note: Temperatures must ALWAYS be in Kelvin)
6. Carnot's Theorem and The Second Law
The derivation of the Carnot cycle efficiency leads to two of the most profound statements in classical thermodynamics, collectively known as Carnot's Theorem:
- No engine working between two given heat reservoirs can be more efficient than a reversible Carnot engine working between those exact same two reservoirs.
- All reversible heat engines operating between the same two temperatures have exactly the same efficiency, regardless of the working substance used (gas, liquid, or solid).
Why can't efficiency be 100%? For $\eta$ to equal 1 (100%), the sink temperature $T_C$ must be $0 \text{ K}$ (Absolute Zero). Since Absolute Zero is physically unattainable (per the Third Law of Thermodynamics), a 100% efficient heat engine is a physical impossibility. This is a direct mathematical manifestation of the Kelvin-Planck statement of the Second Law of Thermodynamics.
7. Extensive Solved Problems (JEE / NEET Level)
Basic Efficiency and Work Calculation
Question:
A Carnot engine operates between a source at $227^\circ \text{C}$ and a sink at $27^\circ \text{C}$. If it absorbs $500 \text{ J}$ of heat from the source per cycle, calculate: (a) The efficiency of the engine, and (b) The work done per cycle.
Solution:
Step 1: Convert temperatures to Kelvin (CRITICAL STEP).
- $T_H = 227 + 273 = 500 \text{ K}$
- $T_C = 27 + 273 = 300 \text{ K}$
Step 2: Calculate Efficiency ($\eta$).
Efficiency = 40%
Step 3: Calculate Work Done ($W$).
We know that $\eta = \frac{W}{Q_H}$.
Work Done = 200 J
Increasing the Efficiency
Question:
The efficiency of a Carnot cycle is $1/6$. By lowering the temperature of the sink by $65 \text{ K}$, the efficiency increases to $1/3$. Find the initial and final temperatures of the source and the sink.
Solution:
Step 1: Set up the initial condition.
Step 2: Set up the second condition (sink temp lowered by 65 K).
Step 3: Solve the simultaneous equations.
Substitute $T_C$ from Eq 1 into the second equation:
Step 4: Find $T_C$.
Answer: Source $T_H = 390 \text{ K}$, Initial Sink $T_C = 325 \text{ K}$.
Two Carnot Engines in Series
Question:
Two Carnot engines A and B are operated in series. Engine A receives heat from a source at $T_1 = 800 \text{ K}$ and rejects heat to a sink at temperature $T_2$. Engine B receives the heat rejected by Engine A and in turn rejects heat to a sink at $T_3 = 200 \text{ K}$. Calculate the intermediate temperature $T_2$ if:
- The work outputs of the two engines are equal.
- The efficiencies of the two engines are equal.
Solution:
Let $Q_1$ be heat absorbed by A, $Q_2$ be heat rejected by A (and absorbed by B), and $Q_3$ be heat rejected by B.
Case (a): Work outputs are equal ($W_A = W_B$)
We know $W_A = Q_1 - Q_2$ and $W_B = Q_2 - Q_3$.
Given $W_A = W_B \implies Q_1 - Q_2 = Q_2 - Q_3 \implies Q_1 + Q_3 = 2Q_2$.
Divide the entire equation by $Q_2$:
From Carnot principles, heat ratios equal absolute temperature ratios ($\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$ and $\frac{Q_3}{Q_2} = \frac{T_3}{T_2}$). Substituting these:
(Notice that $T_2$ is the arithmetic mean of $T_1$ and $T_3$ when work is equal).
Case (b): Efficiencies are equal ($\eta_A = \eta_B$)
Efficiency of A: $\eta_A = 1 - \frac{T_2}{T_1}$
Efficiency of B: $\eta_B = 1 - \frac{T_3}{T_2}$
Equating them:
(Notice that $T_2$ is the geometric mean of $T_1$ and $T_3$ when efficiencies are equal).
Conclusion
The Carnot Cycle is a magnificent theoretical construct that bridges the gap between pure mathematics and mechanical engineering. While no real engine can perfectly execute a Carnot cycle due to inevitable friction, turbulence, and heat losses, it serves as the ultimate yardstick. Understanding the deep connections between the isothermal and adiabatic expansions—and how they dictate efficiency strictly through temperatures—is a critical milestone in mastering physical chemistry and classical thermodynamics.
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