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Thermal Physics
Carnot Cycle and Carnot Theorem
The Carnot cycle is an idealized reversible thermodynamic cycle composed of two isothermal and two adiabatic processes operating between two thermal reservoirs. It establishes the theoretical upper limit for the efficiency of any heat engine operating between the same two temperatures.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Understanding the four stages of the Carnot cycle on a P-V diagram, efficiency formula in terms of absolute temperatures (1 - T₂/T₁), and working of a Carnot heat engine and refrigerator.
PGT extension
Derivation of work done, efficiency of Carnot engine with an ideal gas working substance, Carnot's theorem, thermodynamic temperature scale, and coefficient of performance for refrigerators and heat pumps.
Prerequisites
Concept of Temperature and Zeroth Law of Thermodynamics
First and Second Laws of Thermodynamics
Equation of state for ideal gases (PV = nRT) and adiabatic processes (PV^γ = constant)
Stages of the Carnot Cycle
Stage 1: Isothermal expansion at source temperature T₁ where the working substance absorbs heat Q₁ from the reservoir at constant temperature.
Stage 2: Adiabatic expansion where the system expands without heat exchange, causing the temperature to drop from T₁ to sink temperature T₂.
Stage 3: Isothermal compression at sink temperature T₂ where heat Q₂ is rejected to the cold reservoir.
Stage 4: Adiabatic compression back to the initial state, raising the temperature from T₂ to T₁.
Efficiency of a Carnot Engine
Thermal efficiency η is defined as the ratio of net work done (W) to the heat absorbed from the hot reservoir (Q₁).
For a complete cycle, η = W / Q₁ = (Q₁ - Q₂) / Q₁ = 1 - (Q₂ / Q₁).
Since heat transfer in isothermal processes is proportional to absolute temperatures, Q₂ / Q₁ = T₂ / T₁, yielding η = 1 - (T₂ / T₁).
Carnot's Theorem
Statement 1: No heat engine operating between two given thermal reservoirs can be more efficient than a reversible Carnot engine operating between the same two reservoirs.
Statement 2: All reversible heat engines operating between the same two thermal reservoirs have the same efficiency, regardless of the working substance.
The efficiency depends solely on the temperatures of the source and the sink: η_ma$x = 1 - (T₂ / T₁)$.
Thermodynamic Temperature Scale
Kelvin temperature scale is independent of the physical properties of any particular working substance.
The ratio of heats absorbed and rejected by a Carnot engine is equal to the ratio of absolute temperatures of the reservoirs: Q₁ / Q₂ = T₁ / T₂.
Absolute zero (0 K) is the temperature at which the efficiency of a Carnot engine becomes 100%, which is unattainable in practice.
Derivations
Efficiency of Carnot Engine using an Ideal Gas
Conditions: One mole of an ideal gas as the working substance undergoing a quasi-static Carnot cycle.
Step 1: For Isothermal Expansion (1 to 2 at T₁), work done W₁₂ = RT₁ ln(V₂ / V₁). Heat absorbed Q₁ = W₁₂.
Step 2: For Adiabatic Expansion (2 to 3), T₁V₂^(γ-1) = T₂V₃^(γ-1). Work done W₂₃ = R(T₁ - T₂)/(γ - 1).
Step 3: For Isothermal Compression (3 to 4 at T₂), work done W₃₄ = RT₂ ln(V₄ / V₃) = -RT₂ ln(V₃ / V₄). Heat rejected Q₂ = -W₃₄.
Step 4: For Adiabatic Compression (4 to 1), T₂V₄^(γ-1) = T₁V₁^(γ-1). Using V₂/V₁ = V₃/V₄ from adiabatic relations, net work $W = Q₁ - Q₂ = RT₁ \ln(V₂/V₁) - RT₂ \ln(V₂/V₁)$.
R = Universal gas constant, V₁, V₂ = Initial and final volumes in isothermal expansion Joules (J)
1 mole of ideal gas
Coefficient of Performance (COP) of Refrigerator
$$\beta = \frac{T_2}{T_1 - T_2}$$
T₁ = Temperature of hot body, T₂ = Temperature of cold body Dimensionless
Ideal Carnot refrigerator
Relation between COP and Efficiency
$$\beta = \frac{1 - \eta}{\eta} = \frac{Q_2}{W}$$
η = Efficiency of corresponding Carnot engine, β = COP of refrigerator Dimensionless
Operating between same two temperature limits
Adiabatic Process Relation for Ideal Gas
$$TV^{\gamma-1} = \text{constant}$$
T = Absolute temperature, V = Volume, γ = Ratio of specific heats (C_p/C_v) K m^(3γ-3)
Reversible adiabatic process
Clausius Theorem Inequality
$$\oint \frac{dQ}{T} \le 0$$
dQ = Infinitesimal heat transfer, T = Temperature J K^{-1}
Any thermodynamic cycle (equality holds for reversible cycles)
Carnot Power Output
$$P = \frac{W}{t} = \frac{Q_1 - Q_2}{t}$$
P = Power, W = Work done per cycle, t = Time period of cycle Watts (W)
Cyclic operation
Graphs and experiments
P-V Diagram of Carnot Cycle
Draw: A closed loop on a P-V plane consisting of two isothermal curves (steeper or flatter depending on coordinates, connecting (P₁,V₁) to (P₂,V₂)) and two adiabatic curves (steeper than isotherms, connecting (P₂,V₂) to (P₃,V₃) and (P₄,V₄) to (P₁,V₁)).
Interpret: The area enclosed by the P-V loop represents the net mechanical work done by the Carnot engine in one complete cycle.
Exam use: Direct questions on identifying processes (isothermal vs. adiabatic) and calculating net work or efficiency from graphical coordinates.
T-S Diagram of Carnot Cycle
Draw: A rectangular loop on the Temperature-Entropy (T-S) plane, bounded by T = T₁, T = T₂, S = S₁, and S = S₂.
Interpret: The rectangular area on the T-S diagram represents the net heat absorbed by the engine, which equals the net work done per cycle ($Q_net = W = (T₁ - T₂)(S₂ - S₁))$.
Exam use: Used to test conceptual understanding of entropy changes during isothermal (constant T, varying S) and adiabatic (constant S, isentropic) processes.
Carnot Engine Efficiency vs. Source Temperature
Draw: A non-linear curve plotting efficiency η on the y-axis against source temperature T₁ on the x-axis, keeping sink temperature T₂ constant.
Interpret: Efficiency increases non-linearly with an increase in source temperature T₁, asymptotically approaching 1 as T₁ approaches infinity.
Exam use: Conceptual ranking of engine efficiencies and finding percentage changes in efficiency when source or sink temperatures are altered.
Solved numericals
TGT core
A Carnot engine operates between reservoirs at 500 K and 300 K. If it absorbs 1200 J of heat from the source in each cycle, calculate the work done by the engine per cycle.
Step 1: Identify given parameters: T₁ = 500 K, T₂ = 300 K, Q₁ = 1200 J.
Step 2: Calculate efficiency using η = 1 - (T₂ / T₁).
Step 3: η = 1 - (300 / 500) = 1 - 0.6 = 0.4.
Step 4: Use relation η = W / Q₁ to find work done: $W = η * Q₁ = 0.4 * 1200 J = 480 J$.
Answer: 480 J
PGT extension
A Carnot refrigerator absorbs heat from a freezer at -23 °C and rejects heat to the room at 27 °C. If 500 J of work is done per cycle, find the coefficient of performance and the heat rejected to the room.
By how much must the source temperature of a Carnot engine be increased to raise its efficiency from 40% to 50%, if the sink temperature remains constant at 300 K?
Step 5: Change in source temperature ΔT₁ = T₁₂ - T₁₁ = 600 K - 500 K = 100 K.
Answer: 100 K
Assumptions and limitations
Assumption: All processes in the Carnot cycle are strictly reversible and quasi-static, implying zero friction and infinitely slow operation.
Limitation: Reversible processes take infinite time, meaning a real Carnot engine would deliver zero practical power output.
Limitation: No real working fluid behaves as an ideal gas across extreme temperatures and pressures without intermolecular forces.
Common misconceptions
Common error: A Carnot engine can have 100% efficiency if the working substance is chosen correctly.
Correct: No heat engine can have 100% efficiency unless the sink temperature T₂ = 0 K, which is thermodynamically impossible by the Third Law of Thermodynamics.
Efficiency is strictly governed by the temperature ratio 1 - (T₂/T₁), and T₂ can never be absolute zero.
Common error: The efficiency of a Carnot engine depends on the nature of the working gas (like monatomic vs. diatomic).
Correct: The efficiency of a Carnot engine is independent of the working substance.
Carnot's theorem proves that efficiency depends solely on the temperatures of the source and the sink.
Quick revision
Carnot cycle consists of two isothermal and two adiabatic processes.
Efficiency depends only on absolute temperatures: η = 1 - (T₂ / T₁).
No engine can be more efficient than a reversible Carnot engine operating between the same two temperatures.
Area of P-V loop equals net work done, and area of T-S loop equals net heat absorbed.
Must remember
The Carnot cycle consists of four reversible processes: two isothermal and two adiabatic.
The thermal efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ and $T_2$ are absolute temperatures.
Carnot's theorem states that no heat engine can be more efficient than a reversible Carnot engine operating between the same two temperatures.
The efficiency of a Carnot engine is completely independent of the nature of the working substance.
The Coefficient of Performance (COP) of an ideal refrigerator is given by $\beta = \frac{T_2}{T_1 - T_2} = \frac{1 - \eta}{\eta}$.
Glossary
Source Temperature ($T_1$)The higher temperature of the thermal reservoir from which heat is absorbed by the working substance.
Sink Temperature ($T_2$)The lower temperature of the thermal reservoir to which unused heat is rejected.
Coefficient of Performance ($\beta$)The ratio of heat extracted from the cold reservoir to the work done on the refrigerator.
Reversible ProcessA thermodynamic process that can be reversed without leaving any change in the system or surroundings.
Isothermal vs Adiabatic Processes in Carnot Cycle
Feature
Isothermal Process
Adiabatic Process
Temperature
Constant ($T_1$ or $T_2$)
Varies ($T_1 arrow T_2$ or $T_2 arrow T_1$)
Heat Exchange
Heat absorbed ($Q_1$) or rejected ($Q_2$)
No heat exchange ($dQ = 0$)
Governing Law
$PV = \text{constant}$
$PV^\gamma = \text{constant}$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the sequence of processes in a standard Carnot cycle?
Explanation: A Carnot cycle consists of isothermal expansion, followed by adiabatic expansion, isothermal compression, and finally adiabatic compression.
2. The thermal efficiency of a Carnot engine depends upon:
Explanation: According to Carnot's theorem and formula $\eta = 1 - \frac{T_2}{T_1}$, efficiency depends solely on the source and sink temperatures.
3. If the temperature of the source is increased, the efficiency of a Carnot engine:
Explanation: Since $\eta = 1 - \frac{T_2}{T_1}$, increasing $T_1$ decreases the fraction $\frac{T_2}{T_1}$, thereby increasing the efficiency $\eta$.
4. A Carnot engine absorbs $2000\text{ J}$ of heat from a source at $500\text{ K}$ and rejects heat to a sink at $300\text{ K}$. The work done per cycle is:
Explanation: Using $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$, we get $Q_2 = Q_1 \left(\frac{T_2}{T_1}\right) = 2000 \times \frac{300}{500} = 1200\text{ J}$. Work done $W = Q_1 - Q_2 = 2000 - 1200 = 800\text{ J}$.
5. Can a Carnot engine have 100% efficiency?
Explanation: Efficiency $\eta = 1 - \frac{T_2}{T_1}$ is 1 only when $T_2 = 0\text{ K}$, which is unattainable in practice according to the Third Law of Thermodynamics.
6. What is the Coefficient of Performance ($\beta$) of a Carnot refrigerator operating between $250\text{ K}$ and $300\text{ K}$?
7. Which of the following relations is correct for an adiabatic process involving an ideal gas?
Explanation: For a reversible adiabatic process of an ideal gas, $TV^{\gamma-1} = \text{constant}$ holds true.
8. According to Clausius Theorem, for any thermodynamic cycle, the cyclic integral of $\frac{dQ}{T}$ satisfies:
Explanation: Clausius inequality states that $\oint \frac{dQ}{T} \le 0$, where equality holds for reversible cycles.
9. Two Carnot engines $A$ and $B$ are operated in series. Engine $A$ absorbs heat at $T_1$ and rejects heat to an intermediate reservoir at $T$. Engine $B$ absorbs heat at $T$ and rejects to $T_2$. For equivalent work output or efficiency, what is $T$ in terms of $T_1$ and $T_2$ for maximum overall efficiency?
Explanation: For two engines in series to have equal efficiency, the intermediate temperature is the geometric mean of the two temperatures: $T = \sqrt{T_1 T_2}$.
10. What is the efficiency of a Carnot engine operating between $400\text{ K}$ and $300\text{ K}$?
11. The working substance in a Carnot engine is typically considered to be:
Explanation: In standard theoretical derivations, an ideal gas is used as the working substance in a Carnot engine.
12. The ratio of heat absorbed ($Q_1$) to heat rejected ($Q_2$) in a Carnot cycle is equal to:
Explanation: For isothermal heat transfers in a Carnot cycle, $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.
13. If the Coefficient of Performance of a refrigerator is 4, and it consumes $100\text{ W}$ of power, the heat extracted per second from the cold reservoir is:
14. Which cycle represents the maximum possible efficiency for a heat engine operating between two given temperatures?
Explanation: Carnot's theorem proves that no engine can be more efficient than a reversible Carnot cycle operating between the same temperature limits.
15. In a P-V diagram, what represents the net work done during one complete Carnot cycle?
Explanation: The net work done per cycle in any closed thermodynamic cycle on a P-V diagram is equal to the area enclosed by the cycle path.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A Carnot engine has an efficiency of $50%$ when its sink is at $300\text{ K}$. To increase the efficiency to $60%$, keeping the source temperature constant, the source temperature must be increased by:
2. The efficiency of a Carnot engine working between temperatures $T_1$ and $T_2$ is $1/4$. When the sink temperature $T_2$ is reduced by $65\text{ K}$, the efficiency becomes $1/3$. The temperatures $T_1$ and $T_2$ are respectively:
3. A Carnot engine whose sink is at $300\text{ K}$ has an efficiency of $40%$. By how much should the temperature of the source be increased to raise its efficiency to $50%$?
4. The work done by $1\text{ mole}$ of an ideal gas during isothermal expansion from volume $V_1$ to $V_2$ at temperature $T_1$ is given by:
Explanation: For an isothermal expansion of 1 mole of an ideal gas, work done $W = \int_{V_1}^{V_2} P dV = \int_{V_1}^{V_2} \frac{RT_1}{V} dV = RT_1 \ln\left(\frac{V_2}{V_1}\right)$.
5. If the Coefficient of Performance of a Carnot refrigerator is $5$, then the ratio of the temperature of the source to that of the sink ($T_1 / T_2$) is:
6. Which of the following statements is true regarding Carnot's theorem?
Explanation: Carnot's theorem explicitly establishes that all reversible heat engines operating between the same two thermal reservoirs possess identical efficiency, independent of the working substance.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. During which stages of the Carnot cycle is heat exchanged with the external reservoirs?
Explanation: Heat is absorbed during isothermal expansion ($Q_1$) and rejected during isothermal compression ($Q_2$). Adiabatic stages involve zero heat exchange.
2. A reversible engine takes in $3000\text{ cal}$ of heat from a source at $400\text{ K}$ and gives $2400\text{ cal}$ to the sink. The temperature of the sink is:
3. The efficiency of a Carnot engine is $50%$ when the sink temperature is $500\text{ K}$. If the source temperature is kept constant and efficiency is to be $60%$, the new sink temperature should be:
5. Which of the following P-V curve slopes is steeper for an ideal gas?
Explanation: The slope of an adiabatic curve ($-\gamma P / V$) is $\gamma$ times steeper than the slope of an isothermal curve ($-P / V$). Since $\gamma > 1$, adiabatic is steeper.
6. Two identical Carnot engines are connected in series. The first engine absorbs heat at $T_1$ and rejects to $T$, while the second absorbs at $T$ and rejects to $T_2$. If both engines develop the same work output, the intermediate temperature $T$ is given by:
Explanation: For equal work output in series, $W_1 = W_2 \implies Q_1 - Q = Q - Q_2 \implies \frac{T_1 + T_2}{2} = T$.
7. The Clausius statement of the Second Law of Thermodynamics is related to:
Explanation: Clausius statement asserts that heat cannot spontaneously flow from a body at a lower temperature to a body at a higher temperature without external work.
8. What happens to the entropy of the universe during a reversible Carnot cycle?
Explanation: For any reversible cycle, total entropy change of the system plus surroundings is zero ($\Delta S_{\text{univ}} = 0$).
9. A Carnot engine has power output $P$, absorbing $Q_1$ per cycle with a time period $t$. The heat rejected per cycle $Q_2$ is equal to: