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Thermal Physics
Carnot Engine and Reversible Thermodynamic Cycles
A Carnot engine is an idealized thermal engine operating on a reversible thermodynamic cycle consisting of two isothermal and two adiabatic processes. It establishes the theoretical upper limit for the conversion of heat into mechanical work between two thermal reservoirs.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of efficiency, dependence on source and sink temperatures, calculation of work done, and direct application of Carnot efficiency formula.
PGT extension
Coefficient of performance of refrigerators, Carnot theorem proofs, entropy changes in reversible and irreversible cycles, and T-S diagram analysis.
Prerequisites
Concept of Temperature
First and Second Laws of Thermodynamics
Adiabatic and Isothermal Processes
Cycle Structure and Processes
The Carnot cycle comprises four successive reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
During isothermal stages, the working substance is in thermal contact with a heat reservoir maintaining constant temperature T1 or T2.
During adiabatic stages, the system is thermally isolated, and temperature changes from T1	circ; to T2 without heat exchange.
Efficiency of a Carnot Engine
Efficiency (η) is defined as the ratio of net mechanical work done by the engine to the total heat absorbed from the high-temperature source.
It depends exclusively on the absolute temperatures of the source (T1) and the sink (T2), expressed as η = 1 - (T2 / T1).
Efficiency is always less than unity unless the sink temperature T2 is absolute zero, which is unattainable by the Third Law of Thermodynamics.
Carnot's Theorem
No heat engine operating between two given thermal reservoirs can be more efficient than a reversible Carnot engine operating between the same two reservoirs.
All reversible engines operating between the same two temperatures have the same efficiency, regardless of the working substance used.
The theorem provides the fundamental basis for defining the thermodynamic absolute temperature scale.
Carnot Refrigerator and Heat Pump
Running a Carnot cycle in reverse converts it into a refrigerator or heat pump, transferring heat from a low-temperature sink to a high-temperature source by consuming external work.
The coefficient of performance (COP) for a refrigerator is given by COPref = T2 / (T1 - T2).
For a heat pump, COPhp = T1 / (T1 - T2) = COPref + 1.
Derivations
Derivation of Carnot Engine Efficiency
Conditions: Ideal gas working substance operating in a complete cyclic process.
Step 1: Write heat absorbed during isothermal expansion at temperature T1 from volume V1 to V2: Q1 = nRT1 ln(V2/V1).
Step 2: Write heat rejected during isothermal compression at temperature T2 from volume V4 to V3: Q2 = nRT2 ln(V3/V4).
Step 3: Relate volumes during intermediate adiabatic processes: T1V2γ-1 = T2V3γ-1 and T1V1γ-1 = T2V4γ-1, yielding V2/V1 = V3/V4.
P = Power, W = Work done, t = Time period of cycle Watts (W)
Cyclic operation rate
Graphs and experiments
P-V Diagram of Carnot Cycle
Draw: A closed loop on a pressure-volume plane consisting of two hyperbolic isothermal curves (T1 and T2) connected by two steeper adiabatic curves.
Interpret: The area enclosed by the cycle represents the net mechanical work done by the engine during one complete cycle.
Exam use: Frequently tested in identifying cycle directions, calculating work via area integration, and distinguishing isothermal from adiabatic slopes.
T-S (Temperature-Entropy) Diagram of Carnot Cycle
Draw: A rectangle on the temperature-entropy plane bounded by temperatures T1 and T2 and entropy values SA and SB.
Interpret: The area of the rectangle represents the total heat absorbed (Q1), and the lower inner area represents heat rejected (Q2). The difference represents net work.
Exam use: Used in advanced PGT questions to evaluate entropy transformations and simplify graphical efficiency calculations.
Efficiency vs Sink Temperature Graph
Draw: A downward sloping curve (or straight line depending on variable representation) showing η decreasing as T2 increases for a fixed T1.
Interpret: Demonstrates that efficiency increases as the temperature difference (T1 - T2) increases.
Exam use: Tested in conceptual multiple-choice questions regarding thermal plant optimization and limits of efficiency.
Solved numericals
TGT core
A Carnot engine operates between a source at 500 K and a sink at 300 K. If it absorbs 1200 J of heat from the source, calculate the work done by the engine.
Step 1: Identify given parameters: T1 = 500 K, T2 = 300 K, Q1 = 1200 J.
Step 3: Calculate work done using $W = \eta * Q_{1} = 0.4 * 1200 J = 480 J$.
Answer: 480 J
PGT extension
A Carnot refrigerator absorbs heat from a freezer maintained at -23°C and rejects heat to the surrounding room at 27°C. If 500 J of work is consumed per cycle, find the heat rejected to the room.
By what percentage does the efficiency of a Carnot engine increase if the source temperature is raised from 500 K to 600 K, while the sink temperature remains constant at 300 K?
Assumption: Working parts are frictionless and thermal insulation is perfect, which is practically impossible to achieve in real engineering systems.
Limitation: The Carnot cycle is quasistatic and infinitely slow, rendering it useless for practical high-power output engines operating at finite speeds.
Common misconceptions
Common error: A Carnot engine can achieve 100% efficiency if the sink is kept at room temperature.
Correct: 100% efficiency is impossible unless the sink is at absolute zero (0 K), which violates the Third Law of Thermodynamics.
Efficiency depends on the ratio T2/T1; as long as T2 is above zero, some heat must be rejected.
Common error: The efficiency of a Carnot engine depends on the nature of the working fluid (e.g., whether it is a monoatomic or diatomic gas).
Correct: The efficiency depends solely on the temperatures of the source and sink.
Carnot's theorem proves that efficiency is independent of the working substance.
Quick revision
Carnot efficiency depends only on source and sink absolute temperatures: η = 1 - (T2 / T1).
Work Done in Carnot Cycle: $W = Q_1 - Q_2 = nR(T_1 - T_2)\ln\left(\frac{V_2}{V_1}\right)$
Must remember
The Carnot cycle consists of two isothermal and two adiabatic processes.
The efficiency of a Carnot engine depends exclusively on the absolute temperatures of the source ($T_1$) and sink ($T_2$): $\eta = 1 - \frac{T_2}{T_1}$.
Efficiency is always less than 1 (100%) unless $T_2 = 0\text{ K}$, which is unattainable due to the Third Law of Thermodynamics.
No engine can be more efficient than a reversible Carnot engine operating between the same two temperatures.
Coefficient of performance for a refrigerator is given by $\text{COP}_{\text{ref}} = \frac{T_2}{T_1 - T_2}$.
Coefficient of performance for a heat pump is $\text{COP}_{\text{hp}} = \frac{T_1}{T_1 - T_2} = \text{COP}_{\text{ref}} + 1$.
Net entropy change for a complete reversible Carnot cycle is zero ($\Delta S = 0$).
Efficiency of a Carnot engine is independent of the nature of the working substance.
Glossary
Carnot EngineAn idealized reversible heat engine operating between two thermal reservoirs that achieves the maximum possible efficiency.
Source ($T_1$)A high-temperature thermal reservoir that supplies heat $Q_1$ to the engine at a constant temperature.
Sink ($T_2$)A low-temperature thermal reservoir that absorbs rejected heat $Q_2$ from the engine at a constant temperature.
Coefficient of Performance (COP)The ratio of desired output (heat removed or delivered) to the work input for a refrigerator or heat pump.
Adiabatic ProcessA thermodynamic process in which no heat is exchanged with the surroundings ($Q = 0$).
Isothermal ProcessA thermodynamic process that occurs at a constant temperature ($\Delta T = 0$).
Carnot Engine vs Carnot Refrigerator
Parameter
Carnot Engine
Carnot Refrigerator
Primary Purpose
To convert heat into mechanical work
To transfer heat from low to high temperature using work
Direction of Cycle
Clockwise on $P-V$ diagram
Counter-clockwise on $P-V$ diagram
Performance Measure
Efficiency ($\eta = 1 - T_2/T_1$)
Coefficient of Performance ($\text{COP} = T_2 / (T_1 - T_2)$)
Energy Flow
Heat absorbed from source, work extracted, heat rejected to sink
Heat absorbed from sink, work supplied, heat rejected to source
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following processes constitute a complete Carnot cycle?
Explanation: The Carnot cycle comprises four successive reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
2. The efficiency ($\eta$) of a Carnot engine operating between temperatures $T_1$ and $T_2$ depends on:
Explanation: Efficiency depends exclusively on the absolute temperatures of the source ($T_1$) and the sink ($T_2$), expressed as $\eta = 1 - (T_2 / T_1)$.
3. What is the efficiency of a Carnot engine if the temperature of the sink is $0\text{ K}$?
Explanation: If $T_2 = 0\text{ K}$, then $\eta = 1 - (0 / T_1) = 1$, which corresponds to $100\%$ efficiency, though unattainable in practice due to the Third Law.
4. According to Carnot's Theorem, among all engines operating between the same two temperatures:
Explanation: Carnot's theorem states that no heat engine operating between two given thermal reservoirs can be more efficient than a reversible Carnot engine operating between the same two reservoirs.
5. The coefficient of performance ($\text{COP}_{\text{ref}}$) of a Carnot refrigerator working between temperatures $T_1$ and $T_2$ is given by:
Explanation: For a Carnot refrigerator, $\text{COP}_{\text{ref}} = \frac{T_2}{T_1 - T_2}$.
6. How is the coefficient of performance of a heat pump ($\text{COP}_{\text{hp}}$) related to that of a refrigerator ($\text{COP}_{\text{ref}}$)?
Explanation: For a heat pump, $\text{COP}_{\text{hp}} = \frac{T_1}{T_1 - T_2} = \text{COP}_{\text{ref}} + 1$.
7. During the adiabatic stages of a Carnot cycle, what happens to the temperature of the working substance?
Explanation: During adiabatic stages, the system is thermally isolated, and temperature changes from $T_1$ to $T_2$ without heat exchange.
8. What is the net entropy change of the universe for a complete reversible Carnot cycle?
Explanation: For a complete reversible cycle, $\Delta S = \frac{Q_1}{T_1} - \frac{Q_2}{T_2} = 0$, meaning no net entropy change for the universe.
9. If a Carnot engine absorbs $2000\text{ J}$ of heat from a source at $500\text{ K}$ and rejects heat to a sink at $300\text{ K}$, what is the work done by the engine?
10. Which law of thermodynamics states that absolute zero temperature cannot be attained in a finite number of steps, thus limiting Carnot efficiency to less than 100%?
Explanation: The Third Law of Thermodynamics establishes that absolute zero is unattainable, making $100\%$ efficiency impossible.
11. For an ideal gas working substance undergoing an adiabatic stroke in a Carnot cycle, which relation holds true between initial and final states of the stroke?
Explanation: The adiabatic relation for the Carnot stroke is $T_1 V_2^{\gamma-1} = T_2 V_3^{\gamma-1}$.
12. Can two different reversible Carnot engines operating between the same source and sink temperatures have different efficiencies?
Explanation: Carnot's theorem proves that all reversible engines operating between the same two temperatures have the same efficiency, regardless of the working substance used.
13. In a Carnot cycle, heat is absorbed during:
Explanation: Heat is absorbed from the high-temperature reservoir during the isothermal expansion process.
14. What is the unit of the Coefficient of Performance (COP) of a refrigerator?
Explanation: COP is a ratio of energies (heat removed to work input), making it a dimensionless quantity.
15. The efficiency of a Carnot engine is $50\%$ and the sink temperature is $300\text{ K}$. What is the source temperature?
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 $25\%$ when operating between a source and a sink at temperature $T_2$. If the temperature of the source is increased by $50\text{ K}$, the efficiency becomes $50\%$. The original temperatures of the source ($T_1$) and sink ($T_2$) are:
2. If the ratio of heat absorbed to heat rejected by a Carnot engine is $\frac{5}{3}$, and the sink temperature is $27^{\circ}\text{C}$ ($300\text{ K}$), what is the temperature of the source?
Explanation: For a Carnot engine, $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$. Given $\frac{Q_1}{Q_2} = \frac{5}{3}$ and $T_2 = 300\text{ K}$, $\frac{T_1}{300} = \frac{5}{3} \implies T_1 = \frac{5}{3} \times 300 = 500\text{ K}$.
3. A Carnot refrigerator extracts heat from a cold storage at $200\text{ K}$ and rejects it to the atmosphere at $300\text{ K}$. If $600\text{ J}$ of work is done on the refrigerator per cycle, the heat delivered to the atmosphere is:
4. Two Carnot engines $A$ and $B$ are operated in series. Engine $A$ receives heat at $T_1$ and rejects heat to an intermediate reservoir at temperature $T$. Engine $B$ receives heat at $T$ and rejects heat to a sink at $T_2$. For the efficiencies of both engines to be equal, the intermediate temperature $T$ must be equal to:
5. In a Carnot cycle, the working gas undergoes isothermal expansion at temperature $T_1$ from volume $V_1$ to $V_2$. The work done during this process is:
Explanation: Work done in an isothermal expansion for $n$ moles of an ideal gas at temperature $T_1$ from $V_1$ to $V_2$ is $W = n R T_1 \ln\left(\frac{V_2}{V_1}\right)$.
6. What is the power output of a Carnot engine that absorbs $5000\text{ J}$ of heat per cycle from a source at $500\text{ K}$, rejects heat to a sink at $300\text{ K}$, and completes one cycle in $2\text{ seconds}$?
Attempt these questions under a ten-minute limit using a separate timer.
1. The efficiency of a Carnot engine is $40\%$ when operating between source and sink temperatures $T_1$ and $T_2$. If both temperatures are increased by $100\text{ K}$ each, the new efficiency will be:
Explanation: Let $\eta = 1 - \frac{T_2}{T_1} = 0.4 \implies \frac{T_2}{T_1} = 0.6 \implies T_2 = 0.6 T_1$. With both increased by $100\text{ K}$, $\eta' = 1 - \frac{T_2 + 100}{T_1 + 100}$. Since $\frac{T_2 + 100}{T_1 + 100} > \frac{T_2}{T_1}$ (when $T_1 > T_2$), $\eta'$ will be less than $40\%$.
2. A reversible engine takes in heat from a reservoir at $527^{\circ}\text{C}$ and gives out heat to a sink at $127^{\circ}\text{C}$. The work done by the engine for $1000\text{ J}$ of absorbed heat is:
5. In a Carnot cycle, during which processes is there no change in entropy of the system?
Explanation: Adiabatic processes are isentropic ($\Delta S = 0$) because they are reversible and involve no heat transfer ($dQ = 0$).
6. A Carnot engine operates between $200\text{ K}$ and $400\text{ K}$. If it absorbs $100\text{ kJ}$ of heat from the source, how much heat is rejected to the sink?
7. Why is a practical realization of a Carnot engine impossible?
Explanation: Frictionless movement and perfect thermal insulation are idealizations that cannot be achieved in the real world, making Carnot engines theoretical constructs.
8. If the efficiency of a Carnot engine is $\eta$, the ratio of heat rejected to heat absorbed ($Q_2 / Q_1$) is equal to:
9. An ideal heat pump is used to heat a building. If the outside temperature is $-3^{\circ}\text{C}$ ($270\text{ K}$) and the indoor temperature is $27^{\circ}\text{C}$ ($300\text{ K}$), what is the coefficient of performance of the heat pump?
10. On a Temperature-Entropy ($T-S$) diagram, what geometric shape does a reversible Carnot cycle form?
Explanation: On a $T-S$ diagram, isothermal processes are horizontal lines (constant $T$) and adiabatic processes are vertical lines (constant $S$), forming a rectangle.