Welcome to Lecture 97 of our comprehensive JEE Physics Master Series. In this lecture, we explore the foundational bedrock of thermal physics: the Second Law of Thermodynamics. While the First Law ensures energy conservation across all thermodynamic transformations, it fails to explain why spontaneous processes possess a distinct directional arrow—why heat never spontaneously flows from cold to hot, or why mechanical work can be converted completely into heat via friction while the reverse conversion is strictly forbidden. We systematically analyze the Kelvin-Planck and Clausius formulations, study the operating principles of Heat Engines and Refrigerators/Heat Pumps, rigorously prove the logical equivalence of both statements, and master multi-reservoir systems and JEE Advanced problem archetypes.
Detailed Concept Card: The Second Law of Thermodynamics
1. Inadequacy of the First Law & The Need for a Directional Principle
The First Law of Thermodynamics ($Q = \Delta U + W$) is a purely quantitative statement of energy conservation. It asserts that in any closed cycle, the net heat absorbed equals the net work delivered ($\oint \delta Q = \oint \delta W$). However, the First Law imposes no restriction whatsoever on the direction in which energy transformation can spontaneously proceed:
- Spontaneous Heat Conduction: When a hot body is placed in contact with a cold body, heat spontaneously flows from the hot body to the cold body until thermal equilibrium is established. A spontaneous process where heat flows from the cold body to the hot body would conserve total energy and satisfy the First Law, yet it never occurs in nature.
- Complete Work-to-Heat vs. Heat-to-Work Asymmetry: Work can be completely and continuously converted into heat (e.g., stirring a liquid with a paddle wheel, or braking a moving vehicle). However, no device has ever been constructed that can absorb heat from a single thermal reservoir and convert it completely into continuous mechanical work with zero thermal dissipation.
- Gas Free Expansion: When an ideal gas expands into a vacuum through a partition, it does not do work and its internal energy remains unchanged ($Q = 0, W = 0 \implies \Delta U = 0$). Yet the gas never spontaneously compresses itself back into the initial half of the vessel.
The Second Law of Thermodynamics provides the indispensable criterion governing the directionality, spontaneity, and fundamental limits of physical and engineering processes.
2. Thermal Energy Reservoirs
To analyze cyclic devices, we define a Thermal Reservoir (or heat reservoir) as a hypothetical thermodynamic system of infinite thermal capacity ($C \to \infty$) whose temperature remains strictly constant regardless of how much heat is added or extracted from it:
- Heat Source (High-Temperature Reservoir, $T_H$): A reservoir that supplies heat to a cyclic operating system without experiencing any temperature drop.
- Heat Sink (Low-Temperature Reservoir, $T_C$): A reservoir that absorbs rejected heat from a cyclic operating system without experiencing any temperature rise ($T_H > T_C$).
3. Heat Engines & The Kelvin-Planck Statement
A Heat Engine is a cyclically operating thermodynamic device that converts thermal energy absorbed from a high-temperature reservoir into net mechanical work delivered to the surroundings while rejecting residual heat to a low-temperature sink.
Because the working substance operates in a complete thermodynamic cycle, its state returns to the initial state, meaning the net change in internal energy over a complete cycle is zero ($\Delta U_{\text{cycle}} = 0$). By the First Law:
$$W_{\text{net}} = Q_H – Q_C$$
where $Q_H$ is the heat absorbed from the hot source at $T_H$, $Q_C$ is the heat rejected to the cold sink at $T_C$, and $W_{\text{net}}$ is the net work performed by the engine. The thermal efficiency ($\eta$) of the heat engine is defined as the fraction of heat absorbed that is converted into useful work:
$$\mathbf{\eta = \frac{\text{Net Work Output}}{\text{Heat Absorbed from Source}} = \frac{W_{\text{net}}}{Q_H} = \frac{Q_H – Q_C}{Q_H} = 1 – \frac{Q_C}{Q_H}}$$
The Kelvin-Planck Statement of the Second Law:
“It is impossible for any device operating in a thermodynamic cycle to absorb heat from a single thermal reservoir and convert it entirely into a net amount of mechanical work delivered to the surroundings.”
Critical Deductions from the Kelvin-Planck Statement:
- No cyclic heat engine can have a thermal efficiency of $100\%$ ($\eta < 1$).
- A heat engine must necessarily reject heat to a lower-temperature reservoir ($Q_C > 0$). Heat rejection is not an engineering flaw or frictional defect; it is a fundamental law of nature.
- A hypothetical machine that operates in a cycle and converts all absorbed heat completely into work without rejecting any heat ($Q_H = W, Q_C = 0, \eta = 1$) is termed a Perpetual Motion Machine of the Second Kind (PMM-2). A PMM-2 satisfies the First Law (energy is conserved) but strictly violates the Second Law.
4. Refrigerators, Heat Pumps & The Clausius Statement
A Refrigerator or Heat Pump is a cyclically operating device that extracts heat from a low-temperature cold space ($T_C$), receives an external mechanical work input ($W_{\text{ext}}$), and discharges heat to a high-temperature sink ($T_H$). Over a complete cycle ($\Delta U = 0$):
$$Q_H = Q_C + W_{\text{ext}}$$
The performance of such devices is quantified by the Coefficient of Performance ($\text{COP}$), defined as the ratio of the desired thermal effect to the required work input:
| Device | Primary Objective (Desired Effect) | Coefficient of Performance Formula |
|---|---|---|
| Refrigerator ($\beta$ or $\text{COP}_{\text{ref}}$) | Cooling / Heat extraction from cold body ($Q_C$) | $$\beta = \frac{Q_C}{W_{\text{ext}}} = \frac{Q_C}{Q_H – Q_C}$$ |
| Heat Pump ($\gamma$ or $\text{COP}_{\text{HP}}$) | Heating / Heat rejected into warm room ($Q_H$) | $$\gamma = \frac{Q_H}{W_{\text{ext}}} = \frac{Q_H}{Q_H – Q_C}$$ |
Universal Relationship between Heat Pump and Refrigerator COP:
$$\mathbf{\text{COP}_{\text{HP}} = \frac{Q_H}{W_{\text{ext}}} = \frac{Q_C + W_{\text{ext}}}{W_{\text{ext}}} = \frac{Q_C}{W_{\text{ext}}} + 1 = \text{COP}_{\text{ref}} + 1}$$
Because $\text{COP}_{\text{ref}} > 0$, the COP of a heat pump is always strictly greater than 1 ($\text{COP}_{\text{HP}} > 1$).
The Clausius Statement of the Second Law:
“It is impossible to construct a device operating in a thermodynamic cycle whose sole effect is the transfer of heat from a cooler body to a hotter body without any external work input.”
Critical Deductions from the Clausius Statement:
- Heat cannot spontaneously flow uphill against a thermal gradient from a cold body to a hot body.
- To transfer heat from a cold region to a warm region, external mechanical or electrical work must be supplied ($W_{\text{ext}} > 0$).
- The coefficient of performance of a refrigerator can never be infinite ($\beta < \infty$); a refrigerator with $W_{\text{ext}} = 0$ is physically impossible.
5. Rigorous Proof of Equivalence of Kelvin-Planck and Clausius Statements
Although the Kelvin-Planck statement addresses heat engines and the Clausius statement addresses refrigerators, they are not two independent laws. They are completely equivalent logical statements: a violation of one statement inevitably leads to a violation of the other.
Part A: Violation of Kelvin-Planck Implies Violation of Clausius
1. Suppose a hypothetical engine $E_{\text{hyp}}$ violates the Kelvin-Planck statement: it absorbs heat $Q_H$ from a single reservoir at $T_H$ and converts it entirely into mechanical work $W = Q_H$ without rejecting any heat ($Q_C = 0$).
2. Now, let this work output $W$ be used to drive an ordinary, standard refrigerator $R$ operating between a cold reservoir at $T_C$ and the hot reservoir at $T_H$.
3. Refrigerator $R$ absorbs heat $Q_C$ from the cold reservoir at $T_C$, consumes work input $W = Q_H$, and discharges heat $Q_H’ = Q_C + W = Q_C + Q_H$ into the hot reservoir at $T_H$.
4. Consider the composite system $(E_{\text{hyp}} + R)$ acting as a single cyclic device:
– Net work exchanged with surroundings: $W_{\text{net}} = W – W = 0$.
– Net heat extracted from cold reservoir: $Q_C$.
– Net heat delivered to hot reservoir: $Q_H’ – Q_H = (Q_C + Q_H) – Q_H = Q_C$.
5. The combined device operates in a complete cycle and produces no effect other than transferring heat $Q_C$ from a cooler body ($T_C$) to a hotter body ($T_H$) with zero work input. This directly violates the Clausius statement!
Part B: Violation of Clausius Implies Violation of Kelvin-Planck
1. Suppose a hypothetical refrigerator $R_{\text{hyp}}$ violates the Clausius statement: it transfers heat $Q_C$ from a cold reservoir at $T_C$ to a hot reservoir at $T_H$ without requiring any external work ($W_{\text{ext}} = 0$).
2. Now, couple this with a standard heat engine $E$ operating between $T_H$ and $T_C$. Let engine $E$ absorb heat $Q_H$ from the hot reservoir ($Q_H > Q_C$), deliver net mechanical work $W = Q_H – Q_C$, and reject heat $Q_C$ to the cold reservoir at $T_C$.
3. Consider the composite system $(R_{\text{hyp}} + E)$ acting as a single cyclic device:
– Net heat extracted from cold reservoir: $Q_C – Q_C = 0$. The cold reservoir experiences zero net thermal exchange!
– Net heat extracted from hot reservoir: $Q_H – Q_C$.
– Net mechanical work delivered to surroundings: $W = Q_H – Q_C$.
4. The combined device operates in a complete cycle, extracts net heat $(Q_H – Q_C)$ from a single thermal reservoir at $T_H$, and converts it entirely into mechanical work $W = Q_H – Q_C$ without rejecting any heat to any other reservoir. This directly violates the Kelvin-Planck statement!
Conclusion: The Kelvin-Planck and Clausius statements are dual expressions of the identical fundamental physical reality.
Four Graded Solved Examples
Example 1 (JEE Main Level: Coupled Heat Engine and Refrigerator System)
Problem Statement: A cyclic heat engine absorbs $Q_H = 1200\text{ J}$ of heat from a high-temperature source and rejects $Q_C = 750\text{ J}$ of heat to a low-temperature sink per cycle. The entire mechanical work output of this engine is used to drive a refrigerator operating with a coefficient of performance $\beta = 4.0$. Determine:
- The work output $W$ and the thermal efficiency $\eta$ of the heat engine.
- The heat $Q_C’$ extracted from the cold refrigerated space per cycle.
- The total heat $Q_H’$ dumped into the surroundings by the refrigerator per cycle.
Solution:
Step 1: Work Output and Efficiency of the Heat Engine.
Over a complete cycle, the change in internal energy is zero ($\Delta U = 0$). By First Law:
$$W = Q_H – Q_C = 1200\text{ J} – 750\text{ J} = \mathbf{450\text{ J}}$$
The thermal efficiency is:
$$\eta = \frac{W}{Q_H} = \frac{450\text{ J}}{1200\text{ J}} = 0.375 = \mathbf{37.5\%}$$
Step 2: Heat Extracted by the Refrigerator ($Q_C’$).
The refrigerator receives work input $W_{\text{in}} = W = 450\text{ J}$. By definition of $\text{COP}_{\text{ref}}$:
$$\beta = \frac{Q_C’}{W_{\text{in}}} \implies Q_C’ = \beta \cdot W_{\text{in}} = 4.0 \times 450\text{ J} = \mathbf{1800\text{ J}}$$
Step 3: Total Heat Dumped into Surroundings ($Q_H’$).
Applying the First Law to the refrigerator cycle:
$$Q_H’ = Q_C’ + W_{\text{in}} = 1800\text{ J} + 450\text{ J} = \mathbf{2250\text{ J}}$$
Takeaway: Even though the engine efficiency is only $37.5\%$, the refrigerator can extract significantly more heat ($1800\text{ J}$) from the cold compartment than the work supplied ($450\text{ J}$) because the COP is greater than 1.
Example 2 (Conceptual & Mathematical: Perpetual Motion Machines & Statements of Second Law)
Problem Statement: Three separate inventors submit patent proposals for thermal machines:
– Inventor A proposes a cyclic device that absorbs $1000\text{ J}$ of heat from ocean water at $300\text{ K}$, converts $1000\text{ J}$ into mechanical work to propel a ship, and discharges $0\text{ J}$ of heat.
– Inventor B proposes a cyclic device that requires no external electrical or mechanical power, yet continuously transfers $500\text{ J}$ of heat per minute from inside a domestic deep freezer at $-10^\circ\text{C}$ to room air at $+25^\circ\text{C}$.
– Inventor C proposes a heat pump that maintains a room at $20^\circ\text{C}$ by absorbing $4000\text{ J}$ of heat from cold outdoor air at $0^\circ\text{C}$ while consuming $1000\text{ J}$ of electric work.
Classify each proposal: state whether it is physically admissible, and if not, specify the exact law of thermodynamics it violates.
Solution:
Analysis of Proposal A:
– Total energy absorbed: $Q = 1000\text{ J}$. Work delivered: $W = 1000\text{ J}$. Net change: $\Delta U = Q – W = 0$.
– First Law: Satisfied (energy is conserved).
– Second Law: The engine absorbs heat from a single thermal reservoir (ocean water at $300\text{ K}$) and converts it completely into mechanical work with $Q_C = 0$ ($\eta = 100\%$).
– Conclusion: Impossible. Proposal A violates the Kelvin-Planck statement of the Second Law and is a Perpetual Motion Machine of the Second Kind (PMM-2).
Analysis of Proposal B:
– The device transfers heat from a colder body ($-10^\circ\text{C}$) to a warmer body ($+25^\circ\text{C}$) with zero work input ($W = 0$).
– First Law: Satisfied (heat in = heat out = $500\text{ J/min}$).
– Second Law: Violates the Clausius statement, which establishes that heat cannot spontaneously flow from cold to hot without external work.
– Conclusion: Impossible. Violates the Second Law of Thermodynamics.
Analysis of Proposal C:
– The heat pump absorbs $Q_C = 4000\text{ J}$ and consumes $W = 1000\text{ J}$. Total heat delivered to the room is $Q_H = Q_C + W = 4000 + 1000 = 5000\text{ J}$.
– $\text{COP}_{\text{HP}} = \frac{Q_H}{W} = \frac{5000}{1000} = 5.0$.
– Maximum possible Carnot COP between $T_C = 273.15\text{ K}$ and $T_H = 293.15\text{ K}$:
$$\text{COP}_{\text{HP, max}} = \frac{T_H}{T_H – T_C} = \frac{293.15}{20} \approx 14.66$$
– Since the proposed COP ($5.0$) is strictly less than the theoretical reversible maximum ($14.66$), the proposal satisfies both the First and Second Laws.
– Conclusion: Physically admissible and viable.
Takeaway: A machine can satisfy energy conservation (First Law) completely, yet be physically impossible if it violates the Second Law (PMM-2).
Example 3 (JEE Advanced Level: Two Reversible Heat Engines Operating in Series)
Problem Statement: Two reversible heat engines, $A$ and $B$, operate in series between three thermal reservoirs at absolute temperatures $T_1$, $T_2$, and $T_3$ (where $T_1 > T_2 > T_3$). Engine $A$ absorbs heat $Q_1$ from the reservoir at $T_1$, delivers work $W_A$, and rejects heat $Q_2$ to the intermediate reservoir at $T_2$. Engine $B$ absorbs the entire heat $Q_2$ rejected by Engine $A$, delivers work $W_B$, and rejects heat $Q_3$ to the reservoir at $T_3$. Given $T_1 = 900\text{ K}$ and $T_3 = 400\text{ K}$, determine the intermediate temperature $T_2$ in the following two independent cases:
- The two engines produce equal mechanical work outputs ($W_A = W_B$).
- The two engines operate with identical thermal efficiencies ($\eta_A = \eta_B$).
Solution:
For reversible engines operating between two thermal reservoirs: $$\frac{Q_2}{Q_1} = \frac{T_2}{T_1} \quad \text{and} \quad \frac{Q_3}{Q_2} = \frac{T_3}{T_2}$$ The work outputs of the two engines are: $$W_A = Q_1 – Q_2 = Q_1 \left(1 – \frac{T_2}{T_1}\right) = Q_1 \left(\frac{T_1 – T_2}{T_1}\right)$$ $$W_B = Q_2 – Q_3 = Q_2 \left(1 – \frac{T_3}{T_2}\right) = \left[Q_1 \frac{T_2}{T_1}\right] \left(\frac{T_2 – T_3}{T_2}\right) = Q_1 \left(\frac{T_2 – T_3}{T_1}\right)$$
Case 1: Equal Work Outputs ($W_A = W_B$).
Equating $W_A$ and $W_B$:
$$Q_1 \left(\frac{T_1 – T_2}{T_1}\right) = Q_1 \left(\frac{T_2 – T_3}{T_1}\right) \implies T_1 – T_2 = T_2 – T_3$$
$$2 T_2 = T_1 + T_3 \implies \mathbf{T_2 = \frac{T_1 + T_3}{2}}$$
$T_2$ is the arithmetic mean of $T_1$ and $T_3$. Substituting $T_1 = 900\text{ K}$ and $T_3 = 400\text{ K}$:
$$T_2 = \frac{900 + 400}{2} = \mathbf{650\text{ K}}$$
Case 2: Equal Thermal Efficiencies ($\eta_A = \eta_B$).
The efficiencies are:
$$\eta_A = 1 – \frac{T_2}{T_1}, \quad \eta_B = 1 – \frac{T_3}{T_2}$$
Equating $\eta_A = \eta_B$:
$$1 – \frac{T_2}{T_1} = 1 – \frac{T_3}{T_2} \implies \frac{T_2}{T_1} = \frac{T_3}{T_2} \implies T_2^2 = T_1 T_3 \implies \mathbf{T_2 = \sqrt{T_1 T_3}}$$
$T_2$ is the geometric mean of $T_1$ and $T_3$. Substituting $T_1 = 900\text{ K}$ and $T_3 = 400\text{ K}$:
$$T_2 = \sqrt{900 \times 400} = \sqrt{360000} = \mathbf{600\text{ K}}$$
Takeaway: In series heat engines: – Equal work output $\implies T_2$ is the arithmetic mean ($\text{AM}$). – Equal efficiency $\implies T_2$ is the geometric mean ($\text{GM}$). Since $\text{AM} > \text{GM}$ for distinct positive values, $T_{2,\text{equal work}} (650\text{ K}) > T_{2,\text{equal }\eta} (600\text{ K})$.
Example 4 (JEE Advanced Level: Quantitative Proof of Clausius Inequality & Spontaneous Entropy Production)
Problem Statement: A finite metal block of heat capacity $C = 500\text{ J/K}$ initially at temperature $T_1 = 600\text{ K}$ is brought into direct thermal contact with a large heat reservoir maintained at constant temperature $T_0 = 300\text{ K}$ inside an insulated chamber until the block cools down to $300\text{ K}$. Show by direct calculation that:
1. Total energy is conserved (First Law).
2. The heat transfer process cannot spontaneously reverse itself without external work, by computing the entropy changes of the block ($\Delta S_{\text{block}}$), the reservoir ($\Delta S_{\text{res}}$), and the universe ($\Delta S_{\text{univ}}$). Take $\ln 2 = 0.693$.
Solution:
Step 1: First Law Energy Balance.
Heat lost by the block during cooling from $600\text{ K}$ to $300\text{ K}$:
$$Q_{\text{lost}} = C (T_1 – T_0) = 500\text{ J/K} \times (600 – 300)\text{ K} = 150000\text{ J} = 150\text{ kJ}$$
Heat absorbed by the reservoir at $T_0 = 300\text{ K}$:
$$Q_{\text{res}} = +150\text{ kJ}$$
Since heat lost by the block equals heat gained by the reservoir, $\Delta E_{\text{universe}} = 0$. The First Law is satisfied.
Step 2: Entropy Change of the Metal Block ($\Delta S_{\text{block}}$).
Because entropy is a state function, we evaluate $\Delta S_{\text{block}}$ along a reversible path between the same initial and final states:
$$\Delta S_{\text{block}} = \int_{T_1}^{T_0} \frac{C \, dT}{T} = C \ln\left(\frac{T_0}{T_1}\right) = 500 \ln\left(\frac{300}{600}\right) = -500 \ln 2$$
$$\Delta S_{\text{block}} = -500 \times 0.693 = \mathbf{-346.5\text{ J/K}}$$
Step 3: Entropy Change of the Reservoir ($\Delta S_{\text{res}}$).
The reservoir remains at constant temperature $T_0 = 300\text{ K}$ while absorbing $Q_{\text{res}} = 150000\text{ J}$:
$$\Delta S_{\text{res}} = +\frac{Q_{\text{res}}}{T_0} = \frac{150000\text{ J}}{300\text{ K}} = \mathbf{+500\text{ J/K}}$$
Step 4: Total Entropy Generation of the Universe ($\Delta S_{\text{univ}}$).
$$\Delta S_{\text{univ}} = \Delta S_{\text{block}} + \Delta S_{\text{res}} = -346.5\text{ J/K} + 500\text{ J/K} = \mathbf{+153.5\text{ J/K} > 0}$$
Takeaway: Because $\Delta S_{\text{univ}} > 0$, the process is intrinsically irreversible. For the process to spontaneously reverse (the block heating from $300\text{ K}$ to $600\text{ K}$ while the reservoir loses heat), $\Delta S_{\text{univ}}$ would equal $-153.5\text{ J/K} < 0$, which strictly violates the Second Law of Thermodynamics.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. Which of the following fundamental physical statements best highlights the primary difference between the First and Second Laws of Thermodynamics?
(A) The First Law deals with ideal gases, whereas the Second Law deals only with real gases.
(B) The First Law establishes the conservation of energy, whereas the Second Law dictates the direction and feasibility of spontaneous energy transfers.
(C) The First Law forbids the creation of energy, whereas the Second Law forbids the destruction of energy.
(D) The First Law applies only to reversible processes, whereas the Second Law applies only to irreversible processes.
Q2. A hypothetical cyclic heat engine operates by absorbing heat from a single thermal reservoir at temperature $T$ and converting $100\%$ of this absorbed heat into net mechanical work delivered to the surroundings without rejecting any heat to any other body. Such a machine is known as:
(A) A Perpetual Motion Machine of the First Kind (PMM-1), violating the First Law.
(B) A Perpetual Motion Machine of the Second Kind (PMM-2), violating the Second Law.
(C) A Carnot Engine operating at absolute zero.
(D) An ideal heat pump operating with zero COP.
Q3. A domestic refrigerator extracts heat $Q_C$ from its interior freezing compartment while consuming external electrical work $W$, and discharges heat $Q_H$ into the surrounding kitchen air. If $\text{COP}_{\text{ref}}$ is the coefficient of performance of this refrigerator and $\text{COP}_{\text{HP}}$ is the coefficient of performance when operated as a heat pump, the universal relation between them is:
(A) $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} – 1$
(B) $\text{COP}_{\text{HP}} \times \text{COP}_{\text{ref}} = 1$
(C) $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} + 1$
(D) $\text{COP}_{\text{HP}} = \frac{1}{\text{COP}_{\text{ref}} + 1}$
Q4. A heat engine operating between two thermal reservoirs has a thermal efficiency of $\eta = 25\%$. During each cycle, the engine delivers $W = 500\text{ J}$ of mechanical work. The amount of heat rejected to the low-temperature sink per cycle is:
(A) $2000\text{ J}$
(B) $1500\text{ J}$
(C) $125\text{ J}$
(D) $375\text{ J}$
Q5. According to the Clausius statement of the Second Law of Thermodynamics, it is impossible for any cyclic device to:
(A) Convert mechanical work entirely into thermal energy.
(B) Absorb heat from a hot body and reject heat to a cold body.
(C) Transfer heat from a colder body to a hotter body without any external work input.
(D) Operate with an efficiency strictly less than $100\%$.
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding the Kelvin-Planck and Clausius statements of the Second Law of Thermodynamics is/are correct?
(A) The Kelvin-Planck statement is concerned with the conversion of heat into work, whereas the Clausius statement is concerned with the direction of heat transfer.
(B) A violation of the Kelvin-Planck statement directly implies a violation of the Clausius statement.
(C) A violation of the Clausius statement directly implies a violation of the Kelvin-Planck statement.
(D) The Kelvin-Planck and Clausius statements are mutually exclusive and apply to completely unrelated thermodynamic systems.
Q7. Which of the following consequences logically follow from the Second Law of Thermodynamics?
(A) No cyclic heat engine can have a thermal efficiency of $\eta = 1$.
(B) A cyclic heat engine requires at least two thermal reservoirs at different temperatures to produce continuous net work.
(C) The coefficient of performance of any real or reversible refrigerator must be strictly finite ($\text{COP} < \infty$).
(D) Spontaneous heat transfer across a finite temperature difference is always an irreversible process accompanied by an increase in total entropy.
Q8. Two reversible heat engines, $E_1$ and $E_2$, operate in series between three thermal reservoirs at temperatures $T_1$, $T_2$, and $T_3$ (with $T_1 > T_2 > T_3$). Engine $E_1$ absorbs heat $Q_1$ from reservoir $T_1$ and rejects heat $Q_2$ to reservoir $T_2$. Engine $E_2$ absorbs heat $Q_2$ from reservoir $T_2$ and rejects heat $Q_3$ to reservoir $T_3$. Which of the following statements is/are correct?
(A) If the work outputs of both engines are equal ($W_1 = W_2$), then $T_2 = \frac{T_1 + T_3}{2}$.
(B) If the thermal efficiencies of both engines are equal ($\eta_1 = \eta_2$), then $T_2 = \sqrt{T_1 T_3}$.
(C) The combined efficiency $\eta_{\text{comb}}$ of the two engines operating in series is $\eta_{\text{comb}} = \eta_1 + \eta_2 – \eta_1 \eta_2$.
(D) If $T_1 > T_3$ with $T_1 \neq T_3$, the intermediate temperature for equal work is always strictly greater than that for equal efficiency ($T_{2, \text{equal work}} > T_{2, \text{equal }\eta}$).
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A reversible heat engine absorbs $Q_H = 1500\text{ J}$ of heat from a hot reservoir and delivers $W = 600\text{ J}$ of work per cycle. If this work $W$ is directly used to drive a refrigerator having a coefficient of performance $\beta = 3.5$, determine the heat extracted from the cold refrigerated body per cycle (in $\text{J}$).
Q10. Two reversible engines $A$ and $B$ are connected in series between three thermal reservoirs at temperatures $T_1 = 1200\text{ K}$, $T_2$, and $T_3 = 300\text{ K}$. Engine $A$ absorbs heat from $T_1$ and rejects heat to $T_2$, which is entirely absorbed by Engine $B$, which in turn rejects heat to $T_3$. If both engines produce equal mechanical work ($W_A = W_B$), calculate the intermediate temperature $T_2$ (in $\text{K}$).
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (B) Energy conservation vs. Directionality | First Law vs. Second Law Distinctions |
| Q2 | (B) PMM-2 | Perpetual Motion Machine of the Second Kind |
| Q3 | (C) $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} + 1$ | Heat Pump and Refrigerator COP Relation |
| Q4 | (B) $1500\text{ J}$ | Heat Engine Work & Heat Rejection |
| Q5 | (C) Transfer heat cold to hot with $W = 0$ | Clausius Statement Formulation |
| Q6 | (A), (B), (C) | Equivalence of Kelvin-Planck & Clausius |
| Q7 | (A), (B), (C), (D) | Physical Deductions of the Second Law |
| Q8 | (A), (B), (C), (D) | Series Engines Arithmetic & Geometric Means |
| Q9 | 2100 | Coupled Refrigerator Heat Extraction ($3.5 \times 600\text{ J}$) |
| Q10 | 750 | Series Equal Work Temperature ($(1200 + 300)/2\text{ K}$) |
Complete Step-by-Step Solutions
Solution 1:
The First Law is an energy accounting principle: it states that energy cannot be created or destroyed, requiring $Q = \Delta U + W$. However, the First Law does not forbid spontaneous heat transfer from a cooler body to a hotter body, nor does it forbid $100\%$ conversion of heat to work without rejection, provided energy is balanced.
The Second Law governs the directionality and spontaneity of natural processes and sets the thermodynamic boundary on conversion of heat into work.
Correct Option: (B)
Solution 2:
– A machine that produces work without any energy input ($W > 0$ with $Q = 0$) creates energy out of nothing, violating the First Law; it is a Perpetual Motion Machine of the First Kind (PMM-1).
– A machine that converts all absorbed heat from a single thermal reservoir into work without heat rejection ($Q_H = W, Q_C = 0, \eta = 1$) satisfies the First Law but violates the Kelvin-Planck statement of the Second Law; it is a Perpetual Motion Machine of the Second Kind (PMM-2).
Correct Option: (B)
Solution 3:
For a refrigerator, $\text{COP}_{\text{ref}} = \frac{Q_C}{W}$.
For a heat pump, $\text{COP}_{\text{HP}} = \frac{Q_H}{W}$.
By First Law for a cycle, $Q_H = Q_C + W$.
Substituting into the heat pump COP:
$$\text{COP}_{\text{HP}} = \frac{Q_C + W}{W} = \frac{Q_C}{W} + \frac{W}{W} = \text{COP}_{\text{ref}} + 1$$
Correct Option: (C)
Solution 4:
Thermal efficiency is given by:
$$\eta = \frac{W}{Q_H} \implies Q_H = \frac{W}{\eta} = \frac{500\text{ J}}{0.25} = 2000\text{ J}$$
Applying the First Law over the complete cycle:
$$W = Q_H – Q_C \implies Q_C = Q_H – W = 2000\text{ J} – 500\text{ J} = \mathbf{1500\text{ J}}$$
Correct Option: (B)
Solution 5:
The Clausius statement explicitly states: “It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a cooler body to a hotter body.” External work must be supplied to move heat against a thermal gradient.
Correct Option: (C)
Solution 6:
(A) Correct: Kelvin-Planck concerns heat engine limits ($\eta < 1$), whereas Clausius concerns refrigerator limits ($W_{\text{ext}} > 0$).
(B) Correct: Coupling a PMM-2 engine to a standard refrigerator creates a combined device that transfers heat from cold to hot with zero net work, violating Clausius.
(C) Correct: Coupling a Clausius-violating refrigerator to a standard heat engine yields a combined device that converts heat from a single reservoir completely into work, violating Kelvin-Planck.
(D) Incorrect: They are completely equivalent, not mutually exclusive.
Correct Options: (A), (B), (C)
Solution 7:
(A) Correct: Kelvin-Planck requires $Q_C > 0 \implies \eta = 1 – Q_C/Q_H < 1$.
(B) Correct: Without a second reservoir at lower temperature to absorb $Q_C$, continuous cyclic work cannot be produced.
(C) Correct: Since $W_{\text{ext}} > 0$ by Clausius, $\text{COP} = Q_C / W_{\text{ext}} < \infty$.
(D) Correct: Heat transfer across a finite temperature difference generates positive entropy ($\Delta S_{\text{univ}} > 0$), making it irreversible.
Correct Options: (A), (B), (C), (D)
Solution 8:
For two reversible engines in series:
$$W_1 = Q_1 \left(1 – \frac{T_2}{T_1}\right) = Q_1 \left(\frac{T_1 – T_2}{T_1}\right)$$
$$W_2 = Q_2 \left(1 – \frac{T_3}{T_2}\right) = Q_1 \left(\frac{T_2}{T_1}\right) \left(\frac{T_2 – T_3}{T_2}\right) = Q_1 \left(\frac{T_2 – T_3}{T_1}\right)$$
(A) Correct: $W_1 = W_2 \implies T_1 – T_2 = T_2 – T_3 \implies T_2 = \frac{T_1 + T_3}{2}$.
(B) Correct: $\eta_1 = \eta_2 \implies 1 – \frac{T_2}{T_1} = 1 – \frac{T_3}{T_2} \implies T_2 = \sqrt{T_1 T_3}$.
(C) Correct: Total work $W_{\text{net}} = W_1 + W_2 = \eta_1 Q_1 + \eta_2 Q_2 = \eta_1 Q_1 + \eta_2 (1 – \eta_1) Q_1 = [\eta_1 + \eta_2 – \eta_1 \eta_2] Q_1 \implies \eta_{\text{comb}} = \eta_1 + \eta_2 – \eta_1 \eta_2$.
(D) Correct: By the AM-GM inequality, for distinct positive temperatures $T_1 \neq T_3$, the arithmetic mean is strictly greater than the geometric mean: $\frac{T_1 + T_3}{2} > \sqrt{T_1 T_3}$.
Correct Options: (A), (B), (C), (D)
Solution 9:
The work delivered by the heat engine is $W = 600\text{ J}$.
This work is completely supplied to the refrigerator as work input: $W_{\text{in}} = 600\text{ J}$.
Given coefficient of performance $\beta = 3.5$:
$$\beta = \frac{Q_C’}{W_{\text{in}}} \implies Q_C’ = \beta \cdot W_{\text{in}} = 3.5 \times 600\text{ J} = \mathbf{2100\text{ J}}$$
Answer: 2100
Solution 10:
For two reversible engines in series producing equal mechanical work ($W_A = W_B$), the intermediate temperature $T_2$ is the arithmetic mean of the source and sink temperatures:
$$T_2 = \frac{T_1 + T_3}{2}$$
Given $T_1 = 1200\text{ K}$ and $T_3 = 300\text{ K}$:
$$T_2 = \frac{1200 + 300}{2} = \frac{1500}{2} = \mathbf{750\text{ K}}$$
Answer: 750