Welcome to Lecture 99 of our comprehensive JEE Physics Master Series. In this lecture, we undertake an in-depth, rigorous exploration of Heat Engines and Refrigerators. We formulate the operational principles and thermodynamic cycles governing both forward and reverse cyclic devices, derive the fundamental expressions for Thermal Efficiency ($\eta$) and Coefficient of Performance ($\text{COP}$), master the critical technique of dissecting closed indicator diagrams ($P$-$V$ loops) to isolate positive heat input zones ($dQ > 0$) from heat rejection zones, and solve standard JEE Advanced problem archetypes including triangular and rectangular cycles, combined power-plant/refrigerator systems, and multi-reservoir thermal networks.
Detailed Concept Card: Heat Engines & Refrigerators
1. Operating Principles of a Heat Engine
A Heat Engine is a thermodynamic system operating in a closed thermodynamic cycle designed to continuously convert thermal energy absorbed from a high-temperature source into useful mechanical work while rejecting residual heat to a low-temperature sink.
The standard heat engine configuration consists of three essential components:
- Heat Source ($T_H$): A high-temperature thermal reservoir of infinite heat capacity maintaining constant temperature $T_H$, from which the engine absorbs heat $Q_H$ during each cycle.
- Working Substance: The medium (such as an ideal gas, steam, or air-fuel mixture) that undergoes cyclic expansion, compression, heating, and cooling. Because the working substance completes a closed cycle returning to its original state, its net internal energy change over each complete cycle is strictly zero ($\Delta U_{\text{cycle}} = 0$).
- Heat Sink ($T_C$): A low-temperature thermal reservoir maintaining constant temperature $T_C$ ($T_C < T_H$), into which the engine discharges waste heat $Q_C$ during each cycle.
Applying the First Law of Thermodynamics over one full cycle:
$$Q_{\text{net}} = \Delta U_{\text{cycle}} + W_{\text{net}} \implies (Q_H – Q_C) = 0 + W_{\text{net}} \implies \mathbf{W_{\text{net}} = Q_H – Q_C}$$
2. Thermal Efficiency ($\eta$) & Indicator Diagram Geometry
The thermal efficiency ($\eta$) measures the fraction of thermal energy absorbed from the high-temperature source that is successfully converted into net mechanical work:
$$\mathbf{\eta = \frac{\text{Useful Work Output}}{\text{Heat Absorbed from Source}} = \frac{W_{\text{net}}}{Q_H} = \frac{Q_H – Q_C}{Q_H} = 1 – \frac{Q_C}{Q_H}}$$
Crucial Distinction between $Q_H$ and $Q_{\text{net}}$:
In the denominator of the efficiency expression, $Q_H$ is NOT the net heat $Q_{\text{net}}$! $Q_{\text{net}} = Q_H – Q_C = W_{\text{net}}$. The term $Q_H$ represents the total heat absorbed by the gas along all segments of the cycle where heat is supplied ($dQ > 0$). Substituting $Q_{\text{net}}$ instead of $Q_H$ in $\eta$ is a classic JEE trap that erroneously yields $\eta = W/W = 1$.
Work on $P$-$V$ Indicator Diagrams:
- On a $P$-$V$ indicator diagram, the net work done during one cycle equals the area enclosed by the cyclic path: $$W_{\text{net}} = \oint P \, dV = \text{Enclosed Area}$$
- Clockwise Cycle ($W_{\text{net}} > 0$): Expansion occurs at higher average pressures than compression. The net boundary work is positive, functioning as a Heat Engine.
- Counter-Clockwise Cycle ($W_{\text{net}} < 0$): Compression occurs at higher average pressures than expansion. Net work is done on the gas, functioning as a Refrigerator or Heat Pump.
3. Refrigerators and Heat Pumps (Reversed Cycles)
A Refrigerator or Heat Pump is a cyclic thermodynamic device that consumes external mechanical work ($W_{\text{in}}$) to extract heat ($Q_C$) from a cold space and discharge heat ($Q_H$) to a hotter reservoir. By the First Law over a complete cycle:
$$\mathbf{Q_H = Q_C + W_{\text{in}}}$$
The efficiency of a reversed device is evaluated using the Coefficient of Performance ($\text{COP}$), defined as:
$$\text{COP} = \frac{\text{Desired Output (Target Energy Effect)}}{\text{Required Input (Work Consumed)}}$$
| Device Type | Primary Functional Objective | Coefficient of Performance Expression |
|---|---|---|
| Refrigerator ($\beta$ or $\text{COP}_{\text{ref}}$) | Cooling: Extract heat $Q_C$ from a refrigerated chamber | $$\beta = \frac{Q_C}{W_{\text{in}}} = \frac{Q_C}{Q_H – Q_C}$$ |
| Heat Pump ($\gamma$ or $\text{COP}_{\text{HP}}$) | Heating: Deliver heat $Q_H$ to a warm living space | $$\gamma = \frac{Q_H}{W_{\text{in}}} = \frac{Q_H}{Q_H – Q_C}$$ |
Universal Relationship Connecting Heat Pump & Refrigerator COP:
$$\mathbf{\text{COP}_{\text{HP}} = \frac{Q_H}{W_{\text{in}}} = \frac{Q_C + W_{\text{in}}}{W_{\text{in}}} = \frac{Q_C}{W_{\text{in}}} + 1 = \text{COP}_{\text{ref}} + 1}$$
Because $\text{COP}_{\text{ref}} > 0$ for any functioning cooling device, the coefficient of performance of a heat pump is always strictly greater than unity ($\text{COP}_{\text{HP}} > 1$).
4. Advanced Method for Dissecting P-V Cycles in JEE Problems
In standard JEE Advanced problems, cycles are often presented as geometric shapes (triangles, rectangles, ellipses, or linear chords) on $P$-$V$ diagrams. To calculate the thermal efficiency $\eta = \frac{W_{\text{net}}}{Q_{\text{in}}}$:
- Compute $W_{\text{net}}$: Evaluate the geometric area enclosed by the closed curve.
- Formulate $dQ$ along every path: Using the First Law for an ideal gas: $$dQ = dU + dW = n C_v dT + P dV = \frac{C_v}{R} (P dV + V dP) + P dV = \left(\frac{C_v}{R} + 1\right) P dV + \frac{C_v}{R} V dP$$ For a monoatomic gas ($C_v = \frac{3}{2}R$): $$dQ = \frac{5}{2} P dV + \frac{3}{2} V dP$$
- Identify the Zero-Heat Boundary ($dQ = 0$): Along slanted straight-line paths $P(V) = -aV + b$, the pressure decreases as volume increases ($dP = -a \, dV$). Consequently: $$dQ = \left[\frac{5}{2}(-aV + b) – \frac{3}{2}aV\right] dV = \left[\frac{5}{2}b – 4aV\right] dV$$ The condition $dQ = 0$ identifies the exact inflection volume $V^* = \frac{5b}{8a}$. Heat is absorbed ($dQ > 0$) only on the sub-segment where $V < V^*$. Integrating $dQ$ only over this absorbing sub-segment gives the true heat input $Q_{\text{in}}$.
Four Graded Solved Examples
Example 1 (JEE Main Level: Coupled Heat Engine and Refrigerator Energy Balance)
Problem Statement: A heat engine operating between two thermal reservoirs absorbs $Q_H = 2000\text{ J}$ of heat from the high-temperature source per cycle and operates with a thermal efficiency of $\eta = 30\%$. The net mechanical work delivered by this engine is used directly to drive a domestic refrigerator that operates with a coefficient of performance $\beta = 4.5$. Determine:
- The net work output $W$ and the heat rejected $Q_C$ to the sink by the heat engine per cycle.
- The heat $Q_C’$ extracted from the freezer compartment by the refrigerator per cycle.
- The total heat $Q_H’$ discharged into the surrounding room by the refrigerator per cycle, and verify the heat pump relation $\text{COP}_{\text{HP}} = \beta + 1$.
Solution:
Step 1: Work Output and Heat Rejection of the Engine.
From the definition of thermal efficiency:
$$W = \eta \, Q_H = 0.30 \times 2000\text{ J} = \mathbf{600\text{ J}}$$
Applying the First Law to the engine cycle ($\Delta U_{\text{cycle}} = 0$):
$$Q_C = Q_H – W = 2000\text{ J} – 600\text{ J} = \mathbf{1400\text{ J}}$$
Step 2: Heat Extracted by the Refrigerator ($Q_C’$).
The work input consumed by the refrigerator is $W_{\text{in}} = W = 600\text{ J}$. By definition of $\text{COP}_{\text{ref}}$:
$$\beta = \frac{Q_C’}{W_{\text{in}}} \implies Q_C’ = \beta \cdot W_{\text{in}} = 4.5 \times 600\text{ J} = \mathbf{2700\text{ J}}$$
Step 3: Heat Discharged into the Surroundings ($Q_H’$).
By the First Law for the refrigerator:
$$Q_H’ = Q_C’ + W_{\text{in}} = 2700\text{ J} + 600\text{ J} = \mathbf{3300\text{ J}}$$
Verifying the Heat Pump COP:
$$\text{COP}_{\text{HP}} = \frac{Q_H’}{W_{\text{in}}} = \frac{3300\text{ J}}{600\text{ J}} = 5.5$$
$$\text{COP}_{\text{HP}} = \beta + 1 = 4.5 + 1 = 5.5$$
The relationship holds identically.
Takeaway: Because the refrigerator coefficient of performance is greater than 1 ($\beta = 4.5$), the heat extracted from the food compartment ($2700\text{ J}$) is 4.5 times larger than the mechanical work supplied ($600\text{ J}$).
Example 2 (Conceptual & Graphical: Thermal Efficiency of a Triangular P-V Cycle)
Problem Statement: One mole of an ideal monoatomic gas ($C_v = \frac{3}{2}R, C_p = \frac{5}{2}R$) undergoes a closed cyclic process $A \to B \to C \to A$ represented on a $P$-$V$ indicator diagram where:
– State $A$ has coordinates $(P_0, V_0)$.
– Path $A \to B$ is an isochoric pressure rise to $(3P_0, V_0)$.
– Path $B \to C$ is a straight-line expansion to $(P_0, 3V_0)$.
– Path $C \to A$ is an isobaric compression at constant pressure $P_0$ back to $(P_0, V_0)$.
Determine the net work done $W_{\text{net}}$, the total heat absorbed $Q_{\text{in}}$ by identifying the exact sub-segment of $B \to C$ where $dQ > 0$, and the exact thermal efficiency $\eta$ of the cycle.
Solution:
Step 1: Net Work Done ($W_{\text{net}}$).
The cycle is executed clockwise; hence $W_{\text{net}} > 0$. The net work equals the area of right triangle $ABC$:
$$W_{\text{net}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3V_0 – V_0) \times (3P_0 – P_0) = \frac{1}{2} \times (2V_0) \times (2P_0) = \mathbf{2 P_0 V_0}$$
Step 2: Heat Exchange on Path $A \to B$ (Isochoric Heating).
Since volume is constant ($V = V_0, dV = 0$):
$$Q_{AB} = n C_v (T_B – T_A) = \frac{3}{2} (P_B V_B – P_A V_A) = \frac{3}{2} (3P_0 V_0 – P_0 V_0) = \frac{3}{2} (2 P_0 V_0) = \mathbf{3 P_0 V_0 > 0}$$
Heat is absorbed along the entire path $A \to B$.
Step 3: Heat Exchange on Path $B \to C$ (Straight Line Path).
Equation of the line passing through $(V_0, 3P_0)$ and $(3V_0, P_0)$:
$$P(V) – 3P_0 = \frac{P_0 – 3P_0}{3V_0 – V_0} (V – V_0) = -\frac{P_0}{V_0} (V – V_0) \implies P(V) = 4P_0 – \frac{P_0}{V_0} V$$
Differential heat transfer for a monoatomic gas:
$$dQ = dU + dW = \frac{3}{2} d(PV) + P dV = \frac{5}{2} P dV + \frac{3}{2} V dP$$
Since $dP = -\frac{P_0}{V_0} dV$:
$$dQ = \left[\frac{5}{2} \left(4P_0 – \frac{P_0}{V_0} V\right) – \frac{3}{2} \frac{P_0}{V_0} V\right] dV = \left[10P_0 – 4\frac{P_0}{V_0} V\right] dV$$
Heat absorption occurs ($dQ > 0$) as long as:
$$10P_0 – 4\frac{P_0}{V_0} V > 0 \implies V < 2.5 V_0$$
At $V = 2.5 V_0$, $dQ = 0$. Beyond $V = 2.5 V_0$, $dQ < 0$ (the gas starts rejecting heat!).
Therefore, heat is absorbed on path $BC$ only from $V = V_0$ to $V = 2.5 V_0$:
$$Q_{BC, \text{absorbed}} = \int_{V_0}^{2.5V_0} \left(10P_0 – 4\frac{P_0}{V_0} V\right) dV = \left[10P_0 V – 2\frac{P_0}{V_0} V^2\right]_{V_0}^{2.5V_0}$$
$$= P_0 \left[(10 \times 2.5 – 2 \times 6.25) – (10 \times 1 – 2 \times 1)\right] = P_0 [(25 – 12.5) – (10 – 2)] = P_0 [12.5 – 8] = \mathbf{4.5 P_0 V_0}$$
Step 4: Total Heat Absorbed ($Q_{\text{in}}$) and Thermal Efficiency ($\eta$).
Path $C \to A$ is isobaric compression where $dV < 0$ and $dT < 0$, so $Q_{CA} < 0$ (heat rejection).
Thus, total heat absorbed in the cycle is:
$$Q_{\text{in}} = Q_{AB} + Q_{BC, \text{absorbed}} = 3 P_0 V_0 + 4.5 P_0 V_0 = \mathbf{7.5 P_0 V_0}$$
The thermal efficiency is:
$$\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{2 P_0 V_0}{7.5 P_0 V_0} = \frac{2}{15/2} = \mathbf{\frac{4}{15} \approx 26.67\%}$$
Takeaway: Never assume that heat is absorbed over an entire linear path! Always set $dQ = 0$ to find the transition point between heat absorption and heat rejection.
Example 3 (JEE Advanced Level: Power Plant Driving a Commercial Refrigeration System)
Problem Statement: A thermal power plant operates as a cyclic heat engine with a thermal efficiency of $\eta_{\text{engine}} = 40\%$. The mechanical work delivered by the plant drives an industrial refrigeration unit operating with a coefficient of performance $\beta = 3.0$. The refrigeration plant maintains a cold-storage room at a constant low temperature by removing heat at a rate of $\dot{Q}_C = 120\text{ kW}$. Determine:
- The mechanical power $P_{\text{input}}$ required to drive the refrigeration plant.
- The thermal power input $\dot{Q}_H$ required from fuel combustion in the power plant.
- The total rate of heat $\dot{Q}_{\text{total, rejected}}$ dumped into the environment by both the power plant and the refrigeration unit combined.
Solution:
Step 1: Mechanical Power Required by the Refrigerator ($P_{\text{input}}$).
By definition of the refrigerator coefficient of performance:
$$\beta = \frac{\dot{Q}_C}{P_{\text{input}}} \implies P_{\text{input}} = \frac{\dot{Q}_C}{\beta} = \frac{120\text{ kW}}{3.0} = \mathbf{40\text{ kW}}$$
Step 2: Fuel Thermal Power Required ($\dot{Q}_H$).
The power plant must deliver mechanical output $P_{\text{net}} = P_{\text{input}} = 40\text{ kW}$. With efficiency $\eta_{\text{engine}} = 0.40$:
$$\eta_{\text{engine}} = \frac{P_{\text{net}}}{\dot{Q}_H} \implies \dot{Q}_H = \frac{P_{\text{net}}}{\eta_{\text{engine}}} = \frac{40\text{ kW}}{0.40} = \mathbf{100\text{ kW}}$$
Step 3: Total Heat Rejected to Environment ($\dot{Q}_{\text{total, rejected}}$).
1. Heat rejected by the power plant:
$$\dot{Q}_{\text{plant, rej}} = \dot{Q}_H – P_{\text{net}} = 100\text{ kW} – 40\text{ kW} = 60\text{ kW}$$
2. Heat rejected by the refrigeration plant:
$$\dot{Q}_{\text{ref, rej}} = \dot{Q}_C + P_{\text{input}} = 120\text{ kW} + 40\text{ kW} = 160\text{ kW}$$
3. Total heat rate rejected to the ambient surroundings:
$$\dot{Q}_{\text{total, rejected}} = \dot{Q}_{\text{plant, rej}} + \dot{Q}_{\text{ref, rej}} = 60\text{ kW} + 160\text{ kW} = \mathbf{220\text{ kW}}$$
Energy Conservation Check:
Total energy entering the combined system: $\dot{Q}_H + \dot{Q}_C = 100\text{ kW} + 120\text{ kW} = 220\text{ kW}$.
Total energy leaving into the environment: $220\text{ kW}$. The First Law of Thermodynamics is satisfied.
Example 4 (JEE Advanced Level: Reversible Heat Engine Driving a Reversible Heat Pump across Three Thermal Reservoirs)
Problem Statement: A reversible heat engine operates between a high-temperature reservoir at $T_1 = 800\text{ K}$ and an intermediate reservoir at $T_2 = 400\text{ K}$. The entire work output of this engine is used to drive a reversible heat pump that extracts heat from a low-temperature reservoir at $T_3 = 200\text{ K}$ and delivers heat to the same intermediate reservoir at $T_2 = 400\text{ K}$. If the heat absorbed by the engine from the reservoir at $T_1$ is $Q_1 = 1600\text{ J}$ per cycle, determine:
- The mechanical work output $W$ of the engine.
- The heat $Q_3$ extracted from the low-temperature reservoir at $T_3$.
- The total heat $Q_{2,\text{total}}$ delivered to the intermediate reservoir at $T_2$ by the combined engine-heat pump system.
Solution:
Step 1: Work Output of the Reversible Engine.
For a reversible engine operating between $T_1 = 800\text{ K}$ and $T_2 = 400\text{ K}$:
$$\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{400}{800} = 1 – 0.50 = 0.50$$
$$W = \eta \, Q_1 = 0.50 \times 1600\text{ J} = \mathbf{800\text{ J}}$$
Heat rejected by the engine to the intermediate reservoir at $T_2$:
$$Q_2′ = Q_1 – W = 1600\text{ J} – 800\text{ J} = 800\text{ J}$$
Step 2: Heat Pump Operation & Heat Extracted ($Q_3$).
For a reversible heat pump operating between $T_3 = 200\text{ K}$ and $T_2 = 400\text{ K}$:
$$\text{COP}_{\text{HP}} = \frac{T_2}{T_2 – T_3} = \frac{400\text{ K}}{400\text{ K} – 200\text{ K}} = \frac{400}{200} = 2.0$$
The heat delivered by the heat pump to reservoir $T_2$ is:
$$Q_2” = \text{COP}_{\text{HP}} \times W = 2.0 \times 800\text{ J} = 1600\text{ J}$$
Applying the First Law to the heat pump cycle:
$$Q_2” = Q_3 + W \implies Q_3 = Q_2” – W = 1600\text{ J} – 800\text{ J} = \mathbf{800\text{ J}}$$
Step 3: Total Heat Delivered to Reservoir $T_2$.
Both the engine and the heat pump discharge heat into reservoir $T_2$:
$$Q_{2,\text{total}} = Q_2′ + Q_2” = 800\text{ J} + 1600\text{ J} = \mathbf{2400\text{ J}}$$
Takeaway: By utilizing the heat engine’s work to drive a heat pump rather than directly dumping heat, the system absorbs $1600\text{ J}$ at $800\text{ K}$ and $800\text{ J}$ at $200\text{ K}$, delivering a massive $2400\text{ J}$ of thermal energy at $400\text{ K}$ with zero net mechanical work consumed.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. The thermal efficiency $\eta$ of a heat engine that absorbs heat $Q_H$ from a high-temperature source and rejects heat $Q_C$ to a low-temperature sink is given by:
(A) $\eta = 1 + \frac{Q_C}{Q_H}$
(B) $\eta = \frac{Q_H – Q_C}{Q_C}$
(C) $\eta = 1 – \frac{Q_C}{Q_H}$
(D) $\eta = \frac{Q_C}{Q_H}$
Q2. If $\text{COP}_{\text{ref}}$ represents the coefficient of performance of a refrigerator and $\text{COP}_{\text{HP}}$ represents the coefficient of performance of the same device operated as a heat pump between the same two thermal reservoirs, then:
(A) $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} – 1$
(B) $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} + 1$
(C) $\text{COP}_{\text{HP}} \times \text{COP}_{\text{ref}} = 1$
(D) $\text{COP}_{\text{HP}} = \frac{1}{\text{COP}_{\text{ref}} + 1}$
Q3. A thermodynamic cycle plotted on a $P$-$V$ indicator diagram is traced in a clockwise direction. Which of the following statements is correct?
(A) Net work done by the system is negative, and the cycle represents a refrigerator.
(B) Net work done by the system is positive, and the cycle represents a heat engine.
(C) Net heat exchanged over the cycle is zero.
(D) Change in internal energy over the complete cycle is strictly positive.
Q4. A heat engine operating with an efficiency of $\eta = 0.40$ delivers $W = 1200\text{ J}$ of mechanical work per cycle. The quantity of heat rejected to the low-temperature sink per cycle is:
(A) $3000\text{ J}$
(B) $1800\text{ J}$
(C) $480\text{ J}$
(D) $720\text{ J}$
Q5. A refrigerator with a coefficient of performance $\beta = 5.0$ is used to freeze $1\text{ kg}$ of water initially at $0^\circ\text{C}$ into ice at $0^\circ\text{C}$. Taking the latent heat of fusion of water as $L_f = 3.36 \times 10^5\text{ J/kg}$, the electrical work input required by the compressor is:
(A) $1.68 \times 10^6\text{ J}$
(B) $6.72 \times 10^4\text{ J}$
(C) $3.36 \times 10^5\text{ J}$
(D) $8.40 \times 10^4\text{ J}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. A fixed mass of an ideal gas undergoes a closed cyclic process represented on a $P$-$V$ diagram. Which of the following statements is/are universally true for any complete cycle?
(A) The net change in internal energy over the cycle is zero ($\Delta U_{\text{cycle}} = 0$).
(B) The net heat absorbed by the gas equals the net work done by the gas ($Q_{\text{net}} = W_{\text{net}}$).
(C) The thermal efficiency is equal to $W_{\text{net}} / Q_{\text{net}}$.
(D) The net work done is positive if the cycle is traversed clockwise and negative if traversed counter-clockwise.
Q7. A domestic refrigerator is placed in a closed, thermally insulated room with its door left completely open, and its power cord is plugged into an electrical outlet. Which of the following statements is/are correct?
(A) The room temperature will gradually decrease because the refrigerator cools the surrounding air.
(B) The room temperature will gradually increase because electrical energy consumed by the motor is ultimately dissipated as heat into the room.
(C) The heat dumped into the room by the condenser coils exceeds the heat extracted from the room air by the evaporator coils by an amount equal to the electrical work consumed ($Q_H = Q_C + W$).
(D) The refrigerator acts as a cyclic heater with an effective heat delivery rate equal to the electrical power rating of the compressor.
Q8. An ideal gas is taken along a straight-line path $A \to B$ on a $P$-$V$ indicator diagram where pressure decreases linearly as volume increases from $(P_1, V_1)$ to $(P_2, V_2)$. Which of the following statements is/are correct?
(A) The temperature of the gas increases to a maximum value and then decreases along the straight-line path.
(B) The point of maximum temperature occurs where the product $PV$ attains its maximum value.
(C) Along the path from $A$ to $B$, heat may be absorbed during the initial phase and rejected during the final phase.
(D) Molar heat capacity along this path is constant throughout the expansion.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A cyclic heat engine absorbs $Q_H = 3600\text{ J}$ of thermal energy from a furnace and delivers $W = 1200\text{ J}$ of mechanical work per cycle. The entire work output of this engine is directly coupled to drive a refrigerator operating with a coefficient of performance $\beta = 4.0$. Calculate the heat extracted from the refrigerated space per cycle (in $\text{J}$).
Q10. An ideal monoatomic gas undergoes a rectangular cyclic process on a $P$-$V$ diagram consisting of: an isobaric expansion from $(P_0, V_0)$ to $(P_0, 2V_0)$, an isochoric depressurization to $(P_0/2, 2V_0)$, an isobaric compression to $(P_0/2, V_0)$, and an isochoric pressurization back to $(P_0, V_0)$. If the thermal efficiency of this cycle is expressed in simplest fractional form as $\eta = \frac{a}{b}$ where $a$ and $b$ are co-prime positive integers, find the value of $(a + b)$.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (C) $\eta = 1 – Q_C / Q_H$ | Thermal Efficiency Definition |
| Q2 | (B) $\text{COP}_{\text{HP}} = \text{COP}_{\text{ref}} + 1$ | Heat Pump vs. Refrigerator COP Relation |
| Q3 | (B) Clockwise cycle is a Heat Engine ($W > 0$) | Indicator Diagram Directionality |
| Q4 | (B) $1800\text{ J}$ | Engine Heat Rejection Calculation |
| Q5 | (B) $6.72 \times 10^4\text{ J}$ | Freezer Compressor Work Input |
| Q6 | (A), (B), (D) | Cyclic Thermodynamics Principles |
| Q7 | (B), (C), (D) | Open Refrigerator in an Insulated Room |
| Q8 | (A), (B), (C) | Linear Path on $P$-$V$ Diagram Analysis |
| Q9 | 4800 | Coupled Refrigerator Heat Extraction ($4.0 \times 1200\text{ J}$) |
| Q10 | 15 | Rectangular Cycle Efficiency $\eta = 2/13 \implies a+b=15$ |
Complete Step-by-Step Solutions
Solution 1:
Thermal efficiency is the ratio of net work delivered to heat absorbed from the high-temperature reservoir:
$$\eta = \frac{W}{Q_H} = \frac{Q_H – Q_C}{Q_H} = 1 – \frac{Q_C}{Q_H}$$
Correct Option: (C)
Solution 2:
By definition: $\text{COP}_{\text{ref}} = \frac{Q_C}{W}$ and $\text{COP}_{\text{HP}} = \frac{Q_H}{W}$.
Using the First Law energy balance $Q_H = Q_C + W$:
$$\text{COP}_{\text{HP}} = \frac{Q_C + W}{W} = \frac{Q_C}{W} + 1 = \text{COP}_{\text{ref}} + 1$$
Correct Option: (B)
Solution 3:
On a $P$-$V$ diagram, when a cycle is traversed clockwise, the expansion path lies at higher pressures than the compression path. Consequently, the positive expansion work exceeds the negative compression work, resulting in net positive work ($W_{\text{net}} > 0$). This describes the operation of a Heat Engine.
Correct Option: (B)
Solution 4:
Given $\eta = 0.40$ and $W = 1200\text{ J}$:
$$Q_H = \frac{W}{\eta} = \frac{1200\text{ J}}{0.40} = 3000\text{ J}$$
Applying the First Law to the cycle:
$$Q_C = Q_H – W = 3000\text{ J} – 1200\text{ J} = \mathbf{1800\text{ J}}$$
Correct Option: (B)
Solution 5:
Heat that must be extracted to freeze $1\text{ kg}$ of water at $0^\circ\text{C}$:
$$Q_C = m L_f = 1\text{ kg} \times 3.36 \times 10^5\text{ J/kg} = 3.36 \times 10^5\text{ J}$$
Given coefficient of performance $\beta = 5.0$:
$$\beta = \frac{Q_C}{W} \implies W = \frac{Q_C}{\beta} = \frac{3.36 \times 10^5\text{ J}}{5.0} = \mathbf{6.72 \times 10^4\text{ J}}$$
Correct Option: (B)
Solution 6:
(A) Correct: Internal energy is a state function; after any complete cycle, $\Delta U_{\text{cycle}} = 0$.
(B) Correct: By the First Law, $Q_{\text{net}} = \Delta U + W_{\text{net}} = 0 + W_{\text{net}} = W_{\text{net}}$.
(C) Incorrect: Thermal efficiency is $\eta = W_{\text{net}} / Q_H$ where $Q_H$ is the total heat absorbed, not $Q_{\text{net}}$.
(D) Correct: Clockwise cycles enclose positive work (heat engines), while counter-clockwise cycles enclose negative work (refrigerators/heat pumps).
Correct Options: (A), (B), (D)
Solution 7:
(A) Incorrect: The refrigerator cannot act as a net heat sink for an insulated room.
(B) Correct: Electrical energy supplied from the wall outlet is converted into internal energy and dissipated into the room air.
(C) Correct: By the First Law, heat rejected at the condenser is $Q_H = Q_C + W$. Since both the evaporator and condenser are inside the same room, the net heat added to the room per cycle is $Q_H – Q_C = W > 0$.
(D) Correct: The room gains thermal energy at a rate exactly equal to the electrical power consumed by the motor.
Correct Options: (B), (C), (D)
Solution 8:
(A) Correct: Equation of a downward-sloping line is $P = -aV + b$. Then $T \propto PV = -aV^2 + bV$, which is an inverted parabola opening downward.
(B) Correct: Since $T = PV/(nR)$, temperature is directly proportional to $PV$, attaining its maximum at $V = b/(2a)$.
(C) Correct: For an ideal gas, $dQ = n C_v dT + P dV$. As derived in the concept card, $dQ > 0$ during the initial phase and becomes negative ($dQ < 0$) in the latter phase.
(D) Incorrect: Molar heat capacity $C = \frac{dQ}{n \, dT}$ varies from positive values to infinity (where $dT = 0$) and then becomes negative.
Correct Options: (A), (B), (C)
Solution 9:
The work output delivered by the heat engine is $W = 1200\text{ J}$.
This work is entirely supplied as the input work to the refrigerator: $W_{\text{in}} = 1200\text{ J}$.
Given coefficient of performance $\beta = 4.0$:
$$\beta = \frac{Q_C’}{W_{\text{in}}} \implies Q_C’ = \beta \cdot W_{\text{in}} = 4.0 \times 1200\text{ J} = \mathbf{4800\text{ J}}$$
Answer: 4800
Solution 10:
1. Net Work Done ($W_{\text{net}}$):
The cycle is rectangular with width $\Delta V = 2V_0 – V_0 = V_0$ and height $\Delta P = P_0 – P_0/2 = P_0/2$:
$$W_{\text{net}} = \Delta P \times \Delta V = \left(\frac{P_0}{2}\right) V_0 = \frac{1}{2} P_0 V_0$$
2. Heat Absorbed ($Q_{\text{in}}$):
– Path 1 (Isobaric expansion at $P_0$ from $V_0$ to $2V_0$):
$$Q_1 = n C_p \Delta T = \frac{5}{2} P_0 (2V_0 – V_0) = \frac{5}{2} P_0 V_0$$
– Path 2 (Isochoric cooling at $2V_0$ from $P_0$ to $P_0/2$): $Q_2 = n C_v \Delta T < 0$ (heat rejected).
– Path 3 (Isobaric compression at $P_0/2$ from $2V_0$ to $V_0$): $Q_3 = n C_p \Delta T < 0$ (heat rejected).
– Path 4 (Isochoric pressurization at $V_0$ from $P_0/2$ to $P_0$):
$$Q_4 = n C_v \Delta T = \frac{3}{2} \left(P_0 – \frac{P_0}{2}\right) V_0 = \frac{3}{4} P_0 V_0$$
Total heat absorbed:
$$Q_{\text{in}} = Q_1 + Q_4 = \frac{5}{2} P_0 V_0 + \frac{3}{4} P_0 V_0 = \frac{10 + 3}{4} P_0 V_0 = \frac{13}{4} P_0 V_0$$
3. Thermal Efficiency ($\eta$):
$$\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{\frac{1}{2} P_0 V_0}{\frac{13}{4} P_0 V_0} = \frac{1}{2} \times \frac{4}{13} = \frac{2}{13}$$
Here $a = 2$ and $b = 13$, which are co-prime positive integers ($\gcd(2, 13) = 1$).
$$a + b = 2 + 13 = \mathbf{15}$$
Answer: 15