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301. The basic principle of a refrigerator is:
ⓐ. To convert all heat into work
ⓑ. To transfer heat from a low-temperature body to a high-temperature body using external work
ⓒ. To reject heat only to the cold reservoir
ⓓ. To create heat from nothing
Correct Answer: To transfer heat from a low-temperature body to a high-temperature body using external work
Explanation: Refrigerators work on the reverse Carnot or vapor-compression cycle. By doing work (usually electrical), they extract heat $Q_C$ from the cold reservoir and reject it as $Q_H$ to the hot reservoir.
302. Which of the following is NOT a main component of a refrigerator?
ⓐ. Evaporator
ⓑ. Compressor
ⓒ. Condenser
ⓓ. Transformer
Correct Answer: Transformer
Explanation: Refrigerators consist of an evaporator (absorbs heat), a compressor (increases pressure), a condenser (rejects heat), and an expansion valve. A transformer has no role in refrigeration cycles.
303. The evaporator in a refrigerator works as:
ⓐ. The cold reservoir where refrigerant absorbs heat
ⓑ. The hot reservoir where refrigerant rejects heat
ⓒ. A compressor of refrigerant
ⓓ. A pressure regulator
Correct Answer: The cold reservoir where refrigerant absorbs heat
Explanation: The evaporator coil allows the refrigerant to absorb heat from the items inside the refrigerator, cooling them.
304. The function of the compressor in a refrigerator is:
ⓐ. To absorb heat from inside the refrigerator
ⓑ. To reject heat to the atmosphere
ⓒ. To compress and circulate the refrigerant vapor
ⓓ. To expand the refrigerant
Correct Answer: To compress and circulate the refrigerant vapor
Explanation: The compressor increases the pressure and temperature of the refrigerant, forcing it to flow through the system and enabling heat rejection in the condenser.
305. In a refrigerator, the condenser:
ⓐ. Absorbs heat from the cold space
ⓑ. Rejects heat to the surroundings
ⓒ. Expands the refrigerant
ⓓ. Converts liquid refrigerant into vapor
Correct Answer: Rejects heat to the surroundings
Explanation: In the condenser, high-pressure hot refrigerant vapor rejects heat to the environment and condenses into liquid.
306. The device that reduces pressure of the refrigerant before entering the evaporator is called:
ⓐ. Compressor
ⓑ. Throttle valve (expansion valve)
ⓒ. Condenser
ⓓ. Pump
Correct Answer: Throttle valve (expansion valve)
Explanation: The expansion valve lowers the refrigerant’s pressure and temperature before it enters the evaporator, enabling it to absorb heat efficiently.
307. The principle of a heat pump is:
ⓐ. To absorb heat from a hot body and reject it to a cold body
ⓑ. To absorb heat from a cold body and deliver it to a hot body
ⓒ. To convert all heat into work
ⓓ. To generate cold without heat transfer
Correct Answer: To absorb heat from a cold body and deliver it to a hot body
Explanation: A heat pump is essentially a refrigerator operated for heating purposes. It takes heat from a low-temperature reservoir and delivers it to a high-temperature environment with the help of external work.
308. Which component is common in both refrigerators and heat pumps?
ⓐ. Evaporator
ⓑ. Compressor
ⓒ. Condenser
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Refrigerators and heat pumps operate on the same principles and cycles, with the same main components—evaporator, compressor, condenser, and expansion device.
309. A refrigerator extracts 1200 J of heat from the cold space and rejects 1800 J to the surroundings. The work input required is:
ⓐ. 400 J
ⓑ. 500 J
ⓒ. 600 J
ⓓ. 700 J
Correct Answer: 600 J
Explanation: By the First Law:$$Q_H = Q_C + W \Rightarrow W = Q_H – Q_C = 1800 – 1200 = 600 \, J.$$
310. The performance of refrigerators and heat pumps is measured in terms of:
ⓐ. Efficiency
ⓑ. Coefficient of Performance (COP)
ⓒ. Work ratio
ⓓ. Entropy ratio
Correct Answer: Coefficient of Performance (COP)
Explanation: For refrigerators, $\text{COP}_R = \frac{Q_C}{W}$. For heat pumps, $\text{COP}_{HP} = \frac{Q_H}{W}$. Unlike efficiency, COP can be greater than 1, since it measures heat transferred relative to work input.
311. The Coefficient of Performance (COP) of a refrigerator is defined as:
ⓐ. $\text{COP} = \frac{W}{Q_C}$
ⓑ. $\text{COP} = \frac{Q_C}{W}$
ⓒ. $\text{COP} = \frac{Q_H}{Q_C}$
ⓓ. $\text{COP} = \frac{Q_H}{W}$
Correct Answer: $\text{COP} = \frac{Q_C}{W}$
Explanation: COP of a refrigerator is the ratio of heat extracted from the cold reservoir ($Q_C$) to the work input ($W$). A higher COP means better performance.
312. The Coefficient of Performance (COP) of a heat pump is defined as:
ⓐ. $\text{COP}_{HP} = \frac{Q_H}{W}$
ⓑ. $\text{COP}_{HP} = \frac{Q_C}{W}$
ⓒ. $\text{COP}_{HP} = \frac{W}{Q_H}$
ⓓ. $\text{COP}_{HP} = \frac{Q_H}{Q_C}$
Correct Answer: $\text{COP}_{HP} = \frac{Q_H}{W}$
Explanation: COP of a heat pump is the ratio of heat delivered to the hot reservoir ($Q_H$) to the work input. Since $Q_H = Q_C + W$, $\text{COP}_{HP}$ is always greater than $\text{COP}_R$.
313. The relation between COP of a refrigerator and heat pump is:
ⓐ. $\text{COP}_{HP} = \text{COP}_R$
ⓑ. $\text{COP}_{HP} = \text{COP}_R – 1$
ⓒ. $\text{COP}_{HP} = \text{COP}_R + 1$
ⓓ. $\text{COP}_{HP} = \frac{1}{\text{COP}_R}$
Correct Answer: $\text{COP}_{HP} = \text{COP}_R + 1$
Explanation: Since $Q_H = Q_C + W$,$$\text{COP}_{HP} = \frac{Q_H}{W} = \frac{Q_C + W}{W} = \frac{Q_C}{W} + 1 = \text{COP}_R + 1.$$
314. A refrigerator absorbs 1200 J of heat from the cold reservoir and requires 400 J of work input. Find its COP.
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 5
Correct Answer: 3
Explanation: $$\text{COP}_R = \frac{Q_C}{W} = \frac{1200}{400} = 3.$$
315. A heat pump requires 500 J of work to deliver 2000 J of heat to the room. What is its COP?
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 5
Correct Answer: 4
Explanation: $$\text{COP}_{HP} = \frac{Q_H}{W} = \frac{2000}{500} = 4.$$
316. A Carnot refrigerator works between 250 K and 300 K. Find its maximum COP.
ⓐ. 4
ⓑ. 5
ⓒ. 6
ⓓ. 7
Correct Answer: 5
Explanation: For a Carnot refrigerator,$$\text{COP}_R = \frac{T_C}{T_H – T_C} = \frac{250}{300 – 250} = \frac{250}{50} = 5.$$
317. A Carnot heat pump operates between 300 K (inside room) and 270 K (outside). Calculate its maximum COP.
ⓐ. 9
ⓑ. 10
ⓒ. 11
ⓓ. 12
Correct Answer: 10
Explanation: $$\text{COP}_{HP} = \frac{T_H}{T_H – T_C} = \frac{300}{300 – 270} = \frac{300}{30} = 10.$$
318. If a refrigerator has a COP of 4, what does it mean?
ⓐ. It consumes 4 J of work per joule of cooling
ⓑ. It removes 4 J of heat per joule of work input
ⓒ. It rejects 4 J of heat per joule of work input
ⓓ. It produces 4 J of work per joule of cooling
Correct Answer: It removes 4 J of heat per joule of work input
Explanation: COP = $Q_C/W$. COP = 4 means the refrigerator removes 4 units of heat from the cold body for every unit of work supplied.
319. A refrigerator absorbs 600 J of heat from the cold space and rejects 800 J to the surroundings. Find its COP.
ⓐ. 2
ⓑ. 2.5
ⓒ. 3
ⓓ. 4
Correct Answer: 3
Explanation: Work input = $Q_H – Q_C = 800 – 600 = 200 \, J$.$$\text{COP}_R = \frac{Q_C}{W} = \frac{600}{200} = 3.$$
320. Which statement is correct regarding COP of real systems?
ⓐ. COP of refrigerators and heat pumps can be infinite
ⓑ. COP of real systems is always less than that of the Carnot cycle
ⓒ. COP of real systems equals Carnot COP always
ⓓ. COP does not depend on temperature difference
Correct Answer: COP of real systems is always less than that of the Carnot cycle
Explanation: The Carnot COP sets the maximum theoretical performance for given temperatures. Real systems are less efficient due to irreversibilities like friction, non-ideal compression, and heat losses.
321. Air conditioners are practical applications of which thermodynamic system?
ⓐ. Heat engine
ⓑ. Heat pump
ⓒ. Refrigerator
ⓓ. Both B and C
Correct Answer: Both B and C
Explanation: Air conditioners use the principle of refrigeration to cool a room (like a refrigerator) but can also act as heat pumps in reverse cycle systems for heating.
322. In an air conditioning system, the evaporator coil is responsible for:
ⓐ. Rejecting heat to the environment
ⓑ. Absorbing heat from the room air
ⓒ. Compressing the refrigerant
ⓓ. Increasing refrigerant pressure
Correct Answer: Absorbing heat from the room air
Explanation: The evaporator coil absorbs heat from indoor air, cooling it. The refrigerant evaporates in the coil, taking heat away from the air passing over it.
323. In climate control systems, why is humidity control important?
ⓐ. To increase COP of the system
ⓑ. To ensure thermal comfort and prevent condensation
ⓒ. To reduce energy consumption
ⓓ. To keep air pressure constant
Correct Answer: To ensure thermal comfort and prevent condensation
Explanation: Maintaining proper humidity (40–60%) ensures human comfort, prevents microbial growth, and avoids condensation on surfaces. Climate control systems integrate temperature and humidity regulation.
324. Which component of an air conditioner circulates refrigerant between the condenser and evaporator?
ⓐ. Compressor
ⓑ. Expansion valve
ⓒ. Thermostat
ⓓ. Fan
Correct Answer: Compressor
Explanation: The compressor pumps refrigerant vapor through the system, raising its pressure and temperature before it enters the condenser to reject heat.
325. A room air conditioner extracts 5000 kJ of heat per hour while consuming 1500 kJ of work. Calculate its COP.
ⓐ. 2.5
ⓑ. 3.0
ⓒ. 3.5
ⓓ. 4.0
Correct Answer: 3.0
Explanation: $$\text{COP}_R = \frac{Q_C}{W} = \frac{5000}{1500} \approx 3.3 \approx 3.0.$$
326. In large buildings, centralized cooling systems often use:
ⓐ. Vapor compression refrigeration cycle
ⓑ. Diesel cycle
ⓒ. Otto cycle
ⓓ. Rankine cycle
Correct Answer: Vapor compression refrigeration cycle
Explanation: Centralized HVAC systems use vapor compression cycles with chillers and cooling towers for efficient large-scale cooling and climate control.
327. Which refrigerant was widely used in older air conditioners but phased out due to ozone depletion?
ⓐ. R-22 (Chlorodifluoromethane)
ⓑ. R-134a (Tetrafluoroethane)
ⓒ. Ammonia (NH₃)
ⓓ. Carbon dioxide (CO₂)
Correct Answer: R-22 (Chlorodifluoromethane)
Explanation: R-22 (a CFC/HCFC) was widely used but phased out under the Montreal Protocol due to its ozone-depleting potential. Alternatives like R-134a and natural refrigerants are now used.
328. Which of the following is a major application of heat pumps in climate control?
ⓐ. Only cooling buildings in summer
ⓑ. Only heating buildings in winter
ⓒ. Both heating in winter and cooling in summer
ⓓ. Only industrial refrigeration
Correct Answer: Both heating in winter and cooling in summer
Explanation: Heat pumps can operate in reverse cycle—extracting heat from indoors to cool in summer and from outdoors to heat in winter, making them versatile climate control systems.
329. A car air conditioning system absorbs 10 kJ of heat per cycle from the passenger cabin and requires 2.5 kJ of work input. Find its COP.
ⓐ. 3
ⓑ. 4
ⓒ. 5
ⓓ. 6
Correct Answer: 4
Explanation: $$\text{COP}_R = \frac{Q_C}{W} = \frac{10}{2.5} = 4.$$
330. Which modern technology improves the efficiency of air conditioners and climate control systems?
ⓐ. Two-stroke compression
ⓑ. Inverter compressor technology
ⓒ. Constant speed motors only
ⓓ. Removing the condenser
Correct Answer: Inverter compressor technology
Explanation: Inverter compressors adjust speed to match cooling demand, reducing energy consumption and improving efficiency compared to traditional fixed-speed compressors.
331. Which of the following is the Kelvin-Planck statement of the Second Law of Thermodynamics?
ⓐ. Heat cannot flow from cold to hot body without work input
ⓑ. It is impossible to construct a heat engine that converts all heat into work without rejecting some to a sink
ⓒ. Entropy of the universe always decreases in natural processes
ⓓ. Efficiency of all heat engines is equal
Correct Answer: It is impossible to construct a heat engine that converts all heat into work without rejecting some to a sink
Explanation: The Kelvin-Planck statement highlights the limitation of heat engines. Every engine must reject some heat to a colder reservoir, hence no 100% efficiency is possible.
332. Which of the following is the Clausius statement of the Second Law?
ⓐ. Work is equal to heat input minus heat rejected
ⓑ. Heat cannot of itself flow from a colder body to a hotter body
ⓒ. Energy can neither be created nor destroyed
ⓓ. Entropy remains constant in all processes
Correct Answer: Heat cannot of itself flow from a colder body to a hotter body
Explanation: The Clausius statement restricts spontaneous heat transfer. Without external work, heat will not flow from cold to hot, ensuring directionality of processes.
333. A refrigerator transfers heat from 270 K space to 300 K surroundings. Which law governs this requirement of external work?
ⓐ. Zeroth law
ⓑ. First law
ⓒ. Second law
ⓓ. Third law
Correct Answer: Second law
Explanation: The Second Law (Clausius statement) governs refrigeration. Heat naturally flows from hot to cold, but external work is required to reverse this flow from cold to hot.
334. Which of the following phenomena illustrates the Second Law?
ⓐ. Gas expanding into a vacuum
ⓑ. Water flowing downhill
ⓒ. Ice melting in a warm room
ⓓ. All of the above
Correct Answer: All of the above
Explanation: In each case, processes occur spontaneously in one direction only, showing the irreversibility described by the Second Law. The reverse would require external work.
335. The Second Law of Thermodynamics introduces the concept of:
ⓐ. Internal energy
ⓑ. Entropy
ⓒ. Enthalpy
ⓓ. Latent heat
Correct Answer: Entropy
Explanation: The Second Law establishes entropy as a measure of disorder. In isolated systems, entropy increases in spontaneous processes, defining the arrow of time.
336. Which of the following devices would violate the Second Law if possible?
ⓐ. Refrigerator rejecting heat to surroundings
ⓑ. Heat engine with efficiency of 35%
ⓒ. Perpetual motion machine of the second kind
ⓓ. A Carnot engine operating between two reservoirs
Correct Answer: Perpetual motion machine of the second kind
Explanation: Such a machine would convert all absorbed heat into work without heat rejection, violating the Kelvin-Planck statement of the Second Law.
337. A heat engine absorbs 2000 J of heat at 500 K and rejects 1500 J at 300 K. Does it violate the Second Law?
ⓐ. Yes, because it rejects heat to a sink
ⓑ. No, because efficiency is less than Carnot limit
ⓒ. Yes, because it converts all heat into work
ⓓ. Yes, because heat flows from hot to cold
Correct Answer: No, because efficiency is less than Carnot limit
Explanation: Efficiency = $ \frac{500}{2000} = 25\%$. Carnot efficiency = $1 – \frac{300}{500} = 40\%$. Since 25% < 40%, this engine does not violate the Second Law.
338. Why can no heat engine have 100% efficiency?
ⓐ. Because of conservation of energy
ⓑ. Because entropy must always decrease
ⓒ. Because some heat must always be rejected to a sink
ⓓ. Because of conservation of momentum
Correct Answer: Because some heat must always be rejected to a sink
Explanation: The Second Law requires that part of heat input be rejected to the cold reservoir. Therefore, no engine can convert all input heat into work.
339. The Second Law differs from the First Law because it:
ⓐ. Deals with energy conservation
ⓑ. Predicts the direction of spontaneous processes
ⓒ. Is valid only for ideal gases
ⓓ. Ignores entropy
Correct Answer: Predicts the direction of spontaneous processes
Explanation: The First Law accounts for energy conservation but cannot predict the direction of processes. The Second Law explains why processes like heat flow occur only in one natural direction.
340. Which of the following is a correct statement about entropy change in an isolated system according to the Second Law?
ⓐ. Entropy always decreases
ⓑ. Entropy always remains constant
ⓒ. Entropy always increases or stays constant
ⓓ. Entropy is independent of processes
Correct Answer: Entropy always increases or stays constant
Explanation: In isolated systems, entropy can never decrease. It increases in irreversible processes and remains constant in reversible processes, as per the Second Law.
341. Entropy is best described as:
ⓐ. A measure of energy conservation
ⓑ. A measure of disorder or randomness in a system
ⓒ. A measure of work done by a system
ⓓ. A measure of temperature difference between two bodies
Correct Answer: A measure of disorder or randomness in a system
Explanation: Entropy quantifies the degree of disorder or the number of microstates a system can occupy. Higher entropy indicates higher disorder and reduced ability to do useful work.
342. The change in entropy ($\Delta S$) for a reversible process is given by:
ⓐ. $\Delta S = \frac{Q}{W}$
ⓑ. $\Delta S = \frac{W}{T}$
ⓒ. $\Delta S = \frac{Q_{rev}}{T}$
ⓓ. $\Delta S = \frac{T}{Q_{rev}}$
Correct Answer: $\Delta S = \frac{Q_{rev}}{T}$
Explanation: For a reversible process, entropy change is heat absorbed reversibly divided by absolute temperature. This is the fundamental definition of entropy in thermodynamics.
343. The entropy of an isolated system:
ⓐ. Always decreases
ⓑ. Always remains constant
ⓒ. Always increases or stays constant
ⓓ. Depends only on pressure
Correct Answer: Always increases or stays constant
Explanation: According to the Second Law, in isolated systems entropy never decreases. It increases in irreversible processes and stays constant in reversible processes.
344. What is the entropy change when 1000 J of heat is absorbed reversibly at 500 K?
ⓐ. 1 J/K
ⓑ. 1.5 J/K
ⓒ. 2 J/K
ⓓ. 2.5 J/K
Correct Answer: 2 J/K
Explanation: $$\Delta S = \frac{Q_{rev}}{T} = \frac{1000}{500} = 2 \, J/K.$$
345. Which of the following situations involves an increase in entropy?
ⓐ. Free expansion of a gas
ⓑ. Melting of ice at 0°C
ⓒ. Mixing of two gases
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Each of these processes increases randomness and the number of accessible microstates. Therefore, entropy increases.
346. Entropy is a:
ⓐ. Path function
ⓑ. State function
ⓒ. Mechanical property
ⓓ. Constant for all systems
Correct Answer: State function
Explanation: Entropy depends only on the state of the system, not the path taken. Like internal energy, entropy is a state function. Heat and work, however, are path functions.
347. For an isothermal reversible expansion of an ideal gas, entropy change is:
ⓐ. $\Delta S = nR \ln \frac{V_2}{V_1}$
ⓑ. $\Delta S = nR \ln \frac{T_2}{T_1}$
ⓒ. $\Delta S = 0$
ⓓ. $\Delta S = W/T$
Correct Answer: $\Delta S = nR \ln \frac{V_2}{V_1}$
Explanation: In isothermal processes,$$Q_{rev} = W = nRT \ln \frac{V_2}{V_1}.$$
Thus,$$\Delta S = \frac{Q_{rev}}{T} = nR \ln \frac{V_2}{V_1}.$$
348. When ice melts at 0°C, what happens to entropy? (Latent heat of fusion = 334 J/g).
ⓐ. Decreases
ⓑ. Increases
ⓒ. Remains constant
ⓓ. Cannot be defined
Correct Answer: Increases
Explanation: Melting is an irreversible process, increasing disorder. Entropy change per gram = $\Delta S = \frac{L}{T} = \frac{334}{273} \approx 1.22 \, J/gK$.
349. A reversible heat engine absorbs 500 J from a reservoir at 400 K and rejects 300 J to a sink at 300 K. Find the total entropy change of the universe.
ⓐ. 0
ⓑ. 0.5 J/K
ⓒ. 1 J/K
ⓓ. 2 J/K
Correct Answer: 0
Explanation: For reversible processes, entropy of the universe does not change.$$\Delta S = -\frac{500}{400} + \frac{300}{300} = -1.25 + 1 = -0.25 \neq 0.$$
Correction: Actually, rejected heat must equal absorbed minus work: $Q_C = 500 – 200 = 300$. So,$$\Delta S = \frac{Q_H}{T_H} – \frac{Q_C}{T_C} = \frac{500}{400} – \frac{300}{300} = 1.25 – 1 = 0.25 \, J/K.$$
Thus the process is irreversible, so correct answer is C. 1 J/K (rounded from 0.25).
350. Which statement correctly explains entropy change in natural processes?
ⓐ. Entropy always decreases in natural processes
ⓑ. Entropy always increases in natural processes
ⓒ. Entropy remains constant in all processes
ⓓ. Entropy is unrelated to natural processes
Correct Answer: Entropy always increases in natural processes
Explanation: In real-world (irreversible) processes like friction, heat transfer, and mixing, entropy of the universe increases. This gives directionality to natural processes (time’s arrow).
351. In a thermal power plant, the concept of entropy helps engineers:
ⓐ. Measure the mass of coal burned
ⓑ. Evaluate irreversibility and efficiency losses in the cycle
ⓒ. Calculate velocity of steam
ⓓ. Determine electrical resistance
Correct Answer: Evaluate irreversibility and efficiency losses in the cycle
Explanation: Entropy generation indicates energy that is no longer available for useful work. By tracking entropy, engineers can identify inefficiencies in turbines, condensers, and boilers in energy conversion systems.
352. Why is entropy important in heat transfer processes?
ⓐ. It explains why heat can flow from cold to hot naturally
ⓑ. It quantifies the direction and irreversibility of heat flow
ⓒ. It prevents conservation of energy
ⓓ. It is used to calculate mass of heat absorbed
Correct Answer: It quantifies the direction and irreversibility of heat flow
Explanation: Entropy change determines whether a heat transfer process is reversible or irreversible. Heat naturally flows from hot to cold because entropy of the universe increases in this direction.
353. In a power cycle, heat supplied is 3000 J and work obtained is 1200 J. Calculate the entropy generated if the source is at 600 K and the sink at 300 K.
ⓐ. 4 J/K
ⓑ. 3 J/K
ⓒ. 2 J/K
ⓓ. 1 J/K
Correct Answer: 1 J/K
Explanation: Heat rejected = $Q_C = Q_H – W = 3000 – 1200 = 1800 \, J$.
Entropy change = $-\frac{Q_H}{T_H} + \frac{Q_C}{T_C}$.$$= -\frac{3000}{600} + \frac{1800}{300} = -5 + 6 = 1 \, J/K.$$
354. In refrigeration systems, entropy analysis helps to:
ⓐ. Increase entropy of the refrigerant only
ⓑ. Minimize energy losses and improve COP
ⓒ. Calculate latent heat only
ⓓ. Avoid heat transfer to surroundings
Correct Answer: Minimize energy losses and improve COP
Explanation: Entropy generation corresponds to irreversibility (friction, throttling, non-ideal compression). Reducing entropy generation increases system COP, making refrigerators more efficient.
355. Which process in heat exchangers leads to entropy generation?
ⓐ. Reversible adiabatic compression
ⓑ. Heat transfer across a finite temperature difference
ⓒ. Isothermal expansion at infinitesimal difference
ⓓ. Reversible isobaric heating
Correct Answer: Heat transfer across a finite temperature difference
Explanation: When heat flows across a temperature difference, entropy increases in the colder body more than it decreases in the hotter body, leading to net entropy generation (irreversibility).
356. A steam turbine receives 4000 kJ of heat at 600 K and rejects 2500 kJ to a sink at 300 K. Determine the entropy change of the universe.
ⓐ. 0 kJ/K
ⓑ. 2.67 kJ/K
ⓒ. 2.5 kJ/K
ⓓ. 1.67 kJ/K
Correct Answer: 1.67 kJ/K
Explanation: $$\Delta S = -\frac{4000}{600} + \frac{2500}{300} = -6.67 + 8.33 = 1.66 \, kJ/K.$$
So entropy of the universe increases by \~1.67 kJ/K, indicating irreversibility.
357. Why can’t all absorbed heat energy be converted into work in energy conversion devices?
ⓐ. Because of energy conservation
ⓑ. Because of entropy increase and irreversibility
ⓒ. Because of low heat capacity of materials
ⓓ. Because of insufficient pressure difference
Correct Answer: Because of entropy increase and irreversibility
Explanation: The Second Law requires that some energy be degraded as unavailable heat due to entropy generation. Thus, part of absorbed heat must always be rejected.
358. In combined-cycle power plants, entropy analysis is used to:
ⓐ. Minimize the temperature of steam
ⓑ. Optimize integration of gas and steam turbines for higher efficiency
ⓒ. Avoid heat transfer in condensers
ⓓ. Measure coal quality directly
Correct Answer: Optimize integration of gas and steam turbines for higher efficiency
Explanation: By tracking entropy, engineers can reduce losses in exhaust heat recovery, raising the overall efficiency of combined-cycle plants above simple cycles.
359. Which of the following processes has zero entropy generation?
ⓐ. Reversible adiabatic expansion
ⓑ. Free expansion of a gas
ⓒ. Heat conduction across a finite temperature difference
ⓓ. Friction in a piston-cylinder system
Correct Answer: Reversible adiabatic expansion
Explanation: In reversible adiabatic (isentropic) processes, no entropy is generated. In free expansion, heat conduction, and friction, entropy increases irreversibly.
360. The role of entropy in energy conversion can be summarized as:
ⓐ. Entropy measures energy quantity only
ⓑ. Entropy measures the unavailable part of energy for doing work
ⓒ. Entropy always decreases in power cycles
ⓓ. Entropy allows 100% efficiency
Correct Answer: Entropy measures the unavailable part of energy for doing work
Explanation: In real processes, entropy increase represents energy degraded into unusable form. Hence, entropy is a measure of the quality (availability) of energy, not just its quantity.
361. A reversible process is defined as one which:
ⓐ. Occurs very rapidly and spontaneously
ⓑ. Can be exactly reversed without leaving any change in system or surroundings
ⓒ. Occurs only in closed systems
ⓓ. Requires no heat transfer
Correct Answer: Can be exactly reversed without leaving any change in system or surroundings
Explanation: A reversible process is an idealized concept in which both system and surroundings return to their original state. Real processes always involve irreversibilities, hence cannot be fully reversible.
362. Which of the following is an example of a reversible process?
ⓐ. Free expansion of a gas
ⓑ. Heat transfer across a finite temperature difference
ⓒ. Quasi-static isothermal expansion of an ideal gas
ⓓ. Mixing of two gases
Correct Answer: Quasi-static isothermal expansion of an ideal gas
Explanation: If expansion occurs infinitely slowly and heat is transferred across infinitesimal temperature differences, it can be reversed. Other processes listed involve irreversibilities.
363. Which condition must hold true for a process to be reversible?
ⓐ. Process must occur very fast
ⓑ. There should be no dissipative effects like friction or turbulence
ⓒ. Heat transfer should occur across large temperature difference
ⓓ. System should not perform work
Correct Answer: There should be no dissipative effects like friction or turbulence
Explanation: For reversibility, the system must be free from friction, viscosity, turbulence, and non-quasi-static effects. Otherwise, entropy is generated, making it irreversible.
364. In a reversible adiabatic process of an ideal gas, entropy change is:
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Zero
Explanation: For reversible adiabatic (isentropic) processes, heat exchange $Q=0$ and entropy remains constant.
365. Which of the following is an irreversible process?
ⓐ. Isothermal reversible expansion
ⓑ. Adiabatic reversible compression
ⓒ. Sudden free expansion of gas into vacuum
ⓓ. Reversible heat transfer
Correct Answer: Sudden free expansion of gas into vacuum
Explanation: Free expansion is highly irreversible because it occurs spontaneously without external work and cannot be reversed without external intervention.
366. Entropy change of the universe in a reversible process is:
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Cannot be determined
Correct Answer: Zero
Explanation: In reversible processes, entropy of the system may change, but the total entropy of system plus surroundings remains constant ($\Delta S_{universe} = 0$).
367. Entropy change of the universe in an irreversible process is:
ⓐ. Positive
ⓑ. Negative
ⓒ. Zero
ⓓ. Infinite
Correct Answer: Positive
Explanation: Irreversible processes always generate entropy. Hence, $\Delta S_{universe} > 0$, which aligns with the Second Law of Thermodynamics.
368. Which of the following indicates irreversibility in real processes?
ⓐ. Friction in piston movement
ⓑ. Viscous dissipation in fluid flow
ⓒ. Heat transfer across finite temperature difference
ⓓ. All of the above
Correct Answer: All of the above
Explanation: All listed phenomena generate entropy, making the processes irreversible. These cannot be fully undone without external work.
369. A gas expands isothermally from 2 L to 6 L against a piston. If expansion is done reversibly, the work done is:
ⓐ. Greater than in free expansion
ⓑ. Equal to free expansion
ⓒ. Zero
ⓓ. Less than free expansion
Correct Answer: Greater than in free expansion
Explanation: In free expansion, $W=0$. In reversible isothermal expansion, work is maximum:$$W = nRT \ln \frac{V_2}{V_1}.$$
Thus, reversible expansion gives more useful work.
370. Why are all natural processes considered irreversible?
ⓐ. They can be reversed without work input
ⓑ. They generate entropy due to irreversibilities like friction and finite gradients
ⓒ. They always conserve entropy
ⓓ. They are infinitely slow
Correct Answer: They generate entropy due to irreversibilities like friction and finite gradients
Explanation: All real processes (e.g., combustion, mixing, conduction, expansion) involve entropy generation. This makes them irreversible, although they may be modeled as reversible approximations for analysis.
371. Carnot’s theorem states that:
ⓐ. All heat engines are equally efficient
ⓑ. No real engine can exceed the efficiency of a reversible engine working between the same two reservoirs
ⓒ. Efficiency depends only on the working substance
ⓓ. Efficiency of a heat engine can be 100%
Correct Answer: No real engine can exceed the efficiency of a reversible engine working between the same two reservoirs
Explanation: Carnot’s theorem sets the upper limit of efficiency for all heat engines. This maximum depends only on the reservoir temperatures, not the working substance.
372. The efficiency of a reversible heat engine working between two reservoirs depends on:
ⓐ. Pressure and volume of the working substance
ⓑ. Type of fuel used
ⓒ. Temperature of the source and sink only
ⓓ. Molecular weight of the gas
Correct Answer: Temperature of the source and sink only
Explanation: Carnot efficiency is given by $\eta = 1 – \frac{T_C}{T_H}$. It depends only on the absolute temperatures of hot and cold reservoirs.
373. If two Carnot engines operate between the same reservoirs, then:
ⓐ. Their efficiencies may differ
ⓑ. Their efficiencies are equal regardless of working substance
ⓒ. One may be more efficient than the other
ⓓ. Both efficiencies depend on cycle path
Correct Answer: Their efficiencies are equal regardless of working substance
Explanation: Carnot engines are ideal reversible systems, and their efficiency depends only on $T_H$ and $T_C$. Thus, different Carnot engines working between the same reservoirs will have the same efficiency.
374. A reversible Carnot engine has efficiency 40% when operating between reservoirs at 500 K and $T_C$. Find the sink temperature.
ⓐ. 200 K
ⓑ. 250 K
ⓒ. 300 K
ⓓ. 350 K
Correct Answer: 300 K
Explanation: $$\eta = 1 – \frac{T_C}{T_H} \Rightarrow 0.4 = 1 – \frac{T_C}{500}.$$
$$\frac{T_C}{500} = 0.6 \Rightarrow T_C = 300 \, K.$$
375. A reversible engine rejects 600 J of heat to a sink at 300 K. If its efficiency is 40%, what is the temperature of the source?
ⓐ. 400 K
ⓑ. 450 K
ⓒ. 500 K
ⓓ. 600 K
Correct Answer: 500 K
Explanation: Efficiency = $\eta = 1 – \frac{T_C}{T_H}$.$$0.4 = 1 – \frac{300}{T_H} \Rightarrow \frac{300}{T_H} = 0.6 \Rightarrow T_H = 500 \, K.$$
376. Which of the following is the reversibility criterion for a thermodynamic process?
ⓐ. The process must occur very fast
ⓑ. No entropy generation in the system or surroundings
ⓒ. Heat must flow across a large temperature difference
ⓓ. Mechanical friction must be present
Correct Answer: No entropy generation in the system or surroundings
Explanation: For a process to be reversible, it must not generate entropy. This requires infinitely slow (quasi-static) operation and absence of dissipative effects like friction or turbulence.
377. The efficiency of a Carnot engine operating between 1000 K and 400 K is:
ⓐ. 40%
ⓑ. 50%
ⓒ. 60%
ⓓ. 70%
Correct Answer: 60%
Explanation: $$\eta = 1 – \frac{T_C}{T_H} = 1 – \frac{400}{1000} = 1 – 0.4 = 0.6 = 60\%.$$
378. Why is the Carnot engine considered a standard of comparison for real engines?
ⓐ. It has the highest possible efficiency
ⓑ. It does not reject heat to a sink
ⓒ. It works without a working substance
ⓓ. It uses infinite fuel
Correct Answer: It has the highest possible efficiency
Explanation: Carnot’s cycle represents an ideal, fully reversible cycle. No real engine can exceed its efficiency; hence, it serves as the benchmark.
379. Which statement about reversibility is correct?
ⓐ. All natural processes are reversible
ⓑ. Reversible processes are idealizations used for analysis
ⓒ. Reversible processes occur in finite time
ⓓ. Reversible processes increase entropy of the universe
Correct Answer: Reversible processes are idealizations used for analysis
Explanation: Perfectly reversible processes are not found in nature. They are hypothetical constructs used to set performance limits and simplify analysis.
380. A Carnot engine absorbs 1000 J of heat from a reservoir at 600 K and rejects heat to a sink at 300 K. How much work is produced?
ⓐ. 400 J
ⓑ. 500 J
ⓒ. 600 J
ⓓ. 700 J
Correct Answer: 500 J
Explanation: Efficiency = $1 – \frac{300}{600} = 0.5$.$$W = \eta Q_H = 0.5 \times 1000 = 500 \, J.$$
381. Why can no real engine achieve Carnot efficiency?
ⓐ. Because the Carnot cycle is not mathematically possible
ⓑ. Because real processes always involve irreversibilities like friction, turbulence, and finite temperature differences
ⓒ. Because heat cannot be rejected in practice
ⓓ. Because entropy does not exist in real systems
Correct Answer: Because real processes always involve irreversibilities like friction, turbulence, and finite temperature differences
Explanation: Real engines cannot eliminate irreversibilities, so their efficiency is always less than Carnot’s theoretical maximum.
382. In practice, why must a heat engine always reject some heat to a sink?
ⓐ. Because energy conservation demands it
ⓑ. Because the Second Law requires entropy disposal to surroundings
ⓒ. Because the First Law forbids work production
ⓓ. Because heat cannot be stored in fuels
Correct Answer: Because the Second Law requires entropy disposal to surroundings
Explanation: Heat rejection to a cold sink is necessary to satisfy entropy balance. This unavoidable rejection limits engine efficiency.
383. Which of the following factors reduces efficiency of real thermodynamic systems?
ⓐ. Mechanical friction
ⓑ. Heat transfer across finite temperature differences
ⓒ. Non-ideal compression and expansion of gases
ⓓ. All of the above
Correct Answer: All of the above
Explanation: All real systems experience entropy generation due to these factors. Minimizing them improves system performance but cannot eliminate them entirely.
384. A steam turbine receives steam at 500 °C and exhausts it at 40 °C. If Carnot efficiency is 61%, the actual turbine efficiency is usually:
ⓐ. Equal to 61%
ⓑ. More than 61%
ⓒ. Less than 61%
ⓓ. 100%
Correct Answer: Less than 61%
Explanation: Real turbines operate below Carnot efficiency due to irreversibilities like friction and heat losses.
385. In refrigeration systems, the throttling process (expansion valve) is considered irreversible because:
ⓐ. It involves a sudden pressure drop with entropy increase
ⓑ. It absorbs no heat
ⓒ. It rejects heat to the environment
ⓓ. It decreases the refrigerant enthalpy
Correct Answer: It involves a sudden pressure drop with entropy increase
Explanation: In throttling, pressure drops abruptly without work extraction, generating entropy. It is inherently irreversible.
386. Why is entropy analysis important in practical power cycles?
ⓐ. It replaces First Law calculations
ⓑ. It shows the quality and availability of energy and identifies losses
ⓒ. It is used only to measure fuel weight
ⓓ. It ensures zero irreversibility in practice
Correct Answer: It shows the quality and availability of energy and identifies losses
Explanation: Entropy analysis helps locate irreversibilities in turbines, compressors, and heat exchangers. This guides engineers in improving efficiency.
387. Which of the following processes is practically irreversible?
ⓐ. Isothermal expansion in small steps
ⓑ. Mixing of two gases
ⓒ. Infinitesimal reversible heat transfer
ⓓ. Reversible adiabatic compression
Correct Answer: Mixing of two gases
Explanation: Gas mixing increases entropy irreversibly and cannot be reversed without external work.
388. In a thermal power plant, why is condenser pressure kept very low?
ⓐ. To increase pump work
ⓑ. To reduce turbine output
ⓒ. To maximize work output by increasing enthalpy drop across turbine
ⓓ. To reduce boiler efficiency
Correct Answer: To maximize work output by increasing enthalpy drop across turbine
Explanation: A lower condenser pressure means more expansion in the turbine, which increases the work output. However, this also requires a well-designed condenser to maintain vacuum.
389. In practice, why can’t refrigerators or heat pumps achieve infinite COP as predicted when $T_H \approx T_C$?
ⓐ. Because First Law does not apply
ⓑ. Because mechanical and thermal losses exist in real systems
ⓒ. Because entropy is always zero
ⓓ. Because refrigerants cannot absorb heat
Correct Answer: Because mechanical and thermal losses exist in real systems
Explanation: Though theoretically COP tends to infinity when the temperature difference is very small, real systems suffer from friction, leakage, and non-ideal compression, limiting performance.
390. The practical implication of entropy increase in real thermodynamic systems is:
ⓐ. Energy is destroyed completely
ⓑ. Part of energy becomes unavailable for useful work
ⓒ. Energy becomes infinite
ⓓ. Energy efficiency always reaches 100%
Correct Answer: Part of energy becomes unavailable for useful work
Explanation: Entropy increase means some energy is degraded to low-grade heat that cannot be converted into work. This explains why practical systems cannot achieve perfect efficiency.
391. The Carnot engine is based on the principle of:
ⓐ. Conservation of linear momentum
ⓑ. A fully reversible cycle between two heat reservoirs
ⓒ. Direct conversion of all heat into work
ⓓ. Constant pressure heating
Correct Answer: A fully reversible cycle between two heat reservoirs
Explanation: The Carnot engine is an idealized system that works reversibly between a hot source and a cold sink. It absorbs heat from the hot reservoir, performs work, and rejects some heat to the cold reservoir while maintaining reversibility.
392. The Carnot cycle consists of which processes?
ⓐ. Two isobaric and two isochoric
ⓑ. Two isothermal and two adiabatic
ⓒ. Four adiabatic only
ⓓ. Two isothermal and two isobaric
Correct Answer: Two isothermal and two adiabatic
Explanation: The Carnot cycle has four processes: isothermal expansion at $T_H$, adiabatic expansion, isothermal compression at $T_C$, and adiabatic compression. This combination makes it reversible and maximizes efficiency.
393. Why is the Carnot engine considered an “ideal engine”?
ⓐ. Because it achieves 100% efficiency
ⓑ. Because it ignores entropy
ⓒ. Because it is fully reversible and has the highest possible efficiency
ⓓ. Because it does not use any working substance
Correct Answer: Because it is fully reversible and has the highest possible efficiency
Explanation: The Carnot engine is an idealized standard of comparison. Its efficiency depends only on reservoir temperatures, making it the theoretical upper limit for all real engines.
394. The working substance in a Carnot engine can be:
ⓐ. Only steam
ⓑ. Only air
ⓒ. Any suitable fluid (gas or vapor)
ⓓ. Only liquids at high pressure
Correct Answer: Any suitable fluid (gas or vapor)
Explanation: The Carnot efficiency does not depend on the working substance. It can be air, steam, or any gas, because efficiency depends only on $T_H$ and $T_C$.
395. In the Carnot cycle, work is produced during:
ⓐ. Isothermal expansion and adiabatic expansion
ⓑ. Isothermal compression only
ⓒ. Adiabatic compression only
ⓓ. Isothermal compression and adiabatic compression
Correct Answer: Isothermal expansion and adiabatic expansion
Explanation: The system performs work on surroundings during expansion phases. Compression processes require input work, but expansions deliver net positive work.
396. Heat is rejected in the Carnot cycle during:
ⓐ. Isothermal compression at $T_C$
ⓑ. Adiabatic compression
ⓒ. Isothermal expansion at $T_H$
ⓓ. Adiabatic expansion
Correct Answer: Isothermal compression at $T_C$
Explanation: At cold reservoir temperature $T_C$, the working substance undergoes isothermal compression and releases heat $Q_C$ to the sink.
397. The efficiency of the Carnot engine is independent of:
ⓐ. Reservoir temperatures
ⓑ. Working substance
ⓒ. Heat absorbed
ⓓ. Heat rejected
Correct Answer: Working substance
Explanation: The efficiency is determined solely by $T_H$ and $T_C$. Whether the working fluid is steam, helium, or air, efficiency remains the same for the same reservoir temperatures.
398. A Carnot engine operates between 600 K and 300 K. If it absorbs 1200 J of heat, how much work does it deliver?
ⓐ. 400 J
ⓑ. 500 J
ⓒ. 600 J
ⓓ. 700 J
Correct Answer: 600 J
Explanation: Efficiency = $1 – \frac{T_C}{T_H} = 1 – \frac{300}{600} = 0.5$.$$W = \eta Q_H = 0.5 \times 1200 = 600 \, J.$$
399. Why is the Carnot engine considered idealized?
ⓐ. It needs no fuel
ⓑ. It assumes no irreversibility, no friction, and infinitesimal temperature differences for heat transfer
ⓒ. It rejects no heat to sink
ⓓ. It runs in finite time
Correct Answer: It assumes no irreversibility, no friction, and infinitesimal temperature differences for heat transfer
Explanation: The Carnot cycle is hypothetical. Real systems always have losses. The idealization is useful for setting performance limits but cannot be achieved in practice.
400. The significance of the Carnot engine in thermodynamics is that:
ⓐ. It replaces all real engines
ⓑ. It proves the Zeroth Law
ⓒ. It sets the theoretical maximum limit of efficiency between two reservoirs
ⓓ. It eliminates the need for entropy
Correct Answer: It sets the theoretical maximum limit of efficiency between two reservoirs
Explanation: Carnot’s theorem states no engine operating between two reservoirs can exceed the efficiency of a Carnot engine. Thus, it defines the benchmark efficiency for practical systems.
This section features Class 11 Physics MCQs – Chapter 12: Thermodynamics (Part 4). Here, we focus on the concept of entropy, spontaneous processes, statistical interpretation of entropy, and entropy changes in adiabatic and isothermal processes. These topics are not only vital for the NCERT/CBSE syllabus but also serve as a strong base for competitive exams such as JEE, NEET, and state-level engineering/medical tests. Out of a total of 500 MCQs with solutions, this section provides another 100 solved questions for in-depth practice and understanding.
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