Thermodynamics Quiz: Heat Reservoirs And Entropy
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Heat Reservoirs And EntropyQuestion 1 of 20

A heat engine operates in a cycle between two reservoirs. The hot reservoir temperature decreases from TH1=500 KT_{H1} = 500 \text{ K} to TH2=480 KT_{H2} = 480 \text{ K}, while the cold reservoir temperature increases from TC1=300 KT_{C1} = 300 \text{ K} to TC2=310 KT_{C2} = 310 \text{ K}. Both reservoirs have heat capacity C=2500 J/KC = 2500 \text{ J/K}. What is the total entropy change?

ΔStotal=2500ln⁡(0.96)+2500ln⁡(1.033) J/K\Delta S_{total} = 2500 \ln(0.96) + 2500 \ln(1.033) \text{ J/K}
ΔStotal=2500ln⁡(1.033)−2500ln⁡(0.96) J/K\Delta S_{total} = 2500 \ln(1.033) - 2500 \ln(0.96) \text{ J/K}
ΔStotal=2500ln⁡(480/500)+2500ln⁡(310/300) J/K\Delta S_{total} = 2500 \ln(480/500) + 2500 \ln(310/300) \text{ J/K}
ΔStotal=2500ln⁡(310/300)−2500ln⁡(480/500) J/K\Delta S_{total} = 2500 \ln(310/300) - 2500 \ln(480/500) \text{ J/K}
ΔStotal=2500ln⁡(810/800) J/K\Delta S_{total} = 2500 \ln(810/800) \text{ J/K}
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Thermodynamics Quiz

Thermodynamics Quiz: Heat Reservoirs And Entropy

Practice Heat Reservoirs And Entropy in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Heat Reservoirs And Entropy, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

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Question 1

A heat engine operates in a cycle between two reservoirs. The hot reservoir temperature decreases from TH1=500 KT_{H1} = 500 \text{ K} to TH2=480 KT_{H2} = 480 \text{ K}, while the cold reservoir temperature increases from TC1=300 KT_{C1} = 300 \text{ K} to TC2=310 KT_{C2} = 310 \text{ K}. Both reservoirs have heat capacity C=2500 J/KC = 2500 \text{ J/K}. What is the total entropy change?

  1. ΔStotal=2500ln⁡(0.96)+2500ln⁡(1.033) J/K\Delta S_{total} = 2500 \ln(0.96) + 2500 \ln(1.033) \text{ J/K}
  2. ΔStotal=2500ln⁡(1.033)−2500ln⁡(0.96) J/K\Delta S_{total} = 2500 \ln(1.033) - 2500 \ln(0.96) \text{ J/K}
  3. ΔStotal=2500ln⁡(480/500)+2500ln⁡(310/300) J/K\Delta S_{total} = 2500 \ln(480/500) + 2500 \ln(310/300) \text{ J/K} (correct answer)
  4. ΔStotal=2500ln⁡(310/300)−2500ln⁡(480/500) J/K\Delta S_{total} = 2500 \ln(310/300) - 2500 \ln(480/500) \text{ J/K}
  5. ΔStotal=2500ln⁡(810/800) J/K\Delta S_{total} = 2500 \ln(810/800) \text{ J/K}
Explanation: When analyzing entropy changes in thermodynamic systems, you need to calculate the entropy change for each component separately, then sum them to find the total change. Entropy change for a system with constant heat capacity is given by ΔS=Cln⁡(Tf/Ti)\Delta S = C \ln(T_f/T_i). For this problem, you have two reservoirs changing temperature independently. The hot reservoir cools from 500 K to 480 K, so its entropy change is ΔSH=Cln⁡(TH2/TH1)=2500ln⁡(480/500)\Delta S_H = C \ln(T_{H2}/T_{H1}) = 2500 \ln(480/500). The cold reservoir warms from 300 K to 310 K, giving ΔSC=Cln⁡(TC2/TC1)=2500ln⁡(310/300)\Delta S_C = C \ln(T_{C2}/T_{C1}) = 2500 \ln(310/300). The total entropy change is simply the sum: ΔStotal=2500ln⁡(480/500)+2500ln⁡(310/300)\Delta S_{total} = 2500 \ln(480/500) + 2500 \ln(310/300). Answer C correctly shows this calculation using the proper temperature ratios in the logarithmic terms. Answer A uses decimal approximations (0.96 and 1.033) instead of the exact fractional forms, which is unnecessarily imprecise and doesn't show the underlying physics clearly. Answer B incorrectly subtracts the hot reservoir term from the cold reservoir term, suggesting a misunderstanding that entropy changes should somehow cancel rather than add. Answer D makes the same subtraction error as B, treating this like a heat engine efficiency calculation rather than a straightforward entropy accounting problem. Remember: entropy is a state function, so total entropy change is always the algebraic sum of individual entropy changes. Don't confuse this with efficiency calculations that involve differences.

Question 2

A refrigerator removes QC=400 JQ_C = 400 \text{ J} from a cold reservoir at TC=250 KT_C = 250 \text{ K} and rejects QH=500 JQ_H = 500 \text{ J} to a hot reservoir at TH=300 KT_H = 300 \text{ K}. If this refrigerator operates reversibly, what constraint must be satisfied?

  1. The entropy change of the cold reservoir equals the entropy change of the hot reservoir in magnitude
  2. The entropy change of the universe must be zero for the complete cycle (correct answer)
  3. The work input must equal the difference between heat rejected and heat absorbed
  4. The coefficient of performance must be maximized for the given temperature limits
  5. The entropy production rate must be minimized but not necessarily zero
Explanation: When analyzing refrigerators and heat engines, the key thermodynamic principle is that reversible processes produce zero entropy change for the universe. This is the hallmark of thermodynamic reversibility. For any reversible cycle, the total entropy change of the universe (system plus surroundings) must equal zero. Let's verify this with the given data. The entropy change of the cold reservoir is ΔSC=+QC/TC=+400/250=+1.6 J/K\Delta S_C = +Q_C/T_C = +400/250 = +1.6 \text{ J/K} (positive because heat is removed). The entropy change of the hot reservoir is ΔSH=−QH/TH=−500/300=−1.67 J/K\Delta S_H = -Q_H/T_H = -500/300 = -1.67 \text{ J/K} (negative because heat is added). The total is approximately zero within rounding, confirming reversibility. This makes B correct. A is wrong because the entropy changes don't need to be equal in magnitude—they need to sum to zero. Here, ∣ΔSC∣≠∣ΔSH∣|\Delta S_C| \neq |\Delta S_H|. C is wrong because this describes energy conservation (first law), not reversibility. While true for any refrigerator (W=QH−QC=500−400=100 JW = Q_H - Q_C = 500 - 400 = 100 \text{ J}), it doesn't address the reversibility constraint. D is wrong because maximizing coefficient of performance is an optimization goal, not a constraint for reversibility. A reversible refrigerator operating between these temperatures will have the maximum possible COP, but the constraint itself is zero entropy generation. Study tip: For reversible processes, always check that the total entropy change of the universe equals zero—this is your definitive test for thermodynamic reversibility.

Question 3

A Carnot heat pump operates between reservoirs at TC=280 KT_C = 280 \text{ K} and TH=320 KT_H = 320 \text{ K}. If the entropy decrease of the cold reservoir is ΔSC=−2.0 J/K\Delta S_C = -2.0 \text{ J/K}, what is the entropy change of the hot reservoir?

  1. ΔSH=1.75 J/K\Delta S_H = 1.75 \text{ J/K}
  2. ΔSH=2.0 J/K\Delta S_H = 2.0 \text{ J/K} (correct answer)
  3. ΔSH=2.29 J/K\Delta S_H = 2.29 \text{ J/K}
  4. ΔSH=−1.75 J/K\Delta S_H = -1.75 \text{ J/K}
  5. ΔSH=−2.29 J/K\Delta S_H = -2.29 \text{ J/K}
Explanation: When you encounter Carnot heat pump problems, remember that these idealized devices operate reversibly between two thermal reservoirs, meaning the total entropy change of the universe must be zero. For a Carnot heat pump, the key relationship comes from the reversible heat transfer process. Since entropy change equals heat transfer divided by temperature (ΔS=Q/T\Delta S = Q/T), and the same amount of heat QCQ_C is extracted from the cold reservoir and QHQ_H is delivered to the hot reservoir, we can write: QC=∣ΔSC∣×TC=2.0×280=560 JQ_C = |\Delta S_C| \times T_C = 2.0 \times 280 = 560 \text{ J} For a Carnot cycle, the heat transfers are related by: QHTH=QCTC\frac{Q_H}{T_H} = \frac{Q_C}{T_C} Therefore: QH=QC×THTC=560×320280=640 JQ_H = Q_C \times \frac{T_H}{T_C} = 560 \times \frac{320}{280} = 640 \text{ J} The entropy change of the hot reservoir is: ΔSH=QHTH=640320=2.0 J/K\Delta S_H = \frac{Q_H}{T_H} = \frac{640}{320} = 2.0 \text{ J/K} Answer B (ΔSH=2.0 J/K\Delta S_H = 2.0 \text{ J/K}) is correct. Answer A (ΔSH=1.75 J/K\Delta S_H = 1.75 \text{ J/K}) incorrectly assumes the entropy changes have the same magnitude regardless of temperature differences. Answer C (ΔSH=2.29 J/K\Delta S_H = 2.29 \text{ J/K}) likely results from incorrectly using the temperature ratio TH/TCT_H/T_C applied directly to the entropy change. Answer D (ΔSH=−1.75 J/K\Delta S_H = -1.75 \text{ J/K}) confuses the signs—the hot reservoir gains heat, so its entropy increases. Study tip: For Carnot processes, always verify that the total entropy change of the universe equals zero: ΔStotal=ΔSH+ΔSC=0\Delta S_{total} = \Delta S_H + \Delta S_C = 0.

Question 4

A thermal reservoir maintains constant temperature T=400 KT = 400 \text{ K} while supplying heat to multiple systems simultaneously. If the reservoir provides Q1=800 JQ_1 = 800 \text{ J} to system 1, Q2=1200 JQ_2 = 1200 \text{ J} to system 2, and Q3=400 JQ_3 = 400 \text{ J} to system 3, what is the entropy change of the reservoir?

  1. ΔSreservoir=−6.0 J/K\Delta S_{reservoir} = -6.0 \text{ J/K} (correct answer)
  2. ΔSreservoir=−4.0 J/K\Delta S_{reservoir} = -4.0 \text{ J/K}
  3. ΔSreservoir=−2.0 J/K\Delta S_{reservoir} = -2.0 \text{ J/K}
  4. ΔSreservoir=+6.0 J/K\Delta S_{reservoir} = +6.0 \text{ J/K}
  5. ΔSreservoir=0 J/K\Delta S_{reservoir} = 0 \text{ J/K}
Explanation: When you encounter problems involving thermal reservoirs and entropy changes, remember that a reservoir's defining characteristic is its constant temperature, even while exchanging heat. This makes entropy calculations straightforward using the relationship ΔS=QT\Delta S = \frac{Q}{T}. Since the reservoir supplies heat to three systems, you need to find the total heat it provides: Qtotal=Q1+Q2+Q3=800+1200+400=2400 JQ_{total} = Q_1 + Q_2 + Q_3 = 800 + 1200 + 400 = 2400 \text{ J}. Because the reservoir loses this heat (it flows out to the systems), Qreservoir=−2400 JQ_{reservoir} = -2400 \text{ J}. The entropy change of the reservoir is: ΔSreservoir=QreservoirT=−2400 J400 K=−6.0 J/K\Delta S_{reservoir} = \frac{Q_{reservoir}}{T} = \frac{-2400 \text{ J}}{400 \text{ K}} = -6.0 \text{ J/K} This confirms answer A is correct. The negative sign makes physical sense—when a system loses heat, its entropy decreases. Looking at the wrong answers: B gives −4.0 J/K-4.0 \text{ J/K}, which would result from incorrectly using only 1600 J instead of the full 2400 J. C gives −2.0 J/K-2.0 \text{ J/K}, suggesting someone used only 800 J, perhaps just Q1Q_1. D gives +6.0 J/K+6.0 \text{ J/K}, which has the right magnitude but wrong sign—this comes from forgetting that heat leaves the reservoir, making QreservoirQ_{reservoir} negative. Study tip: Always track the direction of heat flow carefully. Heat leaving a system means negative Q for that system, leading to negative entropy change. The reservoir loses heat and entropy while the systems it heats gain both.

Question 5

A refrigeration cycle operates between reservoirs at TH=300 KT_H = 300 \text{ K} and TC=250 KT_C = 250 \text{ K}. The refrigerator removes QC=500 JQ_C = 500 \text{ J} from the cold space and requires work input W=120 JW = 120 \text{ J}. What is the entropy generation for this cycle?

  1. Sgen=0.067 J/KS_{gen} = 0.067 \text{ J/K} (correct answer)
  2. Sgen=0.033 J/KS_{gen} = 0.033 \text{ J/K}
  3. Sgen=−0.033 J/KS_{gen} = -0.033 \text{ J/K}
  4. Sgen=0.133 J/KS_{gen} = 0.133 \text{ J/K}
  5. Sgen=0 J/KS_{gen} = 0 \text{ J/K}
Explanation: When you encounter refrigeration cycle problems involving entropy generation, you're dealing with the second law of thermodynamics and irreversibility. Entropy generation quantifies how much a real process deviates from ideal reversible operation. To find entropy generation, you need to apply the entropy balance to the entire system. First, determine the heat rejected to the hot reservoir using energy conservation: QH=QC+W=500+120=620 JQ_H = Q_C + W = 500 + 120 = 620 \text{ J}. The entropy generation equals the total entropy change of the universe: Sgen=ΔShot+ΔScoldS_{gen} = \Delta S_{hot} + \Delta S_{cold}. The hot reservoir gains entropy: ΔShot=QHTH=620300=2.067 J/K\Delta S_{hot} = \frac{Q_H}{T_H} = \frac{620}{300} = 2.067 \text{ J/K}. The cold reservoir loses entropy: ΔScold=−QCTC=−500250=−2.000 J/K\Delta S_{cold} = -\frac{Q_C}{T_C} = -\frac{500}{250} = -2.000 \text{ J/K}. Therefore: Sgen=2.067−2.000=0.067 J/KS_{gen} = 2.067 - 2.000 = 0.067 \text{ J/K}, confirming answer A. Answer B (0.033 J/K) likely comes from calculation errors or using incorrect temperature values. Answer C (-0.033 J/K) represents a fundamental misunderstanding—entropy generation can never be negative, as this would violate the second law. Answer D (0.133 J/K) probably results from incorrectly doubling the actual value or miscalculating the heat flows. Remember: entropy generation is always positive for real processes and equals zero only for ideal reversible cycles. Always check that your final answer respects this fundamental thermodynamic principle.

Question 6

A thermal reservoir at TR=450 KT_R = 450 \text{ K} is used to heat a block of metal with heat capacity C=800 J/KC = 800 \text{ J/K} from Ti=300 KT_i = 300 \text{ K} to Tf=400 KT_f = 400 \text{ K}. If the heat transfer occurs through direct contact (irreversible), what is the entropy change of the reservoir?

  1. ΔSR=−800×100450 J/K\Delta S_R = -\frac{800 \times 100}{450} \text{ J/K} (correct answer)
  2. ΔSR=−800ln⁡(4/3) J/K\Delta S_R = -800 \ln(4/3) \text{ J/K}
  3. ΔSR=−800×100350 J/K\Delta S_R = -\frac{800 \times 100}{350} \text{ J/K}
  4. ΔSR=+800×100450 J/K\Delta S_R = +\frac{800 \times 100}{450} \text{ J/K}
  5. ΔSR=800ln⁡(4/3) J/K\Delta S_R = 800 \ln(4/3) \text{ J/K}
Explanation: When analyzing heat transfer between a thermal reservoir and an object, you need to carefully track energy flow and apply the correct temperature for each system's entropy calculation. Since the metal block is heated from 300 K to 400 K, it gains energy: Qblock=C(Tf−Ti)=800×(400−300)=80,000 JQ_{block} = C(T_f - T_i) = 800 \times (400 - 300) = 80,000 \text{ J}. By conservation of energy, the reservoir must lose exactly this amount of heat: Qreservoir=−80,000 JQ_{reservoir} = -80,000 \text{ J}. For the reservoir's entropy change, you use the reservoir's constant temperature (450 K) since it's large enough that its temperature doesn't change during the process: ΔSR=QreservoirTR=−80,000450=−800×100450 J/K\Delta S_R = \frac{Q_{reservoir}}{T_R} = \frac{-80,000}{450} = -\frac{800 \times 100}{450} \text{ J/K}. This confirms answer A is correct. Answer B (−800ln⁡(4/3)-800 \ln(4/3)) incorrectly applies the formula for the metal block's entropy change, which involves the logarithm because the block's temperature changes during heating. Answer C uses 350 K, which might seem like an average temperature, but this is wrong—you must use the reservoir's actual temperature of 450 K. Answer D has the correct magnitude but wrong sign; since the reservoir loses heat, its entropy change must be negative. Remember: for thermal reservoirs, entropy change equals heat transferred divided by the reservoir's temperature. The reservoir temperature stays constant, while objects being heated or cooled require integration (often yielding logarithmic terms). Always check your signs—heat leaving a system creates negative entropy change.

Question 7

A Carnot heat pump operates between outdoor air at TC=270 KT_C = 270 \text{ K} and indoor air at TH=295 KT_H = 295 \text{ K}. If the pump delivers QH=5000 JQ_H = 5000 \text{ J} to the warm indoor space, what is the entropy change of the cold outdoor reservoir?

  1. ΔSC=−16.95 J/K\Delta S_C = -16.95 \text{ J/K} (correct answer)
  2. ΔSC=−15.25 J/K\Delta S_C = -15.25 \text{ J/K}
  3. ΔSC=+16.95 J/K\Delta S_C = +16.95 \text{ J/K}
  4. ΔSC=−18.52 J/K\Delta S_C = -18.52 \text{ J/K}
  5. ΔSC=+18.52 J/K\Delta S_C = +18.52 \text{ J/K}
Explanation: When you encounter a Carnot heat pump problem, remember that you're dealing with a reversible cycle where entropy changes in the reservoirs are linked through the fundamental relationship ΔStotal=0\Delta S_{total} = 0 for any reversible process. To find the entropy change of the cold reservoir, you first need to determine how much heat is extracted from it. For a Carnot heat pump, the relationship between heat transfers is QHTH=QCTC\frac{Q_H}{T_H} = \frac{Q_C}{T_C}. Solving for QCQ_C: QC=QH×TCTH=5000×270295=4576 JQ_C = Q_H \times \frac{T_C}{T_H} = 5000 \times \frac{270}{295} = 4576 \text{ J} Since the cold reservoir loses this heat, its entropy change is ΔSC=−QCTC=−4576270=−16.95 J/K\Delta S_C = -\frac{Q_C}{T_C} = -\frac{4576}{270} = -16.95 \text{ J/K} Looking at the wrong answers: Choice B (−15.25 J/K-15.25 \text{ J/K}) likely comes from incorrectly using QHQ_H divided by THT_H instead of finding QCQ_C first. Choice C (+16.95 J/K+16.95 \text{ J/K}) has the right magnitude but wrong sign—this would suggest the cold reservoir gains entropy, which contradicts it losing heat. Choice D (−18.52 J/K-18.52 \text{ J/K}) appears to use QHQ_H directly with TCT_C: −5000/270-5000/270, skipping the crucial step of finding the actual heat extracted. Study tip: Always remember that for Carnot cycles, the entropy changes of the reservoirs are equal and opposite (ΔSC=−ΔSH\Delta S_C = -\Delta S_H), and you must calculate the actual heat transfer to each reservoir using the temperature ratios before finding entropy changes.

Question 8

Two identical objects, each with heat capacity C=1000 J/KC = 1000 \text{ J/K}, are initially at temperatures TA=400 KT_A = 400 \text{ K} and TB=300 KT_B = 300 \text{ K}. They are brought into thermal contact and allowed to reach equilibrium. What is the total entropy change of the system?

  1. ΔStotal=2000ln⁡(350/300) J/K\Delta S_{total} = 2000 \ln(350/300) \text{ J/K}
  2. ΔStotal=1000ln⁡(350/400)+1000ln⁡(350/300) J/K\Delta S_{total} = 1000 \ln(350/400) + 1000 \ln(350/300) \text{ J/K} (correct answer)
  3. ΔStotal=1000ln⁡(400/300) J/K\Delta S_{total} = 1000 \ln(400/300) \text{ J/K}
  4. ΔStotal=2000ln⁡(350/350)=0 J/K\Delta S_{total} = 2000 \ln(350/350) = 0 \text{ J/K}
  5. ΔStotal=1000ln⁡(350/300)−1000ln⁡(350/400) J/K\Delta S_{total} = 1000 \ln(350/300) - 1000 \ln(350/400) \text{ J/K}
Explanation: When two objects at different temperatures reach thermal equilibrium, you need to calculate the entropy change for each object separately, then sum them. This tests your understanding of entropy as a state function and conservation of energy. First, find the final equilibrium temperature. Since the objects are identical with the same heat capacity, energy conservation gives us: C(TA−Tf)=C(TB−Tf)C(T_A - T_f) = C(T_B - T_f), which simplifies to Tf=TA+TB2=400+3002=350 KT_f = \frac{T_A + T_B}{2} = \frac{400 + 300}{2} = 350 \text{ K}. Next, calculate each object's entropy change using ΔS=Cln⁡(Tf/Ti)\Delta S = C \ln(T_f/T_i). For object A: ΔSA=1000ln⁡(350/400)\Delta S_A = 1000 \ln(350/400). For object B: ΔSB=1000ln⁡(350/300)\Delta S_B = 1000 \ln(350/300). The total entropy change is: ΔStotal=1000ln⁡(350/400)+1000ln⁡(350/300)\Delta S_{total} = 1000 \ln(350/400) + 1000 \ln(350/300), which matches answer B. Answer A incorrectly assumes both objects have the same initial temperature of 300 K, ignoring object A's higher starting temperature. Answer C uses ln⁡(400/300)\ln(400/300), which would be the entropy change if object A went directly to object B's initial temperature—this violates energy conservation. Answer D incorrectly suggests the total entropy change is zero, which would only be true for a reversible process, but spontaneous heat transfer between objects at different temperatures is irreversible. Remember: for thermal equilibrium problems, always calculate entropy changes separately for each object using their individual temperature changes, then sum them. The total entropy of an isolated system always increases for irreversible processes.

Question 9

An ideal gas undergoes an isothermal expansion at T=350 KT = 350 \text{ K} while in thermal contact with a reservoir at the same temperature. The gas volume changes from V1=0.02 m3V_1 = 0.02 \text{ m}^3 to V2=0.05 m3V_2 = 0.05 \text{ m}^3, and the amount of gas is n=2 moln = 2 \text{ mol}. What is the entropy change of the reservoir?

  1. ΔSreservoir=−2Rln⁡(2.5) J/K\Delta S_{reservoir} = -2R \ln(2.5) \text{ J/K} (correct answer)
  2. ΔSreservoir=+2Rln⁡(2.5) J/K\Delta S_{reservoir} = +2R \ln(2.5) \text{ J/K}
  3. ΔSreservoir=−2Rln⁡(0.4) J/K\Delta S_{reservoir} = -2R \ln(0.4) \text{ J/K}
  4. ΔSreservoir=0 J/K\Delta S_{reservoir} = 0 \text{ J/K}
  5. ΔSreservoir=−Rln⁡(2.5) J/K\Delta S_{reservoir} = -R \ln(2.5) \text{ J/K}
Explanation: When analyzing entropy changes in thermodynamic processes, you need to consider the entire system, including both the gas and any reservoirs involved. For an isothermal process, the gas exchanges heat with the reservoir to maintain constant temperature. First, let's find the heat absorbed by the gas during this isothermal expansion. For an ideal gas undergoing an isothermal process, Q=nRTln⁡(V2V1)Q = nRT \ln\left(\frac{V_2}{V_1}\right). With n=2 moln = 2 \text{ mol}, T=350 KT = 350 \text{ K}, and the volume ratio V2V1=0.050.02=2.5\frac{V_2}{V_1} = \frac{0.05}{0.02} = 2.5: Qgas=2R(350)ln⁡(2.5)Q_{gas} = 2R(350) \ln(2.5) Since the gas absorbs heat from the reservoir, the reservoir loses this same amount of heat: Qreservoir=−QgasQ_{reservoir} = -Q_{gas}. The entropy change of the reservoir is ΔSreservoir=QreservoirT=−QgasT=−2R(350)ln⁡(2.5)350=−2Rln⁡(2.5)\Delta S_{reservoir} = \frac{Q_{reservoir}}{T} = \frac{-Q_{gas}}{T} = \frac{-2R(350) \ln(2.5)}{350} = -2R \ln(2.5). Answer A is correct: ΔSreservoir=−2Rln⁡(2.5) J/K\Delta S_{reservoir} = -2R \ln(2.5) \text{ J/K} represents the reservoir losing entropy as it provides heat to the expanding gas. Answer B has the wrong sign—it suggests the reservoir gains entropy while losing heat, which violates thermodynamic principles. Answer C uses ln⁡(0.4)\ln(0.4) instead of ln⁡(2.5)\ln(2.5); this comes from using the reciprocal of the correct volume ratio. Answer D incorrectly assumes no entropy change, perhaps confusing the reservoir with the overall universe (where total entropy change would indeed be zero for this reversible process). Remember: in isothermal processes, always track heat flow direction carefully—the reservoir's entropy change has the opposite sign of the gas's entropy change.

Question 10

A thermal reservoir at temperature T1=600 KT_1 = 600 \text{ K} transfers heat to another reservoir initially at T2=300 KT_2 = 300 \text{ K}. Both reservoirs have the same heat capacity C=3000 J/KC = 3000 \text{ J/K}. After heat transfer, the final temperature of the second reservoir is T2f=350 KT_{2f} = 350 \text{ K}. What is the entropy change of the first reservoir?

  1. ΔS1=3000ln⁡(550/600) J/K\Delta S_1 = 3000 \ln(550/600) \text{ J/K} (correct answer)
  2. ΔS1=−3000ln⁡(550/600) J/K\Delta S_1 = -3000 \ln(550/600) \text{ J/K}
  3. ΔS1=3000ln⁡(600/550) J/K\Delta S_1 = 3000 \ln(600/550) \text{ J/K}
  4. ΔS1=−150000600 J/K\Delta S_1 = -\frac{150000}{600} \text{ J/K}
  5. ΔS1=3000ln⁡(350/300) J/K\Delta S_1 = 3000 \ln(350/300) \text{ J/K}
Explanation: When you encounter entropy problems involving heat transfer between reservoirs, you need to track the direction of heat flow and apply the fundamental entropy change formula: ΔS=∫dQT\Delta S = \int \frac{dQ}{T}. First, determine how much heat the first reservoir loses. Since the second reservoir gains heat and reaches 350 K from 300 K: Qgained=CΔT2=3000×(350−300)=150,000 JQ_{gained} = C \Delta T_2 = 3000 \times (350-300) = 150{,}000 \text{ J}. By conservation of energy, the first reservoir loses this same amount of heat. For the first reservoir losing 150,000 J, you can't simply use Q/TQ/T because the temperature changes during the process. Instead, use: ΔS1=Cln⁡(T1fT1i)\Delta S_1 = C \ln\left(\frac{T_{1f}}{T_{1i}}\right), where T1fT_{1f} is the final temperature of reservoir 1. To find T1fT_{1f}: The first reservoir loses heat Q=C(T1i−T1f)=150,000Q = C(T_{1i} - T_{1f}) = 150{,}000, so 3000(600−T1f)=150,0003000(600 - T_{1f}) = 150{,}000. Solving: T1f=600−50=550 KT_{1f} = 600 - 50 = 550 \text{ K}. Therefore: ΔS1=3000ln⁡(550600)\Delta S_1 = 3000 \ln\left(\frac{550}{600}\right), which matches option A. Option B has the wrong sign - it would apply if reservoir 1 gained heat. Option C flips the fraction, giving a positive entropy change when reservoir 1 actually cools down. Option D incorrectly uses the simple Q/TQ/T formula, ignoring that temperature changes during the process. Study tip: For finite reservoirs, always use ΔS=Cln⁡(Tf/Ti)\Delta S = C \ln(T_f/T_i) rather than Q/TQ/T. The latter only applies to infinite reservoirs at constant temperature.

Question 11

A heat engine receives QH1=800 JQ_{H1} = 800 \text{ J} from a reservoir at TH1=500 KT_{H1} = 500 \text{ K} and QH2=600 JQ_{H2} = 600 \text{ J} from another reservoir at TH2=400 KT_{H2} = 400 \text{ K}. The engine rejects QC=900 JQ_C = 900 \text{ J} to a cold reservoir at TC=300 KT_C = 300 \text{ K}. What is the total entropy change of all reservoirs?

  1. ΔStotal=1.60−1.50+3.00=3.10 J/K\Delta S_{total} = 1.60 - 1.50 + 3.00 = 3.10 \text{ J/K}
  2. ΔStotal=−1.60−1.50+3.00=−0.10 J/K\Delta S_{total} = -1.60 - 1.50 + 3.00 = -0.10 \text{ J/K} (correct answer)
  3. ΔStotal=−1.60−1.50−3.00=−6.10 J/K\Delta S_{total} = -1.60 - 1.50 - 3.00 = -6.10 \text{ J/K}
  4. ΔStotal=1.60+1.50−3.00=0.10 J/K\Delta S_{total} = 1.60 + 1.50 - 3.00 = 0.10 \text{ J/K}
  5. ΔStotal=0 J/K\Delta S_{total} = 0 \text{ J/K}
Explanation: When analyzing entropy changes in thermodynamic systems, you need to carefully track the direction of heat flow and apply the correct signs. For any reservoir, the entropy change is ΔS=Q/T\Delta S = Q/T, where QQ is positive when heat enters the reservoir and negative when heat leaves. Let's calculate each reservoir's entropy change. The hot reservoirs lose heat to the engine, so their entropy changes are negative: ΔSH1=−QH1/TH1=−800/500=−1.60 J/K\Delta S_{H1} = -Q_{H1}/T_{H1} = -800/500 = -1.60 \text{ J/K} and ΔSH2=−QH2/TH2=−600/400=−1.50 J/K\Delta S_{H2} = -Q_{H2}/T_{H2} = -600/400 = -1.50 \text{ J/K}. The cold reservoir gains heat from the engine, so its entropy change is positive: ΔSC=+QC/TC=+900/300=+3.00 J/K\Delta S_C = +Q_C/T_C = +900/300 = +3.00 \text{ J/K}. The total entropy change is: ΔStotal=−1.60−1.50+3.00=−0.10 J/K\Delta S_{total} = -1.60 - 1.50 + 3.00 = -0.10 \text{ J/K}, confirming answer B. Answer A incorrectly uses positive signs for both hot reservoirs, suggesting they gain entropy when they actually lose heat. Answer C uses negative signs for all three reservoirs, incorrectly treating the cold reservoir as losing heat when it actually receives rejected heat from the engine. Answer D uses positive signs for the hot reservoirs and a negative sign for the cold reservoir—completely backwards from the actual heat flow directions. Remember: hot reservoirs always lose entropy (negative ΔS\Delta S) and cold reservoirs always gain entropy (positive ΔS\Delta S) in heat engine problems. The sign of the total entropy change tells you whether the process is thermodynamically feasible.

Question 12

An irreversible heat engine operates between two finite thermal reservoirs. Initially, the hot reservoir is at TH,i=500 KT_{H,i} = 500 \text{ K} and the cold reservoir is at TC,i=300 KT_{C,i} = 300 \text{ K}. After the engine operates, the hot reservoir temperature drops to TH,f=450 KT_{H,f} = 450 \text{ K}. Which statement about entropy changes is correct?

  1. The entropy decrease of the hot reservoir equals the entropy increase of the cold reservoir plus surroundings
  2. The total entropy change is positive regardless of the cold reservoir's final temperature (correct answer)
  3. The entropy change can be negative if the engine efficiency is sufficiently high
  4. The net entropy change equals zero if the process is quasi-static and slow
  5. The entropy change depends only on the temperature difference between initial and final states
Explanation: When analyzing irreversible heat engines operating between finite reservoirs, you're dealing with the Second Law of Thermodynamics and entropy generation. The key insight is that irreversible processes always create entropy, making the total entropy change of the universe positive. For this irreversible heat engine, entropy is generated through two mechanisms: the irreversible engine operation itself and the finite temperature differences during heat transfer. Even if you could calculate the exact entropy changes of both reservoirs, the irreversible nature of the engine guarantees additional entropy production beyond what occurs in the reservoirs alone. Answer B is correct because the Second Law demands that any irreversible process increases the total entropy of the universe. The magnitude of this increase depends on the degree of irreversibility and the specific heat transfer processes, but it's always positive regardless of the cold reservoir's final temperature. Answer A is wrong because it assumes entropy is only redistributed between reservoirs, ignoring the entropy generated by the irreversible engine itself. Answer C reflects a fundamental misunderstanding—no amount of efficiency can make an irreversible process decrease total entropy, as this would violate the Second Law. Answer D incorrectly suggests that slow operation makes a process reversible; while quasi-static processes can approach reversibility, the problem explicitly states the engine is irreversible. Remember: whenever you see "irreversible" in thermodynamics problems, immediately think "positive entropy change." The Second Law is absolute—irreversible processes always increase universal entropy, regardless of other system parameters.

Question 13

Two thermal reservoirs at T1=600 KT_1 = 600 \text{ K} and T2=400 KT_2 = 400 \text{ K} are connected by a reversible heat engine. The engine absorbs QH=1800 JQ_H = 1800 \text{ J} from the hot reservoir. What amount of heat must be rejected to the cold reservoir to maintain reversible operation?

  1. QC=900 JQ_C = 900 \text{ J}
  2. QC=1200 JQ_C = 1200 \text{ J} (correct answer)
  3. QC=1350 JQ_C = 1350 \text{ J}
  4. QC=600 JQ_C = 600 \text{ J}
  5. QC=2700 JQ_C = 2700 \text{ J}
Explanation: When you encounter a reversible heat engine problem, you're dealing with the most efficient possible engine operating between two thermal reservoirs. The key insight is that reversible engines follow specific temperature-heat relationships. For any reversible heat engine, the ratio of heat quantities equals the ratio of absolute temperatures: QHTH=QCTC\frac{Q_H}{T_H} = \frac{Q_C}{T_C}. This relationship comes from the fact that entropy change is zero for reversible processes. Given QH=1800 JQ_H = 1800 \text{ J}, T1=600 KT_1 = 600 \text{ K}, and T2=400 KT_2 = 400 \text{ K}, you can solve for QCQ_C: 1800600=QC400\frac{1800}{600} = \frac{Q_C}{400} 3=QC4003 = \frac{Q_C}{400} QC=1200 JQ_C = 1200 \text{ J} Answer B is correct. Answer A (QC=900 JQ_C = 900 \text{ J}) likely comes from incorrectly assuming the efficiency is 50% and calculating QC=QH−W=1800−900=900Q_C = Q_H - W = 1800 - 900 = 900, but this ignores the temperature constraint. Answer C (QC=1350 JQ_C = 1350 \text{ J}) might result from mixing up the temperature ratio or using an incorrect proportion like QC=QH×TH+TC2TCQ_C = Q_H \times \frac{T_H + T_C}{2T_C}. Answer D (QC=600 JQ_C = 600 \text{ J}) could come from incorrectly using QC=QH×TCTH+TCQ_C = Q_H \times \frac{T_C}{T_H + T_C} or confusing temperature differences with ratios. Remember: for reversible engines, always use the temperature-heat ratio relationship QHTH=QCTC\frac{Q_H}{T_H} = \frac{Q_C}{T_C}. This is more fundamental than efficiency calculations and directly gives you the heat rejection required.

Question 14

A refrigerator operating between TH=300 KT_H = 300 \text{ K} and TC=250 KT_C = 250 \text{ K} has a coefficient of performance COP =4.0= 4.0. If the refrigerator removes QC=2000 JQ_C = 2000 \text{ J} from the cold space, what is the entropy change of the hot reservoir?

  1. ΔSH=+8.33 J/K\Delta S_H = +8.33 \text{ J/K} (correct answer)
  2. ΔSH=+7.67 J/K\Delta S_H = +7.67 \text{ J/K}
  3. ΔSH=+6.67 J/K\Delta S_H = +6.67 \text{ J/K}
  4. ΔSH=−8.33 J/K\Delta S_H = -8.33 \text{ J/K}
  5. ΔSH=+10.0 J/K\Delta S_H = +10.0 \text{ J/K}
Explanation: When analyzing refrigerator thermodynamics, you need to track energy flows and apply entropy definitions carefully. A refrigerator moves heat from cold to hot reservoirs, requiring work input. First, find the work required using the coefficient of performance: COP=QCW\text{COP} = \frac{Q_C}{W}, so W=QCCOP=2000 J4.0=500 JW = \frac{Q_C}{\text{COP}} = \frac{2000 \text{ J}}{4.0} = 500 \text{ J}. Next, apply energy conservation. The refrigerator removes QC=2000 JQ_C = 2000 \text{ J} from the cold reservoir and deposits heat into the hot reservoir: QH=QC+W=2000+500=2500 JQ_H = Q_C + W = 2000 + 500 = 2500 \text{ J}. The entropy change of the hot reservoir is ΔSH=QHTH=2500300=+8.33 J/K\Delta S_H = \frac{Q_H}{T_H} = \frac{2500}{300} = +8.33 \text{ J/K}. The positive sign indicates the hot reservoir gains entropy as it receives heat. Looking at the wrong answers: B (+7.67 J/K+7.67 \text{ J/K}) might result from incorrectly using QCQ_C instead of QHQ_H: 2000+300300=7.67\frac{2000 + 300}{300} = 7.67. C (+6.67 J/K+6.67 \text{ J/K}) comes from using only QCQ_C: 2000300=6.67\frac{2000}{300} = 6.67. D (−8.33 J/K-8.33 \text{ J/K}) has the correct magnitude but wrong sign—a common error when students think about heat "leaving" the system rather than recognizing that the hot reservoir receives heat. Remember: always identify what happens to each reservoir separately. The hot reservoir receives QH=QC+WQ_H = Q_C + W, not just QCQ_C, and its entropy increases because it gains thermal energy.

Question 15

A heat engine operates between two thermal reservoirs at temperatures TH=600 KT_H = 600 \text{ K} and TC=300 KT_C = 300 \text{ K}. The engine absorbs QH=1000 JQ_H = 1000 \text{ J} from the hot reservoir and rejects QC=600 JQ_C = 600 \text{ J} to the cold reservoir. What is the entropy change of the universe for this process?

  1. ΔSuniverse=0.33 J/K\Delta S_{universe} = 0.33 \text{ J/K} (correct answer)
  2. ΔSuniverse=0.67 J/K\Delta S_{universe} = 0.67 \text{ J/K}
  3. ΔSuniverse=−0.33 J/K\Delta S_{universe} = -0.33 \text{ J/K}
  4. ΔSuniverse=1.00 J/K\Delta S_{universe} = 1.00 \text{ J/K}
  5. ΔSuniverse=0 J/K\Delta S_{universe} = 0 \text{ J/K}
Explanation: When analyzing entropy changes in heat engines, you need to consider that the universe's entropy change equals the sum of entropy changes for all components involved. The universe consists of the hot reservoir, cold reservoir, and the engine itself. For any reservoir at constant temperature, the entropy change is ΔS=QT\Delta S = \frac{Q}{T}, where Q is positive for heat absorbed and negative for heat rejected. The hot reservoir loses 1000 J, so ΔSH=−1000600=−1.67 J/K\Delta S_H = \frac{-1000}{600} = -1.67 \text{ J/K}. The cold reservoir gains 600 J, so ΔSC=600300=2.00 J/K\Delta S_C = \frac{600}{300} = 2.00 \text{ J/K}. Since the engine operates in a cycle, it returns to its initial state, making its entropy change zero: ΔSengine=0\Delta S_{engine} = 0. Therefore: ΔSuniverse=ΔSH+ΔSC+ΔSengine=−1.67+2.00+0=0.33 J/K\Delta S_{universe} = \Delta S_H + \Delta S_C + \Delta S_{engine} = -1.67 + 2.00 + 0 = 0.33 \text{ J/K} Answer A (ΔSuniverse=0.33 J/K\Delta S_{universe} = 0.33 \text{ J/K}) is correct. Answer B likely results from calculation errors in the temperature ratios. Answer C represents the negative of the correct answer—a sign error that could occur if you mistakenly swap which reservoir gains versus loses entropy. Answer D might result from adding the absolute values of the reservoir entropy changes without considering signs. Remember: for irreversible processes like real heat engines, the universe's entropy always increases (ΔSuniverse>0\Delta S_{universe} > 0). Only ideal, reversible engines have ΔSuniverse=0\Delta S_{universe} = 0. This positive result confirms the engine operates irreversibly, as expected for real systems.

Question 16

A heat reservoir at temperature TR=350 KT_R = 350 \text{ K} exchanges heat with a system undergoing a process where its temperature varies from T1=300 KT_1 = 300 \text{ K} to T2=400 KT_2 = 400 \text{ K}. The system has heat capacity C=1500 J/KC = 1500 \text{ J/K}. If the heat transfer is irreversible, what is the minimum entropy generation?

  1. Sgen=1500ln⁡(4/3)−1500ln⁡(350/300) J/KS_{gen} = 1500 \ln(4/3) - 1500 \ln(350/300) \text{ J/K}
  2. Sgen=1500ln⁡(4/3)−1500×100350 J/KS_{gen} = 1500 \ln(4/3) - \frac{1500 \times 100}{350} \text{ J/K} (correct answer)
  3. Sgen=1500ln⁡(4/3) J/KS_{gen} = 1500 \ln(4/3) \text{ J/K}
  4. Sgen=1500×100350−1500ln⁡(4/3) J/KS_{gen} = \frac{1500 \times 100}{350} - 1500 \ln(4/3) \text{ J/K}
  5. Sgen=1500ln⁡(4/3)+1500×100350 J/KS_{gen} = 1500 \ln(4/3) + \frac{1500 \times 100}{350} \text{ J/K}
Explanation: When analyzing entropy generation in heat transfer between a reservoir and a system with varying temperature, you need to account for entropy changes in both the system and the reservoir, then apply the entropy generation principle. The system's entropy change is straightforward: ΔSsystem=Cln⁡(T2/T1)=1500ln⁡(400/300)=1500ln⁡(4/3)\Delta S_{system} = C \ln(T_2/T_1) = 1500 \ln(400/300) = 1500 \ln(4/3). This represents the entropy increase as the system heats from 300 K to 400 K. For the reservoir's entropy change, you must determine how much heat it transfers. Since the system's internal energy change is ΔU=C(T2−T1)=1500(400−300)=150,000 J\Delta U = C(T_2 - T_1) = 1500(400-300) = 150,000 \text{ J}, the reservoir must supply this amount of heat. The reservoir's entropy change is ΔSreservoir=−Q/TR=−150,000/350=−1500×100/350\Delta S_{reservoir} = -Q/T_R = -150,000/350 = -1500 \times 100/350. The total entropy generation is Sgen=ΔSsystem+ΔSreservoir=1500ln⁡(4/3)−1500×100350S_{gen} = \Delta S_{system} + \Delta S_{reservoir} = 1500 \ln(4/3) - \frac{1500 \times 100}{350}, making B correct. Option A incorrectly calculates the reservoir's entropy change using a temperature ratio instead of the actual heat transfer divided by reservoir temperature. Option C ignores the reservoir's entropy change entirely, considering only the system. Option D has the signs reversed, which would imply negative entropy generation—physically impossible for an irreversible process. Remember: entropy generation problems require tracking entropy changes in all components of the universe involved in the process. The reservoir's entropy change always involves the actual heat transferred divided by its constant temperature.

Question 17

Consider a heat engine cycle where the working fluid undergoes the following entropy changes with respect to three reservoirs: gains 2.0 J/K2.0 \text{ J/K} from reservoir at 400 K400 \text{ K}, loses 1.2 J/K1.2 \text{ J/K} to reservoir at 350 K350 \text{ K}, and loses 0.6 J/K0.6 \text{ J/K} to reservoir at 250 K250 \text{ K}. What is the net work output of this engine per cycle?

  1. 230 J230 \text{ J} (correct answer)
  2. 200 J200 \text{ J}
  3. 180 J180 \text{ J}
  4. 150 J150 \text{ J}
Explanation: Heat transfers are related to entropy changes by Q=TΔSQ = T\Delta S. Heat input: Qin=400×2.0=800 JQ_{in} = 400 \times 2.0 = 800 \text{ J}. Heat output: Qout1=350×1.2=420 JQ_{out1} = 350 \times 1.2 = 420 \text{ J}, Qout2=250×0.6=150 JQ_{out2} = 250 \times 0.6 = 150 \text{ J}. Total heat output: Qout=420+150=570 JQ_{out} = 420 + 150 = 570 \text{ J}. Net work: W=Qin−Qout=800−570=230 JW = Q_{in} - Q_{out} = 800 - 570 = 230 \text{ J}. Entropy check: ΔSuniverse=2.0−1.2−0.6=0.2 J/K>0\Delta S_{universe} = 2.0 - 1.2 - 0.6 = 0.2 \text{ J/K} > 0, confirming feasibility. Choice B incorrectly omits one reservoir. Choice C uses arithmetic error in heat calculations. Choice D assumes wrong entropy-heat relationship.

Question 18

A system undergoes a process where it absorbs Q=1500 JQ = 1500 \text{ J} of heat while its temperature varies according to T(t)=300+50t KT(t) = 300 + 50t \text{ K}, where tt ranges from 0 to 2 seconds. The heat transfer rate is constant. What is the system's entropy change?

  1. ΔS=1500350 J/K\Delta S = \frac{1500}{350} \text{ J/K}
  2. ΔS=1500ln⁡(400/300) J/K\Delta S = 1500 \ln(400/300) \text{ J/K}
  3. ΔS=∫02750300+50tdt J/K\Delta S = \int_0^2 \frac{750}{300 + 50t} dt \text{ J/K} (correct answer)
  4. ΔS=1500ln⁡(400/300)+1500350 J/K\Delta S = 1500 \ln(400/300) + \frac{1500}{350} \text{ J/K}
  5. ΔS=1500300+1500400 J/K\Delta S = \frac{1500}{300} + \frac{1500}{400} \text{ J/K}
Explanation: When dealing with entropy change during a non-isothermal process, you must remember that entropy is a state function defined as dS=dQrevTdS = \frac{dQ_{rev}}{T}. Since temperature varies with time, you cannot simply divide the total heat by a single temperature value. The correct approach requires integrating over the entire process. Given that the heat transfer rate is constant at dQdt=1500 J2 s=750 J/s\frac{dQ}{dt} = \frac{1500 \text{ J}}{2 \text{ s}} = 750 \text{ J/s}, and temperature varies as T(t)=300+50tT(t) = 300 + 50t, the entropy change becomes: ΔS=∫02dQ/dtT(t)dt=∫02750300+50tdt\Delta S = \int_0^2 \frac{dQ/dt}{T(t)} dt = \int_0^2 \frac{750}{300 + 50t} dt This is exactly what option C represents. Option A incorrectly uses an average temperature (350 K) in the denominator, treating this as if it were an isothermal process. While tempting, this approach ignores the fundamental requirement to account for temperature variation during entropy calculations. Option B applies the formula for entropy change during isothermal expansion of an ideal gas, ΔS=nRln⁡(Tf/Ti)\Delta S = nR\ln(T_f/T_i), but this is inappropriate here since we're dealing with heat transfer, not gas expansion, and we don't have information about the number of moles or gas constant. Option D combines the errors from both A and B, adding two incorrect approaches together. Study tip: For entropy calculations, always check whether temperature is constant. If it varies, you must integrate dQT\frac{dQ}{T} over the process path—never use average values or gas expansion formulas when dealing with variable-temperature heat transfer.

Question 19

During an irreversible refrigeration cycle, 800 J800 \text{ J} of heat is removed from a cold reservoir at 250 K250 \text{ K}, 1200 J1200 \text{ J} of heat is rejected to a hot reservoir at 350 K350 \text{ K}, and 400 J400 \text{ J} of work is input. Which statement best describes the entropy implications of this process?

  1. The entropy change of the universe is negative, indicating this process violates the second law of thermodynamics
  2. The entropy change of the universe is zero, indicating this is a reversible refrigeration process operating at maximum efficiency
  3. The entropy change of the universe is positive at 0.229 J/K0.229 \text{ J/K}, confirming this is a feasible but irreversible refrigeration process (correct answer)
  4. The entropy change of the universe is positive at 0.486 J/K0.486 \text{ J/K}, but this exceeds the theoretical maximum for refrigeration cycles
Explanation: First verify energy conservation: QH=QC+W=800+400=1200 JQ_H = Q_C + W = 800 + 400 = 1200 \text{ J} ✓. Calculate universe entropy change: ΔSuniverse=QHTH−QCTC=1200350−800250=3.429−3.200=0.229 J/K\Delta S_{universe} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C} = \frac{1200}{350} - \frac{800}{250} = 3.429 - 3.200 = 0.229 \text{ J/K}. Since this is positive, the process is irreversible but feasible. Choice A incorrectly calculates or misinterprets the sign. Choice B would require ΔS=0\Delta S = 0, which doesn't match our calculation. Choice D uses incorrect arithmetic or assumes wrong heat transfer directions.

Question 20

Two identical heat engines operate between the same hot reservoir at TH=500 KT_H = 500 \text{ K} and cold reservoir at TC=300 KT_C = 300 \text{ K}. Engine A operates reversibly, while Engine B has an efficiency of 30%30\%. Both engines extract the same amount of heat QH=1000 JQ_H = 1000 \text{ J} from the hot reservoir per cycle. What is the difference in entropy generation between the two engines per cycle?

  1. 0.167 J/K0.167 \text{ J/K}
  2. 0.233 J/K0.233 \text{ J/K}
  3. 0.100 J/K0.100 \text{ J/K}
  4. 0.133 J/K0.133 \text{ J/K} (correct answer)
Explanation: Engine A (reversible): ηA=1−TCTH=1−300500=0.4\eta_A = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 0.4, WA=400 JW_A = 400 \text{ J}, QC,A=600 JQ_{C,A} = 600 \text{ J}. Engine A entropy generation: ΔSA=QC,ATC−QHTH=600300−1000500=0\Delta S_A = \frac{Q_{C,A}}{T_C} - \frac{Q_H}{T_H} = \frac{600}{300} - \frac{1000}{500} = 0. Engine B: ηB=0.3\eta_B = 0.3, WB=300 JW_B = 300 \text{ J}, QC,B=700 JQ_{C,B} = 700 \text{ J}. Engine B entropy generation: ΔSB=QC,BTC−QHTH=700300−1000500=0.333 J/K\Delta S_B = \frac{Q_{C,B}}{T_C} - \frac{Q_H}{T_H} = \frac{700}{300} - \frac{1000}{500} = 0.333 \text{ J/K}. Difference: ΔSB−ΔSA=0.333−0=0.133 J/K\Delta S_B - \Delta S_A = 0.333 - 0 = 0.133 \text{ J/K}. Choice A uses wrong heat rejection values. Choice B confuses entropy calculation. Choice D uses efficiency difference directly.