Heat Reservoirs & Entropy - Thermodynamics

1. A power cycle operates between three thermal reservoirs at $T_1 = 800 \text{ K}$, $T_2 = 500 \text{ K}$, and $T_3 = 300 \text{ K}$. In one complete cycle, the working fluid absorbs $1000 \text{ J}$ from reservoir 1, rejects $300 \text{ J}$ to reservoir 2, and rejects $600 \text{ J}$ to reservoir 3. The entropy change of the working fluid over the complete cycle is zero. What can be concluded about this cycle?
   Answer: The cycle is irreversible but possible since the entropy of the universe increases by $1.35 \text{ J/K}$ per cycle First check energy conservation: $W = Q_{in} - Q_{out} = 1000 - (300 + 600) = 100 \text{ J}$ ✓. Calculate entropy change of universe: $\Delta S_{universe} = -\frac{Q_1}{T_1} + \frac{Q_2}{T_2} + \frac{Q_3}{T_3} = -\frac{1000}{800} + \frac{300}{500} + \frac{600}{300} = -1.25 + 0.6 + 2.0 = 1.35 \text{ J/K}$. Since this is positive, the cycle is possible but irreversible. Choice A is wrong because $\Delta S_{universe} \neq 0$. Choice B is wrong because entropy increases, not decreases. Choice D incorrectly calculates maximum work or assumes wrong efficiency limits for three-reservoir cycles.
2. A heat reservoir at temperature $T_R = 350 \text{ K}$ exchanges heat with a system undergoing a process where its temperature varies from $T_1 = 300 \text{ K}$ to $T_2 = 400 \text{ K}$. The system has heat capacity $C = 1500 \text{ J/K}$. If the heat transfer is irreversible, what is the minimum entropy generation?
   Answer: $S_{gen} = 1500 \ln(4/3) - \frac{1500 \times 100}{350} \text{ J/K}$ 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: $\Delta S_{system} = C \ln(T_2/T_1) = 1500 \ln(400/300) = 1500 \ln(4/3)$. 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 $\Delta U = C(T_2 - T_1) = 1500(400-300) = 150,000 \text{ J}$, the reservoir must supply this amount of heat. The reservoir's entropy change is $\Delta S_{reservoir} = -Q/T_R = -150,000/350 = -1500 \times 100/350$. The total entropy generation is $S_{gen} = \Delta S_{system} + \Delta S_{reservoir} = 1500 \ln(4/3) - \frac{1500 \times 100}{350}$, 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.
3. A heat engine receives $Q_{H1} = 800 \text{ J}$ from a reservoir at $T_{H1} = 500 \text{ K}$ and $Q_{H2} = 600 \text{ J}$ from another reservoir at $T_{H2} = 400 \text{ K}$. The engine rejects $Q_C = 900 \text{ J}$ to a cold reservoir at $T_C = 300 \text{ K}$. What is the total entropy change of all reservoirs?
   Answer: $\Delta S_{total} = -1.60 - 1.50 + 3.00 = -0.10 \text{ J/K}$ 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 $\Delta S = Q/T$, where $Q$ 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: $\Delta S_{H1} = -Q_{H1}/T_{H1} = -800/500 = -1.60 \text{ J/K}$ and $\Delta S_{H2} = -Q_{H2}/T_{H2} = -600/400 = -1.50 \text{ J/K}$. The cold reservoir gains heat from the engine, so its entropy change is positive: $\Delta S_C = +Q_C/T_C = +900/300 = +3.00 \text{ J/K}$. The total entropy change is: $\Delta S_{total} = -1.60 - 1.50 + 3.00 = -0.10 \text{ J/K}$, 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 $\Delta S$) and cold reservoirs always gain entropy (positive $\Delta S$) in heat engine problems. The sign of the total entropy change tells you whether the process is thermodynamically feasible.
4. A refrigeration cycle operates between reservoirs at $T_H = 300 \text{ K}$ and $T_C = 250 \text{ K}$. The refrigerator removes $Q_C = 500 \text{ J}$ from the cold space and requires work input $W = 120 \text{ J}$. What is the entropy generation for this cycle?
   Answer: $S_{gen} = 0.067 \text{ J/K}$ 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: $Q_H = Q_C + W = 500 + 120 = 620 \text{ J}$. The entropy generation equals the total entropy change of the universe: $S_{gen} = \Delta S_{hot} + \Delta S_{cold}$. The hot reservoir gains entropy: $\Delta S_{hot} = \frac{Q_H}{T_H} = \frac{620}{300} = 2.067 \text{ J/K}$. The cold reservoir loses entropy: $\Delta S_{cold} = -\frac{Q_C}{T_C} = -\frac{500}{250} = -2.000 \text{ J/K}$. Therefore: $S_{gen} = 2.067 - 2.000 = 0.067 \text{ J/K}$, 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.
5. An ideal gas undergoes an isothermal expansion at $T = 350 \text{ K}$ while in thermal contact with a reservoir at the same temperature. The gas volume changes from $V_1 = 0.02 \text{ m}^3$ to $V_2 = 0.05 \text{ m}^3$, and the amount of gas is $n = 2 \text{ mol}$. What is the entropy change of the reservoir?
   Answer: $\Delta S_{reservoir} = -2R \ln(2.5) \text{ J/K}$ 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, $Q = nRT \ln\left(\frac{V_2}{V_1}\right)$. With $n = 2 \text{ mol}$, $T = 350 \text{ K}$, and the volume ratio $\frac{V_2}{V_1} = \frac{0.05}{0.02} = 2.5$: $Q_{gas} = 2R(350) \ln(2.5)$ Since the gas absorbs heat from the reservoir, the reservoir loses this same amount of heat: $Q_{reservoir} = -Q_{gas}$. The entropy change of the reservoir is $\Delta S_{reservoir} = \frac{Q_{reservoir}}{T} = \frac{-Q_{gas}}{T} = \frac{-2R(350) \ln(2.5)}{350} = -2R \ln(2.5)$. Answer A is correct: $\Delta S_{reservoir} = -2R \ln(2.5) \text{ J/K}$ 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 $\ln(0.4)$ instead of $\ln(2.5)$; 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.
6. A system undergoes a process where it absorbs $Q = 1500 \text{ J}$ of heat while its temperature varies according to $T(t) = 300 + 50t \text{ K}$, where $t$ ranges from 0 to 2 seconds. The heat transfer rate is constant. What is the system's entropy change?
   Answer: $\Delta S = \int_0^2 \frac{750}{300 + 50t} dt \text{ J/K}$ When dealing with entropy change during a non-isothermal process, you must remember that entropy is a state function defined as $dS = \frac{dQ_{rev}}{T}$. 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 $\frac{dQ}{dt} = \frac{1500 \text{ J}}{2 \text{ s}} = 750 \text{ J/s}$, and temperature varies as $T(t) = 300 + 50t$, the entropy change becomes: $\Delta S = \int_0^2 \frac{dQ/dt}{T(t)} dt = \int_0^2 \frac{750}{300 + 50t} dt$ 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, $\Delta S = nR\ln(T_f/T_i)$, 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 $\frac{dQ}{T}$ over the process path—never use average values or gas expansion formulas when dealing with variable-temperature heat transfer.
7. Two identical heat engines operate between the same hot reservoir at $T_H = 500 \text{ K}$ and cold reservoir at $T_C = 300 \text{ K}$. Engine A operates reversibly, while Engine B has an efficiency of $30\%$. Both engines extract the same amount of heat $Q_H = 1000 \text{ J}$ from the hot reservoir per cycle. What is the difference in entropy generation between the two engines per cycle?
   Answer: $0.133 \text{ J/K}$ Engine A (reversible): $\eta_A = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 0.4$, $W_A = 400 \text{ J}$, $Q_{C,A} = 600 \text{ J}$. Engine A entropy generation: $\Delta S_A = \frac{Q_{C,A}}{T_C} - \frac{Q_H}{T_H} = \frac{600}{300} - \frac{1000}{500} = 0$. Engine B: $\eta_B = 0.3$, $W_B = 300 \text{ J}$, $Q_{C,B} = 700 \text{ J}$. Engine B entropy generation: $\Delta S_B = \frac{Q_{C,B}}{T_C} - \frac{Q_H}{T_H} = \frac{700}{300} - \frac{1000}{500} = 0.333 \text{ J/K}$. Difference: $\Delta S_B - \Delta S_A = 0.333 - 0 = 0.133 \text{ J/K}$. Choice A uses wrong heat rejection values. Choice B confuses entropy calculation. Choice D uses efficiency difference directly.
8. Two identical objects, each with heat capacity $C = 1000 \text{ J/K}$, are initially at temperatures $T_A = 400 \text{ K}$ and $T_B = 300 \text{ K}$. They are brought into thermal contact and allowed to reach equilibrium. What is the total entropy change of the system?
   Answer: $\Delta S_{total} = 1000 \ln(350/400) + 1000 \ln(350/300) \text{ J/K}$ 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: $C(T_A - T_f) = C(T_B - T_f)$, which simplifies to $T_f = \frac{T_A + T_B}{2} = \frac{400 + 300}{2} = 350 \text{ K}$. Next, calculate each object's entropy change using $\Delta S = C \ln(T_f/T_i)$. For object A: $\Delta S_A = 1000 \ln(350/400)$. For object B: $\Delta S_B = 1000 \ln(350/300)$. The total entropy change is: $\Delta S_{total} = 1000 \ln(350/400) + 1000 \ln(350/300)$, 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 $\ln(400/300)$, 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.
9. A heat engine operates between two thermal reservoirs at temperatures $T_H = 600 \text{ K}$ and $T_C = 300 \text{ K}$. The engine absorbs $Q_H = 1000 \text{ J}$ from the hot reservoir and rejects $Q_C = 600 \text{ J}$ to the cold reservoir. What is the entropy change of the universe for this process?
   Answer: $\Delta S_{universe} = 0.33 \text{ J/K}$ 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 $\Delta S = \frac{Q}{T}$, where Q is positive for heat absorbed and negative for heat rejected. The hot reservoir loses 1000 J, so $\Delta S_H = \frac{-1000}{600} = -1.67 \text{ J/K}$. The cold reservoir gains 600 J, so $\Delta S_C = \frac{600}{300} = 2.00 \text{ J/K}$. Since the engine operates in a cycle, it returns to its initial state, making its entropy change zero: $\Delta S_{engine} = 0$. Therefore: $\Delta S_{universe} = \Delta S_H + \Delta S_C + \Delta S_{engine} = -1.67 + 2.00 + 0 = 0.33 \text{ J/K}$ Answer A ($\Delta S_{universe} = 0.33 \text{ J/K}$) 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 ($\Delta S_{universe} > 0$). Only ideal, reversible engines have $\Delta S_{universe} = 0$. This positive result confirms the engine operates irreversibly, as expected for real systems.
10. A thermal reservoir at $T_R = 450 \text{ K}$ is used to heat a block of metal with heat capacity $C = 800 \text{ J/K}$ from $T_i = 300 \text{ K}$ to $T_f = 400 \text{ K}$. If the heat transfer occurs through direct contact (irreversible), what is the entropy change of the reservoir?
    Answer: $\Delta S_R = -\frac{800 \times 100}{450} \text{ J/K}$ 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: $Q_{block} = C(T_f - T_i) = 800 \times (400 - 300) = 80,000 \text{ J}$. By conservation of energy, the reservoir must lose exactly this amount of heat: $Q_{reservoir} = -80,000 \text{ J}$. 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: $\Delta S_R = \frac{Q_{reservoir}}{T_R} = \frac{-80,000}{450} = -\frac{800 \times 100}{450} \text{ J/K}$. This confirms answer A is correct. Answer B ($-800 \ln(4/3)$) 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.
11. Consider a heat engine cycle where the working fluid undergoes the following entropy changes with respect to three reservoirs: gains $2.0 \text{ J/K}$ from reservoir at $400 \text{ K}$, loses $1.2 \text{ J/K}$ to reservoir at $350 \text{ K}$, and loses $0.6 \text{ J/K}$ to reservoir at $250 \text{ K}$. What is the net work output of this engine per cycle?
    Answer: $230 \text{ J}$ Heat transfers are related to entropy changes by $Q = T\Delta S$. Heat input: $Q_{in} = 400 \times 2.0 = 800 \text{ J}$. Heat output: $Q_{out1} = 350 \times 1.2 = 420 \text{ J}$, $Q_{out2} = 250 \times 0.6 = 150 \text{ J}$. Total heat output: $Q_{out} = 420 + 150 = 570 \text{ J}$. Net work: $W = Q_{in} - Q_{out} = 800 - 570 = 230 \text{ J}$. Entropy check: $\Delta S_{universe} = 2.0 - 1.2 - 0.6 = 0.2 \text{ J/K} > 0$, confirming feasibility. Choice B incorrectly omits one reservoir. Choice C uses arithmetic error in heat calculations. Choice D assumes wrong entropy-heat relationship.
12. Two thermal reservoirs at $T_1 = 600 \text{ K}$ and $T_2 = 400 \text{ K}$ are connected by a reversible heat engine. The engine absorbs $Q_H = 1800 \text{ J}$ from the hot reservoir. What amount of heat must be rejected to the cold reservoir to maintain reversible operation?
    Answer: $Q_C = 1200 \text{ J}$ 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: $\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$. This relationship comes from the fact that entropy change is zero for reversible processes. Given $Q_H = 1800 \text{ J}$, $T_1 = 600 \text{ K}$, and $T_2 = 400 \text{ K}$, you can solve for $Q_C$: $\frac{1800}{600} = \frac{Q_C}{400}$ $3 = \frac{Q_C}{400}$ $Q_C = 1200 \text{ J}$ Answer B is correct. Answer A ($Q_C = 900 \text{ J}$) likely comes from incorrectly assuming the efficiency is 50% and calculating $Q_C = Q_H - W = 1800 - 900 = 900$, but this ignores the temperature constraint. Answer C ($Q_C = 1350 \text{ J}$) might result from mixing up the temperature ratio or using an incorrect proportion like $Q_C = Q_H \times \frac{T_H + T_C}{2T_C}$. Answer D ($Q_C = 600 \text{ J}$) could come from incorrectly using $Q_C = Q_H \times \frac{T_C}{T_H + T_C}$ or confusing temperature differences with ratios. Remember: for reversible engines, always use the temperature-heat ratio relationship $\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$. This is more fundamental than efficiency calculations and directly gives you the heat rejection required.
13. During an irreversible refrigeration cycle, $800 \text{ J}$ of heat is removed from a cold reservoir at $250 \text{ K}$, $1200 \text{ J}$ of heat is rejected to a hot reservoir at $350 \text{ K}$, and $400 \text{ J}$ of work is input. Which statement best describes the entropy implications of this process?
    Answer: The entropy change of the universe is positive at $0.229 \text{ J/K}$, confirming this is a feasible but irreversible refrigeration process First verify energy conservation: $Q_H = Q_C + W = 800 + 400 = 1200 \text{ J}$ ✓. Calculate universe entropy change: $\Delta S_{universe} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C} = \frac{1200}{350} - \frac{800}{250} = 3.429 - 3.200 = 0.229 \text{ J/K}$. Since this is positive, the process is irreversible but feasible. Choice A incorrectly calculates or misinterprets the sign. Choice B would require $\Delta S = 0$, which doesn't match our calculation. Choice D uses incorrect arithmetic or assumes wrong heat transfer directions.
14. An ideal gas undergoes a Carnot cycle between two reservoirs at $T_H = 600 \text{ K}$ and $T_C = 300 \text{ K}$. During the isothermal expansion at $T_H$, the gas absorbs $800 \text{ J}$ of heat and its entropy increases by $\Delta S$. What is the entropy change of the gas during the isothermal compression at $T_C$?
    Answer: $-\Delta S$, where $\Delta S = 1.33 \text{ J/K}$ For the isothermal expansion at $T_H$: $\Delta S = \frac{Q_H}{T_H} = \frac{800}{600} = 1.33 \text{ J/K}$. In a Carnot cycle, the entropy change during adiabatic processes is zero, and the total entropy change over the complete cycle must be zero. Therefore: $\Delta S_{isothermal,hot} + \Delta S_{isothermal,cold} = 0$. This means $\Delta S_{isothermal,cold} = -\Delta S = -1.33 \text{ J/K}$. The magnitude is the same but opposite in sign because the gas returns to its initial state. Choice B incorrectly assumes the cold reservoir entropy change is twice as large. Choice C uses wrong heat value for entropy calculation. Choice D incorrectly applies temperature ratio to entropy change.
15. Two heat reservoirs, initially at $T_1 = 600 \text{ K}$ and $T_2 = 400 \text{ K}$, each with heat capacity $C = 5000 \text{ J/K}$, are connected through a heat engine that extracts work while allowing them to approach thermal equilibrium. If the process stops when the temperature difference is reduced to $50 \text{ K}$, what is the change in entropy of the universe?
    Answer: $733 \text{ J/K}$ Final temperatures: Since reservoirs have equal heat capacities and reach a 50 K difference, we need to solve: $T_{1f} - T_{2f} = 50$ and $C(T_1 - T_{1f}) = C(T_{2f} - T_2)$ (energy conservation for the reservoirs alone). This gives: $600 - T_{1f} = T_{2f} - 400$, so $T_{1f} + T_{2f} = 1000$. Combined with $T_{1f} - T_{2f} = 50$: $T_{1f} = 525 \text{ K}$, $T_{2f} = 475 \text{ K}$. Entropy change: $\Delta S = C\ln\left(\frac{T_{1f}}{T_1}\right) + C\ln\left(\frac{T_{2f}}{T_2}\right) = 5000\ln\left(\frac{525}{600}\right) + 5000\ln\left(\frac{475}{400}\right) = 5000(-0.133 + 0.173) = 5000(0.040) = 733 \text{ J/K}$. Choice A uses arithmetic mean incorrectly. Choice B assumes complete equilibrium. Choice D neglects one reservoir's entropy change.
16. A refrigerator operating between $T_H = 300 \text{ K}$ and $T_C = 250 \text{ K}$ has a coefficient of performance COP $= 4.0$. If the refrigerator removes $Q_C = 2000 \text{ J}$ from the cold space, what is the entropy change of the hot reservoir?
    Answer: $\Delta S_H = +8.33 \text{ J/K}$ 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: $\text{COP} = \frac{Q_C}{W}$, so $W = \frac{Q_C}{\text{COP}} = \frac{2000 \text{ J}}{4.0} = 500 \text{ J}$. Next, apply energy conservation. The refrigerator removes $Q_C = 2000 \text{ J}$ from the cold reservoir and deposits heat into the hot reservoir: $Q_H = Q_C + W = 2000 + 500 = 2500 \text{ J}$. The entropy change of the hot reservoir is $\Delta S_H = \frac{Q_H}{T_H} = \frac{2500}{300} = +8.33 \text{ J/K}$. The positive sign indicates the hot reservoir gains entropy as it receives heat. Looking at the wrong answers: B ($+7.67 \text{ J/K}$) might result from incorrectly using $Q_C$ instead of $Q_H$: $\frac{2000 + 300}{300} = 7.67$. C ($+6.67 \text{ J/K}$) comes from using only $Q_C$: $\frac{2000}{300} = 6.67$. D ($-8.33 \text{ J/K}$) 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 $Q_H = Q_C + W$, not just $Q_C$, and its entropy increases because it gains thermal energy.
17. A thermal reservoir at temperature $T_1 = 600 \text{ K}$ transfers heat to another reservoir initially at $T_2 = 300 \text{ K}$. Both reservoirs have the same heat capacity $C = 3000 \text{ J/K}$. After heat transfer, the final temperature of the second reservoir is $T_{2f} = 350 \text{ K}$. What is the entropy change of the first reservoir?
    Answer: $\Delta S_1 = 3000 \ln(550/600) \text{ J/K}$ 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: $\Delta S = \int \frac{dQ}{T}$. First, determine how much heat the first reservoir loses. Since the second reservoir gains heat and reaches 350 K from 300 K: $Q_{gained} = C \Delta T_2 = 3000 \times (350-300) = 150{,}000 \text{ J}$. 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 $Q/T$ because the temperature changes during the process. Instead, use: $\Delta S_1 = C \ln\left(\frac{T_{1f}}{T_{1i}}\right)$, where $T_{1f}$ is the final temperature of reservoir 1. To find $T_{1f}$: The first reservoir loses heat $Q = C(T_{1i} - T_{1f}) = 150{,}000$, so $3000(600 - T_{1f}) = 150{,}000$. Solving: $T_{1f} = 600 - 50 = 550 \text{ K}$. Therefore: $\Delta S_1 = 3000 \ln\left(\frac{550}{600}\right)$, 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 $Q/T$ formula, ignoring that temperature changes during the process. Study tip: For finite reservoirs, always use $\Delta S = C \ln(T_f/T_i)$ rather than $Q/T$. The latter only applies to infinite reservoirs at constant temperature.
18. A thermal reservoir maintains constant temperature $T = 400 \text{ K}$ while supplying heat to multiple systems simultaneously. If the reservoir provides $Q_1 = 800 \text{ J}$ to system 1, $Q_2 = 1200 \text{ J}$ to system 2, and $Q_3 = 400 \text{ J}$ to system 3, what is the entropy change of the reservoir?
    Answer: $\Delta S_{reservoir} = -6.0 \text{ J/K}$ 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 $\Delta S = \frac{Q}{T}$. Since the reservoir supplies heat to three systems, you need to find the total heat it provides: $Q_{total} = Q_1 + Q_2 + Q_3 = 800 + 1200 + 400 = 2400 \text{ J}$. Because the reservoir loses this heat (it flows out to the systems), $Q_{reservoir} = -2400 \text{ J}$. The entropy change of the reservoir is: $\Delta S_{reservoir} = \frac{Q_{reservoir}}{T} = \frac{-2400 \text{ J}}{400 \text{ K}} = -6.0 \text{ J/K}$ 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 $-4.0 \text{ J/K}$, which would result from incorrectly using only 1600 J instead of the full 2400 J. C gives $-2.0 \text{ J/K}$, suggesting someone used only 800 J, perhaps just $Q_1$. D gives $+6.0 \text{ J/K}$, which has the right magnitude but wrong sign—this comes from forgetting that heat leaves the reservoir, making $Q_{reservoir}$ 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.
19. A Carnot heat pump operates between outdoor air at $T_C = 270 \text{ K}$ and indoor air at $T_H = 295 \text{ K}$. If the pump delivers $Q_H = 5000 \text{ J}$ to the warm indoor space, what is the entropy change of the cold outdoor reservoir?
    Answer: $\Delta S_C = -16.95 \text{ J/K}$ 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 $\Delta S_{total} = 0$ 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 $\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$. Solving for $Q_C$: $Q_C = Q_H \times \frac{T_C}{T_H} = 5000 \times \frac{270}{295} = 4576 \text{ J}$ Since the cold reservoir loses this heat, its entropy change is $\Delta S_C = -\frac{Q_C}{T_C} = -\frac{4576}{270} = -16.95 \text{ J/K}$ Looking at the wrong answers: Choice B ($-15.25 \text{ J/K}$) likely comes from incorrectly using $Q_H$ divided by $T_H$ instead of finding $Q_C$ first. Choice C ($+16.95 \text{ J/K}$) has the right magnitude but wrong sign—this would suggest the cold reservoir gains entropy, which contradicts it losing heat. Choice D ($-18.52 \text{ J/K}$) appears to use $Q_H$ directly with $T_C$: $-5000/270$, 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 ($\Delta S_C = -\Delta S_H$), and you must calculate the actual heat transfer to each reservoir using the temperature ratios before finding entropy changes.
20. A refrigerator removes $Q_C = 400 \text{ J}$ from a cold reservoir at $T_C = 250 \text{ K}$ and rejects $Q_H = 500 \text{ J}$ to a hot reservoir at $T_H = 300 \text{ K}$. If this refrigerator operates reversibly, what constraint must be satisfied?
    Answer: The entropy change of the universe must be zero for the complete cycle 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 $\Delta S_C = +Q_C/T_C = +400/250 = +1.6 \text{ J/K}$ (positive because heat is removed). The entropy change of the hot reservoir is $\Delta S_H = -Q_H/T_H = -500/300 = -1.67 \text{ J/K}$ (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, $|\Delta S_C| \neq |\Delta S_H|$. C is wrong because this describes energy conservation (first law), not reversibility. While true for any refrigerator ($W = Q_H - Q_C = 500 - 400 = 100 \text{ J}$), 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.

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