Thermodynamics Quiz: Cycles And Net Work
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Cycles And Net WorkQuestion 1 of 20

A Carnot engine operates between reservoirs at TH=600 KT_H = 600\text{ K} and TC=400 KT_C = 400\text{ K}. In one cycle, the engine absorbs QH=900 JQ_H = 900\text{ J} from the hot reservoir. What is the net work output for this cycle?

200 J200\text{ J}
300 J300\text{ J}
600 J600\text{ J}
450 J450\text{ J}
150 J150\text{ J}
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Thermodynamics Quiz

Thermodynamics Quiz: Cycles And Net Work

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

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A Carnot engine operates between reservoirs at TH=600 KT_H = 600\text{ K} and TC=400 KT_C = 400\text{ K}. In one cycle, the engine absorbs QH=900 JQ_H = 900\text{ J} from the hot reservoir. What is the net work output for this cycle?

  1. 200 J200\text{ J}
  2. 300 J300\text{ J} (correct answer)
  3. 600 J600\text{ J}
  4. 450 J450\text{ J}
  5. 150 J150\text{ J}
Explanation: When you encounter a Carnot engine problem, you're dealing with the most efficient heat engine possible operating between two thermal reservoirs. The key insight is that Carnot engines follow specific relationships between temperature, heat, and work. For any Carnot engine, the efficiency is determined solely by the reservoir temperatures: η=1−TCTH\eta = 1 - \frac{T_C}{T_H}. Here, η=1−400600=1−23=13\eta = 1 - \frac{400}{600} = 1 - \frac{2}{3} = \frac{1}{3}. The net work output equals the efficiency times the heat absorbed: W=η×QH=13×900 J=300 JW = \eta \times Q_H = \frac{1}{3} \times 900\text{ J} = 300\text{ J}. This confirms answer B is correct. Let's examine why the other options are wrong. Answer A (200 J) would correspond to an efficiency of 200900=0.22\frac{200}{900} = 0.22, which is too low for these temperatures. Answer C (600 J) represents an efficiency of 600900=0.67\frac{600}{900} = 0.67, which is impossibly high—no real engine can exceed the Carnot efficiency. Answer D (450 J) gives an efficiency of 0.5, which might seem reasonable but doesn't match the actual Carnot efficiency for these specific temperatures. Remember this pattern: Carnot engine problems always start with calculating the theoretical maximum efficiency from the temperature ratio. Once you have the efficiency, multiply by the heat input to get the work output. Don't guess at reasonable-looking values—the mathematics of thermodynamics gives you the exact answer every time.

Question 2

A heat engine operates between two thermal reservoirs at temperatures TH=500 KT_H = 500\text{ K} and TC=300 KT_C = 300\text{ K}. In one complete cycle, the engine absorbs QH=1200 JQ_H = 1200\text{ J} from the hot reservoir and rejects QC=800 JQ_C = 800\text{ J} to the cold reservoir. What is the net work output of this cycle?

  1. 400 J400\text{ J} (correct answer)
  2. 800 J800\text{ J}
  3. 1200 J1200\text{ J}
  4. 2000 J2000\text{ J}
  5. 200 J200\text{ J}
Explanation: When you encounter a heat engine problem, you're dealing with the fundamental principle of energy conservation. A heat engine takes thermal energy from a hot reservoir, converts some of it to useful work, and rejects the remainder to a cold reservoir. The first law of thermodynamics tells us that energy cannot be created or destroyed, only converted. For a heat engine, this means: W=QH−QCW = Q_H - Q_C, where WW is the net work output, QHQ_H is heat absorbed from the hot reservoir, and QCQ_C is heat rejected to the cold reservoir. Plugging in the given values: W=1200 J−800 J=400 JW = 1200\text{ J} - 800\text{ J} = 400\text{ J}. This confirms that answer A is correct. Let's examine why the other options are wrong. Answer B (800 J800\text{ J}) represents the heat rejected to the cold reservoir, not the work output. Answer C (1200 J1200\text{ J}) is the heat input from the hot reservoir—this would violate conservation of energy since it suggests all input heat becomes work with nothing rejected. Answer D (2000 J2000\text{ J}) appears to be the sum of heat input and output (QH+QCQ_H + Q_C), which has no physical meaning in this context and also violates energy conservation. Remember this key relationship: W=QH−QCW = Q_H - Q_C. The work output is always the difference between heat absorbed and heat rejected. Never confuse the individual heat transfers with the net work—they're fundamentally different quantities in thermodynamic cycles.

Question 3

A refrigerator operates in a cycle between a cold reservoir at TC=250 KT_C = 250\text{ K} and a hot reservoir at TH=350 KT_H = 350\text{ K}. During one cycle, the refrigerator removes QC=600 JQ_C = 600\text{ J} from the cold reservoir and requires W=200 JW = 200\text{ J} of work input. What is the heat rejected to the hot reservoir?

  1. 400 J400\text{ J}
  2. 600 J600\text{ J}
  3. 800 J800\text{ J} (correct answer)
  4. 200 J200\text{ J}
  5. 1000 J1000\text{ J}
Explanation: When you encounter refrigerator problems, remember that these devices move heat from cold to hot reservoirs using work input. The key principle is energy conservation: all energy must be accounted for. For any refrigerator cycle, the first law of thermodynamics requires that the work input plus the heat removed from the cold reservoir equals the heat rejected to the hot reservoir: W+QC=QHW + Q_C = Q_H. This makes physical sense—the refrigerator takes energy from the cold space (QC=600 JQ_C = 600\text{ J}) and adds work energy (W=200 JW = 200\text{ J}), then dumps this combined energy into the hot reservoir. Calculating: QH=W+QC=200 J+600 J=800 JQ_H = W + Q_C = 200\text{ J} + 600\text{ J} = 800\text{ J} Looking at the wrong answers: Choice A (400 J400\text{ J}) represents the common error of subtracting work from heat removed (QC−WQ_C - W), which violates energy conservation. Choice B (600 J600\text{ J}) incorrectly assumes the heat rejected equals the heat removed, ignoring the work input entirely. Choice D (200 J200\text{ J}) confuses the heat rejected with the work input, missing that the refrigerator must reject more energy than it consumes as work. The correct answer is C: 800 J800\text{ J}. Study tip: For refrigerator problems, always start with energy conservation: W+QC=QHW + Q_C = Q_H. The heat rejected to the hot reservoir is always the largest quantity because it includes both the heat removed from the cold reservoir and the work input. This relationship holds for all refrigeration cycles, regardless of efficiency.

Question 4

A cyclic process returns a system to its initial state. During the cycle, the system absorbs Q1=800 JQ_1 = 800\text{ J} in process 1-2, rejects Q2=300 JQ_2 = 300\text{ J} in process 2-3, and absorbs Q3=200 JQ_3 = 200\text{ J} in process 3-1. What is the net work done by the system during this complete cycle?

  1. 500 J500\text{ J}
  2. 700 J700\text{ J} (correct answer)
  3. 1000 J1000\text{ J}
  4. 300 J300\text{ J}
  5. 100 J100\text{ J}
Explanation: When you encounter a cyclic process problem, remember that the system returns to its initial state, making the change in internal energy zero (ΔU=0\Delta U = 0). This is the key insight that unlocks these problems. Since ΔU=0\Delta U = 0 for any complete cycle, the first law of thermodynamics (ΔU=Q−W\Delta U = Q - W) simplifies to Qnet=WnetQ_{net} = W_{net}. The net heat absorbed equals the net work done by the system. To find the net heat, you need to carefully account for signs. Heat absorbed by the system is positive, while heat rejected is negative. Process 1-2: +800 J+800\text{ J} (absorbed), Process 2-3: −300 J-300\text{ J} (rejected), Process 3-1: +200 J+200\text{ J} (absorbed). Therefore: Qnet=800−300+200=700 JQ_{net} = 800 - 300 + 200 = 700\text{ J} Since Wnet=QnetW_{net} = Q_{net}, the net work done by the system is 700 J700\text{ J}, confirming answer B. The wrong answers represent common mistakes: A (500 J500\text{ J}) likely comes from incorrectly subtracting the last heat term instead of adding it. C (1000 J1000\text{ J}) results from adding the absolute values of all heat transfers without considering signs. D (300 J300\text{ J}) might come from confusing which heat transfer to use or making arithmetic errors. Remember: for cyclic processes, always use ΔU=0\Delta U = 0 to connect net heat and net work. Pay careful attention to the signs of heat transfers—absorbed is positive, rejected is negative.

Question 5

A heat engine completes 50 cycles per minute. In each cycle, it absorbs QH=240 JQ_H = 240\text{ J} and rejects QC=160 JQ_C = 160\text{ J}. What is the power output of this engine?

  1. 66.7 W66.7\text{ W} (correct answer)
  2. 80 W80\text{ W}
  3. 200 W200\text{ W}
  4. 133 W133\text{ W}
  5. 40 W40\text{ W}
Explanation: When you encounter heat engine problems, focus on the fundamental relationship between energy input, output, and power. Heat engines convert thermal energy into mechanical work, and power measures how quickly this energy conversion occurs. To find the power output, you need two key pieces: the work done per cycle and how frequently cycles occur. The work done per cycle comes from the first law of thermodynamics: W=QH−QC=240 J−160 J=80 JW = Q_H - Q_C = 240\text{ J} - 160\text{ J} = 80\text{ J}. This represents the net energy converted to useful work in each cycle. Next, convert the frequency to standard units. The engine completes 50 cycles per minute, which equals 5060=0.833\frac{50}{60} = 0.833 cycles per second. Power is work per unit time, so: P=W×frequency=80 J×0.833 s−1=66.7 WP = W \times \text{frequency} = 80\text{ J} \times 0.833\text{ s}^{-1} = 66.7\text{ W}. This confirms answer A is correct. The wrong answers represent common calculation errors. Answer B (80 W80\text{ W}) occurs if you forget to convert from cycles per minute to cycles per second—essentially using the work per cycle as if it were power. Answer C (200 W200\text{ W}) results from incorrectly multiplying QHQ_H by the wrong frequency conversion. Answer D (133 W133\text{ W}) comes from doubling the correct answer, possibly from a unit conversion mistake. Remember this pattern: for heat engine power problems, always calculate work per cycle first using W=QH−QCW = Q_H - Q_C, then multiply by the cycle frequency in proper units (cycles per second, not per minute).

Question 6

A heat pump delivers QH=2400 JQ_H = 2400\text{ J} to a building while extracting QC=1800 JQ_C = 1800\text{ J} from the outdoor environment in one cycle. What is the coefficient of performance of this heat pump?

  1. 3.03.0
  2. 4.04.0 (correct answer)
  3. 1.331.33
  4. 0.750.75
  5. 6.06.0
Explanation: When analyzing heat pump problems, focus on what the device accomplishes: moving heat from a cold reservoir to a hot reservoir. The coefficient of performance (COP) measures how effectively it does this job. For a heat pump, the COP is defined as the ratio of heat delivered to the desired location (usually a warm building) to the work input required. Since energy is conserved, the work input equals the difference between heat delivered and heat extracted: W=QH−QC=2400 J−1800 J=600 JW = Q_H - Q_C = 2400\text{ J} - 1800\text{ J} = 600\text{ J}. Therefore: COP=QHW=2400 J600 J=4.0\text{COP} = \frac{Q_H}{W} = \frac{2400\text{ J}}{600\text{ J}} = 4.0 Looking at the wrong answers: Choice A (3.0) likely comes from incorrectly using QCW=1800600=3.0\frac{Q_C}{W} = \frac{1800}{600} = 3.0, which would be the COP for a refrigerator operating between the same reservoirs, not a heat pump. Choice C (1.33) results from the flawed calculation QHQC=24001800=1.33\frac{Q_H}{Q_C} = \frac{2400}{1800} = 1.33, but this ratio has no physical meaning in thermodynamics. Choice D (0.75) comes from inverting the correct formula: WQH=6002400=0.25\frac{W}{Q_H} = \frac{600}{2400} = 0.25 or possibly QCQH=0.75\frac{Q_C}{Q_H} = 0.75. Remember: Heat pump COP always uses the heat delivered to the desired space in the numerator and work input in the denominator. For refrigerators, you'd use heat removed from the cold space instead. Keep these definitions straight to avoid mixing up the formulas.

Question 7

An engine operates between two reservoirs with temperatures in the ratio TH:TC=5:3T_H:T_C = 5:3. If the maximum theoretical efficiency is achieved and the engine absorbs QH=1000 JQ_H = 1000\text{ J} per cycle, what work is produced?

  1. 200 J200\text{ J}
  2. 400 J400\text{ J} (correct answer)
  3. 600 J600\text{ J}
  4. 333 J333\text{ J}
  5. 167 J167\text{ J}
Explanation: When you encounter a heat engine problem asking for "maximum theoretical efficiency," you're dealing with the Carnot cycle - the most efficient possible heat engine operating between two thermal reservoirs. The Carnot efficiency is given by η=1−TCTH\eta = 1 - \frac{T_C}{T_H}, where temperatures must be in Kelvin. Since you're given the ratio TH:TC=5:3T_H:T_C = 5:3, you can write TCTH=35\frac{T_C}{T_H} = \frac{3}{5}. Therefore: η=1−35=25=0.4=40%\eta = 1 - \frac{3}{5} = \frac{2}{5} = 0.4 = 40\% The work output is simply the efficiency times the heat input: W=η×QH=0.4×1000 J=400 JW = \eta \times Q_H = 0.4 \times 1000\text{ J} = 400\text{ J}. This confirms answer B. Let's examine why the other options are incorrect. Choice A (200 J) would correspond to an efficiency of 20%, which might result from incorrectly using η=TCTH−1\eta = \frac{T_C}{T_H} - 1 or making an arithmetic error. Choice C (600 J) represents 60% efficiency, which could come from mistakenly using η=TH−TCTC\eta = \frac{T_H - T_C}{T_C} instead of the correct Carnot formula. Choice D (333 J) corresponds to about 33% efficiency, which has no clear thermodynamic basis but might arise from calculation errors involving the given ratio. Key strategy: Always remember that Carnot efficiency problems require absolute temperatures and the formula η=1−TCTH\eta = 1 - \frac{T_C}{T_H}. When given temperature ratios, you can work directly with the ratio without converting to specific Kelvin values, making calculations simpler and reducing errors.

Question 8

A cyclic heat engine operates between reservoirs at 400 K400\text{ K} and 300 K300\text{ K}. The engine absorbs 1200 J1200\text{ J} from the hot reservoir and produces 240 J240\text{ J} of work per cycle. How does this engine's efficiency compare to the maximum possible efficiency?

  1. The engine operates at 80% of maximum efficiency (correct answer)
  2. The engine operates at 60% of maximum efficiency
  3. The engine operates at 75% of maximum efficiency
  4. The engine operates at 90% of maximum efficiency
  5. The engine operates at 50% of maximum efficiency
Explanation: When you encounter heat engine problems, you need to compare the actual efficiency to the theoretical maximum (Carnot) efficiency. This tests your understanding of both real and ideal thermodynamic cycles. First, calculate the actual efficiency. Efficiency equals work output divided by heat input: ηactual=WQH=240 J1200 J=0.20=20%\eta_{actual} = \frac{W}{Q_H} = \frac{240\text{ J}}{1200\text{ J}} = 0.20 = 20\% Next, find the maximum possible efficiency using the Carnot formula: ηCarnot=1−TCTH=1−300 K400 K=1−0.75=0.25=25%\eta_{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{300\text{ K}}{400\text{ K}} = 1 - 0.75 = 0.25 = 25\% The ratio of actual to maximum efficiency is: 20%25%=0.80=80%\frac{20\%}{25\%} = 0.80 = 80\% Choice A is correct because the engine operates at 80% of its theoretical maximum efficiency. Choice B (60%) results from incorrectly calculating the actual efficiency as 15% instead of 20%, possibly from arithmetic errors in the work-to-heat ratio. Choice C (75%) occurs if you mistakenly use the cold reservoir fraction (300/400 = 0.75) as the efficiency ratio. Choice D (90%) suggests confusion between the actual efficiency calculation and the temperature ratio, possibly mixing up which values to divide. Remember: always calculate both efficiencies separately using the correct formulas, then find their ratio. The Carnot efficiency depends only on reservoir temperatures, while actual efficiency uses the given work and heat values. Double-check your arithmetic, as these problems often involve simple ratios that are easy to miscalculate.

Question 9

A reversible heat engine operates in a cycle and produces W=350 JW = 350\text{ J} of work while rejecting QC=650 JQ_C = 650\text{ J} to a reservoir at TC=300 KT_C = 300\text{ K}. What is the temperature of the hot reservoir?

  1. 461 K461\text{ K} (correct answer)
  2. 515 K515\text{ K}
  3. 300 K300\text{ K}
  4. 600 K600\text{ K}
  5. 400 K400\text{ K}
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 the Carnot cycle relationships. For any heat engine, energy conservation requires that the heat absorbed from the hot reservoir equals the work produced plus the heat rejected: QH=W+QC=350 J+650 J=1000 JQ_H = W + Q_C = 350\text{ J} + 650\text{ J} = 1000\text{ J}. For a reversible (Carnot) engine specifically, the temperatures and heat transfers are related by QHTH=QCTC\frac{Q_H}{T_H} = \frac{Q_C}{T_C}. Solving for the hot reservoir temperature: TH=TC×QHQC=300 K×1000 J650 J=300×1.538=461 KT_H = T_C \times \frac{Q_H}{Q_C} = 300\text{ K} \times \frac{1000\text{ J}}{650\text{ J}} = 300 \times 1.538 = 461\text{ K} Answer A (461 K461\text{ K}) is correct using this proper Carnot relationship. Answer B (515 K515\text{ K}) might result from incorrectly using efficiency formulas or computational errors. Answer C (300 K300\text{ K}) represents the cold reservoir temperature—a common mistake when students confuse which temperature they're solving for. Answer D (600 K600\text{ K}) could come from misapplying ratios, perhaps using TH=TC×QHWT_H = T_C \times \frac{Q_H}{W} instead of the correct Carnot relation. Remember: for reversible engines, always use the Carnot relationships linking temperature ratios to heat ratios. First find the missing heat quantity using energy conservation, then apply QHTH=QCTC\frac{Q_H}{T_H} = \frac{Q_C}{T_C} to solve for the unknown temperature.

Question 10

An ideal gas undergoes a thermodynamic cycle consisting of two isothermal and two adiabatic processes. In the isothermal expansion at high temperature, the gas does W1=500 JW_1 = 500\text{ J} of work. In the isothermal compression at low temperature, W2=300 JW_2 = 300\text{ J} of work is done on the gas. What is the net work output of this cycle?

  1. 200 J200\text{ J} (correct answer)
  2. 800 J800\text{ J}
  3. 500 J500\text{ J}
  4. 300 J300\text{ J}
  5. 150 J150\text{ J}
Explanation: When analyzing thermodynamic cycles, focus on the sign conventions for work and the net effect over the complete cycle. Work done by the gas is positive, while work done on the gas is negative. In this cycle, during isothermal expansion at high temperature, the gas does W1=+500 JW_1 = +500\text{ J} of work (positive because the gas does work on its surroundings). During isothermal compression at low temperature, W2=300 JW_2 = 300\text{ J} of work is done on the gas, so this contributes −300 J-300\text{ J} to the net work. The adiabatic processes contribute zero net work over the complete cycle since they form a closed loop on the P-V diagram. The net work output is: Wnet=(+500 J)+(−300 J)=200 JW_{net} = (+500\text{ J}) + (-300\text{ J}) = 200\text{ J} This confirms answer A is correct. Answer B (800 J800\text{ J}) incorrectly adds the magnitudes: 500+300500 + 300, ignoring that compression work opposes expansion work. Answer C (500 J500\text{ J}) mistakenly considers only the expansion work, forgetting about the compression phase entirely. Answer D (300 J300\text{ J}) incorrectly treats only the compression work, perhaps misunderstanding which process dominates. Study tip: For any thermodynamic cycle, always track the sign of work carefully—expansion gives positive work output, compression requires negative work input. The net work is the algebraic sum of all work terms, and for heat engines, this should be positive (net work output).

Question 11

An engine operates in a cycle with three processes. In process A→B, the system absorbs Q1=400 JQ_1 = 400\text{ J} and does W1=150 JW_1 = 150\text{ J} of work. In process B→C, the system rejects Q2=200 JQ_2 = 200\text{ J} and W2=100 JW_2 = 100\text{ J} of work is done on it. In process C→A, the system does W3=80 JW_3 = 80\text{ J} of work. What is the net heat absorbed by the system in the complete cycle?

  1. 130 J130\text{ J} (correct answer)
  2. 200 J200\text{ J}
  3. 330 J330\text{ J}
  4. 400 J400\text{ J}
  5. 30 J30\text{ J}
Explanation: When you encounter cyclic thermodynamic processes, remember that the first law of thermodynamics (ΔU=Q−W\Delta U = Q - W) applies to each step, and for a complete cycle, the internal energy change must be zero since the system returns to its initial state. Let's analyze each process systematically. In process A→B: ΔU1=Q1−W1=400 J−150 J=250 J\Delta U_1 = Q_1 - W_1 = 400\text{ J} - 150\text{ J} = 250\text{ J}. In process B→C: the system rejects 200 J (so Q2=−200 JQ_2 = -200\text{ J}) and has work done on it (so W2=−100 JW_2 = -100\text{ J}), giving ΔU2=−200 J−(−100 J)=−100 J\Delta U_2 = -200\text{ J} - (-100\text{ J}) = -100\text{ J}. In process C→A: ΔU3=Q3−80 J\Delta U_3 = Q_3 - 80\text{ J}. Since ΔUtotal=0\Delta U_{total} = 0 for the complete cycle: 250 J+(−100 J)+ΔU3=0250\text{ J} + (-100\text{ J}) + \Delta U_3 = 0, so ΔU3=−150 J\Delta U_3 = -150\text{ J}. Therefore: Q3=ΔU3+W3=−150 J+80 J=−70 JQ_3 = \Delta U_3 + W_3 = -150\text{ J} + 80\text{ J} = -70\text{ J}. The net heat absorbed is: Qnet=Q1+Q2+Q3=400 J+(−200 J)+(−70 J)=130 JQ_{net} = Q_1 + Q_2 + Q_3 = 400\text{ J} + (-200\text{ J}) + (-70\text{ J}) = 130\text{ J}. This confirms answer A. Answer B (200 J) incorrectly uses only the difference Q1−Q2Q_1 - Q_2, ignoring process C→A. Answer C (330 J) mistakenly adds Q1Q_1 and Q2Q_2 as both positive values, forgetting that heat rejection means negative QQ. Answer D (400 J) only considers the heat input from process A→B. Remember: in cyclic processes, always account for all three quantities (QQ, WW, and ΔU\Delta U) in each step, and use the constraint that ΔUcycle=0\Delta U_{cycle} = 0.

Question 12

A Carnot refrigerator operates between reservoirs at TH=320 KT_H = 320\text{ K} and TC=280 KT_C = 280\text{ K}. If the refrigerator removes QC=2100 JQ_C = 2100\text{ J} from the cold reservoir per cycle, what is the minimum work input required?

  1. 300 J300\text{ J} (correct answer)
  2. 525 J525\text{ J}
  3. 400 J400\text{ J}
  4. 240 J240\text{ J}
  5. 350 J350\text{ J}
Explanation: When you encounter Carnot refrigerator problems, you're dealing with the most efficient possible refrigerator operating between two thermal reservoirs. The key insight is that Carnot devices have a specific relationship between heat transfers and temperatures. For a Carnot refrigerator, the heat transfers are related by the temperature ratio: QHQC=THTC\frac{Q_H}{Q_C} = \frac{T_H}{T_C}. Since QC=2100 JQ_C = 2100\text{ J}, TH=320 KT_H = 320\text{ K}, and TC=280 KT_C = 280\text{ K}, you can find: QH=QC×THTC=2100×320280=2400 JQ_H = Q_C \times \frac{T_H}{T_C} = 2100 \times \frac{320}{280} = 2400\text{ J} The work input equals the difference between heat rejected to the hot reservoir and heat absorbed from the cold reservoir: W=QH−QC=2400−2100=300 JW = Q_H - Q_C = 2400 - 2100 = 300\text{ J}. This confirms answer A is correct. Looking at the wrong answers: B (525 J) might result from incorrectly adding the heat values or misapplying the temperature ratio. C (400 J) could come from calculation errors in the temperature ratio or confusing the relationship between heat transfers. D (240 J) might arise from using an incorrect formula or mixing up which temperatures correspond to which reservoirs. Remember that Carnot problems always involve temperature ratios in Kelvin, and for refrigerators, work input is always the difference QH−QCQ_H - Q_C. The coefficient of performance formula COP=TCTH−TCCOP = \frac{T_C}{T_H - T_C} can also verify your answer: COP=28040=7COP = \frac{280}{40} = 7, so W=QCCOP=21007=300 JW = \frac{Q_C}{COP} = \frac{2100}{7} = 300\text{ J}.

Question 13

A heat engine cycle produces a net work output of W=250 JW = 250\text{ J} while rejecting QC=750 JQ_C = 750\text{ J} to a cold reservoir. If this engine were operated in reverse as a heat pump between the same reservoirs, and the same amount of work W=250 JW = 250\text{ J} were input, how much heat would be delivered to the hot reservoir?

  1. 1000 J1000\text{ J} (correct answer)
  2. 750 J750\text{ J}
  3. 500 J500\text{ J}
  4. 1250 J1250\text{ J}
  5. 875 J875\text{ J}
Explanation: This problem tests your understanding of reversible thermodynamic cycles and how the same device can function as either a heat engine or heat pump. When analyzing any heat engine or heat pump, always apply the first law of thermodynamics: energy must be conserved. For the original heat engine, you can find the heat absorbed from the hot reservoir using energy conservation: QH=W+QC=250+750=1000 JQ_H = W + Q_C = 250 + 750 = 1000\text{ J}. This establishes that the cycle transfers 1000 J from hot to cold reservoir while producing 250 J of work. When operated in reverse as a heat pump, the same cycle runs backward with the same energy quantities, just flowing in opposite directions. The 250 J of work input drives 750 J of heat from the cold reservoir up to the hot reservoir. By energy conservation, the total heat delivered to the hot reservoir is QH=W+QC=250+750=1000 JQ_H = W + Q_C = 250 + 750 = 1000\text{ J}. Choice B (750 J) incorrectly assumes only the heat extracted from the cold reservoir is delivered to the hot side, ignoring the work input. Choice C (500 J) mistakenly subtracts the cold reservoir heat from work instead of adding. Choice D (1250 J) incorrectly adds the original engine's hot reservoir heat to the work input, double-counting energy. Remember this key principle: in reversible cycles, the same energy quantities appear whether running as an engine or heat pump—they just flow in opposite directions. Always apply QH=W+QCQ_H = W + Q_C for heat pumps to account for both work input and heat extracted from the cold side.

Question 14

A heat engine cycle consists of three processes. The system does W1=200 JW_1 = 200\text{ J} of work in process 1, work W2=−150 JW_2 = -150\text{ J} is done on the system in process 2, and the system does W3=100 JW_3 = 100\text{ J} of work in process 3. What is the net work done by the system in this complete cycle?

  1. 150 J150\text{ J} (correct answer)
  2. 250 J250\text{ J}
  3. 450 J450\text{ J}
  4. 50 J50\text{ J}
  5. 350 J350\text{ J}
Explanation: When you encounter a heat engine cycle problem, you're dealing with the fundamental principle that work is additive over multiple processes. The key is understanding the sign convention: positive work means the system does work on the surroundings, while negative work means work is done on the system. To find the net work, you simply add all the work values algebraically, respecting their signs. Here, the system does W1=200 JW_1 = 200\text{ J} of work (positive), has W2=−150 JW_2 = -150\text{ J} of work done on it (negative), and does W3=100 JW_3 = 100\text{ J} of work (positive). The net work is: Wnet=200+(−150)+100=150 JW_{net} = 200 + (-150) + 100 = 150\text{ J} Answer A (150 J150\text{ J}) is correct because it properly accounts for the signs and adds all three work terms. Answer B (250 J250\text{ J}) results from incorrectly treating the −150 J-150\text{ J} as positive, giving 200+150+100=450 J200 + 150 + 100 = 450\text{ J}, then somehow getting 250 J250\text{ J} through calculation error. Answer C (450 J450\text{ J}) comes from ignoring the negative sign entirely and adding 200+150+100=450 J200 + 150 + 100 = 450\text{ J}. Answer D (50 J50\text{ J}) might result from incorrectly subtracting instead of adding one of the positive work values, such as 200−150−100=−50 J200 - 150 - 100 = -50\text{ J} and taking the absolute value. Remember: always pay careful attention to the signs in thermodynamics problems. Work done by the system is positive, work done on the system is negative, and net quantities are always algebraic sums.

Question 15

A heat pump operates between an outdoor temperature of TC=280 KT_C = 280\text{ K} and an indoor temperature of TH=295 KT_H = 295\text{ K}. The heat pump has an actual coefficient of performance that is 60% of the ideal value. If W=600 JW = 600\text{ J} of work is input per cycle, how much heat is delivered to the house?

  1. 7080 J7080\text{ J} (correct answer)
  2. 4248 J4248\text{ J}
  3. 6720 J6720\text{ J}
  4. 2360 J2360\text{ J}
  5. 11800 J11800\text{ J}
Explanation: Heat pump problems test your understanding of the coefficient of performance (COP) and how real devices compare to ideal ones. When you see a heat pump question, remember that COP relates the useful heat output to the work input. For an ideal heat pump operating between two thermal reservoirs, the coefficient of performance is COPideal=THTH−TC\text{COP}_{\text{ideal}} = \frac{T_H}{T_H - T_C}. Substituting the given temperatures: COPideal=295295−280=29515=19.67\text{COP}_{\text{ideal}} = \frac{295}{295 - 280} = \frac{295}{15} = 19.67. Since the actual COP is 60% of the ideal value: COPactual=0.60×19.67=11.8\text{COP}_{\text{actual}} = 0.60 \times 19.67 = 11.8. The coefficient of performance for a heat pump is defined as COP=QHW\text{COP} = \frac{Q_H}{W}, where QHQ_H is the heat delivered to the house and WW is the work input. Solving for the heat delivered: QH=COPactual×W=11.8×600=7080 JQ_H = \text{COP}_{\text{actual}} \times W = 11.8 \times 600 = 7080\text{ J}. Looking at the wrong answers: B (4248 J) likely results from using the refrigerator COP formula (QC/WQ_C/W) instead of the heat pump formula. C (6720 J) might come from calculation errors in the ideal COP or efficiency factor. D (2360 J) appears to involve fundamental misunderstanding of the COP relationship, possibly confusing it with efficiency. Study tip: Always distinguish between heat pump COP (QH/WQ_H/W) and refrigerator COP (QC/WQ_C/W). Heat pumps move heat to the warm reservoir (heating), while refrigerators move heat from the cold reservoir (cooling). The question's context—delivering heat to a house—clearly indicates a heat pump scenario.

Question 16

A refrigerator has a coefficient of performance COPR=4.0COP_R = 4.0. If the refrigerator requires W=150 JW = 150\text{ J} of work input per cycle, what is the net heat transfer to the surroundings during one complete cycle?

  1. 450 J450\text{ J}
  2. 600 J600\text{ J}
  3. 750 J750\text{ J} (correct answer)
  4. 150 J150\text{ J}
  5. 300 J300\text{ J}
Explanation: When you encounter refrigerator problems, remember that refrigerators move heat from a cold reservoir to a hot reservoir using work input. The coefficient of performance tells you how efficiently this heat transfer occurs. The coefficient of performance for a refrigerator is defined as COPR=QCWCOP_R = \frac{Q_C}{W}, where QCQ_C is the heat removed from the cold reservoir and WW is the work input. With COPR=4.0COP_R = 4.0 and W=150 JW = 150\text{ J}, you can find: QC=COPR×W=4.0×150 J=600 JQ_C = COP_R \times W = 4.0 \times 150\text{ J} = 600\text{ J} Now apply the first law of thermodynamics to the complete cycle. Energy is conserved, so the work input plus the heat removed from the cold reservoir equals the heat delivered to the hot reservoir (surroundings): QH=W+QC=150 J+600 J=750 JQ_H = W + Q_C = 150\text{ J} + 600\text{ J} = 750\text{ J} Answer A (450 J450\text{ J}) incorrectly subtracts the work from the heat removed from the cold reservoir instead of adding them. Answer B (600 J600\text{ J}) gives you only QCQ_C, the heat removed from the cold reservoir, not the net heat delivered to the surroundings. Answer D (150 J150\text{ J}) represents just the work input, ignoring the heat transfer entirely. Remember: in refrigerator problems, the "net heat transfer to surroundings" means QHQ_H, which always equals the sum of work input and heat removed from the cold space. The refrigerator doesn't destroy energy—it moves it from cold to hot using additional work energy.

Question 17

A heat pump operates between outdoor air at TC=280 KT_C = 280\text{ K} and indoor air at TH=295 KT_H = 295\text{ K}. If the heat pump delivers QH=3000 JQ_H = 3000\text{ J} to the house and requires W=500 JW = 500\text{ J} of electrical work input per cycle, what is the heat extracted from the outdoor air?

  1. 2500 J2500\text{ J} (correct answer)
  2. 3000 J3000\text{ J}
  3. 3500 J3500\text{ J}
  4. 500 J500\text{ J}
  5. 1500 J1500\text{ J}
Explanation: When analyzing heat pump problems, you need to apply the first law of thermodynamics to the complete cycle. A heat pump extracts heat from a cold reservoir, adds work, and delivers heat to a warm reservoir. The energy conservation principle for any heat pump states: QH=QC+WQ_H = Q_C + W, where QHQ_H is heat delivered to the house, QCQ_C is heat extracted from outdoor air, and WW is work input. Rearranging this equation: QC=QH−WQ_C = Q_H - W. Substituting the given values: QC=3000 J−500 J=2500 JQ_C = 3000\text{ J} - 500\text{ J} = 2500\text{ J}. This confirms that answer A is correct. Looking at the wrong answers: B (3000 J3000\text{ J}) represents the heat delivered to the house, not extracted from outdoors—this confuses input and output. C (3500 J3500\text{ J}) incorrectly adds work to the delivered heat (QH+WQ_H + W), violating energy conservation by suggesting the heat pump extracts more energy than it delivers. D (500 J500\text{ J}) is simply the work input, showing confusion between the electrical energy supplied and thermal energy extracted. The temperatures given are actually unnecessary for this calculation—they would be needed to determine efficiency or coefficient of performance, but not for basic energy balance. Remember: Heat pump problems often provide extra information like temperatures. Focus on what's being asked and apply energy conservation systematically: the heat extracted plus work input must equal the heat delivered.

Question 18

An ideal heat engine operates in a cycle with efficiency η=0.30\eta = 0.30. If the engine performs W=450 JW = 450\text{ J} of work in one cycle, what is the heat input from the hot reservoir?

  1. 1350 J1350\text{ J}
  2. 1500 J1500\text{ J} (correct answer)
  3. 450 J450\text{ J}
  4. 135 J135\text{ J}
  5. 1050 J1050\text{ J}
Explanation: When you encounter heat engine problems, focus on the fundamental relationship between efficiency, work output, and heat input. Efficiency tells you what fraction of the input energy becomes useful work. The efficiency of a heat engine is defined as η=WQH\eta = \frac{W}{Q_H}, where WW is the work output and QHQ_H is the heat input from the hot reservoir. Rearranging this equation to solve for heat input: QH=WηQ_H = \frac{W}{\eta}. Substituting the given values: QH=450 J0.30=1500 JQ_H = \frac{450\text{ J}}{0.30} = 1500\text{ J}. This confirms answer B is correct. Let's examine why the other options are wrong. Answer A (1350 J) likely comes from incorrectly calculating QH=W+η×W=450+0.30×450=1350 JQ_H = W + \eta \times W = 450 + 0.30 \times 450 = 1350\text{ J}, which misapplies the efficiency formula. Answer C (450 J) suggests confusing work output with heat input—a fundamental misunderstanding since the engine must always take in more energy than it outputs as work. Answer D (135 J) results from multiplying work by efficiency (450×0.30=135450 \times 0.30 = 135), which would give you the heat rejected to the cold reservoir, not the heat input. Remember this key relationship: in any real heat engine, QH>W>QCQ_H > W > Q_C due to the second law of thermodynamics. The heat input must always exceed the work output, and efficiency tells you the conversion ratio. When solving these problems, always check that your heat input is greater than the work output.

Question 19

A refrigerator removes heat from a freezer at TC=250 KT_C = 250\text{ K} and rejects heat to a room at TH=300 KT_H = 300\text{ K}. If the refrigerator operates with 75% of the ideal coefficient of performance and removes QC=1800 JQ_C = 1800\text{ J} per cycle, what work input is required?

  1. 480 J480\text{ J} (correct answer)
  2. 360 J360\text{ J}
  3. 600 J600\text{ J}
  4. 240 J240\text{ J}
  5. 720 J720\text{ J}
Explanation: When you encounter refrigerator problems, remember that refrigerators are heat pumps operating in reverse - they use work to move heat from a cold reservoir to a hot reservoir. The key relationship is the coefficient of performance (COP). For an ideal refrigerator, the coefficient of performance is COPideal=TCTH−TCCOP_{ideal} = \frac{T_C}{T_H - T_C}. With your given temperatures: COPideal=250300−250=25050=5.0COP_{ideal} = \frac{250}{300 - 250} = \frac{250}{50} = 5.0 Since this refrigerator operates at 75% efficiency, the actual COP is: COPactual=0.75×5.0=3.75COP_{actual} = 0.75 \times 5.0 = 3.75 The coefficient of performance relates the heat removed to the work input: COP=QCWCOP = \frac{Q_C}{W} Solving for work: W=QCCOPactual=18003.75=480 JW = \frac{Q_C}{COP_{actual}} = \frac{1800}{3.75} = 480 \text{ J} Looking at the wrong answers: Choice B (360 J) likely comes from incorrectly using W=QC×0.75×TH−TCTHW = Q_C \times 0.75 \times \frac{T_H - T_C}{T_H}, mixing up efficiency relationships. Choice C (600 J) might result from using the temperature ratio directly as work: W=QC×TH−TCTC=1800×50250W = Q_C \times \frac{T_H - T_C}{T_C} = 1800 \times \frac{50}{250}. Choice D (240 J) probably comes from using the ideal COP without accounting for the 75% efficiency: W=18005.0×23W = \frac{1800}{5.0} \times \frac{2}{3}. The correct answer is A) 480 J. Study tip: Always distinguish between ideal and actual COP for refrigerators. Real devices never achieve theoretical maximum efficiency, so apply the efficiency factor to the ideal COP first, then use that to find work requirements.

Question 20

Three identical reversible heat engines operate in series between thermal reservoirs at T1=900 KT_1 = 900\text{ K}, T2=600 KT_2 = 600\text{ K}, T3=400 KT_3 = 400\text{ K}, and T4=200 KT_4 = 200\text{ K}. Each engine operates between consecutive temperature levels. If the first engine absorbs Q1=3600 JQ_1 = 3600\text{ J} from the 900 K900\text{ K} reservoir, and all heat rejected by one engine is absorbed by the next engine, what is the total work output of all three engines combined?

  1. 2400 J2400\text{ J} because each engine has the same efficiency
  2. 2700 J2700\text{ J} based on the overall temperature difference
  3. 2800 J2800\text{ J} from summing individual engine work outputs (correct answer)
  4. 3200 J3200\text{ J} since the engines operate in series
Explanation: For reversible engines: Engine 1 (900K to 600K): η1=1−600900=13\eta_1 = 1 - \frac{600}{900} = \frac{1}{3}, so W1=13×3600=1200 JW_1 = \frac{1}{3} \times 3600 = 1200\text{ J} and Q2=2400 JQ_2 = 2400\text{ J}. Engine 2 (600K to 400K): η2=1−400600=13\eta_2 = 1 - \frac{400}{600} = \frac{1}{3}, so W2=13×2400=800 JW_2 = \frac{1}{3} \times 2400 = 800\text{ J} and Q3=1600 JQ_3 = 1600\text{ J}. Engine 3 (400K to 200K): η3=1−200400=12\eta_3 = 1 - \frac{200}{400} = \frac{1}{2}, so W3=12×1600=800 JW_3 = \frac{1}{2} \times 1600 = 800\text{ J}. Total work: 1200+800+800=2800 J1200 + 800 + 800 = 2800\text{ J}. Choice A incorrectly assumes all engines have the same work output. Choice B uses overall efficiency 1−200900=791 - \frac{200}{900} = \frac{7}{9} giving 2800 J2800\text{ J} but with wrong reasoning. Choice D is arbitrary.