Cycles & Net Work - Thermodynamics

1. A refrigeration cycle operates between reservoirs at $T_H = 320\text{ K}$ and $T_C = 280\text{ K}$. The cycle consists of four processes, and during one complete cycle, the net work input is $100\text{ J}$ and the heat extracted from the cold reservoir is $700\text{ J}$. If this cycle is reversed to operate as a heat engine between the same reservoirs, what would be the heat absorbed from the hot reservoir?
   Answer: $800\text{ J}$ using energy conservation from the refrigeration cycle In the refrigeration cycle: $W_{in} = 100\text{ J}$, $Q_C = 700\text{ J}$ (extracted from cold), so $Q_H = Q_C + W_{in} = 700 + 100 = 800\text{ J}$ (rejected to hot). When the cycle is reversed to operate as a heat engine, the energy transfers reverse direction but maintain the same magnitudes for a reversible cycle. The engine would absorb $800\text{ J}$ from the hot reservoir, reject $700\text{ J}$ to the cold reservoir, and produce $100\text{ J}$ of work. Choice A incorrectly assumes the heat absorbed equals what was previously extracted. Choice C might come from misapplying COP calculations. Choice D incorrectly applies Carnot efficiency: $\eta = 1 - \frac{280}{320} = 0.125$, giving $W = 0.125 \times Q_H$, so $Q_H = \frac{100}{0.125} = 800\text{ J}$, but this reasoning assumes Carnot operation when the actual cycle performance is given.
2. A heat engine operates on a cycle where the working substance undergoes the following processes: (1) adiabatic compression with work input of $300\text{ J}$, (2) constant volume heat addition of $800\text{ J}$, (3) adiabatic expansion with work output of $500\text{ J}$, and (4) constant volume heat rejection. What is the efficiency of this heat engine?
   Answer: $25\%$ based on the ratio of net work to heat input Net work output: $W_{net} = W_{out} - W_{in} = 500 - 300 = 200\text{ J}$. Heat input occurs only during process 2: $Q_{in} = 800\text{ J}$. For the complete cycle, $\Delta U = 0$, so $Q_{net} = W_{net} = 200\text{ J}$. This means $Q_{out} = Q_{in} - Q_{net} = 800 - 200 = 600\text{ J}$ (rejected during process 4). Efficiency: $\eta = \frac{W_{net}}{Q_{in}} = \frac{200}{800} = 0.25 = 25\%$. Choice B incorrectly uses only expansion work: $\frac{500}{800} \times 100\% = 62.5\%$ but ignores compression work. Choice C might come from incorrect calculation. Choice D uses $\frac{500}{800} = 62.5\%$ which ignores the work input during compression.
3. A heat engine cycle produces a net work output of $W = 250\text{ J}$ while rejecting $Q_C = 750\text{ J}$ 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 $W = 250\text{ J}$ were input, how much heat would be delivered to the hot reservoir?
   Answer: $1000\text{ J}$ 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: $Q_H = W + Q_C = 250 + 750 = 1000\text{ J}$. 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 $Q_H = W + Q_C = 250 + 750 = 1000\text{ J}$. 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 $Q_H = W + Q_C$ for heat pumps to account for both work input and heat extracted from the cold side.
4. A refrigerator operates in a cycle between a cold reservoir at $T_C = 250\text{ K}$ and a hot reservoir at $T_H = 350\text{ K}$. During one cycle, the refrigerator removes $Q_C = 600\text{ J}$ from the cold reservoir and requires $W = 200\text{ J}$ of work input. What is the heat rejected to the hot reservoir?
   Answer: $800\text{ J}$ 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: $W + Q_C = Q_H$. This makes physical sense—the refrigerator takes energy from the cold space ($Q_C = 600\text{ J}$) and adds work energy ($W = 200\text{ J}$), then dumps this combined energy into the hot reservoir. Calculating: $Q_H = W + Q_C = 200\text{ J} + 600\text{ J} = 800\text{ J}$ Looking at the wrong answers: Choice A ($400\text{ J}$) represents the common error of subtracting work from heat removed ($Q_C - W$), which violates energy conservation. Choice B ($600\text{ J}$) incorrectly assumes the heat rejected equals the heat removed, ignoring the work input entirely. Choice D ($200\text{ J}$) 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: $800\text{ J}$. Study tip: For refrigerator problems, always start with energy conservation: $W + Q_C = Q_H$. 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.
5. An engine operates in a cycle with three processes. In process A→B, the system absorbs $Q_1 = 400\text{ J}$ and does $W_1 = 150\text{ J}$ of work. In process B→C, the system rejects $Q_2 = 200\text{ J}$ and $W_2 = 100\text{ J}$ of work is done on it. In process C→A, the system does $W_3 = 80\text{ J}$ of work. What is the net heat absorbed by the system in the complete cycle?
   Answer: $130\text{ J}$ When you encounter cyclic thermodynamic processes, remember that the first law of thermodynamics ($\Delta U = Q - W$) 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: $\Delta U_1 = Q_1 - W_1 = 400\text{ J} - 150\text{ J} = 250\text{ J}$. In process B→C: the system rejects 200 J (so $Q_2 = -200\text{ J}$) and has work done on it (so $W_2 = -100\text{ J}$), giving $\Delta U_2 = -200\text{ J} - (-100\text{ J}) = -100\text{ J}$. In process C→A: $\Delta U_3 = Q_3 - 80\text{ J}$. Since $\Delta U_{total} = 0$ for the complete cycle: $250\text{ J} + (-100\text{ J}) + \Delta U_3 = 0$, so $\Delta U_3 = -150\text{ J}$. Therefore: $Q_3 = \Delta U_3 + W_3 = -150\text{ J} + 80\text{ J} = -70\text{ J}$. The net heat absorbed is: $Q_{net} = Q_1 + Q_2 + Q_3 = 400\text{ J} + (-200\text{ J}) + (-70\text{ J}) = 130\text{ J}$. This confirms answer A. Answer B (200 J) incorrectly uses only the difference $Q_1 - Q_2$, ignoring process C→A. Answer C (330 J) mistakenly adds $Q_1$ and $Q_2$ as both positive values, forgetting that heat rejection means negative $Q$. Answer D (400 J) only considers the heat input from process A→B. Remember: in cyclic processes, always account for all three quantities ($Q$, $W$, and $\Delta U$) in each step, and use the constraint that $\Delta U_{cycle} = 0$.
6. A cyclic heat engine operates between reservoirs at $400\text{ K}$ and $300\text{ K}$. The engine absorbs $1200\text{ J}$ from the hot reservoir and produces $240\text{ J}$ of work per cycle. How does this engine's efficiency compare to the maximum possible efficiency?
   Answer: The engine operates at 80% of maximum efficiency 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: $\eta_{actual} = \frac{W}{Q_H} = \frac{240\text{ J}}{1200\text{ J}} = 0.20 = 20\%$ Next, find the maximum possible efficiency using the Carnot formula: $\eta_{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{300\text{ K}}{400\text{ K}} = 1 - 0.75 = 0.25 = 25\%$ The ratio of actual to maximum efficiency is: $\frac{20\%}{25\%} = 0.80 = 80\%$ 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.
7. An ideal heat engine operates in a cycle with efficiency $\eta = 0.30$. If the engine performs $W = 450\text{ J}$ of work in one cycle, what is the heat input from the hot reservoir?
   Answer: $1500\text{ J}$ 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 $\eta = \frac{W}{Q_H}$, where $W$ is the work output and $Q_H$ is the heat input from the hot reservoir. Rearranging this equation to solve for heat input: $Q_H = \frac{W}{\eta}$. Substituting the given values: $Q_H = \frac{450\text{ J}}{0.30} = 1500\text{ J}$. This confirms answer B is correct. Let's examine why the other options are wrong. Answer A (1350 J) likely comes from incorrectly calculating $Q_H = W + \eta \times W = 450 + 0.30 \times 450 = 1350\text{ J}$, 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 ($450 \times 0.30 = 135$), which would give you the heat rejected to the cold reservoir, not the heat input. Remember this key relationship: in any real heat engine, $Q_H > W > Q_C$ 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.
8. A heat engine operates between two thermal reservoirs at temperatures $T_H = 500\text{ K}$ and $T_C = 300\text{ K}$. In one complete cycle, the engine absorbs $Q_H = 1200\text{ J}$ from the hot reservoir and rejects $Q_C = 800\text{ J}$ to the cold reservoir. What is the net work output of this cycle?
   Answer: $400\text{ J}$ 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: $W = Q_H - Q_C$, where $W$ is the net work output, $Q_H$ is heat absorbed from the hot reservoir, and $Q_C$ is heat rejected to the cold reservoir. Plugging in the given values: $W = 1200\text{ J} - 800\text{ J} = 400\text{ J}$. This confirms that answer A is correct. Let's examine why the other options are wrong. Answer B ($800\text{ J}$) represents the heat rejected to the cold reservoir, not the work output. Answer C ($1200\text{ J}$) 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 ($2000\text{ J}$) appears to be the sum of heat input and output ($Q_H + Q_C$), which has no physical meaning in this context and also violates energy conservation. Remember this key relationship: $W = Q_H - Q_C$. 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.
9. Three identical reversible heat engines operate in series between thermal reservoirs at $T_1 = 900\text{ K}$, $T_2 = 600\text{ K}$, $T_3 = 400\text{ K}$, and $T_4 = 200\text{ K}$. Each engine operates between consecutive temperature levels. If the first engine absorbs $Q_1 = 3600\text{ J}$ from the $900\text{ K}$ 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?
   Answer: $2800\text{ J}$ from summing individual engine work outputs For reversible engines: Engine 1 (900K to 600K): $\eta_1 = 1 - \frac{600}{900} = \frac{1}{3}$, so $W_1 = \frac{1}{3} \times 3600 = 1200\text{ J}$ and $Q_2 = 2400\text{ J}$. Engine 2 (600K to 400K): $\eta_2 = 1 - \frac{400}{600} = \frac{1}{3}$, so $W_2 = \frac{1}{3} \times 2400 = 800\text{ J}$ and $Q_3 = 1600\text{ J}$. Engine 3 (400K to 200K): $\eta_3 = 1 - \frac{200}{400} = \frac{1}{2}$, so $W_3 = \frac{1}{2} \times 1600 = 800\text{ J}$. Total work: $1200 + 800 + 800 = 2800\text{ J}$. Choice A incorrectly assumes all engines have the same work output. Choice B uses overall efficiency $1 - \frac{200}{900} = \frac{7}{9}$ giving $2800\text{ J}$ but with wrong reasoning. Choice D is arbitrary.
10. A refrigerator removes heat from a freezer at $T_C = 250\text{ K}$ and rejects heat to a room at $T_H = 300\text{ K}$. If the refrigerator operates with 75% of the ideal coefficient of performance and removes $Q_C = 1800\text{ J}$ per cycle, what work input is required?
    Answer: $480\text{ J}$ 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 $COP_{ideal} = \frac{T_C}{T_H - T_C}$. With your given temperatures: $COP_{ideal} = \frac{250}{300 - 250} = \frac{250}{50} = 5.0$ Since this refrigerator operates at 75% efficiency, the actual COP is: $COP_{actual} = 0.75 \times 5.0 = 3.75$ The coefficient of performance relates the heat removed to the work input: $COP = \frac{Q_C}{W}$ Solving for work: $W = \frac{Q_C}{COP_{actual}} = \frac{1800}{3.75} = 480 \text{ J}$ Looking at the wrong answers: Choice B (360 J) likely comes from incorrectly using $W = Q_C \times 0.75 \times \frac{T_H - T_C}{T_H}$, mixing up efficiency relationships. Choice C (600 J) might result from using the temperature ratio directly as work: $W = Q_C \times \frac{T_H - T_C}{T_C} = 1800 \times \frac{50}{250}$. Choice D (240 J) probably comes from using the ideal COP without accounting for the 75% efficiency: $W = \frac{1800}{5.0} \times \frac{2}{3}$. 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.
11. A heat pump operates between outdoor air at $T_C = 280\text{ K}$ and indoor air at $T_H = 295\text{ K}$. If the heat pump delivers $Q_H = 3000\text{ J}$ to the house and requires $W = 500\text{ J}$ of electrical work input per cycle, what is the heat extracted from the outdoor air?
    Answer: $2500\text{ J}$ 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: $Q_H = Q_C + W$, where $Q_H$ is heat delivered to the house, $Q_C$ is heat extracted from outdoor air, and $W$ is work input. Rearranging this equation: $Q_C = Q_H - W$. Substituting the given values: $Q_C = 3000\text{ J} - 500\text{ J} = 2500\text{ J}$. This confirms that answer A is correct. Looking at the wrong answers: B ($3000\text{ J}$) represents the heat delivered to the house, not extracted from outdoors—this confuses input and output. C ($3500\text{ J}$) incorrectly adds work to the delivered heat ($Q_H + W$), violating energy conservation by suggesting the heat pump extracts more energy than it delivers. D ($500\text{ J}$) 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.
12. A gas power cycle consists of four processes: (1-2) adiabatic compression, (2-3) constant pressure heat addition, (3-4) adiabatic expansion, and (4-1) constant pressure heat rejection. The temperatures at the four states are $T_1 = 300\text{ K}$, $T_2 = 600\text{ K}$, $T_3 = 900\text{ K}$, and $T_4 = 450\text{ K}$. If the constant pressure heat addition process adds $1200\text{ J}$, what is the net work output of this cycle?
    Answer: $600\text{ J}$ from the difference between heat addition and rejection For constant pressure processes with an ideal gas: $Q = nC_p\Delta T$. Process 2-3 (heat addition): $Q_{23} = nC_p(T_3 - T_2) = nC_p(900 - 600) = 300nC_p = 1200\text{ J}$, so $nC_p = 4\text{ J/K}$. Process 4-1 (heat rejection): $Q_{41} = nC_p(T_1 - T_4) = 4(300 - 450) = 4(-150) = -600\text{ J}$. So $600\text{ J}$ is rejected. Net heat: $Q_{net} = 1200 - 600 = 600\text{ J}$. For a complete cycle, $\Delta U = 0$, so $W_{net} = Q_{net} = 600\text{ J}$. Choice A might come from incorrect temperature difference calculations. Choice C ignores the constant pressure processes. Choice D is arbitrary. The correct approach recognizes that for any cycle, net work equals net heat transfer.
13. A gas undergoes a thermodynamic cycle consisting of: (1) constant pressure expansion from $V_1 = 2\text{ L}$ to $V_2 = 6\text{ L}$ at $P = 3\text{ atm}$, (2) constant volume cooling where pressure drops to $1\text{ atm}$, and (3) isothermal compression back to the initial state. If the internal energy change during the constant volume process is $-800\text{ J}$, what is the net work done by the gas during the complete cycle?
    Answer: $993\text{ J}$ accounting for the complete thermodynamic cycle Converting to SI: $P = 3\text{ atm} = 303,975\text{ Pa}$, $V_1 = 0.002\text{ m}^3$, $V_2 = 0.006\text{ m}^3$. Process 1 (isobaric expansion): $W_1 = P(V_2 - V_1) = 303,975 \times 0.004 = 1216\text{ J}$. Process 2 (isochoric): $W_2 = 0\text{ J}$. Process 3 (isothermal compression): From the pressure change in process 2 and ideal gas law, the temperature ratio gives us the compression work. Using $W_3 = nRT_3\ln(V_1/V_2)$ and the state relationships, $W_3 = -223\text{ J}$. Net work: $W_{net} = 1216 + 0 + (-223) = 993\text{ J}$.
14. A Carnot refrigerator operates between reservoirs at $T_H = 320\text{ K}$ and $T_C = 280\text{ K}$. If the refrigerator removes $Q_C = 2100\text{ J}$ from the cold reservoir per cycle, what is the minimum work input required?
    Answer: $300\text{ J}$ 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: $\frac{Q_H}{Q_C} = \frac{T_H}{T_C}$. Since $Q_C = 2100\text{ J}$, $T_H = 320\text{ K}$, and $T_C = 280\text{ K}$, you can find: $Q_H = Q_C \times \frac{T_H}{T_C} = 2100 \times \frac{320}{280} = 2400\text{ J}$ The work input equals the difference between heat rejected to the hot reservoir and heat absorbed from the cold reservoir: $W = Q_H - Q_C = 2400 - 2100 = 300\text{ J}$. 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 $Q_H - Q_C$. The coefficient of performance formula $COP = \frac{T_C}{T_H - T_C}$ can also verify your answer: $COP = \frac{280}{40} = 7$, so $W = \frac{Q_C}{COP} = \frac{2100}{7} = 300\text{ J}$.
15. A heat engine cycle consists of three processes. The system does $W_1 = 200\text{ J}$ of work in process 1, work $W_2 = -150\text{ J}$ is done on the system in process 2, and the system does $W_3 = 100\text{ J}$ of work in process 3. What is the net work done by the system in this complete cycle?
    Answer: $150\text{ J}$ 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 $W_1 = 200\text{ J}$ of work (positive), has $W_2 = -150\text{ J}$ of work done on it (negative), and does $W_3 = 100\text{ J}$ of work (positive). The net work is: $W_{net} = 200 + (-150) + 100 = 150\text{ J}$ Answer A ($150\text{ J}$) is correct because it properly accounts for the signs and adds all three work terms. Answer B ($250\text{ J}$) results from incorrectly treating the $-150\text{ J}$ as positive, giving $200 + 150 + 100 = 450\text{ J}$, then somehow getting $250\text{ J}$ through calculation error. Answer C ($450\text{ J}$) comes from ignoring the negative sign entirely and adding $200 + 150 + 100 = 450\text{ J}$. Answer D ($50\text{ J}$) might result from incorrectly subtracting instead of adding one of the positive work values, such as $200 - 150 - 100 = -50\text{ J}$ 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.
16. A refrigerator has a coefficient of performance $COP_R = 4.0$. If the refrigerator requires $W = 150\text{ J}$ of work input per cycle, what is the net heat transfer to the surroundings during one complete cycle?
    Answer: $750\text{ J}$ 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 $COP_R = \frac{Q_C}{W}$, where $Q_C$ is the heat removed from the cold reservoir and $W$ is the work input. With $COP_R = 4.0$ and $W = 150\text{ J}$, you can find: $Q_C = COP_R \times W = 4.0 \times 150\text{ J} = 600\text{ J}$ 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): $Q_H = W + Q_C = 150\text{ J} + 600\text{ J} = 750\text{ J}$ Answer A ($450\text{ J}$) incorrectly subtracts the work from the heat removed from the cold reservoir instead of adding them. Answer B ($600\text{ J}$) gives you only $Q_C$, the heat removed from the cold reservoir, not the net heat delivered to the surroundings. Answer D ($150\text{ J}$) represents just the work input, ignoring the heat transfer entirely. Remember: in refrigerator problems, the "net heat transfer to surroundings" means $Q_H$, 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.
17. A cyclic process returns a system to its initial state. During the cycle, the system absorbs $Q_1 = 800\text{ J}$ in process 1-2, rejects $Q_2 = 300\text{ J}$ in process 2-3, and absorbs $Q_3 = 200\text{ J}$ in process 3-1. What is the net work done by the system during this complete cycle?
    Answer: $700\text{ J}$ When you encounter a cyclic process problem, remember that the system returns to its initial state, making the change in internal energy zero ($\Delta U = 0$). This is the key insight that unlocks these problems. Since $\Delta U = 0$ for any complete cycle, the first law of thermodynamics ($\Delta U = Q - W$) simplifies to $Q_{net} = W_{net}$. 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: $+800\text{ J}$ (absorbed), Process 2-3: $-300\text{ J}$ (rejected), Process 3-1: $+200\text{ J}$ (absorbed). Therefore: $Q_{net} = 800 - 300 + 200 = 700\text{ J}$ Since $W_{net} = Q_{net}$, the net work done by the system is $700\text{ J}$, confirming answer B. The wrong answers represent common mistakes: A ($500\text{ J}$) likely comes from incorrectly subtracting the last heat term instead of adding it. C ($1000\text{ J}$) results from adding the absolute values of all heat transfers without considering signs. D ($300\text{ J}$) might come from confusing which heat transfer to use or making arithmetic errors. Remember: for cyclic processes, always use $\Delta U = 0$ to connect net heat and net work. Pay careful attention to the signs of heat transfers—absorbed is positive, rejected is negative.
18. A heat engine completes 50 cycles per minute. In each cycle, it absorbs $Q_H = 240\text{ J}$ and rejects $Q_C = 160\text{ J}$. What is the power output of this engine?
    Answer: $66.7\text{ W}$ 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: $W = Q_H - Q_C = 240\text{ J} - 160\text{ J} = 80\text{ J}$. 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 $\frac{50}{60} = 0.833$ cycles per second. Power is work per unit time, so: $P = W \times \text{frequency} = 80\text{ J} \times 0.833\text{ s}^{-1} = 66.7\text{ W}$. This confirms answer A is correct. The wrong answers represent common calculation errors. Answer B ($80\text{ W}$) 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 ($200\text{ W}$) results from incorrectly multiplying $Q_H$ by the wrong frequency conversion. Answer D ($133\text{ W}$) 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 $W = Q_H - Q_C$, then multiply by the cycle frequency in proper units (cycles per second, not per minute).
19. A heat pump operates between an outdoor temperature of $T_C = 280\text{ K}$ and an indoor temperature of $T_H = 295\text{ K}$. The heat pump has an actual coefficient of performance that is 60% of the ideal value. If $W = 600\text{ J}$ of work is input per cycle, how much heat is delivered to the house?
    Answer: $7080\text{ J}$ 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 $\text{COP}_{\text{ideal}} = \frac{T_H}{T_H - T_C}$. Substituting the given temperatures: $\text{COP}_{\text{ideal}} = \frac{295}{295 - 280} = \frac{295}{15} = 19.67$. Since the actual COP is 60% of the ideal value: $\text{COP}_{\text{actual}} = 0.60 \times 19.67 = 11.8$. The coefficient of performance for a heat pump is defined as $\text{COP} = \frac{Q_H}{W}$, where $Q_H$ is the heat delivered to the house and $W$ is the work input. Solving for the heat delivered: $Q_H = \text{COP}_{\text{actual}} \times W = 11.8 \times 600 = 7080\text{ J}$. Looking at the wrong answers: B (4248 J) likely results from using the refrigerator COP formula ($Q_C/W$) 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 ($Q_H/W$) and refrigerator COP ($Q_C/W$). 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.
20. A Carnot engine operates between reservoirs at $T_H = 600\text{ K}$ and $T_C = 400\text{ K}$. In one cycle, the engine absorbs $Q_H = 900\text{ J}$ from the hot reservoir. What is the net work output for this cycle?
    Answer: $300\text{ J}$ 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: $\eta = 1 - \frac{T_C}{T_H}$. Here, $\eta = 1 - \frac{400}{600} = 1 - \frac{2}{3} = \frac{1}{3}$. The net work output equals the efficiency times the heat absorbed: $W = \eta \times Q_H = \frac{1}{3} \times 900\text{ J} = 300\text{ J}$. 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 $\frac{200}{900} = 0.22$, which is too low for these temperatures. Answer C (600 J) represents an efficiency of $\frac{600}{900} = 0.67$, 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.

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