AP Physics 2 Quiz: Entropy And Second Law Of Thermodynamics

What this quiz covers

This quiz focuses on Entropy And Second Law Of Thermodynamics, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

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.

Question 1

1. The total entropy decreases because energy flows from hot to cold, reducing temperature differences.
2. The total entropy is zero because the energy lost by one block equals the energy gained by the other.
3. The total entropy increases because the magnitude of $\Delta S$ for the cooler block exceeds that of the warmer block. (correct answer)
4. The total entropy increases only if the blocks melt, since entropy change requires a phase change.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When two blocks at different temperatures equilibrate in isolation, heat flows from hot to cold spontaneously. The second law requires total entropy to increase for this spontaneous process. The cooler block gains heat Q at lower temperature (300 K), while the hotter block loses the same heat Q at higher temperature (350 K), so |ΔS_cold| = Q/300 > |ΔS_hot| = Q/350, making the total entropy change positive. Choice A incorrectly assumes reducing temperature differences decreases entropy, confusing uniformity with total entropy. The principle is: when equal amounts of heat transfer occur, the entropy gain at lower temperature exceeds the entropy loss at higher temperature.

Question 2

1. The device is possible because it conserves energy and cycles back to its initial state.
2. The device is possible if its internal entropy decreases each cycle.
3. The device is impossible because a cyclic engine cannot convert heat from one reservoir entirely into work. (correct answer)
4. The device is possible if the reservoir is large enough to supply constant temperature.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. The proposed device violates the Kelvin-Planck statement of the second law: no cyclic engine can convert heat from a single reservoir entirely into work. Such a device would decrease the entropy of the universe (removing heat/entropy from the reservoir without adding it elsewhere), making it impossible. All cyclic heat engines require at least two reservoirs at different temperatures, rejecting some heat to the cold reservoir. The Carnot efficiency limit shows that even ideal engines cannot achieve 100% conversion when operating between finite temperatures. Choice A incorrectly focuses only on energy conservation, ignoring entropy constraints—the second law imposes additional restrictions beyond the first law. When evaluating proposed devices, check both energy conservation and entropy requirements.

Question 3

1. It fails because it would require the entropy of the universe to decrease for a complete cycle. (correct answer)
2. It fails because work is a form of heat, so $Q$ cannot be fully converted into $W$.
3. It fails because the reservoir's temperature would have to increase as it loses heat.
4. It fails because the efficiency of any engine must exceed $100\%$ to be useful.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. The proposed device attempts to convert heat entirely into work using a single reservoir, which violates the Kelvin-Planck statement of the second law. For a cyclic process extracting heat Q from a single reservoir, the entropy change would be ΔS_universe = -Q/T < 0, decreasing the total entropy of the universe, which is forbidden for spontaneous processes. Choice B incorrectly claims work is a form of heat, confusing different energy transfer mechanisms. The key principle is: no cyclic device can convert heat entirely to work while operating with a single thermal reservoir.

Question 4

1. The total entropy decreases because the system becomes more uniform after melting.
2. The total entropy increases because heat transfer from warmer air to colder ice-water yields net positive $\Delta S$. (correct answer)
3. The total entropy is zero because melting is a phase change at constant temperature.
4. The total entropy increases only if the ice appears more disordered than liquid water.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When ice melts in a warm room, heat flows spontaneously from the warmer air to the colder ice-water mixture. The second law requires that total entropy increases for this spontaneous process. The room loses heat at 293 K while the ice-water gains heat at 273 K, so the entropy increase of the cold system exceeds the entropy decrease of the warm room, yielding net positive ΔS. Choice C incorrectly assumes phase changes have zero entropy change, ignoring that melting increases entropy due to increased molecular freedom. The strategy is: for spontaneous heat flow, entropy gain of the cold object exceeds entropy loss of the hot object.

Question 5

1. The total entropy of the isolated block–water system increases as energy disperses from hot to cold. (correct answer)
2. The entropy of the isolated block–water system must decrease because the final temperature is uniform.
3. The process occurs because entropy measures disorder, and the water becomes more disordered than the metal.
4. The process occurs because the thermal energy transferred to the water exceeds the energy lost by the block.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When a hot metal block contacts cooler water in an isolated system, heat flows spontaneously from hot to cold until thermal equilibrium is reached. The second law states that the total entropy of an isolated system must increase for any spontaneous process. During this heat transfer, the entropy decrease of the cooling block is smaller in magnitude than the entropy increase of the warming water (since ΔS = Q/T and the water receives heat at a lower temperature). Choice B incorrectly assumes uniform temperature means lower entropy, confusing equilibrium with disorder. The key strategy is: for spontaneous processes in isolated systems, total entropy always increases.

Question 6

1. The total entropy of the isolated block–water system increases during the heat transfer. (correct answer)
2. The total entropy of the isolated block–water system decreases because the system becomes more ordered.
3. The total entropy of the isolated block–water system remains zero because energy is conserved.
4. The total entropy of the isolated block–water system must decrease if the final temperature is uniform.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When a hot metal block contacts cold water in an isolated system, heat flows spontaneously from hot to cold until thermal equilibrium is reached. During this irreversible process, the total entropy of the isolated system must increase according to the second law. The entropy decrease of the cooling block is smaller in magnitude than the entropy increase of the warming water because entropy changes are inversely proportional to temperature (ΔS = Q/T). Choice B incorrectly assumes that uniform temperature means more order and less entropy, but this confuses microscopic disorder with macroscopic uniformity. Remember: in any spontaneous process in an isolated system, total entropy always increases or remains constant (reversible case only).

Question 7

1. The total entropy of the block-plus-water system increases because energy flows from higher to lower temperature. (correct answer)
2. The total entropy of the block-plus-water system decreases because the system becomes more ordered as it equilibrates.
3. The total entropy of the block-plus-water system must decrease so that the final temperature is between $300\,\text{K}$ and $400\,\text{K}$.
4. The total entropy of the block-plus-water system remains constant because the container is insulated.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When a hot metal block is placed in contact with cooler water in an insulated container, heat flows spontaneously from the hot block to the cold water until thermal equilibrium is reached. According to the second law, the total entropy of an isolated system must increase for any spontaneous process. The entropy increase occurs because energy disperses from the concentrated high-temperature region to the lower-temperature region, creating more accessible microstates overall. Choice B incorrectly claims entropy decreases due to "ordering," confusing macroscopic uniformity with microscopic disorder. Remember: in any spontaneous heat transfer process, total entropy increases even though energy is conserved.

Question 8

1. Moving heat from cold to hot would decrease total entropy unless work is added. (correct answer)
2. Work is required because entropy must decrease in any cyclic device.
3. Moving heat from cold to hot conserves energy but violates the first law unless work is added.
4. Work is required because the room's entropy must always decrease during cooling.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. A refrigerator moves heat from a cold space (270 K) to a warmer room (300 K), which is the opposite of spontaneous heat flow. This process would decrease total entropy if done alone: removing heat from the cold compartment decreases its entropy more than adding the same heat to the warm room increases the room's entropy (since |ΔS| = |Q|/T is larger at lower T). To make this process possible while still increasing total entropy, work must be added to the system, which ultimately gets converted to additional heat expelled to the room. Choice B incorrectly claims this violates the first law—energy is conserved, but the second law requires work input. Remember that moving heat from cold to hot requires work input to ensure total entropy increases.

Question 9

1. The total entropy of gas plus surroundings increases because irreversibility produces entropy. (correct answer)
2. The total entropy of gas plus surroundings is zero because the gas loses heat.
3. The total entropy of gas plus surroundings decreases because friction creates order.
4. The total entropy of gas plus surroundings increases only if the gas temperature remains constant.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. During rapid compression with friction, work is done on the gas in an irreversible manner, and 300 J of heat flows to the surroundings. Irreversible processes always generate entropy, so even though the gas might lose entropy (due to compression and heat loss), the surroundings gain more entropy than the gas loses. The friction converts organized mechanical work into disorganized thermal energy, creating additional entropy beyond what reversible compression would produce. The total entropy of the gas plus surroundings must increase for any spontaneous, irreversible process. Choice C incorrectly claims friction creates order—friction actually increases disorder by converting mechanical energy to heat. For any irreversible process, total entropy of the universe always increases, regardless of local decreases.

Question 10

1. The device is possible if the reservoir is large enough to keep its temperature constant.
2. The device is possible because energy conservation allows $W=Q$ in a cycle.
3. The device is possible because entropy in an isolated system can decrease during a cycle.
4. The device is impossible because a cycle cannot convert heat from one reservoir entirely into work. (correct answer)

Explanation: This question tests understanding of entropy and the second law of thermodynamics. The proposed device claims to convert all heat from a single reservoir into work in a complete cycle, which would be a perfect heat engine with 100% efficiency. This violates the Kelvin-Planck statement of the second law: no cyclic process can convert heat entirely into work while operating with a single heat reservoir. Such a device would decrease the entropy of the universe (reservoir loses entropy, nothing gains it), making it impossible. All real heat engines must reject some heat to a cold reservoir, limiting their efficiency to less than 100%. Choice B incorrectly suggests energy conservation allows this—while energy is conserved, the second law imposes additional restrictions beyond conservation. Remember that any cyclic heat engine requires at least two reservoirs at different temperatures.

Question 11

1. It is consistent because external work allows heat to be transferred from cold to hot in a cycle. (correct answer)
2. It violates the second law because it moves heat from cold to hot, which can never occur.
3. It violates the second law because the coefficient of performance is greater than $1$.
4. It is consistent only if it converts more than $100\%$ of the work into extracted heat.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. A refrigerator moves heat from cold to hot, which doesn't occur spontaneously, but the second law allows this when external work is supplied. The device must reject Qh = Qc + W = 250 J to the hot side while consuming 50 J of work, maintaining positive entropy production overall. The coefficient of performance (COP = Qc/W = 4) being greater than 1 is perfectly allowed and doesn't violate any laws. Choice B incorrectly claims heat can never move from cold to hot, misunderstanding that work input makes this possible. Remember: non-spontaneous processes can occur when external work compensates for entropy decrease.

Question 12

1. The gas's entropy change is negative because expansion makes the gas more disordered.
2. The gas's entropy change is $\Delta S=\frac{T}{Q_{rev}}=\frac{350}{500}\,\text{J/K}$.
3. The gas's entropy change is zero because the internal energy of an ideal gas is constant at constant $T$.
4. The gas's entropy change is $\Delta S=\frac{Q_{rev}}{T}=\frac{500}{350}\,\text{J/K}$. (correct answer)

Explanation: This question tests understanding of entropy and the second law of thermodynamics. For a reversible isothermal process, the entropy change is calculated using ΔS = Q_rev/T, where Q_rev is the heat absorbed reversibly and T is the constant temperature. When the ideal gas expands isothermally at 350 K while absorbing 500 J of heat, its entropy increases by ΔS = 500/350 = 10/7 ≈ 1.43 J/K. This positive entropy change reflects the increased volume and thus increased number of accessible microstates for the gas molecules. Choice D incorrectly inverts the formula as T/Q_rev—entropy has units of J/K, which requires Q/T not T/Q. For any reversible isothermal process, use ΔS = Q_rev/T to calculate entropy change, with positive Q for heat absorbed.

Question 13

1. The gas's entropy must be zero because $Q\approx 0$ implies $\Delta S=0$ for any path.
2. The gas's entropy increases because the process is irreversible and accessible microstates increase. (correct answer)
3. The gas's entropy increases only if the gas converts internal energy to work with efficiency above $100\%$.
4. The gas's entropy decreases because no heat enters, so $\Delta S<0$ must occur.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When a compressed gas expands rapidly (free expansion) with negligible work and heat exchange, the process is irreversible and increases entropy. Even though Q ≈ 0, entropy increases because the gas molecules can access more spatial configurations in the larger volume, increasing the number of microstates. The second law requires entropy to increase in irreversible processes, regardless of heat transfer. This demonstrates that entropy is fundamentally about the number of accessible microstates, not just heat flow. Choice B incorrectly assumes entropy must decrease without heat input, misunderstanding that entropy can increase through volume changes alone. The key principle is that for irreversible expansion, entropy increases due to increased molecular disorder, even in adiabatic conditions.

Question 14

1. Work is required because the refrigerator must have efficiency greater than $100\%$ to move heat.
2. Work is required because total entropy must decrease in any cyclic process.
3. Work is required because entropy is the same as disorder, and the room must become more ordered.
4. Work is required because transferring heat from cold to hot would otherwise decrease total entropy. (correct answer)

Explanation: This question tests understanding of entropy and the second law of thermodynamics. A refrigerator moves heat from a cold space to a warmer room, which is the opposite of the spontaneous direction of heat flow. According to the second law, this process would decrease the total entropy of the universe if it occurred by itself, since entropy change is ΔS = Q/T and removing heat from the cold side decreases entropy more than adding it to the warm side increases it. To make this non-spontaneous process occur, work must be input to increase the total entropy elsewhere, ensuring the universe's entropy still increases overall. Choice C incorrectly equates entropy with everyday disorder rather than thermodynamic microstates. When analyzing refrigeration cycles, remember that work input is required to move heat against its natural direction while still satisfying the second law.

Question 15

1. The proposal violates the second law because a cyclic device cannot convert heat from one reservoir entirely into work. (correct answer)
2. The proposal is allowed because energy conservation alone permits $100\%$ efficiency.
3. The proposal is allowed because entropy in an isolated system can decrease during a cycle.
4. The proposal is allowed because entropy is only a measure of disorder and can be eliminated by careful design.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. The proposed engine violates the Kelvin-Planck statement of the second law, which states that no cyclic device can convert heat from a single thermal reservoir entirely into work. Such a device would decrease the entropy of the universe: the reservoir loses entropy (ΔS = -Q/T) while no compensating entropy increase occurs elsewhere, since all energy becomes organized work. Real engines must reject some heat to a cold reservoir to ensure total entropy increases. Choice B incorrectly suggests energy conservation alone determines what's possible, ignoring entropy constraints. When evaluating proposed engines, check both energy conservation and entropy increase to determine feasibility.

Question 16

1. Total entropy stays zero because the blocks are identical and end at the average temperature.
2. Total entropy increases because heat transfer between different temperatures is irreversible. (correct answer)
3. Total entropy decreases because disorder decreases when temperatures become more uniform.
4. Total entropy decreases because energy is conserved in the insulated box.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When two blocks at different temperatures come into thermal contact within an insulated system, heat flows spontaneously from hot to cold until they reach the same temperature. This irreversible heat transfer increases the total entropy because the cold block gains more entropy than the hot block loses (since ΔS = Q/T and the same Q is transferred at different temperatures). The final uniform temperature represents a more probable macrostate with higher entropy than the initial state of temperature difference. Choice D incorrectly claims entropy decreases when temperatures equalize, misunderstanding that uniform temperature actually represents higher entropy. To analyze thermal equilibration, calculate entropy changes using ΔS = Q/T and verify total entropy increases.

Question 17

1. The engine can have efficiency greater than $1$ because it is cyclic and returns to its initial state.
2. The engine can convert all absorbed heat to work because entropy is conserved in a cycle.
3. The engine must reject some heat to the cold reservoir; it cannot convert all absorbed heat to work. (correct answer)
4. The engine must absorb heat from the cold reservoir because entropy must decrease in the universe.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. A heat engine operating between two thermal reservoirs must obey the Kelvin-Planck statement of the second law: no cyclic engine can convert heat entirely into work while operating with a single reservoir. Even with two reservoirs, the engine cannot convert all absorbed heat Q_h into work; some heat Q_c must be rejected to the cold reservoir to ensure the total entropy of the universe increases. The maximum theoretical efficiency is limited by the Carnot efficiency η = 1 - T_c/T_h, which is always less than 100%. Choice B incorrectly assumes all heat can be converted to work in a cycle, violating the second law. When analyzing heat engines, remember that some heat rejection is always necessary to satisfy entropy requirements.

Question 18

1. The claim violates the second law because it implies $Q_C=300\,\text{J}$ and an efficiency exceeding the Carnot limit. (correct answer)
2. The claim is allowed because any engine can have efficiency greater than $100\%$ if $T_H>T_C$.
3. The claim is allowed because entropy in an isolated system can decrease during a cycle.
4. The claim is allowed because entropy depends only on disorder, and work output increases disorder.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. The claimed engine absorbs Q_H = 1000 J and produces W = 700 J, implying Q_C = Q_H - W = 300 J by energy conservation. This gives an efficiency η = W/Q_H = 700/1000 = 0.70 = 70%, which exceeds the Carnot limit η_Carnot = 1 - T_C/T_H = 1 - 300/500 = 0.40 = 40%. The second law states no heat engine can exceed the Carnot efficiency between given temperatures, as this would decrease total entropy. Choice B incorrectly claims efficiency can exceed 100%, violating both energy conservation and the second law. The key strategy is to always check if a proposed engine's efficiency exceeds the Carnot limit (1 - T_C/T_H) for the given temperatures.

Question 19

1. The total entropy stays constant because a phase change occurs at constant temperature.
2. The melting requires the room to convert heat to work with efficiency greater than $100\%$.
3. The total entropy of ice plus room increases because heat transfer occurs from higher to lower temperature. (correct answer)
4. The total entropy decreases because solid water is more ordered than liquid water.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. When ice melts by absorbing heat from a warmer room, the total entropy increases because heat flows from higher temperature (room at 298 K) to lower temperature (ice at 273 K). The entropy gain of the ice (ΔS_ice = Q/273 K) exceeds the entropy loss of the room (ΔS_room = -Q/298 K) because the same heat Q is divided by a smaller temperature for the ice. Additionally, the phase transition from ordered solid to disordered liquid further increases the ice's entropy. Choice B incorrectly claims total entropy decreases, focusing only on the ice becoming less ordered while ignoring the larger entropy increase from heat transfer. The strategy is to calculate total entropy change for heat transfer: ΔS_total = Q(1/T_cold - 1/T_hot) > 0 when T_cold < T_hot.

Question 20

1. It can exceed $100\%$ efficiency by converting the removed heat entirely into electrical energy.
2. It must require work input because moving heat from cold to hot would otherwise decrease total entropy. (correct answer)
3. It can run with no work because entropy of the cold region decreases while total entropy must decrease.
4. It needs no work because entropy is only disorder, and the cold interior becomes more ordered.

Explanation: This question tests understanding of entropy and the second law of thermodynamics. A refrigerator moves heat from cold to hot, which would decrease total entropy if done alone (ΔS = -Q_C/T_C + Q_C/T_H < 0 since T_C < T_H). The second law prohibits spontaneous entropy decrease, so work input is required to increase entropy elsewhere, ensuring the total entropy of the universe increases. The work done on the refrigerator is converted to additional heat exhausted to the hot reservoir, making Q_H = Q_C + W, which ensures positive total entropy change. Choice B incorrectly suggests the process can occur without work, violating the Clausius statement of the second law that heat cannot spontaneously flow from cold to hot. The strategy is to recognize that any device moving heat from cold to hot requires work input to satisfy the second law.