IB · PHYSICS HL

Physics: Higher Level

Thermodynamics — HL Theme B

Name: ____________________Date: October 10, 2026

1.
State the first law of thermodynamics, defining each term used.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States the first law: ΔU = Q + W (or ΔU = Q − W depending on sign convention, provided stated consistently). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Develop this part of the answer: Defines ΔU as the change in internal energy of the system. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: Defines Q as the thermal energy transferred to the system, and W as the work done on (or by) the system. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The first law of thermodynamics is simply a statement of conservation of energy applied to a thermodynamic system — always state your sign convention explicitly since both conventions are used in different textbooks.
Marking points
- States the first law: ΔU = Q + W (or ΔU = Q − W depending on sign convention, provided stated consistently).
- Defines ΔU as the change in internal energy of the system.
- Defines Q as the thermal energy transferred to the system, and W as the work done on (or by) the system.
Examiner tip: The first law of thermodynamics is simply a statement of conservation of energy applied to a thermodynamic system — always state your sign convention explicitly since both conventions are used in different textbooks.
2.
Marking analysis: A learner attempts the following task: “State the first law of thermodynamics, defining each term used.” Their response addresses only this point: “States the first law: ΔU = Q + W (or ΔU = Q − W depending on sign convention, provided stated consistently).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States the first law: ΔU = Q + W (or ΔU = Q − W depending on sign convention, provided stated consistently). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Defines ΔU as the change in internal energy of the system. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Defines Q as the thermal energy transferred to the system, and W as the work done on (or by) the system. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States the first law: ΔU = Q + W (or ΔU = Q − W depending on sign convention, provided stated consistently).
- Identifies the missing requirement: Defines ΔU as the change in internal energy of the system.
- Identifies the missing requirement: Defines Q as the thermal energy transferred to the system, and W as the work done on (or by) the system.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
A gas absorbs 500 J of thermal energy and does 200 J of work on its surroundings as it expands. Calculate the change in internal energy of the gas.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses the first law ΔU = Q − W, where W is work done by the gas. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes ΔU = 500 − 200. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains ΔU = 300 J. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Be consistent about whether W represents work done BY the system or ON the system — this determines whether it is added or subtracted in the first law equation.
Marking points
- Uses the first law ΔU = Q − W, where W is work done by the gas.
- Substitutes ΔU = 500 − 200.
- Obtains ΔU = 300 J.
Examiner tip: Be consistent about whether W represents work done BY the system or ON the system — this determines whether it is added or subtracted in the first law equation.
4.
Marking analysis: A learner attempts the following task: “A gas absorbs 500 J of thermal energy and does 200 J of work on its surroundings as it expands. Calculate the change in internal energy of the gas.” Their response addresses only this point: “Uses the first law ΔU = Q − W, where W is work done by the gas.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses the first law ΔU = Q − W, where W is work done by the gas. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes ΔU = 500 − 200. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains ΔU = 300 J. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses the first law ΔU = Q − W, where W is work done by the gas.
- Identifies the missing requirement: Substitutes ΔU = 500 − 200.
- Identifies the missing requirement: Obtains ΔU = 300 J.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Distinguish between an isothermal process and an adiabatic process for an ideal gas.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States that an isothermal process occurs at constant temperature, so the internal energy of an ideal gas does not change (ΔU = 0). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: States that an adiabatic process occurs with no thermal energy transfer to or from the system (Q = 0). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: States that in an adiabatic expansion, since Q = 0, any work done by the gas comes entirely from a decrease in internal energy, so the gas cools. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Isothermal means constant temperature (ΔU = 0 for an ideal gas); adiabatic means no heat transfer (Q = 0) — these are two different constraints and are not opposites of each other.
Marking points
- States that an isothermal process occurs at constant temperature, so the internal energy of an ideal gas does not change (ΔU = 0).
- States that an adiabatic process occurs with no thermal energy transfer to or from the system (Q = 0).
- States that in an adiabatic expansion, since Q = 0, any work done by the gas comes entirely from a decrease in internal energy, so the gas cools.
Examiner tip: Isothermal means constant temperature (ΔU = 0 for an ideal gas); adiabatic means no heat transfer (Q = 0) — these are two different constraints and are not opposites of each other.
6.
Marking analysis: A learner attempts the following task: “Distinguish between an isothermal process and an adiabatic process for an ideal gas.” Their response addresses only this point: “States that an isothermal process occurs at constant temperature, so the internal energy of an ideal gas does not change (ΔU = 0).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that an isothermal process occurs at constant temperature, so the internal energy of an ideal gas does not change (ΔU = 0). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that an adiabatic process occurs with no thermal energy transfer to or from the system (Q = 0). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that in an adiabatic expansion, since Q = 0, any work done by the gas comes entirely from a decrease in internal energy, so the gas cools. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that an isothermal process occurs at constant temperature, so the internal energy of an ideal gas does not change (ΔU = 0).
- Identifies the missing requirement: States that an adiabatic process occurs with no thermal energy transfer to or from the system (Q = 0).
- Identifies the missing requirement: States that in an adiabatic expansion, since Q = 0, any work done by the gas comes entirely from a decrease in internal energy, so the gas cools.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
Define entropy in simple qualitative terms, and state the second law of thermodynamics.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: Defines entropy as a measure of the disorder, or the number of possible microscopic arrangements (microstates), of a system. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States the second law: the total entropy of an isolated system (or the universe) never decreases over time in any spontaneous process. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The second law explains why some processes are irreversible in practice (e.g. heat always flows from hot to cold spontaneously) even though the first law alone would not forbid the reverse.
Marking points
- Defines entropy as a measure of the disorder, or the number of possible microscopic arrangements (microstates), of a system.
- States the second law: the total entropy of an isolated system (or the universe) never decreases over time in any spontaneous process.
Examiner tip: The second law explains why some processes are irreversible in practice (e.g. heat always flows from hot to cold spontaneously) even though the first law alone would not forbid the reverse.
8.
Marking analysis: A learner attempts the following task: “Define entropy in simple qualitative terms, and state the second law of thermodynamics.” Their response addresses only this point: “Defines entropy as a measure of the disorder, or the number of possible microscopic arrangements (microstates), of a system.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Defines entropy as a measure of the disorder, or the number of possible microscopic arrangements (microstates), of a system. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States the second law: the total entropy of an isolated system (or the universe) never decreases over time in any spontaneous process. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Defines entropy as a measure of the disorder, or the number of possible microscopic arrangements (microstates), of a system.
- Identifies the missing requirement: States the second law: the total entropy of an isolated system (or the universe) never decreases over time in any spontaneous process.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
Explain, in terms of entropy, why thermal energy spontaneously flows from a hot object to a cold object and never the reverse.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that when thermal energy transfers from a hot object to a cold object, the entropy decrease of the hot object is smaller in magnitude than the entropy increase of the cold object (since entropy change for a given heat transfer is larger at lower temperature). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that the total entropy of the system therefore increases, consistent with the second law, making this direction of heat flow spontaneous. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that the reverse process (heat flowing from cold to hot) would require a total entropy decrease, which violates the second law and therefore cannot occur spontaneously. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Entropy change for a given quantity of heat transferred is Q/T — the same heat transfer causes a larger entropy change at lower temperature, which is why the net entropy change favours heat flowing from hot to cold.
Marking points
- States that when thermal energy transfers from a hot object to a cold object, the entropy decrease of the hot object is smaller in magnitude than the entropy increase of the cold object (since entropy change for a given heat transfer is larger at lower temperature).
- States that the total entropy of the system therefore increases, consistent with the second law, making this direction of heat flow spontaneous.
- States that the reverse process (heat flowing from cold to hot) would require a total entropy decrease, which violates the second law and therefore cannot occur spontaneously.
Examiner tip: Entropy change for a given quantity of heat transferred is Q/T — the same heat transfer causes a larger entropy change at lower temperature, which is why the net entropy change favours heat flowing from hot to cold.
10.
Marking analysis: A learner attempts the following task: “Explain, in terms of entropy, why thermal energy spontaneously flows from a hot object to a cold object and never the reverse.” Their response addresses only this point: “States that when thermal energy transfers from a hot object to a cold object, the entropy decrease of the hot object is smaller in magnitude than the entropy increase of the cold object (since entropy change for a given heat transfer is larger at lower temperature).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that when thermal energy transfers from a hot object to a cold object, the entropy decrease of the hot object is smaller in magnitude than the entropy increase of the cold object (since entropy change for a given heat transfer is larger at lower temperature). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the total entropy of the system therefore increases, consistent with the second law, making this direction of heat flow spontaneous. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that the reverse process (heat flowing from cold to hot) would require a total entropy decrease, which violates the second law and therefore cannot occur spontaneously. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that when thermal energy transfers from a hot object to a cold object, the entropy decrease of the hot object is smaller in magnitude than the entropy increase of the cold object (since entropy change for a given heat transfer is larger at lower temperature).
- Identifies the missing requirement: States that the total entropy of the system therefore increases, consistent with the second law, making this direction of heat flow spontaneous.
- Identifies the missing requirement: States that the reverse process (heat flowing from cold to hot) would require a total entropy decrease, which violates the second law and therefore cannot occur spontaneously.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
A heat engine absorbs 800 J of thermal energy from a hot reservoir and does 300 J of useful work. Calculate the thermal efficiency of the engine.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses efficiency = useful work output / thermal energy input. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes 300/800. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains efficiency = 0.375, or 37.5%. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Thermal efficiency is always less than 100% for a real heat engine, since some thermal energy must always be expelled to a cold reservoir, as required by the second law of thermodynamics.
Marking points
- Uses efficiency = useful work output / thermal energy input.
- Substitutes 300/800.
- Obtains efficiency = 0.375, or 37.5%.
Examiner tip: Thermal efficiency is always less than 100% for a real heat engine, since some thermal energy must always be expelled to a cold reservoir, as required by the second law of thermodynamics.
12.
Marking analysis: A learner attempts the following task: “A heat engine absorbs 800 J of thermal energy from a hot reservoir and does 300 J of useful work. Calculate the thermal efficiency of the engine.” Their response addresses only this point: “Uses efficiency = useful work output / thermal energy input.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses efficiency = useful work output / thermal energy input. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes 300/800. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains efficiency = 0.375, or 37.5%. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses efficiency = useful work output / thermal energy input.
- Identifies the missing requirement: Substitutes 300/800.
- Identifies the missing requirement: Obtains efficiency = 0.375, or 37.5%.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
State the Carnot theorem, and explain why no real heat engine operating between two given temperatures can be more efficient than a Carnot engine.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that the Carnot engine is a theoretical, idealized (reversible) heat engine that achieves the maximum possible efficiency between two given temperatures. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that all real heat engines involve some irreversible processes (e.g. friction, unrestrained expansion), which generate additional entropy and reduce efficiency below the Carnot (reversible) limit. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The Carnot efficiency, 1 − T_cold/T_hot (with temperatures in kelvin), sets an absolute theoretical upper limit that no real engine can exceed, regardless of its design.
Marking points
- States that the Carnot engine is a theoretical, idealized (reversible) heat engine that achieves the maximum possible efficiency between two given temperatures.
- States that all real heat engines involve some irreversible processes (e.g. friction, unrestrained expansion), which generate additional entropy and reduce efficiency below the Carnot (reversible) limit.
Examiner tip: The Carnot efficiency, 1 − T_cold/T_hot (with temperatures in kelvin), sets an absolute theoretical upper limit that no real engine can exceed, regardless of its design.
14.
Marking analysis: A learner attempts the following task: “State the Carnot theorem, and explain why no real heat engine operating between two given temperatures can be more efficient than a Carnot engine.” Their response addresses only this point: “States that the Carnot engine is a theoretical, idealized (reversible) heat engine that achieves the maximum possible efficiency between two given temperatures.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the Carnot engine is a theoretical, idealized (reversible) heat engine that achieves the maximum possible efficiency between two given temperatures. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that all real heat engines involve some irreversible processes (e.g. friction, unrestrained expansion), which generate additional entropy and reduce efficiency below the Carnot (reversible) limit. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the Carnot engine is a theoretical, idealized (reversible) heat engine that achieves the maximum possible efficiency between two given temperatures.
- Identifies the missing requirement: States that all real heat engines involve some irreversible processes (e.g. friction, unrestrained expansion), which generate additional entropy and reduce efficiency below the Carnot (reversible) limit.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate its maximum possible (Carnot) efficiency.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses Carnot efficiency = 1 − T_cold/T_hot, with temperatures in kelvin. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes 1 − 300/500. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains efficiency = 0.40, or 40%. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The Carnot efficiency formula always requires absolute temperatures in kelvin, never Celsius — a common source of error.
Marking points
- Uses Carnot efficiency = 1 − T_cold/T_hot, with temperatures in kelvin.
- Substitutes 1 − 300/500.
- Obtains efficiency = 0.40, or 40%.
Examiner tip: The Carnot efficiency formula always requires absolute temperatures in kelvin, never Celsius — a common source of error.
16.
Marking analysis: A learner attempts the following task: “A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Calculate its maximum possible (Carnot) efficiency.” Their response addresses only this point: “Uses Carnot efficiency = 1 − T_cold/T_hot, with temperatures in kelvin.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses Carnot efficiency = 1 − T_cold/T_hot, with temperatures in kelvin. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes 1 − 300/500. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains efficiency = 0.40, or 40%. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses Carnot efficiency = 1 − T_cold/T_hot, with temperatures in kelvin.
- Identifies the missing requirement: Substitutes 1 − 300/500.
- Identifies the missing requirement: Obtains efficiency = 0.40, or 40%.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
Explain why increasing the temperature of the hot reservoir, or decreasing the temperature of the cold reservoir, increases the maximum possible efficiency of a heat engine.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States the Carnot efficiency formula: efficiency = 1 − T_cold/T_hot. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Develop this part of the answer: States that increasing T_hot decreases the ratio T_cold/T_hot, increasing the efficiency. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that decreasing T_cold also decreases the same ratio, similarly increasing the efficiency, since a greater temperature difference allows more useful work to be extracted per unit of heat input. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Maximizing the temperature difference between the hot and cold reservoirs is the fundamental strategy for improving any real heat engine's efficiency, motivated directly by the Carnot efficiency formula.
Marking points
- States the Carnot efficiency formula: efficiency = 1 − T_cold/T_hot.
- States that increasing T_hot decreases the ratio T_cold/T_hot, increasing the efficiency.
- States that decreasing T_cold also decreases the same ratio, similarly increasing the efficiency, since a greater temperature difference allows more useful work to be extracted per unit of heat input.
Examiner tip: Maximizing the temperature difference between the hot and cold reservoirs is the fundamental strategy for improving any real heat engine's efficiency, motivated directly by the Carnot efficiency formula.
18.
Marking analysis: A learner attempts the following task: “Explain why increasing the temperature of the hot reservoir, or decreasing the temperature of the cold reservoir, increases the maximum possible efficiency of a heat engine.” Their response addresses only this point: “States the Carnot efficiency formula: efficiency = 1 − T_cold/T_hot.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States the Carnot efficiency formula: efficiency = 1 − T_cold/T_hot. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that increasing T_hot decreases the ratio T_cold/T_hot, increasing the efficiency. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that decreasing T_cold also decreases the same ratio, similarly increasing the efficiency, since a greater temperature difference allows more useful work to be extracted per unit of heat input. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States the Carnot efficiency formula: efficiency = 1 − T_cold/T_hot.
- Identifies the missing requirement: States that increasing T_hot decreases the ratio T_cold/T_hot, increasing the efficiency.
- Identifies the missing requirement: States that decreasing T_cold also decreases the same ratio, similarly increasing the efficiency, since a greater temperature difference allows more useful work to be extracted per unit of heat input.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Outline why a refrigerator requires external work input to transfer thermal energy from a cold interior to a warmer exterior, in terms of the second law of thermodynamics.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that heat naturally (spontaneously) flows only from hot to cold, never from cold to hot, without external influence. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that transferring heat from a cold region to a hotter region requires external work to be done on the system, to compensate for and overcome this natural direction, consistent with the second law. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that this work input increases the total entropy of the surroundings by more than the entropy decrease inside the refrigerator, so the total entropy of the universe still increases overall, satisfying the second law. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A refrigerator does not violate the second law — it locally decreases entropy inside the fridge, but the work input increases entropy elsewhere (the surroundings/motor) by a larger amount, so total entropy still increases.
Marking points
- States that heat naturally (spontaneously) flows only from hot to cold, never from cold to hot, without external influence.
- States that transferring heat from a cold region to a hotter region requires external work to be done on the system, to compensate for and overcome this natural direction, consistent with the second law.
- States that this work input increases the total entropy of the surroundings by more than the entropy decrease inside the refrigerator, so the total entropy of the universe still increases overall, satisfying the second law.
Examiner tip: A refrigerator does not violate the second law — it locally decreases entropy inside the fridge, but the work input increases entropy elsewhere (the surroundings/motor) by a larger amount, so total entropy still increases.
20.
Marking analysis: A learner attempts the following task: “Outline why a refrigerator requires external work input to transfer thermal energy from a cold interior to a warmer exterior, in terms of the second law of thermodynamics.” Their response addresses only this point: “States that heat naturally (spontaneously) flows only from hot to cold, never from cold to hot, without external influence.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that heat naturally (spontaneously) flows only from hot to cold, never from cold to hot, without external influence. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that transferring heat from a cold region to a hotter region requires external work to be done on the system, to compensate for and overcome this natural direction, consistent with the second law. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that this work input increases the total entropy of the surroundings by more than the entropy decrease inside the refrigerator, so the total entropy of the universe still increases overall, satisfying the second law. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that heat naturally (spontaneously) flows only from hot to cold, never from cold to hot, without external influence.
- Identifies the missing requirement: States that transferring heat from a cold region to a hotter region requires external work to be done on the system, to compensate for and overcome this natural direction, consistent with the second law.
- Identifies the missing requirement: States that this work input increases the total entropy of the surroundings by more than the entropy decrease inside the refrigerator, so the total entropy of the universe still increases overall, satisfying the second law.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
A gas expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 0.020 m³ to 0.050 m³. Calculate the work done by the gas.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: States that for a constant-pressure (isobaric) process, work done by the gas = PΔV. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Calculates ΔV = 0.050 − 0.020 = 0.030 m³. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains W = 2.0 × 10⁵ × 0.030 = 6000 J. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: On a pressure-volume (PV) diagram, the work done in any process equals the area under the curve — for a constant-pressure process, this area is simply a rectangle, P × ΔV.
Marking points
- States that for a constant-pressure (isobaric) process, work done by the gas = PΔV.
- Calculates ΔV = 0.050 − 0.020 = 0.030 m³.
- Obtains W = 2.0 × 10⁵ × 0.030 = 6000 J.
Examiner tip: On a pressure-volume (PV) diagram, the work done in any process equals the area under the curve — for a constant-pressure process, this area is simply a rectangle, P × ΔV.
22.
Marking analysis: A learner attempts the following task: “A gas expands at a constant pressure of 2.0 × 10⁵ Pa from a volume of 0.020 m³ to 0.050 m³. Calculate the work done by the gas.” Their response addresses only this point: “States that for a constant-pressure (isobaric) process, work done by the gas = PΔV.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that for a constant-pressure (isobaric) process, work done by the gas = PΔV. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Calculates ΔV = 0.050 − 0.020 = 0.030 m³. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains W = 2.0 × 10⁵ × 0.030 = 6000 J. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that for a constant-pressure (isobaric) process, work done by the gas = PΔV.
- Identifies the missing requirement: Calculates ΔV = 0.050 − 0.020 = 0.030 m³.
- Identifies the missing requirement: Obtains W = 2.0 × 10⁵ × 0.030 = 6000 J.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
Explain why no work is done by or on a gas during an isochoric (constant volume) process, and state what happens to any thermal energy transferred to the gas in this case.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States that work done by a gas = PΔV, and since volume does not change (ΔV = 0) in an isochoric process, no work is done. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: States that by the first law, ΔU = Q − W = Q, since W = 0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Develop this part of the answer: States that all of the thermal energy transferred to the gas therefore goes directly into increasing its internal energy, and hence its temperature. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: An isochoric process is the one case where the first law simplifies to ΔU = Q exactly, since the work term vanishes entirely — every joule of heat added shows up as internal energy.
Marking points
- States that work done by a gas = PΔV, and since volume does not change (ΔV = 0) in an isochoric process, no work is done.
- States that by the first law, ΔU = Q − W = Q, since W = 0.
- States that all of the thermal energy transferred to the gas therefore goes directly into increasing its internal energy, and hence its temperature.
Examiner tip: An isochoric process is the one case where the first law simplifies to ΔU = Q exactly, since the work term vanishes entirely — every joule of heat added shows up as internal energy.
24.
Marking analysis: A learner attempts the following task: “Explain why no work is done by or on a gas during an isochoric (constant volume) process, and state what happens to any thermal energy transferred to the gas in this case.” Their response addresses only this point: “States that work done by a gas = PΔV, and since volume does not change (ΔV = 0) in an isochoric process, no work is done.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that work done by a gas = PΔV, and since volume does not change (ΔV = 0) in an isochoric process, no work is done. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that by the first law, ΔU = Q − W = Q, since W = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that all of the thermal energy transferred to the gas therefore goes directly into increasing its internal energy, and hence its temperature. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that work done by a gas = PΔV, and since volume does not change (ΔV = 0) in an isochoric process, no work is done.
- Identifies the missing requirement: States that by the first law, ΔU = Q − W = Q, since W = 0.
- Identifies the missing requirement: States that all of the thermal energy transferred to the gas therefore goes directly into increasing its internal energy, and hence its temperature.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
1.0 mol of an ideal gas undergoes an isothermal expansion at 300 K from a volume of 0.010 m³ to 0.020 m³. Calculate the work done by the gas, using W = nRT ln(V₂/V₁). (R = 8.31 J K⁻¹ mol⁻¹)
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: States W = nRT ln(V₂/V₁). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Calculates the volume ratio V₂/V₁ = 0.020/0.010 = 2.0. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Substitutes W = 1.0 × 8.31 × 300 × ln(2.0). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Develop this part of the answer: Obtains W ≈ 1730 J. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Since internal energy of an ideal gas depends only on temperature, ΔU = 0 for any isothermal process — so by the first law, all the work done by the gas in this expansion equals the thermal energy absorbed, Q = W.
Marking points
- States W = nRT ln(V₂/V₁).
- Calculates the volume ratio V₂/V₁ = 0.020/0.010 = 2.0.
- Substitutes W = 1.0 × 8.31 × 300 × ln(2.0).
- Obtains W ≈ 1730 J.
Examiner tip: Since internal energy of an ideal gas depends only on temperature, ΔU = 0 for any isothermal process — so by the first law, all the work done by the gas in this expansion equals the thermal energy absorbed, Q = W.
26.
Marking analysis: A learner attempts the following task: “1.0 mol of an ideal gas undergoes an isothermal expansion at 300 K from a volume of 0.010 m³ to 0.020 m³. Calculate the work done by the gas, using W = nRT ln(V₂/V₁). (R = 8.31 J K⁻¹ mol⁻¹)” Their response addresses only this point: “States W = nRT ln(V₂/V₁).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States W = nRT ln(V₂/V₁). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Calculates the volume ratio V₂/V₁ = 0.020/0.010 = 2.0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Substitutes W = 1.0 × 8.31 × 300 × ln(2.0). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains W ≈ 1730 J. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States W = nRT ln(V₂/V₁).
- Identifies the missing requirement: Calculates the volume ratio V₂/V₁ = 0.020/0.010 = 2.0.
- Identifies the missing requirement: Substitutes W = 1.0 × 8.31 × 300 × ln(2.0).
- Identifies the missing requirement: Obtains W ≈ 1730 J.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
0.500 kg of ice at 0°C melts completely and reversibly at constant temperature, absorbing 1.67 × 10⁵ J of thermal energy. Calculate the change in entropy of the ice as it melts.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: States ΔS = Q/T for a reversible process at constant temperature. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes Q = 1.67 × 10⁵ J and T = 273 K. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Develop this part of the answer: Obtains ΔS ≈ 612 J K⁻¹. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The formula ΔS = Q/T for entropy change is only directly valid for a reversible process occurring at a single, constant temperature — melting at the melting point is a standard example that satisfies this.
Marking points
- States ΔS = Q/T for a reversible process at constant temperature.
- Substitutes Q = 1.67 × 10⁵ J and T = 273 K.
- Obtains ΔS ≈ 612 J K⁻¹.
Examiner tip: The formula ΔS = Q/T for entropy change is only directly valid for a reversible process occurring at a single, constant temperature — melting at the melting point is a standard example that satisfies this.
28.
Marking analysis: A learner attempts the following task: “0.500 kg of ice at 0°C melts completely and reversibly at constant temperature, absorbing 1.67 × 10⁵ J of thermal energy. Calculate the change in entropy of the ice as it melts.” Their response addresses only this point: “States ΔS = Q/T for a reversible process at constant temperature.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States ΔS = Q/T for a reversible process at constant temperature. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes Q = 1.67 × 10⁵ J and T = 273 K. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains ΔS ≈ 612 J K⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States ΔS = Q/T for a reversible process at constant temperature.
- Identifies the missing requirement: Substitutes Q = 1.67 × 10⁵ J and T = 273 K.
- Identifies the missing requirement: Obtains ΔS ≈ 612 J K⁻¹.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
State the Kelvin-Planck statement of the second law of thermodynamics, and explain what it implies about the possibility of a heat engine with 100% efficiency.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that it is impossible to construct a heat engine that, operating in a cycle, converts thermal energy completely into work with no other effect (i.e. without exhausting some heat to a cold reservoir). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: Explains that this means a heat engine with 100% efficiency is impossible, since some thermal energy must always be rejected to a cold reservoir. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that this statement is equivalent to the more general principle that entropy cannot decrease in an isolated system undergoing a cyclic process. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The Kelvin-Planck statement and the Clausius statement (heat cannot spontaneously flow from cold to hot) are two different-sounding but logically equivalent ways of expressing the same second law.
Marking points
- States that it is impossible to construct a heat engine that, operating in a cycle, converts thermal energy completely into work with no other effect (i.e. without exhausting some heat to a cold reservoir).
- Explains that this means a heat engine with 100% efficiency is impossible, since some thermal energy must always be rejected to a cold reservoir.
- States that this statement is equivalent to the more general principle that entropy cannot decrease in an isolated system undergoing a cyclic process.
Examiner tip: The Kelvin-Planck statement and the Clausius statement (heat cannot spontaneously flow from cold to hot) are two different-sounding but logically equivalent ways of expressing the same second law.
30.
Marking analysis: A learner attempts the following task: “State the Kelvin-Planck statement of the second law of thermodynamics, and explain what it implies about the possibility of a heat engine with 100% efficiency.” Their response addresses only this point: “States that it is impossible to construct a heat engine that, operating in a cycle, converts thermal energy completely into work with no other effect (i.e. without exhausting some heat to a cold reservoir).” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that it is impossible to construct a heat engine that, operating in a cycle, converts thermal energy completely into work with no other effect (i.e. without exhausting some heat to a cold reservoir). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Explains that this means a heat engine with 100% efficiency is impossible, since some thermal energy must always be rejected to a cold reservoir. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that this statement is equivalent to the more general principle that entropy cannot decrease in an isolated system undergoing a cyclic process. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that it is impossible to construct a heat engine that, operating in a cycle, converts thermal energy completely into work with no other effect (i.e. without exhausting some heat to a cold reservoir).
- Identifies the missing requirement: Explains that this means a heat engine with 100% efficiency is impossible, since some thermal energy must always be rejected to a cold reservoir.
- Identifies the missing requirement: States that this statement is equivalent to the more general principle that entropy cannot decrease in an isolated system undergoing a cyclic process.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.