Second Law with Heat Engines Quiz

  • Grade 11th
Reviewed by Ekaterina Yukhnovich
Ekaterina Yukhnovich, PhD |
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Ekaterina V. is a physicist and mathematics expert with a PhD in Physics and Mathematics and extensive experience working with advanced secondary and undergraduate-level content. She specializes in combinatorics, applied mathematics, and scientific writing, with a strong focus on accuracy and academic rigor.
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| Attempts: 14 | Questions: 20 | Updated: Mar 9, 2026
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1. The first law alone would allow 100% conversion of heat to work, but the second law forbids it for cyclic engines.

Explanation

Concept: first law vs second law roles. The first law only requires energy conservation, so it doesn’t specify direction or limits on conversion.

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About This Quiz
Second Law With Heat Engines Quiz - Quiz

This quiz on the Second Law with Heat Engines features 20 questions designed for students in Grade 11. You will explore important concepts like energy efficiency, entropy, and the principles that govern heat engines, which are essential for understanding how energy is transformed and utilized in various systems. Mastering these... see moreideas not only enhances your knowledge of physics but also prepares you for advanced studies and real-world applications. Dive in and see how well you can grasp these fundamental concepts!
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2.

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2. The main second law message for engines is:

Explanation

Concept: kelvin–planck message in plain language. The second law limits how completely heat can be turned into work during cyclic operation.

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3. Which statements about engines are correct?

Explanation

Concept: basic engine constraints. Engines produce work from the difference between absorbed and rejected heat.

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4. A perfectly reversible engine is an ideal model with no friction and no finite temperature differences during heat transfer.

Explanation

Concept: reversible engine idealization. Reversibility requires no friction and heat transfer occurring with infinitesimally small temperature differences.

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5. An engine has efficiency 0.40 and absorbs 1,500 J. Work output is:

Explanation

Concept: using η=w/q_h. Work is w=η*q_h. Substituting gives w=0.40×1500=600 J.

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6. A heat engine must operate between:

Explanation

Concept: engines require a temperature difference. A heat engine needs a hot source to supply heat and a cold sink to reject some heat. The temperature difference enables net work output in a cycle.

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7. A claim: “My engine converts 1,000 J of heat entirely into 1,000 J of work every cycle, with no heat rejected.” This is:

Explanation

Concept: 100% engine efficiency is forbidden. A cyclic engine must reject some heat to a cold sink.

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8. Which change would most likely increase an engine’s efficiency (without changing q_h)?

Explanation

Concept: efficiency improves when less heat is rejected. Since w=q_h−q_c, lowering q_c increases work output for the same q_h.

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9. Increasing the temperature difference between hot and cold reservoirs can allow higher maximum efficiency (in principle).

Explanation

Concept: theoretical efficiency limit depends on temperature gap. A larger temperature difference generally allows a higher possible maximum efficiency.

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10. The efficiency of the engine in Q5 is:

Explanation

Concept: calculating η. Efficiency is η=w/q_h. Substituting gives η=600/2400=0.25, meaning 25% of input heat becomes work.

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11. An engine absorbs 2,400 J and rejects 1,800 J. Work output is:

Explanation

Concept: work as difference of heats. Work output is w=q_h−q_c. Here w=2400−1800=600 J.

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12. If an engine takes q_h from a hot source and rejects q_c to a cold sink, the work output is:

Explanation

Concept: engine energy balance. Conservation of energy for one cycle gives input heat equals work plus rejected heat. Rearranging gives w=q_h−q_c.

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13. Without a cold reservoir, a cyclic heat engine cannot continuously produce work.

Explanation

Concept: heat rejection requirement (second law). The second law says a cyclic engine cannot convert all absorbed heat into work. A cold reservoir is needed as a place to dump unavoidable rejected heat.

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14. Which factors reduce real engine efficiency?

Explanation

Concept: irreversibility lowers efficiency. Friction, heat leaks, and turbulence create entropy and waste energy in less useful forms.

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15. Engine efficiency is η = ___/q_h.

Explanation

Engine efficiency (η) is defined as the ratio of useful work output (w) to the heat input (q_h) from the fuel. It measures how effectively an engine converts the energy from fuel into work. In this context, w represents the work done by the engine, while q_h signifies the total heat energy supplied. Therefore, the equation η = w/q_h illustrates that efficiency is determined by the amount of work produced for a given amount of heat energy consumed, highlighting the importance of maximizing work output to improve efficiency.

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16. In any real engine, some energy becomes less useful because entropy ______.

Explanation

In any real engine, energy is transformed to perform work, but not all of it can be converted into useful work due to the second law of thermodynamics. As entropy increases, it indicates a rise in disorder and energy dispersion within the system. This means that some energy becomes unavailable for doing work, leading to inefficiencies. Therefore, as entropy increases, the amount of useful energy decreases, making it harder to convert all input energy into productive output.

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17. If an engine rejects 70% of its input heat, its efficiency is:

Explanation

Concept: efficiency as fraction converted to work. If 70% is rejected, only 30% becomes work.

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18. Reducing friction always increases efficiency, but cannot remove the need to reject heat to a cold sink.

Explanation

Concept: improvements vs fundamental limits. Lower friction reduces wasted energy and entropy production, improving efficiency.

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19. For the engine in Q12, rejected heat q_c is:

Explanation

Concept: q_c from q_h and w. Rejected heat is q_c=q_h−w. So q_c=1500−600=900 J.

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20. Which heat transfer situation is more reversible (idealised)?

Explanation

Concept: small gradients reduce irreversibility. Heat transfer across a small temperature difference produces less entropy than across a large difference.

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Ekaterina Yukhnovich |PhD |
Science Expert
Ekaterina V. is a physicist and mathematics expert with a PhD in Physics and Mathematics and extensive experience working with advanced secondary and undergraduate-level content. She specializes in combinatorics, applied mathematics, and scientific writing, with a strong focus on accuracy and academic rigor.
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The first law alone would allow 100% conversion of heat to work, but...
The main second law message for engines is:
Which statements about engines are correct?
A perfectly reversible engine is an ideal model with no friction and...
An engine has efficiency 0.40 and absorbs 1,500 J. Work output is:
A heat engine must operate between:
A claim: “My engine converts 1,000 J of heat entirely into 1,000 J...
Which change would most likely increase an engine’s efficiency...
Increasing the temperature difference between hot and cold reservoirs...
The efficiency of the engine in Q5 is:
An engine absorbs 2,400 J and rejects 1,800 J. Work output is:
If an engine takes q_h from a hot source and rejects q_c to a cold...
Without a cold reservoir, a cyclic heat engine cannot continuously...
Which factors reduce real engine efficiency?
Engine efficiency is η = ___/q_h.
In any real engine, some energy becomes less useful because entropy...
If an engine rejects 70% of its input heat, its efficiency is:
Reducing friction always increases efficiency, but cannot remove the...
For the engine in Q12, rejected heat q_c is:
Which heat transfer situation is more reversible (idealised)?
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