IB · B · extended response
An engine's advertised efficiency exceeds its thermal limit
The particulate nature of matter · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- discriminating
- Marks
- 6
- Topics
- 1
- Answer
- Complete
Reversible efficiency limit
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Maximum efficiency
Horizontal axis (x): Cold reservoir / K
Read the exact data as a table
| Cold reservoir / K | Maximum efficiency |
|---|---|
| 150 | 0.75 |
| 300 | 0.5 |
| 450 | 0.25 |
| 600 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
A cyclic engine absorbs 1200 J per cycle from a 600 K reservoir and rejects heat to a 300 K reservoir. The manufacturer claims 800 J of work per cycle. The supplied graph shows the maximum reversible efficiency 1−Tc/Th as cold-reservoir temperature varies for Th = 600 K. Supplied model: For heat Q received reversibly by a reservoir at absolute temperature T, its entropy change is ΔS = Q/T, with Q negative for energy leaving that reservoir. Supplied data (Cold reservoir / K; Maximum efficiency): (150, 0.75), (300, 0.5), (450, 0.25), (600, 0).
- (a)
Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle.
3 marks - (b)
A real engine instead delivers 450 J per cycle. Find its efficiency and total reservoir entropy change; explain why waste heat cannot all be recycled into work in the same cycle.
3 marks
Ready to self-mark?Reveal the detailed answer guide
(a)
Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
Energy conservation alone allows Qout = 1200−800 = 400 J, so a first-law calculation does not expose the problem.
- 2
The reversible limit is 1−300/600 = 0.50. Maximum work is 0.50(1200) = 600 J, and at least 600 J must be rejected.
- 3
At the claimed output the reservoir entropy change would be −1200/600+400/300 = −2/3 J K⁻¹ per cycle, violating the nondecrease of total entropy.
(b)
A real engine instead delivers 450 J per cycle. Find its efficiency and total reservoir entropy change; explain why waste heat cannot all be recycled into work in the same cycle.
Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.
- 1
The actual rejected heat is 1200−450 = 750 J. Efficiency is 450/1200 = 0.375.
- 2
Reservoir entropy change is −1200/600+750/300 = −2+2.5 = +0.50 J K⁻¹ per cycle, consistent with irreversibility.
- 3
The working substance returns to its initial state each cycle. Converting all heat from a single reservoir into cyclic work would violate the second law; reducing losses cannot remove the need for heat rejection at these temperatures.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- Energy conservation alone allows Qout = 1200−800 = 400 J, so a first-law calculation does not expose the problem.
- The reversible limit is 1−300/600 = 0.50. Maximum work is 0.50(1200) = 600 J, and at least 600 J must be rejected.
- At the claimed output the reservoir entropy change would be −1200/600+400/300 = −2/3 J K⁻¹ per cycle, violating the nondecrease of total entropy.
- The actual rejected heat is 1200−450 = 750 J. Efficiency is 450/1200 = 0.375.
- Reservoir entropy change is −1200/600+750/300 = −2+2.5 = +0.50 J K⁻¹ per cycle, consistent with irreversibility.
- The working substance returns to its initial state each cycle. Converting all heat from a single reservoir into cyclic work would violate the second law; reducing losses cannot remove the need for heat rejection at these temperatures.
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