MYP Physics · topic MYP-18.06
Entropy and direction of change
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
An engine's advertised efficiency exceeds its thermal limit
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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]
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.
Show a hint
Begin with part (a): Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle.
Reveal the worked solution
- 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.
Answer (a) Claimed efficiency = 66.7% exceeds the 50% limit; Wmax = 600 J and Qrejected,min = 600 J. (b) Efficiency = 37.5%; reservoir entropy increase = 0.50 J K⁻¹ per cycle.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.