A heat engine absorbs 1,000 J from a hot reservoir and performs 350 J of work in one cycle. How much heat is rejected to the cold reservoir?
A350 J — the rejected heat equals the work done
B650 J — energy conservation requires Q_C = Q_H − W
C1,350 J — both the input heat and the work must be dissipated
D1,000 J — all absorbed heat is eventually rejected
Energy conservation over a complete cycle (ΔU = 0) gives W = Q_H − Q_C, so Q_C = Q_H − W = 1000 − 350 = 650 J. The thermal efficiency is e = W/Q_H = 350/1000 = 0.35 or 35%. Note that only the work output (350 J) leaves as useful mechanical energy; the remaining 650 J is dumped to the cold reservoir.
Question 2 Multiple Choice
An engineer proposes building a heat engine that absorbs heat from a furnace and converts 100% of it into work, with no heat rejection to a cold reservoir. What is wrong with this proposal?
ANothing is fundamentally wrong — 100% efficiency is achievable with sufficiently advanced materials
BRejecting heat to a cold reservoir is thermodynamically necessary for the working fluid to complete its cycle; eliminating it violates the second law of thermodynamics
CThe proposal would work but would produce less power than a conventional engine
DThe proposal is only impossible for gas-cycle engines; steam engines could achieve it
A heat engine's working fluid must return to its initial state after each cycle (otherwise it could only run once). Rejecting Q_C to the cold reservoir is the mechanism that resets the fluid's state. No engine — regardless of design, materials, or working fluid — can complete a cycle using only a single thermal reservoir and produce net work. This is the Kelvin-Planck statement of the second law. It is not an engineering limitation that clever design can overcome; it is a fundamental law of nature.
Question 3 True / False
An engine with 30% thermal efficiency wastes 70% of its input heat purely due to engineering imperfections like friction and heat loss — a perfect engine could theoretically convert most input heat to work.
TTrue
FFalse
Answer: False
Even a perfect (reversible Carnot) engine rejects heat to a cold reservoir. The second law requires heat rejection; it is not an engineering flaw. Some of the 70% may be due to irreversibilities (friction, heat transfer across finite temperature differences), but a fraction is thermodynamically unavoidable. The Carnot efficiency e = 1 − T_C/T_H sets the maximum achievable efficiency for any engine between those reservoirs — and for typical temperature ratios, this maximum is well below 100%.
Question 4 True / False
The thermal efficiency of a heat engine equals the ratio of useful work output to the heat absorbed from the hot reservoir.
TTrue
FFalse
Answer: True
By definition, e = W/Q_H. This can be rewritten as e = 1 − Q_C/Q_H using W = Q_H − Q_C. Both forms are equivalent. The efficiency tells you what fraction of the heat you 'bought' from the hot reservoir actually became useful mechanical work; the remainder is rejected as waste heat to the cold reservoir.
Question 5 Short Answer
Why can no heat engine — real or ideal — achieve 100% thermal efficiency when operating between two thermal reservoirs? Explain in terms of what must happen for the engine to complete a cycle.
Think about your answer, then reveal below.
Model answer: For an engine to run continuously, its working fluid must return to its initial thermodynamic state after each cycle. Completing the cycle requires rejecting some heat Q_C to a cold reservoir — this is the mechanism by which the fluid is 'reset.' Without a cold reservoir to accept Q_C, the fluid cannot complete the cycle and the engine cannot produce sustained work. Since Q_C > 0 always, W = Q_H − Q_C < Q_H, so the efficiency e = W/Q_H is always strictly less than 1. This is not bad design; it is the second law of thermodynamics.
The Carnot efficiency e = 1 − T_C/T_H shows that the only way to approach 100% efficiency would be to have T_C → 0 K (absolute zero, unachievable) or T_H → ∞ (also unachievable). Real engines also suffer irreversibilities that push them further below the Carnot limit.