Carnot Cycle and Reversible Performance Limit
For reservoirs at absolute temperatures T_H > T_C, a reversible Carnot engine has Q_C/Q_H = T_C/T_H and η_C = 1 − T_C/T_H; every irreversible engine between the same reservoirs has η < η_C.
Why this shows up in the exam
Reservoir-temperature calculations · Cascaded reversible engines · Testing whether claimed engine performance is possible
Learn the idea
No engine between two reservoirs can exceed the efficiency of a reversible Carnot engine. A reversible engine wastes the least possible work opportunity, so its performance depends only on reservoir temperatures.
🧠 Memory hook: Carnot compares kelvin temperatures, cold over hot.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- η_C = 1 - T_C/T_H — maximum engine efficiency between two reservoirs; temperatures are in kelvin
- Q_C/Q_H = T_C/T_H — reversible heat ratio using positive magnitudes
How to approach it
- 1Convert all temperatures to kelvin
- 2Identify hot and cold reservoirs
- 3Apply the Carnot relation and check the result lies from zero to one
Common slip-ups that cost marks
- •Using Celsius temperatures in the ratio
- •Assuming every ideal-gas cycle is a Carnot cycle
- •Treating Carnot efficiency as guaranteed rather than maximal
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A gas absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?
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