Carnot Limit, explained.
The Carnot limit bounds heat-engine efficiency between two reservoirs at absolute temperatures: efficiency cannot exceed 1−Tc/Th.
Why it happens
Each proposed cycle takes 100 heat units from the hot reservoir and rejects 100−W to the cold reservoir. Reservoir entropy change is (100−W)/Tc−100/Th. The game rejects negative entropy change and counts admissible proposals.
The Carnot limit bounds heat-engine efficiency between two reservoirs at absolute temperatures: efficiency cannot exceed 1−Tc/Th.
Read the result
Proposed work and rejected heat must sum to one hundred. Check the entropy sign as well as the efficiency percentage.
A worked example
A proposed engine between two reservoirs
Set the hot reservoir to six hundred kelvin and the cold reservoir to three hundred kelvin. The efficiency ceiling is fifty percent.
Propose sixty work units from one hundred heat units. The game rejects the negative reservoir entropy change.
Propose forty instead. It passes this necessary check with sixty heat units rejected, without proving a practical engine can achieve it.
OPTIONAL DEEPER DETAILGo deeper: inside the model
Inside this model
Each proposed cycle takes 100 heat units from the hot reservoir and rejects 100−W to the cold reservoir. Reservoir entropy change is (100−W)/Tc−100/Th. The game rejects negative entropy change and counts admissible proposals.
Where this idea is useful
A practical use
A proposed heat engine must be evaluated against reservoir temperatures in kelvin, not just an appealing output percentage.
A common misconception
“An ideal efficiency ceiling is a typical real engine efficiency.”
A thermodynamic upper bound assumes ideal conditions. Practical irreversibility and engineering constraints generally reduce achievable performance.
What this explanation leaves out
- Passing this entropy check is necessary, not a construction plan or proof of attainability. The ideal reversible bound does not include practical friction, heat-transfer limits or auxiliary energy costs.
Why must the temperatures be in kelvin?
The efficiency ratio requires absolute thermodynamic temperature. Using Celsius would change the ratio arbitrarily with the chosen zero point.
Have you checked an impressive efficiency claim against the actual hot and cold reservoir temperatures?
Associated thinkers
Associations marked provisional are awaiting source review.
Further reading
Explore the original research or the teaching reference behind this experiment.