B.4.13 (HL)—Carnot efficiency limit
- Syllabus
- First assessment 2025
- Objective
- —
- Level
- HL
Carnot limit
For an ideal reversible engine operating between hot and cold reservoirs,
ηC=1−THTC
Both reservoir temperatures must be in kelvin.
What the limit means
No real engine operating between the same reservoir temperatures can exceed ηC. Real friction, finite temperature differences and other irreversibilities make actual efficiency lower.
Read temperature changes
At fixed TH, lowering TC increases the Carnot efficiency. At fixed TC, increasing TH also increases the limit.
Worked example from local Question Bank row 31493
A plant operates between reservoirs at TH=885K and TC=622K.
ηC=1−THTC=1−885622=0.297≈29.7%
This is the theoretical maximum. A real plant between the same temperatures must have lower efficiency because its processes are irreversible.
Common trap
Carnot efficiency is a maximum, not the efficiency every engine achieves. Never use Celsius in the temperature ratio.
The evidence includes a direct Carnot-cycle calculation and a multiple-choice change in cold-reservoir temperature.
Calculate / Determine
Use ηC=1−TC/TH with both reservoir temperatures in kelvin. For a real engine compare η with ηC and state η≤ηC. In ratio questions, recalculate the temperature ratio rather than assuming efficiency changes linearly.
Using Celsius in the ratio or treating Carnot efficiency as the actual efficiency of every engine.
Representative question
Calculate the efficiency of this Carnot cycle.
TATD=PAVAPDVD=<1−4.3×2.21.7×3.8=>0.32 or 32%
Allow use hot/cold instead A, D.
Award [3] for bald correct answer for interval [0.28, 0.34]