B.4.13 (HL)—Carnot efficiency limit

Syllabus
First assessment 2025
Objective
Level
HL

Apply the Carnot Efficiency Limit

HL only

Carnot limit

For an ideal reversible engine operating between hot and cold reservoirs,

ηC=1TCTH\eta_{\mathrm C}=1-\frac{T_C}{T_H}

Both reservoir temperatures must be in kelvin.

What the limit means

No real engine operating between the same reservoir temperatures can exceed ηC\eta_{\mathrm C}. Real friction, finite temperature differences and other irreversibilities make actual efficiency lower.

Read temperature changes

At fixed THT_H, lowering TCT_C increases the Carnot efficiency. At fixed TCT_C, increasing THT_H also increases the limit.

Worked example from local Question Bank row 31493

A plant operates between reservoirs at TH=885KT_H=885\,\mathrm{K} and TC=622KT_C=622\,\mathrm{K}.

ηC=1TCTH=1622885=0.29729.7%\eta_C=1-\frac{T_C}{T_H}=1-\frac{622}{885}=0.297\approx29.7\%

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.

B.4.13 (HL) Exam Analysis

HL only

Assessment in practice

1–3 marks
How it is assessed

The evidence includes a direct Carnot-cycle calculation and a multiple-choice change in cold-reservoir temperature.

Command terms

Calculate / Determine

What earns marks

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.

Watch for

Using Celsius in the ratio or treating Carnot efficiency as the actual efficiency of every engine.

Representative question

Question 1

[Maximum number: 3]

Calculate the efficiency of this Carnot cycle.