Heat Engines and Refrigerators – Carnot Cycle, Efficiency & COP

Heat Engines & Refrigerators | Advanced Thermal Physics

Advanced Physics → Advanced Thermal Physics → Heat Engines & Refrigerators

Heat engine and refrigerator compared side by side — Heat engines convert heat into work, while refrigerators use work to transfer heat.

Central idea
The second law of thermodynamics limits how efficiently heat can be converted into work.

1. What Is a Heat Engine?

A heat engine is a cyclic device that absorbs heat from a hot reservoir, converts part of it into work, and rejects the remaining heat to a cold reservoir.

Working principle of a heat engine — A heat engine operates between two thermal reservoirs.

2. Efficiency of a Heat Engine

$\eta = \dfrac{W}{Q_h} = 1 - \dfrac{Q_c}{Q_h}$

No heat engine can convert all absorbed heat into work. Some heat must always be rejected.

3. Carnot Engine and Maximum Efficiency

$\eta_{\text{max}} = 1 - \dfrac{T_c}{T_h}$

Carnot cycle shown on PV diagram — The Carnot cycle represents an ideal reversible engine.

Carnot efficiency depends only on temperatures, not on the working substance.

4. Entropy and Heat Engines

Even in ideal engines, entropy must be transferred to the cold reservoir.

Entropy flow in a heat engine — Entropy flow explains why 100% efficiency is impossible.

5. Refrigerators and Heat Pumps

A refrigerator uses external work to extract heat from a cold region and reject it to a hotter environment.

Working principle of a refrigerator — Refrigerators move heat against its natural direction using work.

6. Coefficient of Performance (COP)

$\text{COP}_{\text{ref}} = \dfrac{Q_c}{W}$

Comparison of COP and efficiency — COP can be greater than one, unlike engine efficiency.

Refrigerators are judged by COP, not efficiency.

Practice Problems

Level 1 — Conceptual

Why must every heat engine reject heat to a cold reservoir?

Solution

Because entropy must be transferred to satisfy the second law.

Why is Carnot engine unattainable in practice?

Solution

It requires perfectly reversible processes.

Level 2 — Numerical

An engine absorbs 500 J of heat and rejects 300 J. Find efficiency.

Solution

Efficiency = (500 − 300)/500 = 0.4.

A refrigerator removes 200 J of heat using 50 J of work. Find COP.

Solution

COP = 200/50 = 4.

Level 3 — Advanced

Why does lowering cold reservoir temperature increase engine efficiency?

Solution

It increases the temperature difference, reducing entropy rejection.

Why is perpetual motion of second kind impossible?

Solution

It violates entropy increase of the universe.

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