Heat Engines, Efficiency and COP

Heat Engines, Efficiency and COP — A Complete Guide

Heat engines sit at the meeting point of thermodynamics and engineering, and they form one of the most reliably scoring sub-topics in JEE Main and NEET. Every car engine, every refrigerator in your kitchen, every air conditioner — they all obey the same handful of equations. Once we get the conceptual core right, the numerical questions become almost mechanical.

The chapter has three core characters: the heat engine (converts heat to work), the refrigerator (pumps heat from cold to hot using work), and the heat pump (refrigerator in reverse, used for heating). They share the same first-law bookkeeping but ask different questions.

We’ll work through the definitions, the Carnot cycle as the gold standard, common numerical patterns, and the typical traps that examiners set. By the end, a JEE-style question on Carnot efficiency should take you under a minute.

Key Terms & Definitions

Heat Engine. A cyclic device that absorbs heat $Q_H$ from a hot reservoir at temperature $T_H$ , performs work $W$ , and rejects heat $Q_C$ to a cold reservoir at temperature $T_C$ . By the first law over a complete cycle: $W = Q_H - Q_C$ .

Efficiency ( $\eta$ ). The fraction of input heat converted to useful work:

\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}

Coefficient of Performance (COP). For a refrigerator, the ratio of heat extracted from cold reservoir to work input:

\text{COP}_{\text{ref}} = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C}

For a heat pump, $\text{COP}_{\text{pump}} = Q_H / W$ , which is always greater than $\text{COP}_{\text{ref}}$ by exactly $1$ .

Carnot Engine. An idealised reversible engine operating between two reservoirs. It sets the upper bound on efficiency for any engine working between the same two temperatures.

Methods & Concepts

The Carnot Cycle

The Carnot cycle has four reversible strokes: isothermal expansion at $T_H$ , adiabatic expansion (cooling to $T_C$ ), isothermal compression at $T_C$ , adiabatic compression (heating back to $T_H$ ). The crucial result is that for this cycle:

\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}

Temperatures must be in kelvin. This is the most common slip-up — students plug in Celsius and get nonsense.

Why No Engine Can Beat Carnot

The second law of thermodynamics forbids any engine operating between $T_H$ and $T_C$ from exceeding Carnot efficiency. If such a super-Carnot engine existed, you could couple it backwards to a Carnot engine and pump heat from cold to hot with no work — violating Clausius’s statement of the second law.

Refrigerator and Heat Pump Relations

For a Carnot refrigerator:

\text{COP}_{\text{ref,Carnot}} = \frac{T_C}{T_H - T_C}

Notice that as $T_C \to T_H$ , COP $\to \infty$ — easy to pump heat across a small gap. As $T_C \to 0$ , COP $\to 0$ — extremely hard to extract the last bits of heat.

Solved Examples

Example 1 — Carnot Efficiency (CBSE Boards)

A Carnot engine works between $T_H = 500\ \text{K}$ and $T_C = 300\ \text{K}$ . Find efficiency.

\eta = 1 - \frac{300}{500} = 0.4 = 40\%

Example 2 — Work Output (JEE Main pattern)

The same engine absorbs $Q_H = 1000\ \text{J}$ per cycle. Find work done per cycle.

W = \eta Q_H = 0.4 \times 1000 = 400\ \text{J}

Example 3 — Refrigerator COP (NEET pattern)

A refrigerator removes $300\ \text{J}$ from the cold compartment per second while consuming $100\ \text{J/s}$ of electrical work. Find its COP.

\text{COP} = \frac{Q_C}{W} = \frac{300}{100} = 3

Example 4 — Hard (JEE Advanced pattern)

A Carnot engine has efficiency $\eta_1$ . Both temperatures are increased by $50\ \text{K}$ . Show that the new efficiency $\eta_2 < \eta_1$ .

The ratio $T_C/T_H$ moves closer to $1$ when we add the same amount to both, so $1 - T_C/T_H$ decreases. Quantitatively: if $T_H = 400, T_C = 300$ , then $\eta_1 = 25\%$ , but $\eta_2 = 1 - 350/450 \approx 22.2\%$ . Engineers prefer raising $T_H$ alone — which is why power plants chase higher steam temperatures.

Exam-Specific Tips

JEE Main weightage: Heat engines reliably bring 1–2 marks per paper. Almost always one Carnot efficiency or COP question. NEET weightage: Often combined with thermodynamics process questions; expect 1 question. CBSE Class 11: A 3-mark or 5-mark derivation question on Carnot efficiency is standard.

For COP questions, write out $Q_H$ , $Q_C$ , $W$ explicitly with one missing, then use $W = Q_H - Q_C$ and the COP definition. Two equations, one unknown — done in 30 seconds.

Common Mistakes to Avoid

Using Celsius instead of Kelvin. Always convert. $T(K) = T(°C) + 273$ .

Confusing efficiency and COP. Efficiency is bounded above by $1$ ; COP is often greater than $1$ . They’re not the same scale.

Forgetting that $W$ is input for refrigerators, output for engines. Sign convention switches based on the device.

Assuming all cycles are Carnot. The Otto, Diesel, and Brayton cycles have their own efficiency formulas — only use $1 - T_C/T_H$ when the problem says Carnot or reversible.

Forgetting the second law upper bound. If a question claims an engine has efficiency higher than $1 - T_C/T_H$ , mark it impossible without further calculation.

Practice Questions

Q1. A Carnot engine works between $T_H = 600\ \text{K}$ and $T_C = 400\ \text{K}$ . What is its efficiency?

$\eta = 1 - 400/600 = 1/3 \approx 33.3\%$ .

Q2. A refrigerator with COP $= 4$ extracts $400\ \text{J}$ from cold reservoir. Find work done.

$W = Q_C/\text{COP} = 400/4 = 100\ \text{J}$ .

Q3. An engine claims efficiency of $60\%$ between $T_H = 500\ \text{K}$ and $T_C = 250\ \text{K}$ . Is this possible?

Carnot limit is $1 - 250/500 = 50\%$ . Claimed $60\% > 50\%$ , so this violates the second law. Impossible.

Q4. A Carnot heat pump operates with $T_H = 300\ \text{K}, T_C = 270\ \text{K}$ . Find COP.

$\text{COP}_{\text{pump}} = T_H/(T_H - T_C) = 300/30 = 10$ .

Q5. Engine takes $Q_H = 800\ \text{J}$ , rejects $Q_C = 600\ \text{J}$ . Find efficiency.

$\eta = 1 - 600/800 = 25\%$ .

Q6. If a Carnot engine’s $T_H$ doubles (with $T_C$ fixed), how does efficiency change?

New efficiency $= 1 - T_C/(2T_H)$ . Always larger than the original $1 - T_C/T_H$ . Specific increase depends on initial values.

Q7. A heat pump and refrigerator operate between same temperatures. How are their COPs related?

$\text{COP}_{\text{pump}} = \text{COP}_{\text{ref}} + 1$ — exact relationship.

Q8. Why can’t an engine convert all heat to work?

Second law (Kelvin-Planck statement) forbids it. Some heat must be dumped to a cold reservoir.

FAQs

Why is the Carnot efficiency the maximum? Because Carnot is reversible, and reversible engines have the maximum efficiency between any two reservoirs. This is Carnot’s theorem, a direct consequence of the second law.

Can COP be infinite? Mathematically, yes — when $T_H = T_C$ , COP diverges. Physically, no useful work happens because there’s no temperature gradient to fight against.

Why use kelvin? The thermodynamic temperature scale is defined so that $T_C/T_H = Q_C/Q_H$ for a Carnot cycle. This relation holds only in kelvin.

Does efficiency depend on the working substance? For a Carnot engine, no — only on the two reservoir temperatures. For real engines, yes.

What’s the relationship between entropy and efficiency? Real engines have entropy generation, which always reduces efficiency below the Carnot limit. The second law says total entropy never decreases.

Are heat pumps energy efficient? Yes — a heat pump with COP $= 4$ delivers $4\ \text{kWh}$ of heating per $1\ \text{kWh}$ of electricity. Direct electric heaters have COP $= 1$ .