η = 1 - T₂/T₁. Second law says T₂ can never be 0. Cold reservoir always absorbs some heat.

Carnot Engine Efficiency — Why 100% Efficiency is Impossible

Question

A Carnot engine operates between a source at temperature $T_1 = 500\,\text{K}$ and a sink at temperature $T_2 = 300\,\text{K}$ . Calculate its efficiency. Then explain why no engine — however well-built — can achieve 100% efficiency.

Solution — Step by Step

The efficiency of a Carnot engine is:

\eta = 1 - \frac{T_2}{T_1}

Both temperatures must be in Kelvin. Never use Celsius here — this is a trap that costs marks.

\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4

So $\eta = 0.4$ , or 40%.

The engine converts 40% of the heat absorbed from the source into work. The remaining 60% is rejected to the cold reservoir — not because of poor engineering, but because the Second Law demands it.

Efficiency = 40%

For $\eta = 1$ , we need $T_2 = 0\,\text{K}$ (absolute zero). Let’s check:

\eta = 1 - \frac{0}{T_1} = 1 - 0 = 1 = 100\%

The Third Law of Thermodynamics states that reaching absolute zero is physically impossible. So 100% efficiency is not just an engineering challenge — it’s forbidden by the laws of physics.

Why This Works

The Carnot engine is not a real engine — it’s the theoretical upper limit. Any real engine (petrol, diesel, steam) operating between the same two temperatures will always have lower efficiency than the Carnot engine. This is Carnot’s theorem.

The formula $\eta = 1 - T_2/T_1$ tells us something profound: efficiency depends only on the temperatures, not on the working substance or the design of the engine. A more expensive piston or a better lubricant won’t help — only raising $T_1$ or lowering $T_2$ can improve efficiency.

The Second Law is what makes $T_2 = 0$ unreachable. Heat always flows from hot to cold on its own — you can’t have a cold reservoir that absorbs zero heat unless it’s at absolute zero, which no real object can be.

Alternative Method — Using Heat Absorbed and Rejected

If a problem gives you $Q_1$ (heat absorbed) and $Q_2$ (heat rejected) instead of temperatures, use:

\eta = 1 - \frac{Q_2}{Q_1}

For a Carnot engine specifically, these two ratios are equal:

\frac{Q_2}{Q_1} = \frac{T_2}{T_1}

This equivalence is what makes the Carnot cycle reversible. In JEE problems, you’ll sometimes get $Q_1$ and $Q_2$ and be asked to find the temperatures — just flip this relation.

In JEE Main, questions often give work done $W$ and heat absorbed $Q_1$ , then ask for efficiency. Use $\eta = W/Q_1$ directly. No need to find $Q_2$ first. This appeared in JEE Main 2024 Shift 1 in a slightly disguised form.

Common Mistake

Using Celsius instead of Kelvin is the most penalised error in this topic. If $T_1 = 227°\text{C}$ and $T_2 = 27°\text{C}$ , the efficiency is NOT $1 - 27/227$ . Convert first: $T_1 = 500\,\text{K}$ , $T_2 = 300\,\text{K}$ , then apply the formula. The answer changes from a wrong ~88% to the correct 40%.

A second mistake: students sometimes argue that a “perfect” engine with no friction would give 100% efficiency. Friction is an irreversibility problem — the 100% limit is about the Second Law, not mechanical losses. Even a frictionless Carnot engine can’t cross 100% unless $T_2 = 0\,\text{K}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Heat Absorbed and Rejected

Common Mistake

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