Maximum possible efficiency depends only on temperatures. Why cold reservoir matters.

Carnot Engine Efficiency — η = 1 - T₂/T₁

Question

A Carnot engine operates between a source at 627°C and a sink at 27°C. Find the efficiency of the engine. If the engine absorbs 1000 J of heat per cycle, how much useful work does it deliver?

Solution — Step by Step

The Carnot formula uses absolute temperature — Celsius will give you a wrong answer. Add 273 to each.

T_1 = 627 + 273 = 900 \text{ K}

T_2 = 27 + 273 = 300 \text{ K}

The efficiency of any Carnot engine is fixed entirely by the two reservoir temperatures:

\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}

So $\eta = 0.667$ , or 66.7%.

Efficiency is defined as $\eta = W/Q_1$ , where $Q_1$ is heat absorbed from the source.

W = \eta \times Q_1 = \frac{2}{3} \times 1000 = \textbf{666.7 J}

By energy conservation, whatever wasn’t converted to work was dumped into the sink:

Q_2 = Q_1 - W = 1000 - 666.7 = \textbf{333.3 J}

This also follows directly from $Q_2/Q_1 = T_2/T_1$ .

Why This Works

The Carnot engine is not a real machine — it’s a theoretical upper bound. No engine operating between the same two temperatures can be more efficient. This is a direct consequence of the Second Law of Thermodynamics.

The key insight: efficiency depends only on the temperature ratio $T_2/T_1$ . Making the source hotter helps, but lowering the cold reservoir temperature helps equally. This is why real power plants use cold river water or cooling towers — they’re trying to bring $T_2$ down.

The formula $\eta = 1 - T_2/T_1$ also tells you that 100% efficiency is impossible unless $T_2 = 0$ K (absolute zero), which we can never reach. This is actually another way of stating the Third Law of Thermodynamics.

Alternative Method

Instead of using the efficiency formula directly, use the Carnot cycle’s defining property: the ratio of heats equals the ratio of temperatures.

\frac{Q_2}{Q_1} = \frac{T_2}{T_1} = \frac{300}{900} = \frac{1}{3}

So $Q_2 = \frac{1}{3} \times 1000 = 333.3$ J is rejected. Then:

W = Q_1 - Q_2 = 1000 - 333.3 = 666.7 \text{ J}

Same answer, different route. In JEE, if the question directly gives $Q_1$ and asks for $W$ , this path is faster because you skip computing $\eta$ explicitly.

If temperatures are given as a nice ratio (like 1:3 here), work with fractions throughout. $\frac{300}{900} = \frac{1}{3}$ is cleaner than converting to 0.333… mid-calculation — you avoid rounding errors and the examiner sees exact values.

Common Mistake

Using Celsius instead of Kelvin. Students plug in $\eta = 1 - 27/627 = 1 - 0.043 = 95.7\%$ and wonder why the answer looks suspiciously high. The formula $\eta = 1 - T_2/T_1$ is derived from entropy changes, which are defined using absolute temperature. Celsius has an arbitrary zero — it means nothing thermodynamically. Always convert first, before touching the formula.

A second trap: assuming that a 66.7% efficient Carnot engine is “better” than a real engine with, say, 40% efficiency. The Carnot efficiency is the maximum possible — any real engine will always fall below it due to irreversibilities. The question “what fraction of Carnot efficiency does this engine achieve?” appears in CBSE board papers and JEE as a follow-up.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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