Carnot engine efficiency — calculate when source is 500K and sink is 300K

Question

A Carnot engine operates between a source at $500\,\text{K}$ and a sink at $300\,\text{K}$ . Calculate the efficiency of the engine. If the engine does $200\,\text{J}$ of work per cycle, find the heat absorbed from the source.

(NCERT Class 11, Chapter 12)

Solution — Step by Step

\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}

Temperatures must be in Kelvin.

\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 = \mathbf{40\%}

Efficiency is also defined as:

\eta = \frac{W}{Q_H}

where $W$ is work done and $Q_H$ is heat absorbed from the source.

Q_H = \frac{W}{\eta} = \frac{200}{0.4} = \mathbf{500\,\text{J}}

By energy conservation: $Q_H = W + Q_L$

Q_L = Q_H - W = 500 - 200 = \mathbf{300\,\text{J}}

\boxed{\eta = 40\%, \quad Q_H = 500\,\text{J}, \quad Q_L = 300\,\text{J}}

Why This Works

The Carnot engine is the most efficient engine possible between two temperatures. Its efficiency depends only on the temperatures of the source and sink, not on the working substance.

The formula $\eta = 1 - T_L/T_H$ comes from the second law of thermodynamics. It tells us that no engine can be 100% efficient (that would require $T_L = 0\,\text{K}$ , which is unattainable). The larger the temperature ratio $T_H/T_L$ , the higher the efficiency.

In our example, 40% of the heat absorbed is converted to work, and 60% is rejected to the sink. This “wasted” heat is not due to poor engineering — it’s a fundamental limit imposed by thermodynamics.

Alternative Method — Using the heat ratio

For a Carnot engine: $\frac{Q_L}{Q_H} = \frac{T_L}{T_H}$

\frac{Q_L}{Q_H} = \frac{300}{500} = 0.6

So $Q_L = 0.6 \times Q_H$ and $W = Q_H - Q_L = 0.4 \times Q_H$ .

Given $W = 200\,\text{J}$ : $Q_H = 500\,\text{J}$ , $Q_L = 300\,\text{J}$ .

For NEET, remember: Carnot efficiency sets the upper limit. Any real engine between the same temperatures will have efficiency less than the Carnot value. If a problem gives you an efficiency higher than the Carnot limit, the engine violates the second law — it’s impossible.

Common Mistake

The most frequent error: using temperatures in Celsius instead of Kelvin. If you use $T_H = 227°C$ and $T_L = 27°C$ (the equivalent Celsius values) directly in the formula, you get $\eta = 1 - 27/227 = 88\%$ — completely wrong. The Carnot formula requires absolute temperatures (Kelvin). Always convert: $T(\text{K}) = T(°C) + 273$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the heat ratio

Common Mistake

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