Carnot engine efficiency between 500K and 300K — find work per cycle if Q₁=1000J

Question

A Carnot engine operates between a source temperature of $T_1 = 500$ K and a sink temperature of $T_2 = 300$ K. If the engine absorbs $Q_1 = 1000$ J of heat per cycle from the source, find: (a) The efficiency of the engine (b) The work done per cycle (c) The heat rejected to the sink per cycle

Solution — Step by Step

The efficiency of a Carnot engine depends only on the temperatures of the source and sink:

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

This is the maximum possible efficiency for any heat engine operating between these two temperatures. No real engine can exceed this.

\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4

\boxed{\eta = 40\%}

The engine converts 40% of the heat absorbed from the source into work.

Work done is the fraction of input heat that gets converted:

W = \eta \times Q_1 = 0.4 \times 1000 = 400 \text{ J}

\boxed{W = 400 \text{ J per cycle}}

By the first law of thermodynamics, energy is conserved:

Q_1 = W + Q_2

Q_2 = Q_1 - W = 1000 - 400 = 600 \text{ J}

\boxed{Q_2 = 600 \text{ J per cycle}}

We can verify: $Q_2/Q_1 = 600/1000 = 0.6 = T_2/T_1$ . ✓

Why This Works

The Carnot engine is a theoretical ideal — it operates on a reversible cycle (two isothermal and two adiabatic processes). Its efficiency depends only on temperatures, not on the working substance or any other property.

The formula $\eta = 1 - T_2/T_1$ comes from the Carnot theorem, which is a direct consequence of the second law of thermodynamics. The second law sets an absolute upper limit on how efficiently we can convert heat into work.

The relation $Q_2/Q_1 = T_2/T_1$ is the thermodynamic definition of temperature on the Kelvin scale — this is actually why absolute temperature is defined the way it is.

Alternative Method

You can also find $Q_2$ directly using $Q_2/Q_1 = T_2/T_1$ :

Q_2 = Q_1 \times \frac{T_2}{T_1} = 1000 \times \frac{300}{500} = 600 \text{ J}

Then $W = Q_1 - Q_2 = 1000 - 600 = 400$ J. Same answer, slightly fewer steps.

Common Mistake

Always use absolute temperatures (Kelvin) in the Carnot efficiency formula. If you use Celsius, you get nonsense. Converting: $T(K) = T(°C) + 273$ . If the question gives temperatures as 227°C and 27°C, convert to 500 K and 300 K first. This is the #1 error students make on this type of problem.

Notice that efficiency can never reach 100% unless $T_2 = 0$ K (absolute zero), which is unattainable. And efficiency is zero when $T_1 = T_2$ (no temperature difference to drive the engine). JEE Main sometimes asks: “What must $T_1$ be for 50% efficiency if $T_2 = 400$ K?” — solve $0.5 = 1 - 400/T_1$ to get $T_1 = 800$ K.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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