heat-engine efficiency and Carnot limit

A heat engine absorbs heat $Q_1$ from a hot reservoir, performs work $A$, and rejects heat $Q_2$ to a cold reservoir. Over a complete cycle the internal energy of the working substance returns to its initial value, so the first law of thermodynamics gives $A = Q_1 - Q_2$. Efficiency is defined as the ratio of useful work to heat supplied: $\eta = \frac{A}{Q_1} = 1 - \frac{Q_2}{Q_1}.$ Carnot showed that for fixed hot-reservoir and cold-reservoir temperatures $T_1$ and $T_2$, no cycle can exceed $\eta_{\max} = 1 - \frac{T_2}{T_1},$ and equality is reached only by the reversible Carnot cycle. In this formula the temperatures must be absolute (kelvin), otherwise the ratio $T_2/T_1$ is meaningless. Three consequences follow: efficiency is always less than one, it increases when $T_1$ rises or $T_2$ falls, and $\eta = 1$ would require $T_2 = 0$ K, which is unattainable.

In one cycle the working substance absorbs $Q_1 = 800$ J from the heater and rejects $Q_2 = 600$ J to the cooler. Find the work per cycle and the efficiency of the engine.

An ideal Carnot engine operates between a heater at $T_1 = 500$ K and a cooler at $T_2 = 300$ K. Each cycle absorbs $Q_1 = 1000$ J from the heater. Find the maximum efficiency, the maximum work per cycle, and the heat rejected to the cooler.

An inventor claims that his engine takes $Q_1 = 5000$ J per cycle from a heater at $T_1 = 400$ K and rejects $Q_2 = 2500$ J to a cooler at $T_2 = 300$ K. Is this possible? Justify your answer with a calculation. What is the maximum permissible work under these conditions?

Self-study work with a heat-engine model (safe, no real heaters).

1. Open an interactive heat-engine or Carnot-cycle simulation (for example, a browser-based Carnot cycle simulator). Do not use real boiling water, steam, or open flames: the hot reservoirs are modelled virtually.
2. For 4–5 pairs of temperatures $T_1$ and $T_2$, record the heat supplied $Q_1$, the heat rejected $Q_2$, the work $A$, and the efficiency $\eta$; put the results in a table.
3. For each row compute $\eta_{\max} = 1 - T_2/T_1$ and compare it with the $\eta$ measured in the model.
4. Plot $\eta$ against the ratio $T_2/T_1$ and overlay the straight line $1 - T_2/T_1$.
5. State a conclusion: how efficiency changes as $T_1$ rises and as $T_2$ falls, and why the points never rise above the Carnot line.

To show the energy flows in a heat engine and to make clear why efficiency is bounded by the temperature ratio and can never reach 100%.

Diagram: the hot reservoir at temperature $T_1$ on top, an arrow $Q_1$ down to the working substance, from which an arrow $A$ goes sideways (useful work) and an arrow $Q_2$ goes down to the cold reservoir at temperature $T_2$. Beside it, a graph: horizontal axis shows the ratio $T_2/T_1$ from 0 to 1, vertical axis shows $\eta$; the straight line $\eta_{\max} = 1 - T_2/T_1$ and points lying below it.

A $1.6\ \mathrm{L}$ petrol engine burns its fuel in short bursts; the burning gases reach roughly $2200\ \mathrm{K}$, and the engine is cooled by outside air at about $300\ \mathrm{K}$. Of every $100\ \mathrm{MJ}$ of heat released by the fuel, only about $30\ \mathrm{MJ}$ reach the wheels as useful work — the radiator, the exhaust pipe and friction take the other $70\ \mathrm{MJ}$.

Here $\eta = W/Q_H = 30/100 = 0.30$, i.e. $30\%$. The Carnot limit for the same temperatures is $\eta_{C} = 1 - T_C/T_H = 1 - 300/2200 \approx 0.86$; the real engine sits far below it because heat transfer in a few milliseconds is irreversible, the gas does not expand fully, and part of the heat leaves with the exhaust instead of doing work.

In a thermal power station steam enters the turbine at $550\ ^\circ\mathrm{C}$, that is $823\ \mathrm{K}$, and the used steam condenses in a cooling tower against water at $30\ ^\circ\mathrm{C}$, that is $303\ \mathrm{K}$. The station sends about $400\ \mathrm{MW}$ of electrical power to the grid while the fuel supplies about $1000\ \mathrm{MW}$ of heat.

$\eta = 400/1000 = 0.40$, i.e. $40\%$, while the Carnot limit is $\eta_{C} = 1 - 303/823 \approx 0.63$. The gap of $23$ percentage points is the price of a real cycle: the turbine blades, the boiler's finite temperature difference and the pumps all consume part of the work, and heat is transferred across temperature differences rather than reversibly.

A turbofan cruising at $11\ \mathrm{km}$ burns about $1\ \mathrm{kg}$ of kerosene per second, which releases roughly $43\ \mathrm{MJ}$ of heat each second ($43\ \mathrm{MW}$). The combustion chamber runs at about $1800\ \mathrm{K}$ and the outside air at that altitude is about $220\ \mathrm{K}$. The engine turns that heat into roughly $15\ \mathrm{MW}$ of thrust power.

$\eta = 15/43 \approx 0.35$, i.e. about $35\%$. The Carnot limit $\eta_{C} = 1 - 220/1800 \approx 0.88$ shows the temperature difference is more than wide enough; the loss comes from the jet exhaust still carrying kinetic and thermal energy, from compressor and turbine inefficiency, and from the fact that a heat engine must reject $Q_C = Q_H - W \approx 28\ \mathrm{MW}$ to the cold air.