A car engine, a power plant's turbine and a steam locomotive all do one thing: take heat from something hot, turn part of it into work, and let the rest go. Each runs the kind of cycle you just learned, and its efficiency is what you get out over what you put in.
Why it works
Around a cycle $\Delta U = 0$, so energy in equals energy out: $Q_H = W + Q_C$. Why can't $Q_C$ be zero? To return to its start, the gas must be compressed again. Compressing it while it is hot would cost all the work it just did, so the engine cools it first, and cooling means giving heat $Q_C$ to something colder. This is the second law of thermodynamics: no engine running in a cycle can turn heat completely into work.
Key idea
$$e = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}, \qquad Q_H = W + Q_C$$ Every heat engine gives off some heat, so $e < 1$.
Worked examples
Worked example: One cycle
An engine takes in $2000\,\text{J}$ of heat per cycle and does $500\,\text{J}$ of work. Find $Q_C$ and the efficiency.
- $Q_C = Q_H - W = 2000 - 500 = 1500\,\text{J}$.
- $e = \dfrac{W}{Q_H} = \dfrac{500}{2000} = 0.25 = 25\%$. Check: $1 - \dfrac{1500}{2000} = 0.25$.
Answer: $Q_C = 1500\,\text{J}$, $e = 25\%$
Worked example: A car engine
An engine delivers $60\,\text{kW}$ of mechanical power at $30\%$ efficiency. At what rates does it take in heat and give it off?
- Rates work like amounts: $P_{\text{in}} = \dfrac{P_{\text{out}}}{e} = \dfrac{60\,\text{kW}}{0.30} = 200\,\text{kW}$.
- Heat given off: $200 - 60 = 140\,\text{kW}$, through the exhaust and the radiator.
Answer: $200\,\text{kW}$ in; $140\,\text{kW}$ out
Common mistake
Thinking a well-designed engine could turn all its heat into work. Friction and leaks do waste some, but even a perfect engine must give off heat $Q_C$ to finish its cycle. The next topic finds the best any engine can do. Check: in every engine problem, $Q_C$ must be greater than zero.
The Science Lab's engine chapter also counts $W$ as the work the engine delivers, a positive number, as here.