How is the thermal efficiency ($ ext{eta}$) of a heat engine calculated based on energy inputs and outputs?
Answer
The ratio of the net work output ($W$) to the total heat input ($Q_H$).
Thermal efficiency ($ ext{eta}$) is defined as the ratio of the mechanical work produced ($W$) to the total heat energy absorbed from the high-temperature source ($Q_H$).
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How is the thermal efficiency ($ ext{eta}$) of a heat engine calculated based on energy inputs and outputs?What fundamental requirement dictates that the efficiency of any real heat engine must always be less than 100%?What factors exclusively determine the Carnot efficiency, according to the derived formula?In what temperature scale must $T_H$ and $T_C$ be expressed to correctly calculate Carnot efficiency?What is the fundamental thermodynamic requirement for extracting work from a heat engine?Which principle serves as the theoretical foundation proving that no heat engine can be more efficient than a reversible engine operating between the same two temperatures?Which of the following represents an irreversible process causing real engines to fall short of the Carnot limit?Besides thermodynamic limits, what practical constraint governs the achievable operating temperature ($T_H$) in modern engines?How does the Second Law of Thermodynamics differentiate itself from the First Law regarding energy?If Plant A has $T_H = 900 ext{ K}$ and Plant B has $T_H = 1200 ext{ K}$ (both rejecting at $T_C = 300 ext{ K}$), what does this illustrate about efficiency gains?