The maximum theoretical efficiency of a heat engine depends only on the source and sink temperatures

Efficiency of a heat engine

• The goal of a heat engine is to transfer thermal energy into useful mechanical work as efficiently as possible.
• The efficiency can be calculated using:

• Where:
  • $W$ = useful work output (J)
  • $Q_H$ = energy transferred from the source (J)
  • $Q_C$ = energy transferred to the sink (J)
• Since the efficiency of a heat engine can never be zero (otherwise there would be no work), this means no heat engine can completely convert heat into work.

Maximum theoretical efficiency

• As the efficiency of a thermodynamic system increases, the difference between the temperatures of the source and sink increases.
• The maximum theoretical efficiency of a heat engine is:

• Where:
  • $T_C$ = temperature of the sink (cold reservoir) in kelvin (K)
  • $T_H$ = temperature of the source (hot reservoir) in kelvin (K)
• This equation can be used if an ideal gas is used as the working substance for the engine.
• Fundamentally, to make an engine as efficient as possible, the source temperature must be as high as possible and the sink temperature as low as possible.
• The maximum theoretical efficiency is 100% only if the sink temperature is 0 K, which is physically unattainable.

Using the second law to test proposed engines

• If a proposed engine claims an efficiency greater than the maximum theoretical efficiency for its operating temperatures, then it violates the second law of thermodynamics and is impossible.
• This is a common exam question: calculate the claimed efficiency using $\frac{Q_H - Q_C}{Q_H}$, then calculate the maximum theoretical efficiency using $\frac{T_H - T_C}{T_H}$. If the claimed efficiency exceeds the maximum, the engine is thermodynamically impossible.