Question

A Heat engine receives 5 MJ heat transfer at 995 K, gives out 2 MJ as work and the rest of the heat to the surrondings at 25*C. The first and second law efficiencies are ----- and ----.

Answer 1

The first law efficiency or thermal efficiency of the heat engine is 40%, and the second law efficiency also known as Carnot efficiency is approximately 70%.

Step-by-step explanation:

The student's question concerns the calculation of the first and second law efficiencies of a heat engine that operates between two thermal reservoirs at different temperatures. Initially, the heat engine receives 5 MJ of heat at 995 K and performs 2 MJ of work, ejecting the rest of the heat to the surroundings at 25°C (which translates to 298 K).

The first law efficiency is also known as the thermal efficiency and is calculated using the formula:

• First law efficiency (η) = Work output / Heat input

Therefore, η = (2 MJ) / (5 MJ) = 0.4 or 40%.

The second law efficiency, also known as the Carnot efficiency, is determined by the temperatures of the hot and cold reservoirs and is given by:

• Second law efficiency (ηCarnot) = 1 - (Tc / Th)

Where Tc and Th are absolute temperatures of the cold and hot reservoirs respectively. Thus, ηCarnot = 1 - (298 K / 995 K) ≈ 0.70 or 70%.

The first and second law efficiencies are 40% and approximately 70%, respectively.