The Department of Energy develops a new reversible engine which has a COP of 4 when operated as a refrigerator and a COP of 5 when operated as a heat pump. What is its efficiency when operated as an engine doing work?

Question

Quizplus · Accepted Answer

The efficiency ($\eta$) of an engine is given by $\eta = 1 - \frac{1}{COP_{hp}}$, where $COP_{hp}$ is the Coefficient of Performance as a heat pump. Given $COP_{hp} = 5$, $\eta = 1 - \frac{1}{5} = 0.8$ or 80%. However, the efficiency of an engine is the fraction of heat input converted to work, so we need to convert this to a percentage. The correct calculation for efficiency in terms of work output, not as a heat pump, involves recognizing that the given COPs are not directly used to calculate engine efficiency. Instead, the relationship between COP of a heat pump and the efficiency of an engine doing work is indirect. Given the misunderstanding in the explanation, the correct approach involves recognizing that the efficiency of a reversible engine is not directly derived from the COP values provided for its operation as a refrigerator or a heat pump. The correct answer, based on the information provided and the correct interpretation of the efficiency of an engine, should reflect the fundamental principles of thermodynamics, particularly the second law, and the correct formula should be applied accordingly. My explanation mistakenly applied the formula for efficiency based on the COP of a heat pump, which led to an incorrect conclusion. The correct approach to find the efficiency of an engine doing work involves using the relationship between the heat absorbed and work done, not the COP values given for its operation in different modes.

The Department of Energy Develops a New Reversible Engine Which