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A heat engine absorbs 100 J of heat from the hot reservoir
Chapter , Problem 29(choose chapter or problem)
A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine.
Questions & Answers
QUESTION:
A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine.
ANSWER:Step 1 of 5
heat absorbed by the engine from the hot reservoir during each cycle
heat released by the engine to the cold reservoir during each cycle
The Coefficient of Performance (COP)
A ratio that measures the efficiency of a refrigeration or heat pump system. It is defined as the ratio of the heat transfer rate to the work input rate.
For a refrigeration system, the COP is defined as the ratio of the heat removed from the cold space to the work input required to drive the refrigeration cycle. Mathematically, it can be expressed as
For a heat pump system, the COP is defined as the ratio of the heat supplied to the hot space to the work input required to drive the heat pump cycle.
The COP is always greater than one, since it represents the ratio of useful output (heat transfer) to input (work). The higher the COP, the more efficient the refrigeration or heat pump system is.