Thermodynamics – Reversible devices operating between the same two reservoirs A reversible heat engine operates between a hot and a cold thermal reservoir and has a thermal efficiency of 0.40. If the same temperature limits are used to run a reversible refrigerator and a reversible heat pump, determine the correct pair: coefficient of performance as a refrigerator (COP_R) and coefficient of performance as a heat pump (COP_HP).

Introduction / Context:
This problem tests core relations among reversible heat engines, refrigerators, and heat pumps operating between the same two temperature reservoirs. For Carnot (reversible) devices, efficiency and coefficients of performance are dictated solely by the temperature ratio, so once one quantity is known, all others follow deterministically.

Given Data / Assumptions:

• Reversible (Carnot) behavior between the same hot and cold reservoirs.
• Thermal efficiency of the heat engine, eta = 0.40.
• No irreversibilities or additional losses; all devices see identical Th and Tc.

Concept / Approach:
For a reversible heat engine: eta = 1 - Tc/Th. For a reversible refrigerator: COP_R = Tc/(Th - Tc). For a reversible heat pump: COP_HP = Th/(Th - Tc). These can be written in terms of the temperature ratio r = Th/Tc. Using eta, we compute r, then evaluate COP_R and COP_HP directly.

Step-by-Step Solution:

Verification / Alternative check:
For any reversible cycle between fixed reservoirs, COP_HP = COP_R + 1. Since COP_R ≈ 1.5, COP_HP should be ≈ 2.5, confirming consistency.

Why Other Options Are Wrong:

• (COP)R = (COP)HP = 0.6: Confuses efficiency with COP and violates COP_HP = COP_R + 1.
• (2.5; 1.5): Reverses refrigerator and heat pump values.
• (Equal 2.5): COPs cannot be equal; heat pump COP must exceed refrigerator COP by 1.

Common Pitfalls:
Mixing up efficiency (engine) with COP (refrigerator/heat pump), or forgetting the identity COP_HP = COP_R + 1 for the same temperature limits.

Final Answer:
(COP)R = 1.5; (COP)HP = 2.5