Reversible device reconfigured as a refrigerator A reversible heat engine has an ideal thermal efficiency of 30%. If the same device operates reversibly as a refrigerator between the same two temperature levels, what will be its coefficient of performance (C.O.P.)?

Introduction / Context:
Carnot relations connect the efficiency of a reversible heat engine with the performance of its corresponding reversible refrigerator or heat pump operating between the same two thermal reservoirs.

Given Data / Assumptions:

• Reversible (Carnot) operation between T_H and T_L.
• Thermal efficiency of the engine η = 0.30.
• All other conditions (reservoir temperatures) unchanged when running as a refrigerator.

Concept / Approach:
For a Carnot engine, η = 1 − (T_L / T_H). For a Carnot refrigerator, C.O.P._R = T_L / (T_H − T_L). We first find the temperature ratio from the efficiency, then compute C.O.P.

Step-by-Step Solution:
η = 1 − (T_L / T_H) = 0.30 ⇒ T_L / T_H = 1 − 0.30 = 0.70.C.O.P._R = T_L / (T_H − T_L) = (T_L / T_H) / (1 − T_L / T_H).Substitute: C.O.P._R = 0.70 / (1 − 0.70) = 0.70 / 0.30 = 2.333…Rounded to two decimals, C.O.P. ≈ 2.33.

Verification / Alternative check:
Use the relation (C.O.P.)_HP = (C.O.P.)_R + 1 to see that the heat pump C.O.P. would be 3.33; this is consistent with η and the same temperature pair.

Why Other Options Are Wrong:
1.33 and 4.33 correspond to different temperature ratios; 3.33 is the heat pump value, not the refrigerator; 0.70 is the temperature ratio, not a C.O.P.

Common Pitfalls:
Confusing engine efficiency η with C.O.P., and mixing the refrigerator and heat pump expressions.

Final Answer:
2.33