Reversed Carnot devices — relating COP of heat pump and refrigerator A heat pump operating on a reversed Carnot cycle has a COP (heat pump) of 5. If the same device now operates as a refrigerator with a work input of 1 kW, what is the refrigerating effect (cooling capacity)?

Introduction / Context:
For any reversed cycle, the coefficient of performance (COP) takes different forms depending on whether the device is viewed as a heat pump (delivering heat to a warm space) or a refrigerator (removing heat from a cold space). The two COPs are simply related for the same temperature limits.

Given Data / Assumptions:

• Reversed Carnot cycle with the same temperature limits.
• COP_heat pump = 5.
• As a refrigerator, work input W = 1 kW.

Concept / Approach:
The fundamental relation is COP_HP = COP_R + 1. This derives from energy balance: Q_H = Q_L + W. Dividing by W gives Q_H/W = Q_L/W + 1, or COP_HP = COP_R + 1. Once COP_R is known, the cooling capacity follows as Q_L = COP_R * W.

Step-by-Step Solution:
Given COP_HP = 5 → COP_R = COP_HP - 1 = 4.Work input as refrigerator W = 1 kW.Refrigerating effect Q_L = COP_R * W = 4 * 1 kW = 4 kW.

Verification / Alternative check:
Using Carnot expressions: COP_R = T_L / (T_H - T_L) and COP_HP = T_H / (T_H - T_L). Their difference is 1, confirming the rule used.

Why Other Options Are Wrong:
(a), (b), (c) understate capacity; (e) confuses COP_HP with cooling capacity; COP of 5 is not power directly.

Common Pitfalls:
Forgetting the relation COP_HP = COP_R + 1 and mixing up which heat flow (Q_H or Q_L) a COP refers to.

Final Answer:
4 kW