Refrigerator and Coefficient of Performance - UNSOLVED PRACTICE SET

Q7. Define the coefficient of performance (COP) of a refrigerator. Why is it called a "coefficient of performance" rather than "efficiency"?

Q8. Explain why the COP of a refrigerator can be greater than 1, while the efficiency of a heat engine is always less than 1.

Q9. Draw a schematic diagram of a refrigerator, showing the hot reservoir, cold reservoir, working substance, and the direction of heat flow and work input.

Q10. A refrigerator has a COP of 4. If it extracts 800 J of heat from the cold reservoir, how much work is required? Calculate the heat rejected to the hot reservoir.

Q11. Why does the back of a refrigerator feel warm? Explain using energy conservation principles.

Q12. What happens to the COP of a refrigerator if the temperature difference between the hot and cold reservoirs increases? Explain with reasoning.

Q13. Describe the working principle of a refrigerator with a neat labeled diagram. Define coefficient of performance (COP) and derive its expression: β = Q₂/W = Q₂/(Q₁ – Q₂). For an ideal refrigerator, show that β = T₂/(T₁ – T₂).

Q14. A refrigerator maintains a freezer at –20°C while the room temperature is 30°C.

(i) Calculate the COP of an ideal refrigerator operating between these temperatures.

(ii) If the refrigerator extracts 1000 J of heat per cycle from the freezer, calculate the work required per cycle.

(iii) Calculate the heat rejected to the room per cycle.

(iv) If the actual COP is only 50% of the ideal COP, calculate the actual work required.

Q15. Compare a heat engine and a refrigerator on the following points:

(i) Direction of heat flow

(ii) Purpose

(iii) Performance measure (efficiency vs. COP)

(iv) Relation between Q₁, Q₂, and W

(v) Limitations imposed by the Second Law

Present your answer in a clear, well-organized format.

Q16. A refrigerator operates between –10°C (263 K) and 27°C (300 K). It extracts 600 J of heat from the cold reservoir per cycle.

(i) Calculate the ideal COP of the refrigerator.

(ii) Calculate the minimum work required per cycle.

(iii) Calculate the heat rejected to the room per cycle.

(iv) If the actual COP is 3.0, calculate the actual work required per cycle.

(v) Calculate the heat actually rejected to the room per cycle.

Q17. An air conditioner in a classroom maintains the room at 22°C while the outside temperature is 40°C. The air conditioner removes 2500 J of heat from the room per second.

(i) Calculate the ideal COP of the air conditioner.

(ii) Calculate the minimum power (work per second) required.

(iii) If the actual power consumption is 500 W, calculate the actual COP.

(iv) Calculate the heat rejected to the outside per second.

(v) Why does the actual power consumption exceed the minimum calculated?

Q18. A freezer in a laboratory maintains a temperature of –50°C (223 K) while the room temperature is 27°C (300 K). The freezer extracts 500 J of heat per cycle from its interior.

(i) Calculate the COP of an ideal freezer operating between these temperatures.

(ii) Calculate the work required per cycle for this ideal freezer.

(iii) A student claims that by using a better compressor, the COP can be made infinite. Is this possible? Explain using the Second Law of Thermodynamics.

(iv) If the freezer door is left open in the room, will the room cool down? Explain your answer carefully.