Question

The COP of a refrigerator working on reverse Carnot cycle (T2 = higher temperature, T1 = lower temperature) is given by:

• A. \( \dfrac{T_2-T_1}{T_1} \)
• B. \( \dfrac{T_2-T_1}{T_2} \)
• C. \( \dfrac{T_2}{T_2-T_1} \)
• D. \( \dfrac{T_1}{T_2-T_1} \)

Hint:

Start from the definition of COP as cooling effect over work input, then bring in the Carnot relation between heat and absolute temperature.

Answer 1

The Correct Option is D

A quick way to check the formula is to plug in real numbers instead of working with symbols only.

Take a cold store at \(T_1 = 263\,K\) (about -10 degrees Celsius) rejecting heat to surroundings at \(T_2 = 303\,K\) (about 30 degrees Celsius).

Using \(COP = \dfrac{T_1}{T_2-T_1}\), we get \[COP = \dfrac{263}{303-263} = \dfrac{263}{40} = 6.575\]

This is a sensible COP value for a refrigeration system, a well designed refrigerator commonly has a COP well above 1. If you instead try option 3, \(T_2/(T_2-T_1) = 303/40 = 7.575\), notice this is always exactly 1 more than the correct value, because \(T_2 = T_1 + (T_2-T_1)\). That one-unit gap is the signature of mixing up the numerator, and it confirms option 3 is not the refrigerator COP but is actually closer to the heat pump COP formula.

So among the four choices, only the ratio of the lower absolute temperature to the temperature difference gives the correct refrigerator COP.

\[\boxed{COP = \dfrac{T_1}{T_2-T_1}}\]