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

Step 1: Coefficient of performance (COP) of a refrigerator is defined as the useful cooling effect divided by the work input, \[COP = \dfrac{Q_1}{W}\] where \(Q_1\) is the heat absorbed from the cold space at temperature \(T_1\).

Step 2: The refrigerator rejects heat \(Q_2\) to the surroundings at the higher temperature \(T_2\). By the first law of thermodynamics, the work input is the difference between heat rejected and heat absorbed, \[W = Q_2 - Q_1\]

Step 3: For a reverse Carnot cycle, heat transfer is proportional to absolute temperature, \[\dfrac{Q_1}{T_1} = \dfrac{Q_2}{T_2}\]

Step 4: Substitute \(W = Q_2 - Q_1\) into the COP formula and divide numerator and denominator by \(Q_1\), using \(Q_2/Q_1 = T_2/T_1\): \[COP = \dfrac{Q_1}{Q_2-Q_1} = \dfrac{1}{\frac{Q_2}{Q_1}-1} = \dfrac{1}{\frac{T_2}{T_1}-1} = \dfrac{T_1}{T_2-T_1}\]

This matches option 4. The other expressions either invert this ratio or use the wrong temperature in the numerator, so they do not represent COP of a refrigerator on the reverse Carnot cycle.