Question

The coefficient of performance for a refrigerator can be:
A. greater than 1
B. smaller than 1
C. equal to 1
D. None of these

Answer 1

Hint: The Coefficient of Performance or COP is a proportion of useful heating or cooling provided to the work required in the system. Higher COPs likely bring down working expenses. The COP normally surpasses 1, particularly in heat pumps, in light of the fact that, rather than simply changing over work to heat which, if 100% effective, would be a COP of 1, it pumps extra heat from a heat source to where the energy is required.

Complete step by step answer:
We know that a refrigerator is a power absorbing device.
Coefficient of Performance or COP is nothing but the efficiency in different words.
$\text{COP}=\dfrac{\text{ Output }}{\text{ Input }}=\left( \dfrac{\text{Desired effect}}{\text{Work Input}} \right)$
For a refrigerator working between two temperature limits, $T_1$ (higher) and $T_2$ (lower),
Refrigerators will be throwing the heat taking from the $T_2$ (Storage Space) to $T_1$(Room or any exhaust at higher temperature).
The coefficient of performance is the heat taken from the storage space divided by work input $={\text{Q}}_2/\text{W}$, where W in work input.
as, $Q_{1}=Q_{2}+W$
so, $\mathrm{COP}=\dfrac{\mathrm{Q}_{2}}{\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right)}$
From the $100 \%$ Second Law Efficiency, we know that applying the reversible refrigerator theory, we get that Q can be replaced by temperature.
$\mathrm{COP}=\dfrac{\mathrm{T}_{2}}{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}$
as, $T_{1}>T_{2}$
so, $\mathrm{COP}>1$
Generally, we know that the difference between $T_1$ and $T_2$ is less than $T_2$ so efficiency is always greater than 1.

Hence, the correct answer is Option A.

Note: We must have a clear concept about COPs and where they are required. COPs are usually required in heat consuming devices or heat providing devices and are kept into consideration for calculating the efficiency of the device.