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Hint
We should know that a constant that denotes the efficiency of a refrigerator, expressed as the amount of heat removed from a substance being refrigerated divided by the amount of work necessary to remove the heat. The maximum coefficient of performance is the temperature of the environment to which the heat is dispelled to. is the temperature of the space to cool off. Efficiency is a measure of how much work or energy is conserved in a process. In many processes, work or energy is lost, for example as waste heat or vibration. The efficiency is the energy output, divided by the energy input, and expressed as a percentage. A perfect process would have an efficiency of 100%. Based on this we have to solve this question.
Complete step by step answer
We know that a refrigerator is a power absorbing device.
COP is nothing but the efficiency in different words.
$\text{COP}=\dfrac{\text{Output}}{\text{ Input }}=$ (Desired effect/ Work Input)
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).
COP = (Heat taken from the Storage Space)/ (Work Input) = Q2/W, where $\mathrm{W}$ in work input.
as, $Q_{1}=Q_{2}+W$
$\mathrm{so}, \mathrm{COP}=\dfrac{\mathrm{Q}_{2}}{\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right)}$
applying the reversible refrigerator theory, Q can be replaced by temperature $(100 \%$ Second Law Efficiency).
$\mathrm{COP}=\dfrac{\mathrm{T}_{2}}{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}$
as, $T_{1}>T_{2}$
so, $\mathrm{COP}>1$
Generally, $T_1$ and $T_2$ difference is less than $T_2$ so efficiency is always greater than 1.
Hence the correct answer is option (A).
Note
Thus, we can say that the COP of the refrigerator and air-conditioner can be less than one or greater than one. The above formula also shows that the COP of the heat pump can never be less than one; it is always more than one. As we have seen, the performance of a heat pump can vary wildly depending on a number of factors, but generally speaking ground source heat pumps can regularly be found on the market with a COP greater than 4. Air source heat pumps tend to be less efficient, and a COP over 3 is considered good.
It should be known that work output is always less than work input, so an actual machine cannot be 100% efficient. The effort force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.
We should know that a constant that denotes the efficiency of a refrigerator, expressed as the amount of heat removed from a substance being refrigerated divided by the amount of work necessary to remove the heat. The maximum coefficient of performance is the temperature of the environment to which the heat is dispelled to. is the temperature of the space to cool off. Efficiency is a measure of how much work or energy is conserved in a process. In many processes, work or energy is lost, for example as waste heat or vibration. The efficiency is the energy output, divided by the energy input, and expressed as a percentage. A perfect process would have an efficiency of 100%. Based on this we have to solve this question.
Complete step by step answer
We know that a refrigerator is a power absorbing device.
COP is nothing but the efficiency in different words.
$\text{COP}=\dfrac{\text{Output}}{\text{ Input }}=$ (Desired effect/ Work Input)
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).
COP = (Heat taken from the Storage Space)/ (Work Input) = Q2/W, where $\mathrm{W}$ in work input.
as, $Q_{1}=Q_{2}+W$
$\mathrm{so}, \mathrm{COP}=\dfrac{\mathrm{Q}_{2}}{\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right)}$
applying the reversible refrigerator theory, Q can be replaced by temperature $(100 \%$ Second Law Efficiency).
$\mathrm{COP}=\dfrac{\mathrm{T}_{2}}{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}$
as, $T_{1}>T_{2}$
so, $\mathrm{COP}>1$
Generally, $T_1$ and $T_2$ difference is less than $T_2$ so efficiency is always greater than 1.
Hence the correct answer is option (A).
Note
Thus, we can say that the COP of the refrigerator and air-conditioner can be less than one or greater than one. The above formula also shows that the COP of the heat pump can never be less than one; it is always more than one. As we have seen, the performance of a heat pump can vary wildly depending on a number of factors, but generally speaking ground source heat pumps can regularly be found on the market with a COP greater than 4. Air source heat pumps tend to be less efficient, and a COP over 3 is considered good.
It should be known that work output is always less than work input, so an actual machine cannot be 100% efficient. The effort force is the force applied to a machine. Work input is the work done on a machine. The work input of a machine is equal to the effort force times the distance over which the effort force is exerted.
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