
The relation between efficiency $\eta$ of a heat engine and the coefficient of
performance $\alpha$ of a refrigerator is:
A $\quad \eta=\dfrac{1}{1-\alpha}$
B $\quad \eta=\dfrac{1}{1+\alpha}$
C $\quad \eta=1+\alpha$
D $\quad \eta=1-\alpha$
Answer
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Hint: We know that energy efficiency measures the proportion of energy which is wasted. The rate at which energy is wasted needs to be reduced and there are a number of ways to save energy, especially thermal energy. Efficiency signifies a peak level of performance that uses the least amount of inputs to achieve the highest amount of output. Efficiency requires reducing the number of unnecessary resources used to produce a given output including personal time and energy. Based on this we have to solve this question.
Complete step by step answerThe Efficiency if a heat engine can be given by the following equation:
$\Rightarrow \eta=1-\dfrac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$
$=\dfrac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}$
$=\dfrac{1}{1+\dfrac{\mathrm{T}_{2}}{\mathrm{T}_{1}-\mathrm{T}_{2}}}$
$=\dfrac{1}{1+\alpha}$
Where,
$\Rightarrow \eta=1-\dfrac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$ is the efficiency
${{\text{T}}_{1}}$ and ${{\text{T}}_{2}}$are the temperatures
And, $\alpha $is the coefficient of performance.
Therefore, from the solution, we can conclude that the correct answer is Option B.
Note: It is known that the coefficient of performance or COP of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required. Higher COPs equate to lower operating costs. Coefficient of Performance (COP) is the efficiency ratio of the amount of heating or cooling provided by a heating or cooling unit to the energy consumed by the system. The higher the Coefficient of Performance the more efficient the system.
Thus, we conclude that efficiency relates to the use of all inputs in producing any given output, including personal time and energy. The coefficient of performance or COP (sometimes CP) of a heat pump is the ratio of the heating or cooling provided over the electrical energy consumed.
Complete step by step answerThe Efficiency if a heat engine can be given by the following equation:
$\Rightarrow \eta=1-\dfrac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$
$=\dfrac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{T}_{1}}$
$=\dfrac{1}{1+\dfrac{\mathrm{T}_{2}}{\mathrm{T}_{1}-\mathrm{T}_{2}}}$
$=\dfrac{1}{1+\alpha}$
Where,
$\Rightarrow \eta=1-\dfrac{\mathrm{T}_{2}}{\mathrm{T}_{1}}$ is the efficiency
${{\text{T}}_{1}}$ and ${{\text{T}}_{2}}$are the temperatures
And, $\alpha $is the coefficient of performance.
Therefore, from the solution, we can conclude that the correct answer is Option B.
Note: It is known that the coefficient of performance or COP of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required. Higher COPs equate to lower operating costs. Coefficient of Performance (COP) is the efficiency ratio of the amount of heating or cooling provided by a heating or cooling unit to the energy consumed by the system. The higher the Coefficient of Performance the more efficient the system.
Thus, we conclude that efficiency relates to the use of all inputs in producing any given output, including personal time and energy. The coefficient of performance or COP (sometimes CP) of a heat pump is the ratio of the heating or cooling provided over the electrical energy consumed.
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