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AP EAMCET · PHYSICS · Thermodynamics

If a heat engine and a refrigerator are working between the same two temperatures \(T_1\) and \(T_2\left(T_1>T_2\right)\), then the ratio of efficiency of heat engine to coefficient of performance of refrigerator is

  1. A \(\frac{\left(\mathrm{T}_1-\mathrm{T}_2\right)}{\mathrm{T}_1 \mathrm{~T}_2}\)
  2. B \(\frac{\left(\mathrm{T}_1+\mathrm{T}_2\right)}{\mathrm{T}_1 \mathrm{~T}_2}\)
  3. C \(\frac{\left(\mathrm{T}_1-\mathrm{T}_2\right)^2}{\mathrm{~T}_1 \mathrm{~T}_2}\)
  4. D \(\frac{\left(T_1+T_2\right)^2}{T_1 T_2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\left(\mathrm{T}_1-\mathrm{T}_2\right)^2}{\mathrm{~T}_1 \mathrm{~T}_2}\)

Step-by-step Solution

Detailed explanation

Efficiency of heat engine: \(\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}\) Coefficient of performance of refrigerator: \(\text{COP}_{\text{ref}} = \frac{T_2}{T_1 - T_2}\) Ratio: \(\frac{\eta}{\text{COP}_{\text{ref}}} = \frac{\frac{T_1 - T_2}{T_1}}{\frac{T_2}{T_1 - T_2}}\)…