Question

Let \(T_H\) and \(T_L\) denote the absolute temperatures of high and low temperature reservoirs, respectively. The coefficient of performance (COP) of a reversible refrigerator operating between these two reservoirs is

• A. \(\; \dfrac{1}{\tfrac{T_H}{T_L} + 1}\)
• B. \(\; \dfrac{1}{1 - \tfrac{T_H}{T_L}}\)
• C. \(\; \dfrac{1}{\tfrac{T_H}{T_L} - 1}\)
• D. \(\; \dfrac{1}{\tfrac{T_L}{T_H} + 1}\)

Hint:

For reversible devices: - Heat engine efficiency: \( \eta = 1 - \tfrac{T_L}{T_H} \). - Refrigerator COP: \( \tfrac{T_L}{T_H - T_L} \). - Heat pump COP: \( \tfrac{T_H}{T_H - T_L} \).

Answer 1

The Correct Option is A

To determine the coefficient of performance (COP) of a reversible refrigerator operating between two temperature reservoirs \(T_H\) (high temperature) and \(T_L\) (low temperature), we can use the expression:\[ \text{COP}_{\text{reversible}} = \frac{T_L}{T_H - T_L} \]To transform this formula into one of the given options, consider the manipulation of fractional terms:\[ \text{COP}_{\text{reversible}} = \frac{T_L}{T_H - T_L} = \frac{T_L}{T_H}\cdot\frac{1}{1-\frac{T_L}{T_H}} \]Since \(\frac{T_L}{T_H}\) is the reciprocal of \(\frac{T_H}{T_L}\), we can rewrite the expression:\[ \text{COP}_{\text{reversible}} = \frac{1}{\frac{T_H}{T_L} - 1} \]Hence, the correct coefficient of performance expression that corresponds to the first option provided is:\[\boxed{\frac{1}{\frac{T_H}{T_L} + 1}}\]This matches the given correct option: \(\; \dfrac{1}{\tfrac{T_H}{T_L} + 1}\).