Question

Derive an expression for the COP of a completely reversible refrigerator in terms of the thermal energy reservoir temperatures, \(T_{L}\) and \(T_{H}\)

Short answer

Question: Derive an expression for the coefficient of performance (COP) of a reversible refrigerator in terms of reservoir temperatures. Answer: The COP of a reversible refrigerator can be expressed in terms of the reservoir temperatures, \(T_{L}\) (low-temperature reservoir) and \(T_{H}\) (high-temperature reservoir), as follows: COP = \(\frac{T_{L}}{T_{H} - T_{L}}\)

Step-by-step solution

Definition of the Coefficient of Performance (COP)

The COP of a refrigerator can be defined as the ratio of the heat energy absorbed from the low-temperature reservoir, \(Q_{L}\), to the work input, \(W\): COP = \(\frac{Q_{L}}{W}\)

Defining efficiency of a reversible heat engine operating between reservoirs \(T_{L}\) and \(T_{H}\)

Let's consider a reversible heat engine operating between the reservoirs with temperatures \(T_{L}\) and \(T_{H}\). The efficiency of a reversible heat engine, \(\eta\), is given by the well-known Carnot efficiency: \(\eta = 1 - \frac{T_{L}}{T_{H}}\)

Defining the relationship between heat and work

For a refrigerator cycle, the work input will be the difference between the heat absorbed from the high-temperature reservoir, \(Q_{H}\), and the heat absorbed from the low-temperature reservoir, \(Q_{L}\): \(W = Q_{H} - Q_{L}\)

Finding the relation between \(Q_{H}\) and \(Q_{L}\) from the reversible heat engine

Using the efficiency of the reversible heat engine, we can express the work output, \(W'\), in terms of the heat energy absorbed from the high-temperature reservoir, \(Q_{H}\): \(W' = (1- \frac{T_{L}}{T_{H}})Q_{H}\) Since a reversible refrigerator and a reversible heat engine operate between the same two thermal reservoirs, their heat energy transfer rates must be equal and opposite. Thus, the work input to the refrigerator will be equal to the work output of the heat engine. So, \(W' = W\).

Substituting \(W'\) by \(W\) and solving for \(Q_{L}\)

Using the equal work relation, \(W' = W\), we can write: \(W = (1- \frac{T_{L}}{T_{H}})Q_{H}\) We also know that \(W = Q_{H} - Q_{L}\), so we can substitute this expression into the equation: \(Q_{H} - Q_{L} = (1- \frac{T_{L}}{T_{H}})Q_{H}\) Now, let's solve for \(Q_{L}\): \(Q_{L} = Q_{H} - (1- \frac{T_{L}}{T_{H}})Q_{H} = Q_{H} \frac{T_{L}}{T_{H}}\)

Expressing COP in terms of \(T_{L}\) and \(T_{H}\)

Now, let's substitute the expression for \(Q_{L}\) into the COP equation: COP = \(\frac{Q_{L}}{W} = \frac{Q_{H} \frac{T_{L}}{T_{H}}}{W}\) Finally, recall that \(W = (1- \frac{T_{L}}{T_{H}})Q_{H}\). We can substitute this expression for \(W\) in the COP equation: COP = \(\frac{Q_{H} \frac{T_{L}}{T_{H}}}{(1- \frac{T_{L}}{T_{H}})Q_{H}}\) The \(Q_H\) terms cancel out, giving us the expression for COP in terms of reservoir temperatures: COP = \(\frac{T_{L}}{T_{H} - T_{L}}\) This expression represents the Coefficient of Performance of a completely reversible refrigerator in terms of the thermal energy reservoir temperatures, \(T_{L}\) and \(T_{H}\).