Question

A completely reversible heat pump produces heat at a rate of \(300 \mathrm{kW}\) to warm a house maintained at \(24^{\circ} \mathrm{C}\). The exterior air, which is at \(7^{\circ} \mathrm{C}\), serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle.

Short answer

Answer: Yes, the heat pump satisfies the second law of thermodynamics according to the increase of entropy principle, as the total entropy change is positive (0.21 kW/K > 0).

Step-by-step solution

Calculate the minimum work input for the heat pump

We can use the equation of a reversible heat pump to find the minimum work required to transfer the heat energy: \(COP = \frac{Q_H}{W} = \frac{T_H}{T_H - T_C}\) Rearrange the equation to solve for W: \(W = \frac{Q_H}{COP}\)

Convert the Celsius temperatures to Kelvin

To calculate the entropy change, we need to deal with the absolute temperatures. Therefore, we need to convert the temperatures from Celsius to Kelvin: \(T_C = 7^{\circ}C + 273.15 = 280.15 K\) \(T_H = 24^{\circ}C + 273.15 = 297.15 K\)

Calculate the rate of entropy change in both heat source and heat sink

The rate of entropy change is the ratio of heat transfer to the temperature of each reservoir: For the heat source (outdoors), the rate of entropy change: \(\frac{dS_C}{dt} = -\frac{Q_C}{T_C}\) For the heat pump (indoors), the rate of entropy change: \(\frac{dS_H}{dt} = \frac{Q_H}{T_H}\)

Check if the second law is satisfied

The increase of entropy principle states that the total change in entropy must be equal to or greater than zero for a process to satisfy the second law: \(\frac{dS_H}{dt} + \frac{dS_C}{dt} \ge 0\) Now, we will plug in the values we've calculated, including the heat transfer rate and temperature of each reservoir: \(\frac{300 kW}{297.15K} - \frac{Q_C}{280.15K} \ge 0\) To find \(Q_C\), we can use the heat pump performance equation: \(COP = \frac{Q_H}{Q_H - Q_C}\) Plugging in the values: \(COP = \frac{297.15}{297.15 - 280.15} = \frac{300 kW}{300 kW - Q_C}\) Now, we can solve for \(Q_C\): \(Q_C = 197.5 kW\) Finally, we check if the second law is satisfied: \(\frac{300 kW}{297.15K} - \frac{197.5 kW}{280.15K} = 0.21 kW/K\) Since the total entropy change is positive (\(0.21 kW/K > 0\)), the heat pump does satisfy the second law according to the increase of entropy principle.