A cycle consists of two processes: (1) heating at constant pressure from state 1 to state 2, and (2) cooling at constant volume from state 2 to state 1. For air, assume R = 287 J/kg.K and k = 1.4. Sta

Using the ideal gas law at state 1: $P_1 V_1 = m R T_1$

$$m = \frac{P_1 V_1}{R T_1} = \frac{(1.03 \times 10^5 \, \text{Pa}) \times (1 \, \text{m}^3)}{(287 \, \text{J/kg}\cdot\text{K}) \times (288.15 \, \text{K})} = \frac{103000}{82685.05} \approx 1.2457 \, \text{kg}$$

The process consists of two parts:

1. Constant pressure heating from $T_1$ to $T_2$.
2. Constant volume cooling from $T_2$ to $T_3$, where $T_3 = T_1$.

Heat flow during constant pressure heating ($Q_{12}$):

$$Q_{12} = m c_p (T_2 - T_1)$$ $$Q_{12} = (1.2457 \, \text{kg}) \times (1.005 \, \text{kJ/kg}\cdot\text{K}) \times (573.15 \, \text{K} - 288.15 \, \text{K})$$ $$Q_{12} = (1.2457) \times (1.005) \times (285) = 356.59 \, \text{kJ}$$

Heat flow during constant volume cooling ($Q_{23}$):

$$Q_{23} = m c_v (T_3 - T_2)$$ $$Q_{23} = (1.2457 \, \text{kg}) \times (0.718 \, \text{kJ/kg}\cdot\text{K}) \times (288.15 \, \text{K} - 573.15 \, \text{K})$$ $$Q_{23} = (1.2457) \times (0.718) \times (-285) = -255.06 \, \text{kJ}$$

The net heat flow is the sum of heat flows for both processes.

$$Q_{net} = Q_{12} + Q_{23}$$ $$Q_{net} = 356.59 \, \text{kJ} + (-255.06 \, \text{kJ}) = 101.53 \, \text{kJ}$$

The net heat flow is $\boxed{\text{101.53 kJ}}$.

Change in entropy during constant pressure heating ($\Delta S_{12}$):

$$\Delta S_{12} = m c_p \ln\left(\frac{T_2}{T_1}\right)$$ $$\Delta S_{12} = (1.2457 \, \text{kg}) \times (1.005 \, \text{kJ/kg}\cdot\text{K}) \times \ln\left(\frac{573.15 \, \text{K}}{288.15 \, \text{K}}\right)$$ $$\Delta S_{12} = (1.2519885) \times \ln(1.9891) = (1.2519885) \times (0.6877) = 0.8611 \, \text{kJ/K}$$

Change in entropy during constant volume cooling ($\Delta S_{23}$):

$$\Delta S_{23} = m c_v \ln\left(\frac{T_3}{T_2}\right)$$ $$\Delta S_{23} = (1.2457 \, \text{kg}) \times (0.718 \, \text{kJ/kg}\cdot\text{K}) \times \ln\left(\frac{288.15 \, \text{K}}{573.15 \, \text{K}}\right)$$ $$\Delta S_{23} = (0.8945) \times \ln(0.5027) = (0.8945) \times (-0.6877) = -0.6150 \, \text{kJ/K}$$

The overall change in entropy is the sum of entropy changes for both processes.

$$\Delta S_{overall} = \Delta S_{12} + \Delta S_{23}$$ $$\Delta S_{overall} = 0.8611 \, \text{kJ/K} + (-0.6150 \, \text{kJ/K}) = 0.2461 \, \text{kJ/K}$$

The overall change in entropy is $\boxed{\text{0.2461 kJ/K}}$.