This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 2: Calculate the mass of air (m). Using the ideal gas law at state 1: P_1 V_1 = m R T_1 m = (P_1 V_1)/(R T_1) = (1.03 × 10^5 \, Pa) × (1 \, m^3)(287 \, J/kg·K) × (288.15 \, K) = (103000)/(82685.05) ≈ 1.2457 \, kg Step 3: Calculate the heat flow for each process. 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 \, kg) × (1.005 \, kJ/kg·K) × (573.15 \, K - 288.15 \, K) Q_12 = (1.2457) × (1.005) × (285) = 356.59 \, kJ Heat flow during constant volume cooling (Q_23): Q_23 = m c_v (T_3 - T_2) Q_23 = (1.2457 \, kg) × (0.718 \, kJ/kg·K) × (288.15 \, K - 573.15 \, K) Q_23 = (1.2457) × (0.718) × (-285) = -255.06 \, kJ Step 4: Calculate the net heat flow (i). The net heat flow is the sum of heat flows for both processes. Q_net = Q_12 + Q_23 Q_net = 356.59 \, kJ + (-255.06 \, kJ) = 101.53 \, kJ The net heat flow is 101.53 kJ. Step 5: Calculate the change in entropy for each process. Change in entropy during constant pressure heating ( S_12): S_12 = m c_p ((T_2)/(T_1)) S_12 = (1.2457 \, kg) × (1.005 \, kJ/kg·K) × (573.15 \, K288.15 \, K) S_12 = (1.2519885) × (1.9891) = (1.2519885) × (0.6877) = 0.8611 \, kJ/K Change in entropy during constant volume cooling ( S_23): S_23 = m c_v ((T_3)/(T_2)) S_23 = (1.2457 \, kg) × (0.718 \, kJ/kg·K) × (288.15 \, K573.15 \, K) S_23 = (0.8945) × (0.5027) = (0.8945) × (-0.6877) = -0.6150 \, kJ/K Step 6: Calculate the overall change in entropy (ii). The overall change in entropy is the sum of entropy changes for both processes. S_overall = S_12 + S_23 S_overall = 0.8611 \, kJ/K + (-0.6150 \, kJ/K) = 0.2461 \, kJ/K The overall change in entropy is 0.2461 kJ/K. That's 2 down. 3 left today — send the next one.