Final answer:
The total entropy created in the irreversible process of the engine is 0.32 J/K, calculated by considering the heat exchanged with both hot and cold reservoirs and their respective temperatures.
Step-by-step explanation:
To calculate the entropy created in an irreversible engine process, we can use the second law of thermodynamics which states that the total entropy of an isolated system can never decrease over time. The engine extracts 1200 J of heat from the hot reservoir at 550 K and does 450 J of work. The amount of heat rejected to the cold reservoir at 300 K is the difference between the heat extracted from the hot reservoir and the work output, which equals 1200 J - 450 J = 750 J.
Now, we calculate the change in entropy for the hot and cold reservoirs separately. The entropy change for the hot reservoir is the heat removed divided by its temperature, which gives us ΔS_{hot} = -ΑQ_{hot}/T_{hot} = -1200 J / 550 K. For the cold reservoir, it's the heat added divided by its temperature, which gives us ΔS_{cold} = ΑQ_{cold}/T_{cold} = 750 J / 300 K.
Add these two together to find the total change in entropy of the system:
- ΔS_{total} = ΔS_{hot} + ΔS_{cold}
- ΔS_{total} = (-1200 J / 550 K) + (750 J / 300 K)
- ΔS_{total} = -2.18 J/K + 2.50 J/K
- ΔS_{total} = 0.32 J/K
The total entropy created in this irreversible process is 0.32 J/K, which corresponds to option B.