Final answer:
The total entropy change when 100 kJ of heat is transferred from a 600 K cold reservoir to a 1,400 K hot reservoir is calculated separately for each reservoir. For the cold reservoir, the entropy change is -166.6667 J/K, and for the hot reservoir, it is 71.4286 J/K. The combined entropy change is -95.2381 J/K.
Step-by-step explanation:
To assume that heat in the amount of 100 kJ is transferred from a cold reservoir at 600 K to a hot reservoir at 1,400 K contrary to the Clausius statement of the second law and calculate the total entropy change, we treat this scenario as a hypothetical reversible process for calculation purposes. In such cases, the entropy change of the reservoirs can be calculated separately and then combined to find the total entropy change.
Entropy change ΔS for a heat transfer Q at a constant temperature T is given by ΔS = Q/T. Hence, the entropy decrease for the cold reservoir losing heat is ΔSc = -100,000 J / 600 K = -166.6667 J/K. For the hot reservoir gaining heat, the entropy increase is ΔSh = 100,000 J / 1,400 K = 71.4286 J/K.
The total entropy change ΔStot is the sum of ΔSh and ΔSc: ΔStot = ΔSh + ΔSc = 71.4286 J/K - 166.6667 J/K = -95.2381 J/K. Rounding this to four decimal places, the total entropy change is -95.2381 kJ/K.