Answer: D
The hot reservoir loses entropy −1200/600=−2.0 JK−1 and the cold one gains +1200/400=+3.0 JK−1. ΔSuniv=+1.0 JK−1>0, as required for an irreversible heat flow. Adding the magnitudes gives 5.0 and the reverse direction would give −1.0 (forbidden).