A Carnot engine operating directly between 600 K and 100 K produces 3333 J of work from 4000 J of heat transfer. If that same 4000 J instead passes irreversibly from the 600 K reservoir to a 250 K reservoir before entering a second Carnot engine operating between 250 K and 100 K, the work output drops to 2400 J. How much less work is produced because of this intermediate irreversible step?

1. The direct Carnot engine between 600 K and 100 K produces W = 3333 J from the same 4000 J of heat transfer. 2. The two-step process, which first loses heat irreversibly to an intermediate 250 K reservoir, produces only W = 2400 J. 3. The difference between these two work outputs is 3333 J - 2400 J = 933 J. 4. This result is stated directly: there is 933 J less work from the same heat transfer in the second process, tying it to the entropy increase caused by the irreversible step. 5. Option D, 2400 J, is simply the second work output by itself, not the difference between the two cases. 6. Options A and C, 600 J and 1600 J, do not come from subtracting 2400 J from 3333 J, so neither matches the required comparison. _Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy_