Correct Answer - Option 4 : 2

__Concept:__

The thermal efficiency of the Carnot cycle is given by

\(\eta = 1 - \frac{{{T_L}}}{{{T_H}}}\)

efficiency is also given by

\(\eta = \frac{{{W_{out}}}}{{{Q_S}}}\)

The change of entropy is given by \(dS = \frac{{dQ}}{T}\)

__Calculation:__

__Given:__

T1= 600K and T2 = 300K, W = 600 kJ

therefore

\(\eta = 1 - \frac{{{T_L}}}{{{T_H}}} = 1 - \frac{{300}}{{600}} = \frac{1}{2}\)

Now the \(\frac{1}{2} = \frac{{600}}{{{Q_S}}}\)

⇒ QS = 1200 kJ

Now, change in entropy of working fluid during heat addition process is given by:

\(dS = \frac{{dQ}}{T} = \frac{{1200}}{{600}} = 2\;kJ/K\)