Answer
The choice (b) is correct. Approximately equals to $+5 \times 10^5 J/K$ per second.
Work Step by Step
The power plant does work at the rate of 10MW with the efficiency, $\epsilon = 6.5 \%$
This means with this $6.5 \%$ efficiency, the plant with a certain heat input can give out the power of 10MW.
Now we need to calculate the entropy per second,
$\frac{\Delta S}{t} = \frac{\frac{Q_H}{T_H}}{t}$
$\frac{\Delta S}{t} = \frac{Q_H}{(t)(T_H)}$
But $\frac{W}{t} = \frac{0.065 Q_H}{t}$
So $10 \times 10^6 J/s= \frac{0.065 Q_H}{t}$
$ {\frac{Q_H}{t}} = \frac{10 \times 10^6 J/s}{0.065 } $
$ {\frac{Q_H}{t}} = 1.54 \times 10^8 J/s$
Substitute into the entropy equation
$\frac{\Delta S}{t} = \frac{Q_H}{(t)(T_H)}$
$\frac{\Delta S}{t} = \frac{1.54 \times 10^8 J/s}{300K} $
$\frac{\Delta S}{t} = 5.13 \times 10^5 J/s.K$
The choice (b) is correct. Approximately equals to $+5 \times 10^5 J/K$ per second.
