Answer
$4.86\times10^{-2}\frac{J/K}{s}$.
Work Step by Step
$$\Delta S=\frac{Q}{T}$$
$$\Delta S=\Delta S_H+\Delta S_L=-\frac{Q}{T_H}+\frac{Q}{T_L}$$
All the heat that leaves the high-temperature heat source will be dumped into the low-temperature body of water.
$$ \Delta S =Q(\frac{1}{T_L}-\frac{1}{T_H})$$
Find the rate at which entropy increases.
$$ \frac{\Delta S}{t} =\frac{Q}{t}(\frac{1}{T_L}-\frac{1}{T_H})$$
$$ =(8.40\;cal/s)(\frac{4.186J}{cal})(\frac{1}{(22+273)K}-\frac{1}{(225+273)K })$$
$$=4.86\times10^{-2}\frac{J/K}{s}$$
