Answer
$1.1\frac{J}{K}$.
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 water will be dumped into the low-temperature water.
$$ \Delta S =Q(\frac{1}{T_L}-\frac{1}{T_H})$$
The amount of heat transferred depends on the change in temperature.
$$ \Delta S =(mc\Delta T)(\frac{1}{T_L}-\frac{1}{T_H})$$
The two masses of water are identical, so the equilibrium temperature is halfway between the two initial temperatures, or $40^{\circ}C$. The average temperature of the cold water is $37.5^{\circ}C$ and the average temperature of the hot water is $42.5^{\circ}C$.
$$ \Delta S =(mc\Delta T)(\frac{1}{T_L}-\frac{1}{T_H})$$
$$ =(1.0kg)(4186J/(kg \cdot C^{\circ}))(5^{\circ}C) (\frac{1}{(37.5+273)K}-\frac{1}{(42.5+273)K })$$
$$=1.068\approx1.1\frac{J}{K}$$
