Answer
The final temperature of the water is $~82^{\circ}C$
Work Step by Step
We can find the change in temperature:
$Q = m~c~\Delta T$
$\Delta T = \frac{Q}{m~c}$
$\Delta T = \frac{125.6\times 10^3~J}{(0.500~kg)(4186~J/kg~C^{\circ})}$
$\Delta T = 60~C^{\circ}$
The final temperature of the water is $22^{\circ}C+60~C^{\circ}$ which is $82^{\circ}C$
