Answer
$49^{\circ}C$
Work Step by Step
To find the temperature we are going to use the equation $q= m\times C_{s}\times\Delta T$ where $q$ is the heat in joules, $m$ is the mass in grams, $C_{s}$ is the specific heat (found on table on pg. 257), and $\Delta T$ is the change in temperature in $^{\circ}C$.
Now we plug in the information from the question into our equation. $2.35\ kJ= 2,350J$ as $1\ kJ= 1000J$. The heat capacity of water is $4.18$.
$2,350J= 25g\times 4.18\frac{J}{g \times^{\circ}C}\times\Delta T$
Rearranging we get:
$\frac{2,350}{25\times 4.18}^{\circ}C=\Delta T$
So the change in temperature is $22^{\circ}C$. The temperature is then $22+27.0$ which is $49^{\circ}C$.