Answer
10.7$^\circ$C
Work Step by Step
A kilocalorie raises the temperature of 1 kg of water by 1 degree Celsius.
$$(8200J)(\frac{1kcal}{4186J})\frac{(1kg) 1C^{\circ}}{1kcal}(\frac{1}{3.0kg})=0.653C^{\circ}$$
The final temperature is 10.0+0.653, or 10.7 degrees C.
