Answer
The final temperature of the aluminum and water is $T_{f} = 27.2^{\circ}C$
Work Step by Step
Using the equation $q=mcΔT$ and $-q_{system}, = q_{surroundings}$ we can solve for the final temperature of the system (i.e the aluminum and the water) provided the given information. Also the heat capacity of aluminum is 0.902 which is found online.
$-q_{system}, = q_{surroundings}$
$-(15.7g)(0.902 \frac{J}{g\times^{\circ}C})(T_{f}-53.2^{\circ}C) = (32.5g)((4.184 \frac{J}{g\times^{\circ}C})(T_{f}-24.5^{\circ}C)$
$-14.1614T_{f}+753.38648 = 135.98T_{f}-3331.51$
$150.1414 T_{f} = 4084.89648$
$T_{f} = 27.2^{\circ}C$
