Answer
The final temperature is; $T_{f} = 20.75^{\circ}C$
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 ethanol and the water) provided the given information. First we must convert the volume of ethanol into its mass using its density; $25 mL\times \frac{0.789g}{ml}= 19.725 g$ Also the heat capacity of ethanol is 2.46 which is found online.
$-q_{system}, = q_{surroundings}$
$-(19.725g)(2.46 \frac{J}{g\times^{\circ}C})(T_{f}-7^{\circ}C) = (35g)((4.184 \frac{J}{g\times^{\circ}C})(T_{f}-25.3^{\circ}C)$
$-48.5235T_{f}+339.6645 = 146.44T_{f}-3704.932$
$194.9635T_{f} = 4044.5965$
$T_{f} = 20.75^{\circ}C$
