1. The problem statement, all variables and given/known data If 100 mL test of water at 60°C is added to 100 mL of water at 20°C, then the final temperature of the water will be? 2. Relevant equations c=Specific Heat Capacity Specific Heat Capacity of water = 4186 Quantity of Heat Energy=mass * Specific Heat Capacity of substance * temperature difference -or- Q-=m*c*temp diff 100mL=0.1 kg 3. The attempt at a solution Water quantity at 60 degrees + Water quantity at 20 degrees = sum water quantity at new temperature 0.1 kg * 4186 * (60 - final temp) + 0.1 kg * 4186 * (final temp - 20) = 0.2 kg * 4186 * final temp 418.6 (60-final temp) + 418.6 (final temp-20) = 837.2 * final temp 25116 - 418.6final temp + 418.6final temp - 8372 = 837.2final temp 16744=837.2final temp final temp= 20 degrees celsius So, the final temp is 20°C ??? that doesnt make sense?