If you combine 410.0 mL of water at 25.00 ∘C and 120.0 mL of water at 95.00 ∘C, what is the final temperature of the mixture? ?

Heat loss/gain = m c ΔT

Heat gained by the water at 25.00°C = Heat lost by water at 95.00°C
410.0 × c × (T - 25.00) = 120.0 × c × (95.00 - T)
410.0T - 10250 = 11400 - 120.0T
530.0T = 21650
Final temperature, T = 40.85 °C

Recognize that the heat lost by the hot water is equal to the heat absorbed by the cool water:
q(hot) = -q(cool)
The negative sign is needed since one side will have a positive value and the other a negative value.
m c (T2-T1) = - m c (T2-T1)
120.0 g (4.184 J/gC) (T2-95C) = - 410 g (4.184 J/gC) (T2-25 C)
Solve for T2