If you combine 340mL of water at 25.00 degree C and 130.0 mL of water at 95 degree C, what is the final temperature of the mixture?

Let T°C be the final temperature.

First portion of water :
m₁ = 340 g, ΔT₁ = (T - 25)°C

Second portion of water :
m₂ = 130 g, ΔT₁ = (95 - T)°C

Assume that there is no heat transferred with the surroundings.
Heat gained by the first portion of water = Heat lost by the second portion of water
m₁ c ΔT₁ = m₂ c ΔT₂
m₁ ΔT₁ = m₂ ΔT₂
340 × (T - 25) = 130 × (95 - T)
340T - 8500 = 12350 - 130T
470T = 20850
Final temperature, T = 44.4°C

There is no need for the density. The calculation is a weighted average:

(((340 mL) x (25.00 °C)) + ((130.0 mL) x (95 °C))) / (340 mL + 130.0 mL) = 44 °C