If you combine 260.0 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? Use 1.00 g/mL as the density of water.

Answer 1

The heat lost by the hot water must equal the heat gained by the cooler water

q = mC∆T

q = heat; m = mass; C = specific heat; ∆T = change in temperature

heat lost by hot water = q = (120.0 g)(4.184 J/gº)(95.00º-Tf) where Tf is the final temperature

heat gained by cool water = q = (260.0 g)(4.184 J/gº)(Tf-25.00º) where Tf is the final temperature

(120.0 g)(4.184 J/gº)(95.00º-Tf) = (260.0 g)(4.184 J/gº)(Tf-25.00º)

47698 - 502.1Tf = 1088Tf - 27196

1590Tf = 74894

Tf = 47.10º = final temperature of the mixture