J.R. S. answered 01/26/21
Ph.D. University Professor with 10+ years Tutoring Experience
q = mC∆T where q = heat, m = mass, C = specific heat and ∆T = change in temperature
Heat gained by cooler water = heat lost by hotter water
(270.0 g)(4.184 J/gº)(Tf - 25º) = (100 g)(4.184 J/gº)(95º - Tf) where Tf = final temperature
Since C (specific heat for water) appears on both sides, we can eliminate it and have...
(270 g)(Tf - 25) = (100 g)(95 - Tf)
270Tf - 6750 = 9500 - 100Tf
370Tf = 16,250
Tf = 43.9º = final temperature