When 50.0 g of water at 95.0 C is transferred to 35.0 g of water at 28.0 C, what is the final temperature of the mixture of water?

Answer 1

heat lost by warm water = heat gained by cool water

mC∆T warm = mC∆T cool

(50.0 g)(4.184 J/gº)(95 - Tf) = (35.0 g)(4.184 J/gº)(Tf - 28)

4750 - 50Tf = 35Tf - 980

85Tf = 5730

Tf = 67.4º = final temperature