the sum of the heats gained is zero.
qH2O+qFe=0
25*4.184(Tf-398K)+24*.449*(Tf-298)=0
now solve for Tf in that.
A 25.0 g sample of iron at 398K is placed in a styrofoam cup containing 25.0 mL of water at 298K. Assuming that no heat is lost outside the cup, what will the final temperature of the water be? What will the final temperature of the iron be?
I set them equal to each other to solve for the final temp. but I don't seem to get any reasonable answers.
I figured i should use
qH2O= 25.0 x 4.184 x (_ -25)
and
qFe = 25.0 x 0.449 x (_ -125)
1 answer
1 answer