[chrismb]

If 150. grams of iron at 95.0 C is placed in an insulated container containing 500. grams of water at 25.0 C, what will the final temperature be (assuming no heat lost)? (CH2O = 4.18 J/gC, CFe = 0.444 J/gC)

{use 3 significant figures in answer}

Ok so far i've done all the work

150g x 0.444 J/g C x (Tf - 95.0) C = - [ 25.0g x 4.18J/g C x (Tf - 25.0 C]

im just having troubling on how to complete it past here....someone help please

[magician4]

150g x 0.444 J/g C x (Tf - 95.0) C = - [ 25.0g x 4.18J/g C x (Tf - 25.0 C]

there's one small but very important error in this equation

correct it

after that , resolve the equation for T_f , and calculate

regards

Ingo