1. The problem statement, all variables and given/known data I have been working on a University Physics thermodynamics problem for over a week now, day and night. I've consulted two physics PhDs, posted on Yahoo Answers (nobody took the bait) and I'm still stuck. I just wasted an hour and a half carefully posting my problem and variables, showing every detail of my work, only to have the site dump me onto a login page. So now I'm making it short and dirty. You cool a 100.0-g slug of red-hot iron (temperature 745°C) by dropping it into an insulated cup (of negligible mass) containing 75.0 g of water at 20.0°C. Assuming no heat exchange with the surroundings, what is the final mass of the iron and the remaining water? c(iron) = 0.47 J/gC c(water = 4.19 J/gC L(water) = 2260 J/g 2. Relevant equations Q = mc(Tf-Ti) [tex]\Delta[/tex]Q = mL 3. The attempt at a solution I assume the iron does not change mass as it doesn't change state. I solved for Tf of the system: 0.47 J/g°C * 100 g * (Tf - 745C) + 4.18 J/g°C * 75g * (Tf - 20C) = 0 47.0 J/C * (Tf - 745C) + 314 J/C * (Tf - 20C) = 0 47 J/C*Tf – 35015 J + 314 J/C*Tf - 6280 J= 0 (-35015 J - 6280 J) + (47 J/C *Tf + 314 J/c*Tf) =0 -41295 J + Tf*(47 J/C + 314 J/C) = 0 Tf(361 J/C) = 41295 J Tf = 114°C I suspect this is incorrect because it doesn't take into account the latent heat of vaporization value for steam, which, if that temperature is correct, is surely involved. My other calculations showed that water does evaporate, but there is not enough heat from the iron to evaporate _all_ of the water. ... [tex]\Delta[/tex]Q(avail for steam) = Q(iron to 100C) – Q(water to 100C) = 3.03 x 104 J − 2.51 x 104 J = 5200 J Solving for the mass of the steam, m(steam) = [tex]\Delta[/tex]Q(avail)/L = 5200 J / 2260 J/g = 2.30 g So I estimate that 72.7 g of water remains. Can someone PLEASE, PLEASE, PLEASE verify that answer? And explain where I went wrong with the calculation of the final temperature? Thanks so very much!