Heat exchange involving phase change, hot iron into water ->steam,

[StarryV]

Homework Statement

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

Homework Equations

Q = mc(Tf-Ti)
\DeltaQ = mL

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.
...
\DeltaQ(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) = \DeltaQ(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!

 

[kuruman]

Your "other" calculation is correct. I too got that 2.3 g of water are vaporized. The problem with the first calculation is that, as you said, you didn't take into ccount the latent heat of vaporization. You can't have water at 114 ^oC, can you? (Let's not worry about superheated water and other such exotica for this problem).

 

[StarryV]

Thank you so much for helping! I'm done with my problem and the tutorial, and confidently so! What a relief!

 

[virupaksha]

Did you performed the experiment?
If Yes
Then you can easily got the answer
I performed it differently....
500gm in 500 ml water
in the end i found only 400 ml of water
........
answer why?

 

[rude man]

Homework Statement

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

Homework Equations

Q = mc(Tf-Ti)
\DeltaQ = mL

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.
...
\DeltaQ(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) = \DeltaQ(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!

If the vessel is truly insulating then no vaor can escape. So then the answer would be 175g.

So the problem is poorly worded. It should have specified that steam can escape without heat enetering.

 

[virupaksha]

Even then you have a point there.
Did you get the answer?
Do the experiment, you will get the answer.
bye
virupakshgupta@gmail.com