Jump to content

bartovan

Members
  • Content Count

    6
  • Joined

  • Last visited

Community Reputation

0 Neutral

About bartovan

  • Rank
    Lepton

Profile Information

  • Favorite Area of Science
    neuroscience

Recent Profile Visitors

186 profile views
  1. Interesting diagram, it clearly shows the relation between temperature and pressure, but how does that shed a light on the constant temperature of the system during phase change? Since there's no time dimension? It only shows that with constant pressure, if you raise the temperature from 0 K (or whatever the X origin is), at some point you'll go from solid to liquid, not how long this transition will take? Concerning this: That's not consistent with the premise of the second thought experiment, where heat would be applied perfectly uniformly, thus also at the core of the ice. If you state as a premise (as you do) that the interior of the ice remains colder, you return to the other thought experiment, where different temperatures exist, and thus to the problem of defining/measuring the temperature "of the system", as Studiot also mentioned.
  2. Exactly. But then the question is, what do they mean by "the temperature of the system"? And how would you measure that temperature? (Obviously not by sticking a thermometer somewhere, because that would be the temperature in one point). "Thus the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process (...)." (https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves) Well yes and no. They do take it to the kitchen: "The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils." (https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves)
  3. Thank you for all your answers. J.C.MacSwell makes exactly the point I'm trying to make: according to the theory, 1g of ice in the ocean would make the entire ocean remain at 0°C, which it doesn't of course. And I do believe that in an ocean of pure water this would be exactly the same. I have been thinking it over some more and thought of two extremes, as a thought experiment. 1) Extreme case 1: The heat is applied in one single point at the bottom. In that case it seems inevitable to me that, according to where exactly one measures, different temperatures will be measured. Especially in the case of a very large container with only a tiny fraction of ice remaining (floating at the top of course, to make things worse), when one would measure very close to the heating point, the temperature would be significantly higher than close to the ice. Importantly, no amount of stirring would remedy this (think of the extreme of 1g of ice in 20 000 liters of water, or 1g of ice in the ocean). So, in case 1, you simply have different temperatures in different areas, even if you stir, so the diagram doesn't really apply in a literal sense. However, it does make sense, see below point 3. 2) Extreme case 2: The heat is applied perfectly uniformly (seems impossible in the real world, but it's a thought experiment). (Maybe the microwave experiment of studiot comes close.) So we have a huge chunk of ice at say -50°C and start applying heat in a perfectly uniform manner (meaning: we add kinetic energy to every molecule at exactly the same rate per molecule). In that case the whole block of ice will turn to water in one and the same instant (in other words: at every point, the molecules will reach the necessary kinetic energy to get loose from the solid state), and then, being completely water, will immediately continue to rise in temperature. (Since the phase change happens in one single point in time). In this case, the diagram doesn't apply either. You go in a straight, rising line from -50°C to 0°C, then in one single instant (a point in time) you have the phase change, and you continue with a straight rising line. No horizontal part between B and C (in the diagram I posted at the beginning). 3) So to come back to case 1, I think what the diagram and the theory mean, is that the system, as a whole, remains at 0°C during the phase change in terms of the sum of kinetic energies. Temperature is actually kinetic energy, right? (at least in one of the multiple frameworks for understanding temperature). So during the phase change, some areas will have a rather high kinetic energy (close to the heating point, where they will be excited more), say corresponding to 10°C; while other areas will have a low kinetic energy, corresponding to 0°C or even less in the ice; still others somewhere in between, etc. Probably, during the phase change, the total sum of these kinetic energies is equivalent to the kinetic energy corresponding to 0°C. Does that make sense? (I propose we leave soup and bishops out of the discussion, since the crux of the question is not really in the "purity" part - see the example of a bit of ice in a gigantic quantity of water).
  4. OK, sounds realistic, although I don't think it's mentioned in the phase change description that the mixture has to be stirred. The page mentioned before says: [as long as there's ice in the mixture] "the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together". No mention of stirring anywhere. Still I wonder, if you have a swimming pool full of melting ice, with a heat source below (let's say a heated floor), and you're almost at the end of the melting process, with 1g of ice in 20000 liters of water, even if you stir like mad, I can't imagine the whole swimming pool remaining at 0°C until the last gram of ice melts? So maybe it's actually completely theoretical and idealistic, and not to be taken as such in any real world setting?
  5. Well soup is at least in majority water, in- and outside of Somerset, so it wouldn't be all that weird if it would act like water. So you think it's only due to not being a pure substance? And how much impurity is acceptable then? Tap water? Distilled water - outside of a dust free lab? I might try with tap water and see what that gives...
  6. I was just heating some frozen soup on my electric stove, and then remembered that supposedly as long as there is some ice left the system stays at 0°C until all the ice is melted, and only then does the temperature rise (if heat is continued to be added). I looked it up and this page (https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.3%3A_Heating_Curves - sorry, couldn't find guidelines on how to properly post links) phrases it like this: "the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process". See graph below between B and C. Only, my soup doesn't act like this. There's a big lump of ice in the middle, and 3 cm of melted soup bubbling and boiling around it. How does this work really?
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.