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Homework Help: Latent heat

  1. Nov 28, 2003 #1
    Me again...

    Is it possible to find th etemperature of the steam if the water boils at 96 degrees Celsius? How?

    Another one, the boiling temperature of a liquid vaires with atmospheric pressure. Does the heat of vaporization vary with the boiling temperature of a liquid? Can someone explain to me the "reason".... =)
  2. jcsd
  3. Nov 28, 2003 #2


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    We'll just let you think on this one for a bit.
    Have you discussed Hess's law? Think about two different paths for evaporating a liquid that is initially at a low P, T state: 1) adding heat enough to evaporate it at the initial P, T, and then heating the vapor; 2) heating the liquid to some higher temperature, and adding heat as necessary to maintain a higher vapor pressure while allowing it to evaporate.
  4. Nov 28, 2003 #3
    Hmmm, I haven't encountered Hess' Law yet...
  5. Nov 29, 2003 #4


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    This sounds like thermo for engineers --- hmmm. Okay, we'll talk about cycles, or alternate paths between initial and final states.

    Consider a liquid at {P,T,V}1 and two of the possible paths to convert that liquid to vapor at {P,T,V}2: 1) evaporate the liquid at {P,T}1 from the liquid volume V1,liq to a vapor volume V1,vap, heat the vapor from T1 to T2, isothermally compress the vapor at T2 to P2; and, 2) compress the liquid at T1 to P2, heat the liquid to T2, evaporate the liquid (there are a couple missing details --- this is help, not a "gimme"). Work out the total enthalpy changes for each path: 1) they are equal simple from the definition of state functions; 2) the enthalpies of evaporation at T1 and T2 can be related through this equality of enthalpy change between states; 3) consider what you know about the other steps in the two different paths, heat capacities, isothermal expansions/compressions, to reach a conclusion regarding the behavior of ΔHvap as a function of temperature, or whether it is independent of T.
  6. Nov 29, 2003 #5
    Hello again, Bystander

    Hello again... =)

    Thank you for the "Hess' Law". I will try to understand it well.

    Regarding on the steam, I know that the boiling pt. of the water, which is 100 degrees celsius is at the sea level.

    But what I am worrying about is that how to compute the final
    temperature of the steam without the mass of the water, and without the initial temperature of the water.... Also, will the latent heat of vaporization be affected in this problem?
  7. Nov 30, 2003 #6


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    You are dealing with "equilibrium" between the liquid and vapor phases of a pure single component (water). Several things should have been clearly stated in your text, or in the lecture material, but it appears that they may have been omitted --- I'll restate them here for you:

    1) the Gibbs phase rule states that the number of degrees of freedom of a system is equal to the number of components plus two minus the number of phases present in the system, f = c + 2 - p ; what this means is that the number of independent variables (pressure, temperature, volume, composition, etc.) you have to play with, "degrees of freedom," is fixed; for this case with one component and two phases, it equals 1 (fix T and everything else is fixed, or fix P and T is fixed);

    2) evaporation of a pure component (liquid, in this case) proceeds at constant T if P is fixed, or at constant P if T is fixed --- if your system is completely in the liquid phase at fixed T, and you gradually reduce pressure, the point at which the very first vapor forms is called the bubble point pressure; if you continue to maintain T, the system can then be expanded at constant pressure, the vapor pressure for the liquid, at the fixed T, until the last "drop" of liquid evaporates --- the only thing changing is volume and heat content; enthalpy of vaporization (per gram, per mole, per unit volume of liquid evaporated) at fixed T is constant, no matter what fraction of the liquid has been evaporated (I think this is your question in your last post);

    3) the locii of points at which two phases exist in equilibrium for a single, pure liquid-vapor system plotted on a P,T diagram is the "liquid-vapor coexistence line/curve," or the "vapor pressure curve;" the properties (densities, heat capacities, etc.) of the liquid and vapor phases are different except at the high T,P end of this curve (the critical point);

    4) "equilibrium" means that a system is in mechanical, chemical, and thermal equilibrium with itself and the surroundings --- "boiling" processes are not strictly speaking at equilibrium, but introductory courses seldom make such distinctions, and you have not mentioned any remarks in the problem about effects (superheating) that suggest that you are to consider this as anything other than an equilibrium state; that is, if the problem states "boiling water at some temperature T," it means that the liquid phase is in equilibrium with the vapor phase.

    You have my apologies for my failure to make this easily translatable --- I tried, but I was more concerned with stating things correctly without getting into too many of the complications that go along with real processes.

    Holler if the slightest thing is unclear to you --- I hope this comes a little closer to addressing your questions.
  8. Nov 30, 2003 #7
    Hello again...

    Hello again...

    I have read your "statement".... Unfortunately, I still don't get it that much...

    I thought of using the formlua of heat, Q = mcT (change in temp.) and then the heat of vaporization Q = mL... but I think it's incorrect...

    I'm really sorry... I don't get it yet... Um, could you help me understanding it? =]
  9. Nov 30, 2003 #8
    A temperature which is given in the problem usually reflects the temperature of the water which reflects the distribution of kinetic energy among molecules. Different temperatures reflect different distributions of velocities, at higher temperatures more molecules will have enough kinetic energy to leave the liquid state. The heat of vaporization is needed to transfer the rest of these molecules which do not have enough kinetic energy into the gaseous state. Thus the heat of vap is in a sense independent of temperature, boiling point, and pressure...that is the heat of vap is applicable even when, for example, the water has not even reached its boiling point. The heat of vap is related to the kinetic energy distribution.

    By putting water in a closed container different temperatures of the water will reflect different vapor pressures. As a correlation when the vapor pressure equals the atmospheric pressure, water will start to boil.

    The rational for this can be explained as follows. At higher temperatures water has a greater density; that is it expands against the atmospheric pressure. When the water's density increases until it cannot expand anymore this is when bubbles start to form. The atmospheric pressure determines how difficult it will be for water to expand to this maximum state. In order for water to expand it has to exert a force which is greater than the atmospheric pressure. And thus when we look at a separate experiment, that is the one which involves vapor pressure, one can draw the connection between vapor pressure and atmospheric pressure...at certain temperatures vapor pressure equals the atmospheric pressure, since the vapor pressure is the magnitude of expansion against an environment (in this case the container) vapor pressure which is a measure of expansion against the environment must at least be greater than the atmospheric pressure in order for water to expand, increase its density to the maximum, and then expand in the form of a bubble.
  10. Dec 2, 2003 #9
    Reply... =)

    Hello again!

    Well, I'm not certain of my answer right now... =)

    I believe the temperature of the steam is also 96 degrees celsius because if the boiling pt. of water is 96, before going to the vapor phase, it must absorb a required amount of heat before it turns into a vapor.

    Thank you very much for your time and help with me, Bystander and Meninger. =)
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