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A confusing question - melting ice

  1. Dec 7, 2005 #1
    Question:

    The heat of vaporization of water is 2260 J/g at 100°C. A 100.0 g sample of water condenses and all of the heat liberated is then transferred to a block of ice. How many grams of ice will melt? Assume the condensed water remains at 100°C and the melted ice remains at 0°C. The heat of fusion of ice is 335 J/g.

    I am totally confused by this question, and I don't even know how to start. Can someone explain this question to me, or just give some hints?
     
  2. jcsd
  3. Dec 7, 2005 #2

    Doc Al

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    This means that 100.0 g of steam (at 100°C) condensed to become 100.0 g of liquid water (at 100°C). How much energy was released when the water changed phase? (How much energy is required to vaporize the water into steam?)
     
  4. Dec 7, 2005 #3
    I tried to solve it based on this forumla:
    energy released by the reaction = specific heat capacity X mass of solution X increase in temperature...

    2260 J/g X 100°C X 100g = 335 J/g X 100°C X (M)

    M= 675 g....is this correct?

    I am still very confused...
     
  5. Dec 7, 2005 #4
    what about the ice? " The heat of fusion of ice is 335 J/g."
     
  6. Dec 7, 2005 #5

    Astronuc

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    I can appreciate the confusion because this situation is phyiscally impossible, but the problem is trying to emphasize the change of phase - from vapor (steam) to liquid (water) and solid (ice) to water.

    Now when a substance changes phase, it does so more or less at constant temperature.

    In the problem the ice changes phase at 0°C to liquid water absorbing 335 J/g, and the steam changes phase at 100°C to liquid water giving 2260 J/g.
     
  7. Dec 7, 2005 #6

    Bystander

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    Yes. The latent heat of fusion at 0 C is 6.75 x the latent heat of fusion at 100 C. I've reddened your errors: you have NOT been asked to use heat capacities; there is, therefore, NO temperature change for the two separate systems for which you are calculating "latent heat changes."
    "I can appreciate the confusion because this situation is phyiscally impossible...."​

    Not really --- throw the ice in the upper pot of a double boiler, shove a steam line into the bottom pot rather than filling it and throwing it on the stove, and do the measurement with whatever degree of precision is desired (reducing heat capacities of apparatus (the double boiler) to less than uncertainties in measurement of water masses).
     
  8. Dec 7, 2005 #7
    thank you very much, and there is another question i want to confirm.
    is it "heat of vaporization" = "heat of the formation of vapor" - "heat of formation of liquid"
     
  9. Dec 7, 2005 #8

    Bystander

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    Yup. You got it.
     
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