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What did i do wrong? Am i on the right track?

  1. Mar 31, 2006 #1

    danago

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    Gold Member

    Hey. Heres the question i was given:

    I am studying year 11 physics, and this is one of the harder questions ive had to face so far. Im going to explain my theory of whats happening as i go along. If Any of my theory is incorrect, PLEASE correct me on it. Im not only doing this to get the answer correct, but to further improve my understanding of heat.

    ____________________________________

    I understand that the steam will condense, and the ice will melt, to produce a final mixture of liquid water. Before the ice and steam changes phase, they must reach 0 degrees and 100 degrees respectively.

    I need to calculate the energy it takes for the ice to increase to 0 degrees, and the steam to decrease to 100 degrees, using the Q=mcT formula, and knowing that the specific heat capacity of ice is 2100, and steam is 2000.

    [tex]Q_{Ice}=mc\Delta T[/tex]
    [tex]=0.22\times 2100\times 16[/tex]
    [tex]=7392J[/tex]
    Therefore, 7392J of energy is required to increase the temperature of the ice to 0 degrees.

    [tex]Q_{Steam}=mc\Delta T[/tex]
    [tex]=4\times 2000\times 4[/tex]
    [tex]=32000J[/tex]
    Therefore, 32kJ of energy is required to cool the steam to 100 degrees.

    From what i understand, a hotter substance transfers energy to a cooler substance, so in this case, the steam transfers to the ice. When the ice has recieved 7392J of energy, it is 0 degrees, but the steam isnt yet 100. It still needs to transfer 24608J of energy before it is 100. Since the ice is already at 0 degrees, the extra energy from the steam will begin to cause fusion/melting of the ice. If 24608J of energy is to be used in fusion of ice (latent heat=334000), the following will give how much of the ice melts:

    [tex]Q=mL[/tex]
    [tex]24608=334000m[/tex]
    [tex]m=0.0727kg[/tex]
    Therefore, when the steam becomes 100 degrees, 0.0737kg of ice will have melted.

    So from what i understand, this is what we currently have:
    4kg steam (100 degrees)
    0.0737kg water (0 degrees)
    0.1463kg ice (0 degrees)

    For the final temperature to be calculated, all the substances must be in the same phase. So now i must calculate the energy levels for fusion and condensation to occur. (knowing that the latent heat for vaporization of water is 2250000).

    [tex]Q_{Ice}=mL[/tex]
    [tex]=0.1463\times 334000[/tex]
    [tex]=48864.2J[/tex]
    Therefore, 48864.2J of energy is required to completely melt the remaining ice.

    [tex]Q_{Steam}=mL[/tex]
    [tex]=4\times 2250000[/tex]
    [tex]=9000000J[/tex]
    Therefore, 9MJ of energy is required to completely condense the steam.

    Again, the energy transfers from the steam to the ice. Once the ice recieves 48864.2J of energy, it will be water at 0 degrees celcius. But for the steam to completely condense, it must transfer 8951135.8J more. In doing this, the temperature of the water increases.

    [tex]Q=mc\Delta T[/tex]
    [tex]8951135.8=0.22\times 4200\times T[/tex]
    [tex]T=9687.3764 degrees[/tex]
    Therefore, once the steam has reached full condesation, the water will have become 9687.3764 degrees.

    Now, from what i know, that is impossible, because water wont exceed 100 degrees without vaporizing. This is where i think ive gone wrong, and i dont know where to go from here.

    From here id usually apply a formula to find the final temperature of the 0.22kg water mixed with the 4kg water, but since im pretty sure ive made a mistake, i wont bother going further.

    Thank you to anyone who can put me on the right track for this question, or correct me on any of my understandings on what is happening.

    Dan.
     
    Last edited: Mar 31, 2006
  2. jcsd
  3. Mar 31, 2006 #2

    andrevdh

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    Homework Helper

    The amount of ice is quite small compared to the amount of steam. Maybe the steam cools off just a bit (it stays in the steam phase) and the heat it loses converts all of the ice into steam.
     
  4. Mar 31, 2006 #3

    Doc Al

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    Staff: Mentor

    You are on the right track. I suggest looking at it this way. First find the answers to these questions (some of which you've already answered):

    (A) How much energy is needed to warm the ice to 0 degrees?
    (B) How much energy is needed to melt the ice?
    (C) How much energy would be needed to convert all the liquid water (from the ice) to the boiling point (100 degrees)?

    Now compare these answers with the energy released from the steam:

    (D) How much energy is released if the steam cools to 100 degrees?
    (E) How much energy is released if all the steam changes to liquid water at 100 degrees?

    By comparing these energies you should be able to figure out the final temperature.
    Not so. All you need is for all the substances to be at the same temperature. (For example, if the final temp is 100 degrees, you could have a mix of water and steam.)
     
  5. Apr 1, 2006 #4

    danago

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    Gold Member

    I can get all of those values without any problems at all (unless theyre all wrong lol), but where do i go from there? is this right?:

    When the steam hits 100 degrees, the following remains:
    4kg steam 100 degrees
    0.0737kg water 0 degrees
    0.1463kg ice 0 degrees

    Then i calculate how much energy is required to fully melt the ice and heat all of the water to 100 degrees:
    [tex]Q_{fusion}=mL[/tex]
    [tex]=0.1463\times 3.34\times 10^5[/tex]
    [tex]=48864.2J[/tex]

    [tex]Q_{heating}=mc\Detla T[/tex]
    [tex]=0.22\times 4200\times 100[/tex]
    [tex]=92400J[/tex]

    [tex]Q=Q_{fusion}+Q_{heating}[/tex]
    [tex]=48864.2+92400[/tex]
    [tex]=141264.2J[/tex]

    So 141264.2J later, the ice has fully melted and heated, so there is 0.22kg of water at 100 degrees, but since the energy was tranferred from the steam, some of it will have condensed.

    [tex]Q=mL[/tex]
    [tex]141264.2=m\times 22.5\times 10^5[/tex]
    [tex]m=0.0628kg[/tex]

    So once the ice has fully melted and heated to 100 degrees, 0.0628kg of the steam will have condensed, leaving us with:
    0.2828kg water 100 degrees
    3.9372kg steam 100 degrees

    Since they are the same temperature, they remain as a steam/water mixture at 100 degrees, and 100 is the answer to the question?

    And sorry about the double post, the one which you locked before. And thanks so much for the help.

    Also, my physics teacher was saying something about the ice also transferring energy. From what i previously understood, the hotter substance transfers energy to the colder substance, so i didnt think ice would transfer any energy. he had alot of trouble explaining the concepts and theories, so if you didnt mind, would you maybe be able to help out with explaining the energy transfer processes?
     
    Last edited: Apr 1, 2006
  6. Apr 1, 2006 #5

    Doc Al

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    Staff: Mentor

    Yep. I didn't check your numbers, but it looks like you have a good handle on it.

    You can easily see that D+E is more than enough to bring all the ice to 100 degrees (since D+E > A+B+C). Right there that tells you that the final temp is 100. (And by analyzing it carefully, like you did, you can even figure out the mix of steam and water.)
     
  7. Apr 1, 2006 #6

    danago

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    Gold Member

    ok thanks so much. It makes perfect sense to me now. I dont understand why it didnt hit me as soon as i calculated the energy levels the first time :P
     
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