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Specific Latent Heat of Fusion

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data
    need to work out the Specific Latent Heat of Fusion
    i did the expirment and here are the results
    Results:
    Mass of calorimeter
    Mcal 116.87g
    Mass of H2O
    MH2O = [Mcal + H2O] - Mcal 80.82g (197.69g)
    Mass of Ice
    Mice = [Mcal + Mice + H2O] - MH2O 12.58g (210.27g)
    Room Temperature (Tr) 26°c
    6°c above (Tr) = Ti 32°c
    Aprox. 6°c below (Tr) = Tf 21°c
    Change in Temperature ∆T -11°K
    Specific Latent Head of Ice L (?)


    2. Relevant equations
    Mice l + Mice CH2O ∆T = -[Mcal Ccal ∆T + MH2O CH2O ∆T]

    we already know the following
    Ccal = Ccu = 390 Jkg-1K-1
    C H2O = 4200 Jkg-1K-1
    LH2O = 33x104 Jkg-1


    3. The attempt at a solution
    so there where i have gotten so far with
    Mice L(?) + Mice CH2O ∆T = -[Mcal Ccal ∆T + MH2O CH2O ∆T]

    I need to find out what L(?) is
    12.58g x L (?) +12.58g x 4200 Jkg-1K-1 x -11°K = -[116.87g x 390 Jkg-1K-1 x -11°K + 80.82g x 4200 Jkg-1K-1 x -11°K]

    Convert grams into kilo grams
    0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = 0.11687kg x 390 Jkg-1K-1 x -11°K +0.08082kg x 4200 Jkg-1K-1 x -11°K

    0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = -501.3723 + -3733.884

    0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = -4235.2563 J

    how do i switch over the 0.01258kg + 0.01258kg x 4200 Jkg-1K-1 x -11°K to the other side to have just L (?) on one side

    I need help to find out L(?)
    how do i rearange to find L(?)




    Thais is wat i mean
    [​IMG]


    PS i know that i am not very good at science, i am a International Relations and Politics student but they r making me do this
     
    Last edited: Oct 15, 2007
  2. jcsd
  3. Oct 15, 2007 #2
    Heres a lil correction i made
     
  4. Oct 15, 2007 #3

    learningphysics

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    "0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = 4235.2563 J"

    I'm going to leave out the units and rewrite:

    0.01258L + (0.01258)(4200)(-11) = 4235.2563

    subtract (0.01258)(4200)(-11) from both sides...
    or equivalently add (0.01258)(4200)(11) to both sides...
     
  5. Oct 15, 2007 #4
    =======================

    my attempt at the final answer

    0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = 4235.2563 J
    0.01258kg x L (?) = 4235.2563 J + 581.196 J
    L (?) = (4235.2563 J + 581.196 J) / 0.01258kg
    l (?) = 382865.6041

    which i think is wrong coz i dont know wat i am doing
     
  6. Oct 15, 2007 #5
    0.01258L + (0.01258)(4200)(-11) = 4235.2563
    0.01258L + (-581.196) = 4235.2563

    u mean subtract (-581.196) from 4235.2563
    4235.2563 - (-581.196)
    4816.4523
    ?????
     
  7. Oct 15, 2007 #6

    learningphysics

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    math wise it looks right to me. l = 382865.6041 J/kg.

    I don't know if the answer is right... I'm rusty with my chemistry... but as far as the algebra for this part is concerned, the math is right.
     
  8. Oct 15, 2007 #7
    lol i know it means 382865.6041 Jouels per kilogram
    382865.6041 Jouels is a HELL load of jouels to melt ice in water at room temprature
    382865.6041 Jouels is like wat ur house used in 2 weeks (just guessing)
     
  9. Oct 15, 2007 #8

    learningphysics

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    yeah... it's basically the same as what you did...

    0.01258L + (-581.196) = 4235.2563

    0.01258L + (-581.196) - (581.196) = 4235.2563 - (-581.196)

    0.01258L = 4816.4523

    which gives the same answer you got.
     
  10. Oct 15, 2007 #9
    thanx mate even tho the answer is wrong atleast the math is write so i can put that in my Conclusion but thanx alot for the help

    i still have to do the following experiment
    combustion of fules
    something on electricity
    something on water turbines
    something on solar energy
    biodiesel

    so i will be here much more to get help with my results :P

    then after that i am freeeeeeeeeeeeeeeeeeeeeeeeeeee from anykind of science
     
  11. Oct 15, 2007 #10

    learningphysics

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    I don't know. This webpage:

    http://www.physchem.co.za/Heat/Latent.htm

    gives the latent heat of ice as 334 kJ.kg-1 = 334,000J/kg, which is not that far off from what you have.
     
  12. Oct 15, 2007 #11
    lol by bad wooooohooo i did it
     
  13. Oct 15, 2007 #12

    learningphysics

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    :) good job. seems like it was a tough experiment! I was terrible at chemistry.
     
  14. Oct 15, 2007 #13
    the expirment was easy just melt ice in water in room temp untill u reach a certian temp

    but i find physics hard even the basics like this coz this is ment to be basic and i am not even a science student i am doing my BA in international relations and politics
     
  15. Oct 15, 2007 #14

    learningphysics

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    You'll usually be able to get lots of help here. good luck!
     
  16. Oct 15, 2007 #15

    learningphysics

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    shouldn't the left side of this equation be:

    12.58g x L (?) +12.58g x 4200 Jkg-1K-1 x (21°K) ?

    because the ice is initially at 0°C... and this water that is from the melted ice goes to a final temperature of 21°C from 0°C.
     
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