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Lab Experiment:Calculating Molar Enthalpy of Fusion of Water

  1. Dec 5, 2004 #1
    Hey.
    I am doing a lab experiment in school to find the molar enthalpy of fusion of water. Ice and water are mixed in a calorimeter and the temperatures are recorded. Here are my results. Mass of ice cube: 6.61g. Mass of water: 100g. Initial ice temperature: 0 degrees celsius. Initial water temperature: 21 degrees celsius. Final Temperature of ice and water: 16 degrees celsius.

    To calculate the molar enthaply I used mcT(ice)+nH = mcT(water)
    now I substitute in my values. 6.61(4.19)(16)+(6.61/18.02)H=100(4.19)(5)
    My question is about the change in temperature value for water(T). Should it be 5, or -5? Would it be 5 since heat is being lost? I've calculated using both values and it seems like using -5 will get an answer that is closer to the accepted value of molar enthalpy of fusion of water. Can someone please tell me the correct way to do this? Thanks.
     
  2. jcsd
  3. Dec 5, 2004 #2

    chem_tr

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    In your binary ice-water mixture, ice is the one that receives energy, and water is the one that gives energy. So energy is lost from water, indicating a minus sign before. The minus has a physical meaning, not a mathematical one.
     
  4. Dec 5, 2004 #3
    And that means I dont use it?
     
  5. Dec 5, 2004 #4
    nobb, you use the negative sign in your calculation because it indicates a physical temperature decrease.

    if you were to use the absolute value all of the time it would be hard to determine whether there was a loss of energy or gain.

    also why did you choose to use that formula. I received a value of -6906 j.

    finding the change in enthalpy of water(q=mct)then dividing it by the # of moles of ice (6.61g/18.02g/mol) gives me a value of -5691 J. This value from your data is closer to the molar enthalpy of fusion of ice.
     
  6. Dec 5, 2004 #5

    Gokul43201

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    Nobb. You must use change in T = absolute value of change = +5. Here's the reason :

    [tex]Q_f = Q _i [/tex]

    [tex]Q_f(water) + Q_f(water from ice) = Q _i(water) + Q_i(ice) [/tex]

    [tex]Set~Q(ice~at~0C) = 0 [/tex]

    Now :
    [tex]Q_f(water) = m(water)*C*T_f + n(water)H [/tex]

    [tex]Q_f(water from ice) = m(water~from~ice)*C*T_f + n(water~from~ice)H [/tex]

    [tex]Q_i(water) = m(water)*C*T_i + n(water)H [/tex]

    [tex]Q_i(ice) = 0 [/tex]

    Plugging in these values into the second equation gives:

    [tex] m(water)*C*T_f + n(water)H + m(water~from~ice)*C*T_f + n(water~from~ice)H = m(water)*C*T_i + n(water)H + 0 [/tex]

    Collecting terms, this becomes :

    [tex] m(water)*C*(T_f-T_i) + m(water~from~ice)*C*T_f + n(water~from~ice)H = 0 [/tex]
    Which, upon rearranging, becomes :

    [tex] m(water)*C*(T_i-T_f) = m(water~from~ice)*C*T_f + n(water~from~ice)H [/tex]

    Look at the LHS and you will see that it contains T(i) - T(f) = 21 - 16 = +5, not -5.

    A shorter, but less rigorous explanation of why you use +5 is in looking at the the conservation statement that says "heat lost = heat gained". Now if one side of this equation is negative, and the other side is positive, the two sides can never be equal to each other. So, stated this way, the equation only talks about absolute values of heat lost or gained.

    You will end up with a number like 4.5 KJ/mol, which appears to be lower than the expected 6 KJ/mol. This is because of heat leaking in from the surroundings, raising the final temperature. A more carefully controlled experiment would have resulted in a lower final temperature.
     
    Last edited: Dec 5, 2004
  7. Dec 5, 2004 #6

    Gokul43201

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    No gerald, that's not correct.

    gerald, this is not the correct way to find the enthalpy of fusion. Nodd has used the correct approach. The number you get may be closer to the expected number, but that is not because your approach is correct.
     
  8. Dec 5, 2004 #7
    hmm, so we always use an absolute value for temperature readings ? do we then determine a negative sign for molar enthalpy by looking at whether energy was lost or gained.

    what was wrong with my approach?
     
    Last edited: Dec 5, 2004
  9. Dec 5, 2004 #8
    if we always use an absolute value why is it denoted as delta t which is = t(final) - t(initial)
     
  10. Dec 5, 2004 #9
    alright, i think i understand what was wrong with my approach. the energy involved in decreasing the temperature of water isn't necessarily the amount of energy involved in the fusion of ice right. am i on the right track?

    i'm going to read over my textbook more thoroughly again and do some more problems.
     
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