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Experiement of determining specifc latent heat of vapourisation

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Explain why heat loss is the same in both cases, provided the time is the same.


    2. Relevant equations
    VIt = mL + H


    3. The attempt at a solution

    Is it the same because in both cases I'm asked to boil off 50g of alcohol?
     
  2. jcsd
  3. Dec 30, 2012 #2
    Well I realize what I said really makes no sense... I found something in a book that says H is negligible if we do 2 experiments and eliminate H from the two equations and H is also relatively small due to lagging but I'm confused because the diagram of setup my teacher gave me has NO LAGGING.
     
  4. Dec 30, 2012 #3

    haruspex

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    What 'both cases'? Have you forgoten to attach something?
     
  5. Dec 30, 2012 #4
    umm here's the method of experiment
    method : Set up the apparatus as in the diagram of the experiment. Choose
    a voltage setting by sliding the slider of the potentiometer. Record
    the readings on the voltmeter and the ammeter. Bring the liquid
    to boil. After boiling for 30 seconds, start your stopwatch and note
    the time taken to boil away 50g of the alcohol. Repeat the
    procedure with different values of V and I but with the same
    time, t. Theory states that the thermal energy supplied after the
    liquid has started to boil is equal to the heat required to boil off a
    a mass m of the liquid plus the heat lost to the surroundings,
    H .i.e.

    VIt= mL + H

    where V is the voltmeter reading, I is the ammeter reading,
    m is the mass of the liquid boiled off, t is the time, L is the
    Specific Latent Heat of Vapourisation of alcohol.

    Note , since the same time is used in both parts of the experiment, then
    H may be eliminated by subtraction. Explain why heat loss is the same in both cases, provided the time is the same. Calculate the Specific Latent Heat of Vapourisation of alcohol and state its unit.
     
  6. Dec 30, 2012 #5

    haruspex

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    Why didn't you post all that the first time? There's no chance of making a useful reply to the original post.
    The question could have been better worded. No, you are not expected to boil off 50g in both cases. In fact, you don't need to boil off exactly 50g in either case. it should say about 50g the first time. The second time it's whatever happens to boil off in the same time period. The important thing is that you measure how much boiled off in each case.
    How would varying t affect H?
     
  7. Dec 30, 2012 #6
    t will affect the magnitude of H
     
  8. Dec 30, 2012 #7

    haruspex

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    Right. So if you conduct the measurements (voltage, current, mass lost) over the same time period each time, will H be the same? If not, why not?
     
  9. Dec 30, 2012 #8
    Umm I believe H will be the same, since the liquid is at it's boiling point each time?
    and since it's the same t the amount of energy should practically be the same right?
     
  10. Dec 30, 2012 #9

    haruspex

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    Yes.
    Does that answer all your questions?
     
  11. Dec 30, 2012 #10
    Yes I believe so, thank you so much
     
  12. Dec 31, 2012 #11
    Sorry ignore this post I had two tabs open and posted in the wrong thread.
     
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