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Homework Help: Time for melting and vaporising a wax candle

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Assume that a wax candle with mass m at temperature T1 undergoes constant heat admission dQ/dt=Kp

    T1=25 celsius

    Calculate how long it will take for the wax to melt and vaporise during constant heat admission
    without any loss.

    ρ=791 kg/m3
    M=310 g/mol
    Tm=317.15 K (melting temperature)
    Lf=252 kJ/kg
    Tb=641.8 K (boiling temperature)
    Lv=105 kJ/kg
    Cp(s)=598.1 J/mol K
    Cp(l)=739 J/mol K
    Cp(g)=1193 J/mol K

    2. Relevant equations


    3. The attempt at a solution
    Well the posibilities overwhelmes me.

    I know that it all boils down to using Δt=Q/P the question is though which Q are they asking for, I have only made an attempt at the boiling question though.

    For boiling
    If I apply what I know from heat of transformation:
    Q=0.05kg*252 kJ/kg -> Q=12600J
    Δt=12600J/10W -> Δt=1260s
    In other words it would take 21mins.

    However if that's the case then it would be even faster to boil it away when it's in liquid.

    Am I doing something wrong or is there something blatantly obvious that I don't seem to get?
  2. jcsd
  3. Apr 8, 2013 #2


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    Staff: Mentor

    Is this consistant?

    Before plugging any numbers into equations, can you describe to me what will happen to the candle, from beginning to end?
  4. Apr 8, 2013 #3
    I can see I said I would calculate the boiling, when I mean't I would calculate the melting.

    The candle is in solid phase and it's placed in a isolated room which has the temperature of 25 celsius. I will now light the candle and have to calculate the time it will take the 50g of solid wax to become liquid.

    I tried to calculate the time it would take for the candle to melt, and my guess for that was to use the equation Q=mL.

    I here apply the constant Lf since the candle is going from solid to liquid state. The heat I calculate from that is 12.6 kJ or 12600J.

    From that I then use Δt=Q/P

    What I'm unsure about is whether or not I've done it correctly given the options that I have.
  5. Apr 8, 2013 #4


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    Staff: Mentor


    You are not lighting the candle. You are supplying heat at a uniform rate. It is not the same thing.

    But what happens before it becomes liquid?

    That part is correct. But if you want to get through the problem, you first have to figure out everything that happens from beginning to end.
  6. Apr 8, 2013 #5
    I assume there are two stages then, the first stage is when the wax is heating which is expressed by:
    [itex]Q= \frac{m C_p (s) (T_m-T_1)}{M}[/itex]

    The second stage would be when the wax reaches the melting temperature and begins to melt and so I add that to the equation so that Q will be:
    [itex]Q= \frac{m C_p (s) (T_m-T_1)}{M} +mL_f[/itex]

    From this I can then apply my final formula:
    [itex]\Delta t= \frac{Q}{P}[/itex]

    Would this be correct?
  7. Apr 8, 2013 #6


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    Staff: Mentor

    This is correct but incomplete. The problem asks for the time for complete vaporization.
  8. Apr 8, 2013 #7
    Not entirely. I'm sorry if the translation is terrible but it's in danish. They ask for the time it takes for it to melt and then to vaporize. For vaporization I assume it would be in 4 stages:

    1. Heating to reach the melting temperature.
    2. Melting the wax.
    3. Heating the liquid wax to the boiling temperature.
    4. The wax is vaporizing.

    This would give the heat calculation Q:

    [itex]Q= \frac{m C_p (l) (T_b - T_m)}{M} + mL_v + \frac{m C_p (s) (T_m - T_1)}{M} + mL_f[/itex]

    And then it is just [itex] \Delta t = \frac{Q}{P}[/itex]

    Now I should be home free unless there is more to it.
  9. Apr 8, 2013 #8


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    Staff: Mentor

  10. Apr 8, 2013 #9
    You purposely searched up that word didn't ya? :D, regardless thanks for your help!
  11. Apr 9, 2013 #10


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    Staff: Mentor

    Nej, jag kan svenska.

    You're welcome!
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