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Quantum Physics in the Body: Energy

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A pulsed dye laser emits light of wavelength 576 nm in pulses of 456 µs duration. This light is absorbed by the hemoglobin in the blood and can therefore be used to remove vascular lesions, such as certain blemishes and birthmarks. To get an estimate for the power required for this laser surgery, assume that blood has the same specific heat capacity (4190 J/(kg·K)) and heat of vaporisation (2.256×106 J/kg) as water. Suppose that each pulse must remove 2.5 µg of blood by evaporating it, starting at 33 °C.
    a) How much energy must the pulse deliver to the blemish?
    b) What must be the power output of the laser?
    c) How many photons does each pulse deliver?

    2. Relevant equations

    Q = m*c*delta(t)
    f = c/wavelength
    E = h*f

    3. The attempt at a solution

    I don't even know how to start this... but I'd really appreciate the help, especially an answer to part a.
     
  2. jcsd
  3. Mar 18, 2009 #2
    Okay... I'm starting to make some progress, but I keep getting stuck in my ideas...

    Energy to Vaporize = Q = m*l = (2.5e-9 kg)(2.256×106 J/kg) = 5.64e-03 J...
    so...
    Q = m*c*delta(t) ; delta(t) = Q/(m*c) = 5.64e-03 J/[(2.5e-9 kg)x(4190 J/(kg·K))]
    delta(t) = 538.42 K

    but... I don't understand.... This means there's a difference...
    33°C = 306.15 K, so there's a difference of 232.275 K.

    What can I do with this?
     
  4. Mar 18, 2009 #3
    Your thermal energy for vaporizing blood is correct. But you start out at 33°C and blood might not undergo a phase transition then, right? You might have to add additional heat to get the temperature high enough to reach the temperature at which blood transitions from a liquid to a gas.

    It looks like you attempted to find the temperature at which blood transitions from a liquid to a gas. My guess is is that you're supposed to use the boiling point of water.
     
  5. Mar 18, 2009 #4
    Ah, that's very interesting!
    I never though of using that 100°C boiling point.
    so, with that...
    Energy to Vaporize = Q = m*l = (2.5e-9 kg)(2.256×106 J/kg) = 5.64e-03 J
    and
    Q = m*c*delta(t) = (2.5e-9 kg)(4190 J/(kg·K))(67K) = 7.01825e-4 J

    adding these two values... Q1 +Q2 = 6.341825e-3 J, which makes sense...

    now,
    b) What must be the power output of the laser?
    c) How many photons does each pulse deliver?

    ideas there?
     
  6. Mar 18, 2009 #5
    Well how do we define power?
     
  7. Mar 18, 2009 #6
    Okay, I figured out b...

    I thought, since a J/s is a W...

    Then the energy, (6.341825e-3 J)/0.000456s = 13.9076 W....

    so far, so good.... now, c.
     
  8. Mar 18, 2009 #7
    I figured out c too!!!
    Thank you so much for your help!!
     
  9. Mar 18, 2009 #8
    Looks right so far.

    EDIT: Nice, good job.
     
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