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Incredibly lost on Calorimetry Problem

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    An alcohol rub can rapidly reduce an elevated body temperature in a patient. The heat energy lost by the person is due to the evaporation of alcohol. Find the number of grams of alcohol that must be evaporated from the surface of a 75 kg person to reduce the body temperature by 2.5 degrees C. The heat of vaporization for alcohol is 204 cal/g. The specific heat capacity of the human body is 0.83 cal/g degrees C.


    2. Relevant equations

    It is a change of temperature problem and a change of phase problem so I am unsure as to the formula. Change of temp is Q=(mass)(specific heat)(change of Temp) and change of phase is Q=(mass)(latent heat)

    Can someone please help me determine which formula I am to use to solve? Thanks in advance for any assistance you can provide!
     
  2. jcsd
  3. Jun 28, 2011 #2

    vela

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    As you noted, it's both "a change of temperature problem and a change of phase problem," so you need to use both formulas.

    Energy conservation tells you the amount the heat lost by the body is equal to the amount of heat that goes into evaporating the alcohol.
     
  4. Jun 28, 2011 #3
    Thank you. I was sure it was but he has never given a problem of this difficulty so I was sure that I was reading more into it. I appreciate the confirmation. I'm going to get to work on it and hope I can figure it out. Thanks again for the quick reply.
     
  5. Jun 28, 2011 #4
    I've been working on this problem for hours and have come up with the following thus far. Can someone let me know if I am on the right track?

    Q(heat energy)=m(mass)*c(specific heat)*change of time

    Q=(75 kg)*(2430 J/kg degrees C)*(2.5 degrees C)= 455,625 Joules

    That portion is for the temperature change. Now I believe I need to find the phase change. Does this look like I am heading in the right direction?
     
  6. Jun 28, 2011 #5

    ideasrule

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    Yes, that's exactly what you need to do. However, where did you get 2430 from? 1 calorie is 4.184 J, so 0.86 cal/g should not be 2430 J/kg.
     
  7. Jun 28, 2011 #6
    2430 J/kg degrees C is the specific heat for Ethyl Alcohol listed in my textbook. Is that what I am supposed to have in the equation?
     
  8. Jun 28, 2011 #7
    Oops, nevermind. I just realized that I am working on the patient specific heat not the alcohol.
     
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