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Homework Help: Specific Heat and the body

  1. Dec 22, 2007 #1
    1. The problem statement, all variables and given/known data
    I was having this discussion, and a friend of mine asked if swallowing hot coffee would denature the enzymes of the stomach

    2. Relevant equations
    q= cg(Delta T)

    3. The attempt at a solution
    First thing I did was assume coffee to be around 140 deg F. I then made the assumption that a typical gulp of coffee is around 10ml. Since the density of coffee should be close to water, I computed mass as 1(10ml) = 10 g.

    Using the formula: Q= cg(delta T) I calculated (4.186j/g) (10g) (140 - 98.6) = 1.7 Kj

    Is it safe to say then, that to drop coffee to body temp we would need to release 1.7Kj.

    Now to determine what specific heat is needed to release this heat I calculated as follows: C= Q/(g x Delta T)

    Inserting numbers into equation I get 1.7Kj/10g(40) = 4.1 Does this mean we need a specific heat of 4.1 to cool the coffee?

    The human body specific heat is around 3.5 J/g so then does this mean the coffee is still hot in the stomach?
    I am sure I am missing something here. Any thoughts?
  2. jcsd
  3. Dec 22, 2007 #2


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    You threw your numbers back into the same equation, q = mcdeltaT, which got you the specific heat of water, which is not surprising, since it's what you used earlier. In other words, you went in a circle. :P
  4. Dec 22, 2007 #3
    Ok, but now I am confused. How do I fix my error?
  5. Dec 23, 2007 #4
    I see my circle, but not sure how to proceed.
  6. Dec 24, 2007 #5


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    Staff Emeritus
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    This is not an easy calculation. You want to essentially find the temperature of the coffee when it reaches your stomach, and compare that with the denaturation temperature of the enzyme of interest, is that right? Wouldn't it be easier to worry about the denaturing of salivary amylase (ptyalin)?

    Also, do you care about whether or not the denaturation is reversible? The temperatures needed to permanently denature a protein are typically much higher (and are a function of secondary structure) than the reversible denaturation temperature.
    Last edited: Dec 24, 2007
  7. Dec 24, 2007 #6
    Thanks for the response. I admit it was a silly argument. My goal was to show that the temp of the coffee would be negligable by the time it reached the stomach. I thought I may be able to show this with a few equations. I guess I was mistaken, but thanks all the same.
  8. Sep 18, 2010 #7
    Three years later, never stop guessing! : )
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