- #1
MichaelXY
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Homework Statement
I was having this discussion, and a friend of mine asked if swallowing hot coffee would denature the enzymes of the stomach
Homework Equations
q= cg(Delta T)
M=DV
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?