body butter pats
Entropy change from digesting fat?
digesting fat produces 9.3 food calories per gram of fat, and typically 80% of this energy goes to heat when metabolized. the body then moves all this heat to the surface by a combination of thermal conductivity and motion of the blood. The internal temperature of the body (where digestion occurs) is normally 37 degrees celcius and the surface is usually about 30 degrees celcius. By how much does the digestion and metabolism of a 2.5 gram pat of butter change your body's entropy? does it increase or decrease?
One food calorie is equal to 1000 calories or 4184 Joules.
So the heat transferred to your body by the digestion of 2.5g fat is
Q = 0.8 ∙ 2.5g ∙ 9.3∙4184J/g = 77822.4J
If a system exchanges amount of heat Q at constant absolute temperature T its entropy changes by:
ΔS = Q/T
Q is sign sensitive. If the system absorbs heat q is positive and entropy rises; if systems rejects heat Q is negative and its entropy decreases.
Digestion of fat adds amount of heat calculated above to you body at a temperature of
T = (37+ 273)K = 310K
So the digestion changes entropy of the body by
ΔS₁ = 77822.4J/310K = 251.04J/K
On the other hand your body have to transfer the same amount of heat to the surround to maintain constant body temperature. The heat rejection occurs at the surface temperature of
T = (30+ 273)K = 303K
Due to this heat transfer your body's entropy changes by:
ΔS₂ = -77822.4J/303K = -256.84J/K
The total entropy change of your body is:
ΔS = ΔS₁ + ΔS₂ = 251.04J - 256.84J/K = -5.8J/K
That means your body's entropy decreases.
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