By Meredith Barthold

Since we just had Thanksgiving and we ate a rather large
turkey in my house, I was wondering how much mass the turkey loses as it is
baked in the oven. I know the initial weight, and I could have just weighed the
turkey after being cooked, but instead I decided to use diffusion to calculate
the change in mass after the turkey is cooked. So I measured the turkeys
dimensions before putting it in the oven, kept track of the time it took to
cook, and I used Fick’s Law of Diffusion.

Most variables were
obvious, but the concentration of water inside the turkey is an estimation
based on the pressure required to penetrate human skin. It takes 7 bar to
penetrate the skin with a needle (http://www.ncbi.nlm.nih.gob/pmc/articles/PMC2291478),
and to convert bar to kg/m

^{3}, I multiplied 7 bar times the density of air in an estimated 400°F oven, which is 0.5243 (http://www.engineeringtoolbox.com/air-desity-specific-weight-d_600.ntml). This gives an estimation of 3.6701 kg/m^{3}inside of the turkey.
For the concentration
of water vapor outside the turkey I used the same value as was used in class
for the water vapor outside the sunflower.

All variables are as
follows:

**L**ength of turkey: 35.5 cmà0.355 m

Cross sectional

**A**rea of turkey: (0.29 m)*(0.18 m)= 0.0522 m^{2}**C**oncentration of water vapor outside the turkey: 0.011 kg/m

^{3}

**C**oncentration of water inside the turkey: 3.6701 kg/m

^{3}

**T**ime in the oven to cook: 5 hrs. 33 mins. à 333 minutes à19980 seconds

**D**iffusion constant, water vapor in air: 2.4 x 10

^{-5}m

^{2}/s

Using the following
equation:

Rate of diffusion = Δm/Δt = DA [(C

_{2}-C_{1})/L]
Δm = (2.4 x 10

^{-5}m^{2}/s)(0.0522 m^{2})[(3.6701-0.011)/(0.355 m)](19980 seconds)
Δm= 0.258
kg = 0.569 lbs = 0.6 lbs

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