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# A skier wears a jacket filled with goose down that is 15mmthick. Another skier wears a wool sweater that is 5.0mmthick. Both have the same surface are

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A skier wears a jacket filled with goose down that is 15mmthick. Another skier wears a wool sweater that is 5.0mmthick. Both have the same surface area. Assuming thetemperature difference between the inner and outer surfaces of eachgarment is the same, calculate the ratio (wool./goose down) of theheat lost due to conduction during the same time interval.

2021-03-27
The rate of heat flow through a conductor is:
$$\displaystyle{H}={\left({\frac{{{k}{A}}}{{{L}}}}\right)}\triangle{T}$$ where k is the conductivity of the material, A is the surface area, L is the thickness
You can write this for wool sweater and for down jacket:
$$\displaystyle{H}_{{w}}={\left({\frac{{{k}_{{w}}{A}}}{{{L}_{{w}}}}}\right)}\triangle{T}$$ and $$\displaystyle{H}_{{d}}={\left({\frac{{{k}_{{d}}{A}}}{{{L}_{{d}}}}}\right)}\triangle{T}$$
And take the ratio of these equations:
$$\displaystyle{\left({\frac{{{H}_{{w}}}}{{{H}_{{d}}}}}\right)}={\frac{{{\frac{{{k}_{{w}}}}{{{L}_{{w}}}}}}}{{{\frac{{{k}_{{d}}}}{{{L}_{{d}}}}}}}}={\frac{{{k}_{{w}}{L}_{{d}}}}{{{k}_{{d}}{L}_{{w}}}}}$$
my book doesnt have the k values for wool ordown, but yours must have them if its asking this question. Popthem in and calc the answer. Notice the units drop out and you willjust have a number that represents how much more heat is lost persecond through the sweater than through the down.