College Physics
posted by Becky .
The top layer of your goose down sleeping bag has a thickness of 5.0 cm and a surface area of 1.0 m^2. When the outside temperature is 18 degrees celcius you lose 24 Cal/hr by heat conduction through the bag (which remains at a cozy 34 degrees celcius inside). Assume that you're sleeping on an insulated pad that eliminates heat conduction to the ground beneath you. What is the thermal conductivity of the goose down?

You must have heard of the basic thermal conduction equation.
dQ/dt = k A (deltaT)/(thickness)
In your case, dQ/dt = 24 Cal/hr
(those are kilocalories, by the way)
deltaT = 52 C
thickness = 0.05 m
A = 1 m^2
Solve for k 
is k 0.023076923 ?

the answer is 0.027

I get k = 0.02307.. Cal/hr*degC*m also
Radiation through the sides has been neglected. 
just for others who are looking for this answer:
You have to convert 24Cal/hr into J/s
since 1Cal=4190 Joules
24Cal/hr x 1hr/3600s x 4190 J = Q/t
Then plug into formula stated above to get a
final answer of 0.026858974 = 0.027 (2 sigfigs)
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