Posted by **Becky** on Friday, October 7, 2011 at 11:36pm.

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?

- College Physics -
**drwls**, Friday, October 7, 2011 at 11:54pm
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

- College Physics -
**Becky**, Saturday, October 8, 2011 at 12:21am
is k 0.023076923 ?

- College Physics -
**Becky**, Saturday, October 8, 2011 at 2:07am
the answer is 0.027

- College Physics -
**drwls**, Saturday, October 8, 2011 at 8:48am
I get k = 0.02307.. Cal/hr*degC*m also

Radiation through the sides has been neglected.

- College Physics -
**Daniela**, Friday, November 8, 2013 at 5:21am
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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