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Posted by 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 - , 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 - , Saturday, October 8, 2011 at 12:21am

    is k 0.023076923 ?

  • College Physics - , Saturday, October 8, 2011 at 2:07am

    the answer is 0.027

  • College Physics - , 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 - , 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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