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College Physics

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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 - ,

    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 - ,

    is k 0.023076923 ?

  • College Physics - ,

    the answer is 0.027

  • College Physics - ,

    I get k = 0.02307.. Cal/hr*degC*m also

    Radiation through the sides has been neglected.

  • College Physics - ,

    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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