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Posted by on Monday, April 25, 2011 at 7:54am.

The rate that an object cools is directly proportional to the difference between its temperature ( in Kelvins) at that time and the surrounding temperature (in Kelvins). If an object is initially at 35K, and the surrounding temperature remains constant at 10K, it takes 5 minutes for the object to cool to 25K. How long will it take for the object to cool to 20K?

  • calc - , Monday, April 25, 2011 at 8:14am

    You are assuming Newton's law of cooling. This leads to a differential equation of T(t) that has a simple exponential decay law. I will skip these steps.

    Let k be the cooling rate constant and T' be the surrounding temperature (10 K in this case). t is time in minutes.

    The temperature decay equation is

    T - T' = 25 e^(-kt)

    where the 25 K is the initial temperature difference relative to the surroundings.

    Note that dT/dt = -k*(T-T'), as required by the cooling law, and that T(t=0) = 25 + T'

    15 = 25*e^(-5k)
    -5k = ln(0.6)
    k= 0.102 min^-1

    When T = 20K,
    10 = 25*e^(-0.102 t)
    0.4 = e^(-0.102 t)
    -.102t = ln(0.4)
    t = 9.0 minutes

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