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Math Help - Newton’s law of cooling, difficult question

  1. #1
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    Newton’s law of cooling, difficult question

    By Newton’s law of cooling the temperature T of a cooling object as a function of time is



    TO - initial temperature,
    T1 is ambient temperature (meaning the temperature outside the cooling object)
    k is constant
    W

    -32C was the te,perature when electricity was cut off and heating of a house stopped working. After 2 hours from the power cut the inside temperature was measured to be 19,2 C, and after that the following measurements were done:

    2,5 hours: 18,3 oC
    3,0 hours: 17,5 oC
    3,5 hours: 16,6 oC
    4,0 hours: 15,8 oC
    4,5 hours: 14,9 oC
    5,0 hours: 14,2 oC

    Supposing Newton’s cooling law to be valid in this case,

    Search:

    k
    T0 initial temperature (which is the temperature at moment t=0 when the power was cut off). Try also to predict when the inside temperature will be dropped below zero.


    __________________________________________________ __________________________________________________ __

    my attempt:

    19.2-(-32)=(T0-32)e-k(2)
    14.2-(-32)=(T0-32)e-k(5)

    51.2 (T0-32)e-k(2)
    46.2 (T0-32)e-k(5)

    1.11 e2,5k
    k= 0.041103894
    k= ln(1.11)/2.5


    Substitude k into the first equation

    51.2=(T0-32)e-2k
    51.2=(T0-32)e-2((ln(1,11)/2,5))


    and now ?????????????
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  2. #2
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    Re: Newton’s law of cooling, difficult question

    I have no idea what you are doing. You posted this in "differential equations" but there is no differential equation. You title this "Newton's law of cooling" but never say what that is.
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  3. #3
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    Re: Newton’s law of cooling, difficult question

    Hi Plastique!

    Newton's Law of Cooling is defined as:
    dT/dt=k(T-Tm) where k is some constant and Tm is the ambient temperature.

    Solving the ODE gives:
    T=ce^(kt)+Tm

    We can set Tm= -32
    so, T=ce^(kt)-32

    Using the initial conditions t=2,T=19.2, our equation becomes:
    19.2=ce^(2k)-32 ---> 51.2=ce^(2k) ---> (51.2/c)=e^(2k) ---> ln(51.2/c)/2=k

    We can now substitute k back into the equation:
    T=ce^[ln(51.2/c)t/2]-32

    Now we use the next initial condition (which can be any of the given values by the way) t=5,T=14.2, and our equation becomes:
    14.2=ce^[ln(51.2/c)5/2]-32

    And solving for c gives:
    c= 54.83047 Remember that ln(51.2/c)5/2=ln((51.2/c)^(5/2))

    We can now also find k by substituting c into our equation for k:
    k=ln(51.2/c)/2 ---> k=ln(51.2/54.83047)/2= -0.03425

    So we now have both c=54.82047 and k=-0.03425 which makes our equation:
    T=(54.82047)e^(-0.03425t)-32

    Finding the initial temperature of the house comes from setting t=0 which will give T0= 22.83047 (a reasonable temperature).

    To find how long it takes for the temperature of the house to drop below 0 degrees Celsius, set T=0 and solve for t.
    t=15.72 hours.
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