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Newton's cooling to be applied to heating

  1. Jul 6, 2009 #1
    I have experimented the converse of newton's law of cooling, i.e. the rate of heating of a cold body is directly proportional to the deficient temperature when the difference in temperatures is small and the states of the system remain the same.
     
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  3. Jul 6, 2009 #2

    ideasrule

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    Yes, the heat transferred by conduction is given by H=kA(dT/dx), where dT/dx is the temperature gradient. The equation applies whether an object is being heated up or cooled down.
     
  4. Jul 6, 2009 #3
    Yes, it is given by H/t=kAdT/dx, but that is conduction while my consideration is radiation. The law is given by dT/dt=r(Tenvironment -T) which gives on integration ln (Tenv-T)=rt +constant or Tenv-T=e^(rt+c)
    In this case, however, since Tenv-T is negative, e^(rt +c) results negative which is impossible. For calculus to be applicable, the equation must be dT/dt=r(T-Tenv) which is not acheived by negating the right side, so r is a positive constant(this constant is different from that of integration,c) as before.
    Thus there is a new equation but the question is whether the new constant r would be equal to the corresponding original constant.
     
    Last edited: Jul 6, 2009
  5. Jul 6, 2009 #4

    Mapes

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    If we let

    [tex]\theta = T-T_\mathrm{env}[/tex]

    then we integrate

    [tex]\frac{d\theta}{dt}=-r\theta[/tex]

    to get

    [tex]\ln\theta=-rt+C_1[/tex]

    [tex]\theta=C_2\exp(-rt)[/tex]

    and from the initial conditions,

    [tex]C_2=T_0-T_\mathrm{env}[/tex]

    to produce

    [tex]T=T_\mathrm{env}+(T_0-T_\mathrm{env})\exp(-rt)[/tex]

    Does this help? This is a little easier that your integration, which requires your constant c to be a complex number.
     
  6. Jul 6, 2009 #5
    There are two things I would like to point out I was wrong in using a positive sign in the right side and secondly, c is not a complex constant as you said.
     
  7. Jul 7, 2009 #6

    Mapes

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    Then what is the value of c that makes [itex]\exp(-rt+c)[/itex] negative?
     
  8. Jul 7, 2009 #7
    Since you have already said from the initial conditions,

    [tex]C_2=T_0-T_\mathrm{env}[/tex]
    and C2=e^c
    e^c is directly [tex]T_0-T_\mathrm{env}[/tex]
    which gives
    e^-rt(T0-Tenv)=(T-Tenv)
    which only on reversing signs becomes the expected integrated result of "the law of heating" as it should have been.
    It would be the same result if we had started with theta=Tenv-T,right?
    which in your interpretation yields a positive [itex]\exp(-rt+c)[/tex]
     
    Last edited: Jul 7, 2009
  9. Jul 7, 2009 #8

    HallsofIvy

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    Your integration is wrong. The integral of dx/x is ln|x|+ C, not ln x+ C.

    No. [itex]\int dx/x= ln |x|+ C[/itex]. [itex]ln|T_{env}- T|= rt+ constant[/itex] so [itex]|T_{env}-T|= Ce^{rt}[/itex]
    If [itex]T_{env}-T[/itex] is negative, [itex]|T_{env}- T|= -(T_{env}- T)= Ce^{rt}[/itex]

     
  10. Jul 7, 2009 #9
    r is negative,because d(theta)/dt is negative as the cooling or heating decreases over time and you haven't used a negative sign in the equation. I had mentioned that I was wrong in using a positive sign in the differential equation.
    Even if integral of dx/x is written as ln x +c instead of ln|x| +c the result is the same with a negative sign on both sides of the equation.
     
    Last edited: Jul 8, 2009
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