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Thread: Newton's Law of Cooling?

  1. #1

    Newton's Law of Cooling?

    I think this is a Newton's Law of Cooling question but the formula I have for that isn't working...

    A pizza, heated to a temperature of 350 degrees Fahrenheit, is taken out of an oven and placed in a 75 degrees Fahrenheit room at time t=0 minutes. The temperature of the pizza is changing at a rate of -110e^-0.4t degrees Fahrenheit per minute. To the nearest degree, what is the temperature of the pizza at
    time t=5 minutes?


    The formula I tried to use was T= (To - Ts)e^kt + Ts

    The problem I'm having is with the "changing at a rate of..." I don't know what to do with that information. I'd really appreciate the step by step of how to do this particular question.

    Thank you!
    Chels

  2. #2
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    Re: Newton's Law of Cooling?

    You are given the rate of change of the pizza temperature after removal from the oven.

    A negative rate means the temperature is decreasing as expected.
    Differentiating the temperature function gives the rate of change.

    [tex]\frac{d}{dt}f(T)=-110e^{-0.4t}[/tex]

    [tex]Hence,\ f(T)=\frac{-110}{-0.4}e^{-0.4t}+c=275e^{-0.4t}+c[/tex]

    [tex]f(0)=350,\ therefore\ c=350-275=75.[/tex]

    [tex]f(T)=275e^{-0.4t}+75\ degrees\ fahrenheit.[/tex]

  3. #3
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    Re: Newton's Law of Cooling?

    [tex]Another \ way, \ to \ wit:[/tex]

    [tex]Newton's \ Law \ of \ Cooling \] states \ that \ the \ rate \ of \ change \ of \ the \ temperature \ of \ an \ object \ is \ proportional \ to \ the \ difference \ between \ the \ temperature \ of \ the \ object \ and \ the \ temperature \ of \ the \ surrounding \ medium.[/tex]

    [tex]Ergo, \ \frac{dT(t)}{dt} \ = \ k[T(t)-L][/tex]

    [tex]Given, \ \frac{dT(t)}{dt} \ = \ -110e^{-.4t}, \ L \ = \ 75, \ T(0) \ =350, \ find \ T(5).[/tex]

    [tex]k[T(t)-75] \ = \ -110e^{-.4t}[/tex]

    [tex][T(t)-75] \ = \ \frac{-110e^{-.4t}}{k}[/tex]

    [tex]T(t) \ = \ \frac{-110e^{-.4t}}{k}+75[/tex]

    [tex]T(0) \ = \ 350 \ = \ \frac{-110}{k}+75, \ k \ = \ \frac{-110}{275}[/tex]

    [tex]Hence, \ T(t) \ = \ \frac{-110e^{-.4t}}{\frac{-110}{275}} \ +75 \ = \ 275e^{-.4t}+75[/tex]

    [tex]Therefore, \ T(5) \ = \ about \ 112^{o} \ F.[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

  4. #4
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    Re: Newton's Law of Cooling?

    [tex]chrisr's \ solution \ is \ better \ than \ mine. \ In \ chrisr's \ solution, \ given \ the \ rate \ of \ change, \ the[/tex]

    [tex]temperature \ of \ the \ room \ (75^{o} F) \ is \ a \ red \ herring, \ as \ it \ isn't \ necessary \ (the \ temperature \ of[/tex]

    [tex]\ the \ room) \ to \ find \ the \ temperature \ of \ the \ pizza \ 5 \ minutes \ after \ it \ was \ taken \ out \ of \ the \ oven.[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

  5. #5
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    Re: Newton's Law of Cooling?

    Note: The above problem could have been written this way, to wit:

    A pizza, heated to a temperature of 350 degrees Fahrenheit, is taken out of an oven and placed in a room at time t=0 minutes. The temperature of the pizza is changing at a rate of -110e^-0.4t degrees Fahrenheit per minute. To the nearest degree, (a)what is the temperature of the pizza at time t=5 minutes and (b)what is the temperature of the room (assume the room is at a constant temperature)?

    [tex]\frac{dT(t)}{dt} \ = \ -110e^{-.4t} \ \implies \ T(t) \ = \ 275e^{-.4t}+C[/tex]

    [tex]T(0) \ = \ 350 \ = \ 275+C, \ C \ = \ 75[/tex]

    [tex]a) \ Hence, \ T(t) \ = \ 275e^{-.4t}+75, \ T(5) \ = \ 112^{o}(about).[/tex]

    [tex]b) \ \lim_{t\to\infty}[275e^{-.4t}+75] \ = \ 75, \ equals \ the \ temperature \ of \ the \ room, \ Why?[/tex]
    I am not, therefore I do not think. Contrapositive of Descartes' quip.

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