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Writing a differential equation to describe cooling

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Newton's Law of Cooling states that the temperature of an object changes at a rate proportional to the difference between the temperature of the object itself and the temperature of its environment.

    Suppose the ambient air temperature is 70*F and that the rate constant is 0.05min^-1.

    Write a differential equation for the temperature change the object undergoes.

    2. Relevant equations



    3. The attempt at a solution

    Just starting with DE in preparation for fall.

    I just wrote:

    let q be heat, t be time in minutes.

    (dq/dt) = 0.05(q-70)

    ...

    Is that really it?

    And solving this differential equation, would mean finding a function, q(t), such that its derivative is equal to itself, minus 70, times 0.05, for any value of t?

    Does that mean that modeling and solving differential equations is mainly a method of finding an equation that models a situation by examining the behavior of its rate of change?
     
  2. jcsd
  3. Jun 24, 2012 #2

    Dick

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    Yes, it is that easy, if your 'heat' variable q means temperature.
     
  4. Jun 25, 2012 #3

    HallsofIvy

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    Actually, it is NOT quite that easy! If the initial temperature, q, is greater than the ambient temperature, 70, then the object will cool! That is, q will decrease and dq/dt is negative. If the initial temperature, q, is lower than the ambient temperature, 70, then the object will warm! That is, q will increase and dq/dt is positive.

    dq/dt= 0.05(70- q) or dq/dt= -0.05(q- 70)
     
  5. Jun 25, 2012 #4
    Since it introduced it as the "law of cooling" my instinct was to write a de for the rate of cooling, not heating.
     
  6. Jun 25, 2012 #5

    Dick

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    Halls is right. I missed the sign. You have to think about the physics of the situation to get the right sign. If u is greater than 70 it should be cooling, if u is less the 70 it should be heating.
     
  7. Jun 25, 2012 #6
    I understand what you are saying. I mean that I "chose" a drop in temperature to be positive. I see why this may be considered awkward.
     
  8. Jun 25, 2012 #7

    Dick

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    It's worse than awkward. It's wrong. I was sloppy in overlooking it. Sorry.
     
  9. Jun 25, 2012 #8
    Still easy. :P
     
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