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Newton's cooling law

  1. Jun 11, 2008 #1
    For newton's cooling law
    [tex]
    q = h*a \Delta T
    [/tex]

    q is the rate of energy loss of a body but for what unit time?
    For example if q = 3 does the body lose 3 watts of energy in 1 second?
     
  2. jcsd
  3. Jun 11, 2008 #2

    f95toli

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    Whatever units you want as long as you are consistent (i.e mixing imperial and SI is a bad idea).
    So yes, assuming you are using SI for the constant and the variables the time will be in seconds.
     
  4. Jun 11, 2008 #3
    The differential form is more general
    [tex] \partial{Q}/\partial{t} = -k{\oint}\nabla{T}\vec{dS} [/tex]

    [tex] \partial{Q}/\partial{t}[/tex] is the amount of heat transferred per time unit as long as you are using SI. [W] or [J*s^-1]. So it's J that are transferred in one second not W
    And you have a minus missing

    I may be wrong, feel free to correct me
     
    Last edited: Jun 11, 2008
  5. Jun 11, 2008 #4
    Newton's law of cooling: If you put milk in your coffee then leave it for a minute it will be warmer than if you leave it for a minute then add milk.
     
  6. Jun 11, 2008 #5
    That's an efficient way of applying the Newton's law of cooling :)
     
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