Newton's cooling law

  • Thread starter Ry122
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #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.
 
  • #3
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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:
  • #4
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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.
 
  • #5
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That's an efficient way of applying the Newton's law of cooling :)
 

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