Newton's Cooling Law: q = h*a \Delta T

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    Cooling Law
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Discussion Overview

The discussion revolves around Newton's cooling law, specifically the equation q = h*a ΔT, and seeks to clarify the meaning of the variable q in terms of energy loss over time. Participants explore the implications of units and the generality of the law in different contexts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions the time frame associated with the energy loss rate q, asking if a value of q = 3 implies a loss of 3 watts in one second.
  • Another participant responds that the time unit can be consistent with the chosen system of units, indicating that if SI units are used, time would be in seconds.
  • A third participant introduces a differential form of the law, suggesting it is more general and clarifies that the amount of heat transferred per time unit is in joules, not watts, while also noting a missing minus sign in the equation.
  • One participant provides a practical example of Newton's law of cooling involving milk and coffee to illustrate the concept.
  • A later reply acknowledges the example as an efficient application of the law.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of q and its units, with some agreeing on the use of SI units while others introduce alternative formulations. The discussion remains unresolved regarding the implications of the differential form and the correct interpretation of energy loss.

Contextual Notes

There are limitations regarding the assumptions made about units and the conditions under which the equations apply. The discussion also highlights potential misunderstandings about the relationship between watts and joules in the context of energy transfer.

Ry122
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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?
 
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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.
 
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:
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.
 
That's an efficient way of applying the Newton's law of cooling :)
 

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