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

[Ry122]

For Newton's cooling law
<br /> q = h*a \Delta T<br />

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?

 

[f95toli]

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.

 

[armis]

The differential form is more general
\partial{Q}/\partial{t} = -k{\oint}\nabla{T}\vec{dS}

\partial{Q}/\partial{t} 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

 

[ironhill]

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.

 

[armis]

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