# Newton's cooling law

For newton's cooling law
$$q = h*a \Delta T$$

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
Gold Member
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
$$\partial{Q}/\partial{t} = -k{\oint}\nabla{T}\vec{dS}$$

$$\partial{Q}/\partial{t}$$ 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

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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.

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