- #1

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[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?

- Thread starter Ry122
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- #1

- 565

- 2

[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

f95toli

Science Advisor

Gold Member

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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

[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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- #5

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

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