Talk:Heat conduction

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excuse me, but i'm a freshmen in high school and i have to write about newton's law of cooling. could anyone help me to understand it?...maybe i'm just stupid.....

You probably shouldn't try to get all your information here.. Wikipedia is very much a work in progress. Try learning from resources available at your local library. Using the internet as your sole source of information is a very bad idea.CSTAR 18:45, 2 Jan 2005 (UTC)

Heat is transfer of thermal energy. Isn't "transfer of heat" redundant, then?

This probably isn't the absolute best place to get an answer to this question, but where does "q" come from in these equations? There's no "q" in "heat!" I can't seem to find its origin.

I think it comes from quantity. (to distinguish from temperature)--njh 04:49, 8 March 2006 (UTC)

Contents

[edit] incorrect/vague equations

The equations on this page need to be double checked. The first equation presented originally had the heat equal to the gradient of the temperature -- contrary to what the text stated. I will temporarily change the entry so that \frac{\Delta Q_{i}}{\Delta t}\sim\frac{\Delta T}{\Delta x_{i}}, but the standard definitions of Fourier's law (which is essentially what the above equation is) has some weird notation with the time rate of change of the vector 'heat flux' \underline{q} being proportional to the gradient of the temperature: \frac{d\underline{q}}{dt}\sim\nabla T. The people over at Heat equation have it going-on, mathematically speaking -- this page should be more rigorous and use derivatives, etc.

[edit] equations

I'm pretty sure all the vector does is indicate and denote the direction of the quantity. Let me explain. Draw an arrow from the cold region to the hot region. The arrow points in the direction of the temperature gradient. Now, in which direction does the heat flow? That's right, in the reverse direction, from hot to cold. Draw this arrow for the conducted heat flow, and it will be parallel to the temperature gradient vector, but in the opposite direction. I think that's why there is a negative sign in the equation from the Heat_equation page, which says:

  • The Fourier law states that heat energy flow has the following linear dependence on the temperature gradient
\mathbf{H}(x) = -\mathbf{A}(x) \cdot \nabla u (x)

Note: They use a notation here that is common in textbooks, where bold symbols are vector quantities. Handwritten notation often uses a horizontal line or an arrow, since it's difficult to handwrite in bold. Mikiemike 18:50, 26 April 2006 (UTC)

[edit] More equations

I think it's redundant to say "proportional to a constant times something". At the top of the page where it says this:

\frac{\Delta Q}{\Delta t} \sim k A \frac{\Delta T}{\Delta x}

I think it should be this:

\frac{\Delta Q}{\Delta t} \sim A \frac{\Delta T}{\Delta x}

[edit] Headline text

no good information get more!!!!!!!!!!!!! —The preceding unsigned comment was added by 75.22.15.204 (talk) 15:58, 28 January 2007 (UTC). or,

\frac{\Delta Q}{\Delta t} = k A \frac{\Delta T}{\Delta x}

Mikiemike 18:50, 26 April 2006 (UTC)

Agreed. The third example is the one I'd use. Ojcit 19:59, 2 October 2006 (UTC)

[edit] Fourier's law

This article is a redirect here. Would it make sense to give it its own full article, with maybe a more rigorous mathematical treatment, and leave Heat conduction a more general overview? If Fourier's law doesn't apply to anything else, then the current setup is appropriate. Ojcit 18:37, 19 September 2006 (UTC)

Same question with Newton's law of cooling. It must be less popular than his other three, but arguably each of these laws could support a separate article, rather than being subsections of this one. Ojcit 18:40, 19 September 2006 (UTC)

[edit] figure

I think the first figure is misleading. The ΔT line is perpendicular to the wire, which isn't really what we want to show. Likewise, the hypotenuse is unlabled and its meaning is not clear. I guess the fact that Q is the slope of the line T vs. x (i.e. dt/dx) is what the illustrator was trying to get at. If the reader has no experience with derivatives, the picture in its present state doesn't help. Ojcit 19:57, 2 October 2006 (UTC)

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