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GR9677 #75
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Thermodynamics}Fourier's Law

Recall Fourier's Law , where is the heat flux vector (rate of heat flowing through a unit area) and is the temperature and is the thermal conductivity. (One can also derive it from dimensional analysis, knowing that the energy flux has dimensions of )

Fourier's Law implies the following simplification:

The problem wants the ratio of heat flows , as in choice (D). (The problem gives , , and , .)  Alternate Solutions
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lrichey
2011-11-05 12:08:50
maybe I'm just stupid... but could someone explain why I can keep .8 in terms of meters and 4 in terms of mm?
 titanqq2012-10-12 05:12:23 you get the variables k_A/B and k_B/A to divide one by one. This means, it doesn't matter in what terms you take it, they just must be taken both in 1 terms redmomatt
2011-09-24 06:52:22
The problem can be solved without knowing Fourier's Law specifically.

Since we are given conductivities in watts/meter C, the ratios will be independent of these quantities. Given that the only variable then is the thickness of the pane, we take the limiting case to see how heat flow is affected; a monumentally thicker piece of glass will have a lower heat flow through it, given the same conditions. So, we can see that heat flow is inversely proportional to thickness.

Thus, we can simply divide the conductivity by the thickness, and then divide A by B, obtaining (.8/4)/(.025/2) = .2/.0125 = 80/5 = 16 . walczyk
2011-04-03 06:18:57
why is it that the temperature differentials are equal? I'm not really sure what it signifies. Temperature per length... shouldn't the temperature be larger per distance for one of the walls/windows?
 walczyk2012-09-29 16:30:28 I see, delta T is T_outdoor - T_indoor, which is the assumed the same for each case. carle257
2010-04-05 00:26:17
You can also think of it like a resistance problem. A is 32 times more conductive to heat, but double the length. Thus the thermal resistance is proportional to conductivity*area/length = 32/2 = 16. jw111
2008-11-05 15:07:59
Fourier's Law is very similar to V=IR

for
V ~ temperature difference
I ~ heat flow
R ~ heat resistance = inverse of conductivity

here ra = 1/0.8 and rb = 1/ 0.025, thus

Ra = ra*4, Rb = rb*2

then Ia/Ib = Rb/Ra = 16. bat_pesso
2007-10-31 14:02:21
dimensional analyses:

heat flow A = 0.8/4mm

heat flow B = 0.025/2mm

A/B = (0.8x2)/(0.025x4) = 16/1
 FortranMan2008-09-02 17:10:17 provided you remember how area is involved in Fourier's law, otherwise you might get alarmed with meters^2 popping up      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$