GREPhysics.NET: GR9677 #75

GR9677 #75

Problem

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Thermodynamics $\Rightarrow$ }Fourier's Law

Recall Fourier's Law $q = -k\nabla T$ , where $q$ is the heat flux vector (rate of heat flowing through a unit area) and $T$ is the temperature and $k$ is the thermal conductivity. (One can also derive it from dimensional analysis, knowing that the energy flux has dimensions of $J/(s m^2)$ )

Fourier's Law implies the following simplification: $q = -k \frac{\Delta T}{\Delta l}$

The problem wants the ratio of heat flows $q_A/q_B=\frac{k_A l_B}{k_B l_A}=\frac{0.8 \times 2}{0.025 \times 4}=32/2=16$ , as in choice (D). (The problem gives $l_A = 4$ , $l_B=2$ , and $k_A=0.8$ , $k_B=0.025$ .)

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Comments

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.

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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.

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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

FortranMan
2008-09-02 17:10:17

provided you remember how area is involved in Fourier's law, otherwise you might get alarmed with meters^2 popping up

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