Rate of Heat Flow (EK 1001 Chem, #323)

[zer0crew]

Alright, so I'm struggling to understand question 323 from Exam Krackers 1001 Chemistry. The question asks;

Sorry I don't have a picture of the diagram, but it basically shows a hot and cold plate on either side, with 4 slabs of various sizes (height, width, depth) between them.

Through which slab is the rate of heat flow, Q/t, the greatest?
A.) slab 2
B.) slab 3
C.) slab 4
D.) The rate of heat flow is the same through all slabs.


Given the equation; Q/t = kA[(Th-Tc)/L]

I reasoned that to maximize Q/t you would want the a large A value and a small L value (given that k, Th, and Tc are all fixed in the problem). Apparently that's not the case though. I calculated slab 2 to have the largest A/L ratio, therefore reasoning that it would have the best rate of heat transfer.

The back of the book says "Like fluid flow or electronic current, the rate of heat flow through a slab is constant everywhere." which didn't help me very much. I'm sure I'm missing some key concept here. This answer makes it sound like the above equation is completely irrelevant if the rate of heat flow is the same through all slab, regardless of cross sectional area and length.

Any thoughts? or better explanations?

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[fizzgig]

don't have the problem, but say you have 3 slabs, Thot and Tcold on far sides.

say the outer slabs are the same and the middle one differs. if the middle slab had more Q/t than the previous slab, it would be dumping more heat than it was taking in, and would cool off, eventually becoming cooler than the 3rd slab it's trying to dump heat to, and obviously a cooler object can't dump heat to the hotter one.

if the middle slab had less Q/t than the first, it would heat up, and then the first slab would be cooler than the middle. again, cooler slab can't dump heat to the hotter one.

relating this to circuits, if you have 3 resistors in series and one driving gradient (deltaV here vs deltaT), does each resistor end up with more or less current? no, because there is steady flow of electrons 'downhill'. the current is the same through all, and is determined by the OVERALL resistance to flow. if one resistor has a very low resistance to flow, you end up with a low voltage drop across that one, not having a different current value for it.

so for the slabs, the overall resistance to heat flow, Q/t, determines the rate of flow, and if you have a slab that conducts really well, you just get a low temperature drop across that slab.