as we know
Q = k A d T / dx
for heat flow rate, constant area and temp differences
k is inversely proportional to dx
dx2 / dx1 = k 1/k2
dx2 = 14.8 * ( 34.3 / 217)
dx2 = 93.63 cm
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Constants 1 Periodic Table Two cyindrical metail nods with equal eross section- one lead, the lead...
Constants 1 Periodic Table Two oylindrical metal nods with equal cross section one lead, the lead end of the tods is 230 C,the temperabure al the aluminum Part A Oventhat the temperatro at the leadainium interface is 53.5-c. and that the lead rod is 14.8㎝log.dthelength ofthe aannum rod end is 84.0 C Note the thermal conductivites of lead and aluminum are 343 Witm-K) and 217 Wim-K),nespectively Submit ( Retun to Assigement Provide Feedbadk