Unformatted text preview: determined to be K W/m 24 . 34 2 K 350) + (500 ) K 10 (8.7 + 1 K) W/m 25 ( 2 1 ) ( 141 2 ave ave ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × ⋅ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + β + = = T T k T k k Then the rate of heat conduction through the plate becomes W 30,820 = − × ⋅ = − = m 15 . 0)K 35 (500 m) 0.6 m K)(1.5 W/m 24 . 34 ( 2 1 ave L T T A k Q & Discussion We would obtain the same result if we substituted the given k ( T ) relation into the second part of Eq, 276, and performed the indicated integration. 261...
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This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.
 Fall '10
 Dr.DanielArenas
 Thermodynamics, Conductivity, Mass, Heat

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