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One end of a 20 cm rod is in a furnace at 800 K, with its sides insulated; the other end radiates as a blackbody and sits at 750 K in steady state, with surroundings at 300 K. Treating radiation as the only heat exchange at the open end, find the rod's thermal conductivity. Take Stefan's constant sigma = 6.0 x 10⁻⁸ W/m²-K⁴.

1. 74 W/m-K
2. 148 W/m-K
3. 37 W/m-K
4. 222 W/m-K

Correct answer: 74 W/m-K

Solution

K*(50)/0.20 = sigma*(750⁴ - 300⁴); solving gives K ~ 74 W/m-K.

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