A rod of length 20 cm has one end placed inside a furnace at 800 K . The sides of the rod are covered by an insulating material, whereas the other end behaves as a blackbody and emits radiation. During the steady state

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Quantrex Academy · Accepted Answer

Correct answer: D. 74 W m^ -1 K^ -1

In steady state, the rate of heat conduction through the rod is equal to the net rate of heat radiated by the exposed end.

\dfrac{KA(T_1 - T_2)}{L} = \sigma A (T_2^4 - T_0^4)

K = \dfrac{\sigma L (T_2^4 - T_0^4)}{T_1 - T_2}

Substituting the given values:

\sigma = 6.0 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}

L = 20 \text{ cm} = 0.2 \text{ m}

T_1 = 800 \text{ K}

T_2 = 750 \text{ K}

T_0 = 300 \text{ K}

K = \dfrac{6.0 \times 10^{-8} \times 0.2 \times (750^4 - 300^4)}{800 - 750}

K = \dfrac{1.2 \times 10^{-8} \times (3164.0625 \times 10^8 - 81 \times 10^8)}{50}

K = \dfrac{1.2 \times 3083.0625}{50}

K = \dfrac{3699.675}{50} = 73.9935 \text{ W m}^{-1} \text{ K}^{-1}

Rounding to the nearest integer, we get

K \approx 74 \text{ W m}^{-1} \text{ K}^{-1}

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A rod of length 20 cm has one end placed inside a furnace at 800 K...

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