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A heat source at $T = 10^{3} K$ is connected to another heat reservoir at $T = 10^{2} K$ by a copper slab which is 2 m thick. Given that the thermal conductivity of copper is $0.1 W K^{- 1} m^{- 1}$ , the energy flux through it in the steady state is

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$90 W m^{- 2}$

$45 W m^{- 2}$

$65 W m^{- 2}$

$200 W m^{- 2}$

answer is B.

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Detailed Solution

$(\frac{d Q}{d t}) = \frac{k A Δ T}{l} E n e r g y f l u x, \frac{1}{A} (\frac{d Q}{d t}) = \frac{k Δ T}{l} = \frac{(0.1) (900)}{2} = 45 W / m^{2}$

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