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A heat source atT = 500 k is connected to another heat reservoir att = 100 kby a copper slab which is 1 m long. Given that the thermal conductivity of copper is 0.1 Wk^(-1) m^(-1) ,the energy flux through it in the steady state is: |
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Answer» `4000 Wm^(-2)`  ` (dQ)/(DT) = (KA)/(I) (500 - 100) ` Heat/time ` = (0.1A )/(1) xx 400 xx 40A` energy/flux ` = ("heat current")/(A) = 40` |
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