What is the heat flux (in W/m²) through the slab of 10cm thick if temperature drop across the slab is 40°C and the thermal conductivity of slab is 35 W/m.°C.?
ϕq=−kdT(x)dxϕq=−35∗4010100ϕq=−14000W/m²ϕq=14000W/m²\phi _q=-k\frac{dT(x)}{dx}\\ \phi _q=-35*\frac{40}{\frac{10}{100}}\\ \phi _q=-14000W/m²\\ \phi _q=14000W/m²ϕq=−kdxdT(x)ϕq=−35∗1001040ϕq=−14000W/m²ϕq=14000W/m²
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