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Asked in AIEEE 2009 · Temperature profile along a bar
Idea: in the steady state no heat piles up anywhere, so the same current (dQ)/(dt) crosses every cross-section of the bar.
With the sides insulated, (dQ)/(dt)=-KA(dθ)/(dx) with K, A and (dQ)/(dt) all constant.
So (dθ)/(dx) is a constant, and θ falls along a straight line from the hot end to the cold end.
θ(x)=θₕₒₜ-((θₕₒₜ-θ_cold)/ℓ)x.
If instead the bar were left bare, heat would leak from the sides, the current would fall as x grew, and the graph would bend into the exponential θ-θ₀=(θ₁-θ₀)e^-mx.
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