A rod of length L and uniform cross-sectional | vedclass

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(C) In steady state, the rate of heat flow $H = -kA(dT/dx)$ is constant throughout the rod.Given $k(x) = 2K - (K/L)x = K(2 - x/L)$.Thus, $H = -K(2 - x

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(C) In steady state, the rate of heat flow $H = -kA(dT/dx)$ is constant throughout the rod.Given $k(x) = 2K - (K/L)x = K(2 - x/L)$.Thus, $H = -K(2 - x/L)A(dT/dx) = 	ext{constant} = C$.Rearranging, $dT = -[C / (KA(2 - x/L))] dx$.Integrating both sides, $T(x) = -[C / (KA)] \int [1 / (2 - x/L)] dx$.$T(x) = [CL / (KA)] \ln(2 - x/L) + C_0$.Since $T(x)$ involves a logarithmic function, the slope $dT/dx = -C / [KA(2 - x/L)]$ is not constant.As $x$ increases from $0$ to $L$, the term $(2 - x/L)$ decreases from $2$ to $1$.Therefore, the magnitude of the slope $|dT/dx| = C / [KA(2 - x/L)]$ increases as $x$ increases.$A$ curve where the magnitude of the slope increases as $x$ increases is concave up. Thus, the graph is concave up.

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