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(B) The entropy change for a system or body is given by $\Delta S = \frac{Q}{T}$.For the system to be in a steady state,the heat lost by the reservoir

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(B) The entropy change for a system or body is given by $\Delta S = \frac{Q}{T}$.For the system to be in a steady state,the heat lost by the reservoir at temperature $T_{1}$ must equal the heat gained by the reservoir at temperature $T_{2}$ (let this be $Q$).The total rate of change of entropy for the conduction process is the sum of the entropy changes of the two reservoirs:$\frac{dS}{dt} = \frac{d}{dt} \left( \frac{-Q}{T_{1}} + \frac{Q}{T_{2}} \right) = \left( \frac{dQ}{dt} \right) \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)$From Fourier's law of heat conduction,the rate of heat flow is $\frac{dQ}{dt} = kA \frac{(T_{1} - T_{2})}{L}$,where $k$ is thermal conductivity,$A$ is the cross-sectional area,and $L$ is the length.Substituting this into the entropy rate equation:$\frac{dS}{dt} = \left( kA \frac{(T_{1} - T_{2})}{L} \right) \left( \frac{T_{1} - T_{2}}{T_{1} T_{2}} \right) = \frac{kA}{L} \frac{(T_{1} - T_{2})^2}{T_{1} T_{2}}$We can rewrite this in terms of the ratio $x = T_{1} / T_{2}$:$\frac{dS}{dt} = \frac{kA}{L} \frac{T_{2}^2 (x - 1)^2}{T_{2}^2 x} = \frac{kA}{L} \frac{(x - 1)^2}{x} = \frac{kA}{L} \left( x - 2 + \frac{1}{x} \right)$When $x = 1$ $(T_{1} = T_{2})$,$\frac{dS}{dt} = 0$. For $x > 1$ or $x  0$. The function $f(x) = \frac{(x-1)^2}{x}$ is non-negative and symmetric in its behavior relative to the minimum at $x=1$. Plot $(b)$ correctly represents this relationship.

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