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As we knowSolution.Conductance \sigma=- conductivity \times \frac{\Delta t}{\Delta x}=-k \frac{\Delta t}{\Delta x}Thus, the temperature ws a function of position and time asA(x, t)=\frac{d \sigma}{d x}=-k \frac{d}{d x} \frac{\Delta t}{\Delta x}=-k x-\frac{\Delta t}{\Delta x^{2}}=k \frac{\Delta t}{\Delta x^{2}}Thus for \Delta t=0.15-0=0.15 & \forall x=0.25 \quad \& k=1\lambda(x, t)=1 \times \frac{115}{125^{2}}=2.4000^{\circ} \mathrm{C}& For \Delta t=.025 \quad \Delta x=0.25 \quad \lambda(x, t)=1 x^{.025}\left(\frac{25}{125}=0.4000^{\circ} \mathrm{C}\right. ...