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SolutionGiven data\begin{aligned}k & =400 \mathrm{w} / \mathrm{m} \cdot \mathrm{k} ; \alpha=10^{-4} \mathrm{~m}^{2} / \mathrm{s} \\h & =150 \mathrm{w} / \mathrm{m}^{2} \cdot \mathrm{k} ; \quad T_{\infty}=300^{\circ} \mathrm{C} ; T_{i}=25^{\circ} \mathrm{C} \\t=t_{m} & =400 \mathrm{sec} .\end{aligned}The termal response of transient conduction in a semiinfinite solid is\begin{aligned}\frac{T_{(x, t)}-T_{i}}{T_{\infty}-T_{i}}=\operatorname{erfc}\left[\frac{x}{2(\alpha t)^{1 / 2}}\right]- & {\left[\exp \left[\frac{h x}{k}+\frac{h^{2} \alpha t}{k^{2}}\right]\right] } \\& *\left[\operatorname{erfc}\left[\frac{x}{2(\alpha t)^{1 / 2}}+\frac{h(\alpha t)^{1 / 2}}{k}\right]\right]\end{aligned}when \left.x=0 ; T^{T}(x, t)=T_{(0, t)}\right)=T_{s} and \operatorname{erfc}(0)=1.\begin{array}{l}\frac{T_{s}-T_{i}}{T_{\infty}-T_{i}}=1-\exp \left[\frac{h^{2} \alpha t}{k^{2}}\right] \operatorname{ertc}\left[\frac{h(\alpha t)^{1 / 2}}{k}\right] \\\frac{T_{s}-T_{i}}{T_{\infty}-T_{j}}=1-\exp \left[\frac{(150)^{2}\left(10^{-4}\right)(400)}{(400)^{2}}\right] \operatorname{erfc}\left[\frac{150\left(10^{-4} \times 400\right)^{1 / 2}}{400}\right] \\\frac{T_{s}-25}{300-25}=1-[\exp (0.005625) * \operatorname{erfc}(0.075)] \\T_{s}-25=275[1-(1.00564)(0.9155)]\end{array}\begin{aligned}& { }^{s}-25=21.817 \\\Rightarrow & T_{s}=21.817+25 \\& T_{s}=46.817^{\circ} \mathrm{C}\end{aligned}\therefore The melting point of Coating is T_{S}=46.87^{\circ} \mathrm{C} ...