This is the heat transfer rate, $Q$. We are also given the temperature difference, $\Delta T = T_{\text{inside}} - T_{\text{outside}} = 75^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 55^{\circ} \mathrm{C}$, the total surface area, $A = 3 \mathrm{m}^{2}$, and the
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