Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

Based on the given problem and solution, a short answer question can be created as follows: Question: Two blocks - a cubic block and a cylindrical block made of granite are exposed to hot gases at \(500^{\circ} \mathrm{C}\). The length of the cubic block and the height and diameter of the cylindrical block are \(5 \mathrm{~cm}\). Both blocks have an initial temperature of \(20^{\circ} \mathrm{C}\). Using the lumped capacitance method, determine the center temperature of each block after 10, 20, and 60 minutes. (Given: \(k=2.5\ \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), \(\alpha=1.15 \times 10^{-6}\ \mathrm{m}^{2} / \mathrm{s}\), heat transfer coefficient \(40\ \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), granite density \(\rho\approx 2700\ \mathrm{kg}/\mathrm{m}^3\), and specific heat \(c_p\approx 790\ \mathrm{J}/\mathrm{kg \cdot K}\).) Answer: *Before providing the final answer, make sure to follow the steps provided in the solution, calculate the Biot number for each block, and check if the lumped capacitance method can be used.* For the cubic block: - \(T_{cube}(10\ \mathrm{min}) = ...^{\circ} \mathrm{C}\) - \(T_{cube}(20\ \mathrm{min}) = ...^{\circ} \mathrm{C}\) - \(T_{cube}(60\ \mathrm{min}) = ...^{\circ} \mathrm{C}\) For the cylindrical block: - \(T_{cylinder}(10\ \mathrm{min}) = ...^{\circ} \mathrm{C}\) - \(T_{cylinder}(20\ \mathrm{min}) = ...^{\circ} \mathrm{C}\) - \(T_{cylinder}(60\ \mathrm{min}) = ...^{\circ} \mathrm{C}\)