MgO ceramic has heat capacity \( 940 \mathrm{~J} / \mathrm{kg} . \mathrm{K} \), coefficient of thermal expansion \( 13.5 \times 10^{-6} \) per \( { }^{\circ} \mathrm{C} \), thermal conductivity 37.7 W/m.K and density of \( 3.58 \mathrm{~g} / \mathrm{cm}^{3} \). Then the thermal diffusivity for MgO is \( \mathrm{cm}^{2} / \mathrm{s} \) (round off to two decimal places).

To calculate the thermal diffusivity (\( \alpha \)) of MgO, you can use the formula: \[ \alpha = \frac{k}{\rho c} \] where: - \( k \) is the thermal conductivity (37.7 W/m·K), - \( \rho \) is the density (3.58 g/cm³ or 3580 kg/m³), - \( c \) is the heat capacity (940 J/kg·K). First, convert the density to kg/m³: \[ 3.58 \, \mathrm{g/cm}^3 = 3580 \, \mathrm{kg/m}^3 \] Now substitute the values into the formula: \[ \alpha = \frac{37.7 \, \mathrm{W/m \cdot K}}{3580 \, \mathrm{kg/m}^3 \cdot 940 \, \mathrm{J/kg \cdot K}} \] This becomes: \[ \alpha = \frac{37.7}{3580 \times 940} \] \[ \alpha = \frac{37.7}{3369200} \] \[ \alpha \approx 1.12 \times 10^{-5} \, \mathrm{m}^2/\mathrm{s} \] Now, convert it to cm²/s: \[ \alpha \approx 1.12 \times 10^{-5} \, \mathrm{m}^2/\mathrm{s} = 1.12 \, \mathrm{cm}^2/\mathrm{s} \] So, the thermal diffusivity for MgO is approximately \( 1.12 \, \mathrm{cm}^2/\mathrm{s} \).