Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

The rate of heat transfer from the ball surface to the environment can be calculated using the formula: \(q = hA \Delta T\) where: - \(q\) is the rate of heat transfer - \(h\) is the heat transfer coefficient, which is given as \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - \(A\) is the surface area of the ball, which can be calculated as \(A = 4 \pi r_{1}^{2}\), where \(r_{1}\) is the radius of the ball. Given the diameter of the ball is \(3\ \mathrm{cm}\), the radius is \(r_1 = 1.5\ \mathrm{cm} = 0.015\ \mathrm{m}\). Therefore, \(A = 4 \pi (0.015\ \mathrm{m})^2 = 2.827 \times 10^{-3}\ \mathrm{m}^{2}\). - \(\Delta T\) is the temperature difference between the ball surface and the ambient air. The ball surface temperature is constant, and the ambient air temperature is given as \(26^{\circ} \mathrm{C}\). Now, we can calculate the rate of heat transfer, \(q\): \(q = 7.5 \frac{\mathrm{W}}{ \mathrm{m}^{2} \cdot \mathrm{K}} \times 2.827 \times 10^{-3}\ \mathrm{m}^{2} \times \Delta T = 0.02120\ \mathrm{W}\cdot \mathrm{K}\cdot \Delta T\)

The heat transfer through the insulation material can be calculated using the Fourier's law of heat conduction: \(q = \frac{k \cdot A \cdot \Delta T}{t}\), where - \(q\) is the heat transfer - \(k\) is the thermal conductivity of the insulation material, which is given as \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - \(A\) is the surface area of the ball, which we have calculated in step 1 - \(t\) is the thickness of the insulation material - \(\Delta T\) is the temperature difference across the insulation layer Let's rewrite the equation from step 1 and substitute the value of \(q\): \(\frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot A \cdot \Delta T}{t} = 0.02120\mathrm{W}\cdot\mathrm{K} \cdot \Delta T\)

Now, we will find the optimal thickness, \(t\), that maximizes heat generation while keeping the surface temperature constant. To do this, we will solve the equation obtained in step 2 for \(t\): \(t = \frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot A \cdot \Delta T}{0.02120\mathrm{W}\cdot\mathrm{K} \cdot \Delta T}\) The temperature difference \(\Delta T\) gets canceled out in both the numerator and denominator: \(t = \frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot A}{0.02120\mathrm{W}\cdot\mathrm{K}}\) Now, we can calculate the optimal thickness, \(t\), by substituting the values of \(k\) and \(A\): \(t = \frac{0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 2.827 \times 10^{-3}\ \mathrm{m}^{2}}{0.02120\mathrm{W}\cdot\mathrm{K}} = 0.020\ \mathrm{m} = 2\ \mathrm{cm}\)

We have calculated the optimal thickness of the insulation material to be \(2\ \mathrm{cm}\). Comparing this value with the given options, we find that the correct answer is: (d) \(2.0\ \mathrm{cm}\)