Heat is generated uniformly at a rate of \(4.2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\) in a spherical ball \((k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of diameter \(24 \mathrm{~cm}\). The ball is exposed to iced-water at \(0^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperatures at the center and the surface of the ball.

We can use the following equation for steady-state heat transfer in a spherical coordinate system: $$\frac{d}{dr}\left(r^2 \frac{d T(r)}{dr}\right) = -\frac{q_{\text{gen}}}{k} r^2$$ First, we calculate \(r^2 \frac{dT}{dr}\): $$\int \frac{d}{dr}\left(r^2 \frac{d T(r)}{dr}\right) = \int -\frac{q_{\text{gen}}}{k}r^2 dr$$ $$r^2 \frac{dT}{dr} = -\frac{q_{\text{gen}}}{3k}r^3 + C_1$$ Now, we can find \(\frac{dT}{dr}\): $$\frac{dT}{dr} = -\frac{q_{\text{gen}}}{3k}r + \frac{C_1}{r^2}$$ Integrating this expression will give us the temperature distribution: $$\int dT(r) = \int \left(-\frac{q_{\text{gen}}}{3k}r + \frac{C_1}{r^2}\right) dr$$ $$T(r) = -\frac{q_{\text{gen}}}{6k}r^2 + C_1 \ln{r} + C_2$$

To find the constants \(C_1\) and \(C_2\), we need to apply the boundary conditions of the problem: 1. At the surface of the ball \((r=R)\), the temperature has to satisfy the convective boundary condition: $$h(T_s - T_{\infty}) = k \left. \frac{dT}{dr} \right|_{r=R}$$ 2. The temperature must be finite at the center \((r=0)\), which means that \(C_1\) must be zero to prevent a singularity: $$C_1 = 0$$ The temperature distribution becomes simplified: $$T(r) = -\frac{q_{\text{gen}}}{6k}r^2 + C_2$$ Now we apply the first boundary condition: $$1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}(T_s) = 45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot \frac{d T}{d r} \Big|_{r = R}$$ Calculate the temperature gradient at the surface: $$\frac{dT}{dr} = -\frac{q_{\text{gen}}}{3k}r = -\frac{4.2 \times 10^6 \mathrm{~W} / \mathrm{m}^{3}}{3 \times 45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} \cdot 0.12 \mathrm{~m} = -28 \mathrm{~K} / \mathrm{m}$$ Now we can find \(T_s\): $$T_s = T_{\infty} + \frac{k}{h} \left(\frac{dT}{dr}\right)_{r=R} = 0 + \frac{45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}{1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}} (-28 \mathrm{~K} / \mathrm{m}) = -1.05^{\circ}\mathrm{C}$$ Finally, we can find the temperature at the center of the ball: $$T(0) = -\frac{q_{\text{gen}}}{6k}(0)^2 + C_2 = -1.05^{\circ}\mathrm{C}$$ This leads us to conclude that: 1. The temperature at the surface of the ball is \(T_s = -1.05^{\circ}\mathrm{C}\). 2. The temperature at the center of the ball is \(T_c = -1.05^{\circ}\mathrm{C}\).