To solve this problem, we will use the principles of steady$-$state heat transfer in a solid sphere with uniform heat generation$\mathrm{.}1.$ Rate of Heat Transfer from the Sphere to the Surroundings $($q$)$:The heat transfer through the surface of the sphere can be expressed using Fourier's law of heat conduction. The heat transfer rate $($q$)$ can be calculated using the formula:q $=-$k $*$ A $*($dT$/$dr$)$Where:$-$ k is the thermal conductivity $(400$ J$/$s$.$m$.\backslash$deg C$)-$ A is the surface area of the sphere $(4**$ R$^2)-$ dT$/$dr is the temperature gradient.Since we are at steady state, we can simplify this to:q $=$ k $*$ A $*($T$2-$ T$1)/($r$2-$ r$1)$To find the surface area $($A$)$:A $=4**$ R$^2=4**(0\mathrm{.}05)^2=0\mathrm{.}0314$ m$^2($approximately$)$Now$,$ we need to calculate the temperature gradient. We can assume that the temperature difference is between T$2$ and T$1,$ where T$1$ is the temperature at the outer radius $(0\mathrm{.}04$ m$)$ and T$2$ is the temperature at the inner radius $(0\mathrm{.}02$ m$).$Using the given values:$-$ T$1=100\backslash$deg C$-$ T$2=110\backslash$deg C$-$ r$1=0\mathrm{.}04$ m$-$ r$2=0\mathrm{.}02$ mNow we can substitute these values into the equation for q$.$q $=400*0\mathrm{.}0314*(110-100)/(0\mathrm{.}02-0\mathrm{.}04)$Calculating this gives us:q $=400*0\mathrm{.}0314*10/(-0\mathrm{.}02)$Since we are looking for the heat transfer rate, we need to take the absolute value:q $=400*0\mathrm{.}0314*10/0\mathrm{.}02=6280$ J$/$s$2.$ Temperature at the Center of the Sphere $($T$\_$center$)$:In a sphere with uniform heat generation, the temperature at the center can be determined by using the formula for the temperature distribution in a sphere:T$($r$)=$ T$1+($T$2-$ T$1)*($r$/$R$)+($B$/(6*$k$))*($R$^2-$ r$^2)$At the center $($r $=0),$ we have:T$\_$center $=$ T$1+($T$2-$ T$1)*(0/$R$)+($B$/(6*$k$))*($R$^2)$Since we don't have B $($the rate of heat generation$),$ we cannot calculate T$\_$center directly without additional information. However, we will note that T$\_$center will be influenced by B$\mathrm{.}3.$ Neglecting Heat Generated within the Sphere:If we neglect the heat generated within the sphere $($meaning B $=0),$ the temperature distribution would be linear between T$1$ and T$2.$ In this case, the temperature at the center would be the average of T$1$ and T$2$:T$\_$center $=($T$1+$ T$2)/2=(100+110)/2=105\backslash$deg CIn summary:a$)$ The rate of heat transfer from the sphere to the surroundings is approximately $6280$ J$/$s$.$b$)$ The temperature at the center of the sphere, neglecting heat generation, would be $105\backslash$deg C$.$c$)$ If we neglect the heat generated within the sphere, the temperature at the center would be $105\backslash$deg C$.$