Use the convergence criterion of 10-8 K and make a table to display the iteration process
until the convergence tolerance is met.
 

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Part II (30 pt) Suppose that a hot stainless-steel spherical ball is suspended within a vacuum chamber maintained at a constant temperature of 300 K (Tsur= 300 K). Initially, the stainless-steel ball is at 800 K (T = 800 K), and its radius is 0.006 m (r=0.006 m). Although we neglect conduction and convection heat transfer, the ball's temperature (T) changes over time (t) due to radiation heat exchange between its surface and the inner surface of the chamber. Using the energy balance equation, the relationship between time (t, sec) and temperature (T, K) is derived, and it is expressed as the equation in the red box below. Energy balance: in-out + g = = dEst dt dT dt dT (=pcV. -) => -A(T -T) = pvc- dt dT =>t= pVc 4ATT T +T T +T. T In In + 2 tan T - T - tan Tsur sur SUP A, S dt = S EAST-T pVc sur (In: natural logarithm, tan: inverse function of tan, i.e., atan) Stainless- Steel Ball, T (0) Vacuum Chamber Stefan-Boltzmann constant = 5.6710-8 W/m-K+, Surface emissivity: &= 0.1 (Vacuum chamber) Temperature of the inner surface Tsur = 300 K, (Stainless Steel) Radius = 0.006 m; T-T(1-0)-800 K T (1=3600 sec)=? T-300 K- Surface area A, 4r; Volume V=4r/3; Initial temperature T-800 K; Density p=8000 kg/m; Specific heat capacity c = 480 J/kg-K (a) Plot the temperature of the spherical steel ball (7) and the elapsed time (t). (b) Determine the temperature of the spherical steel ball (7) after one hour of cooling (t = 3600 sec) using the 1) Bisection and 2) Secant methods with MATLAB. Part II (30 pt) Suppose that a hot stainless-steel spherical ball is suspended within a vacuum chamber maintained at a constant temperature of 300 K (Tsur= 300 K). Initially, the stainless-steel ball is at 800 K (T = 800 K), and its radius is 0.006 m (r=0.006 m). Although we neglect conduction and convection heat transfer, the ball's temperature (T) changes over time (t) due to radiation heat exchange between its surface and the inner surface of the chamber. Using the energy balance equation, the relationship between time (t, sec) and temperature (T, K) is derived, and it is expressed as the equation in the red box below. Energy balance: in-out + g = = dEst dt dT dt dT (=pcV. -) => -A(T -T) = pvc- dt dT =>t= pVc 4ATT T +T T +T. T In In + 2 tan T - T - tan Tsur sur SUP A, S dt = S EAST-T pVc sur (In: natural logarithm, tan: inverse function of tan, i.e., atan) Stainless- Steel Ball, T (0) Vacuum Chamber Stefan-Boltzmann constant = 5.6710-8 W/m-K+, Surface emissivity: &= 0.1 (Vacuum chamber) Temperature of the inner surface Tsur = 300 K, (Stainless Steel) Radius = 0.006 m; T-T(1-0)-800 K T (1=3600 sec)=? T-300 K- Surface area A, 4r; Volume V=4r/3; Initial temperature T-800 K; Density p=8000 kg/m; Specific heat capacity c = 480 J/kg-K (a) Plot the temperature of the spherical steel ball (7) and the elapsed time (t). (b) Determine the temperature of the spherical steel ball (7) after one hour of cooling (t = 3600 sec) using the 1) Bisection and 2) Secant methods with MATLAB.