Calculate the Biot Number to determine if the Lumped Parameter Analysis can be used. If the Biot Number is greater than 0.10, use the Temperature-Time charts in Appendix F to calculate an answer and compare the result to the answer based on lumped capacitance. appendix F in this textbook https://drive.google.com/file/d/145Cbs8i465qUq9JhBFhia58fnPkuhU3k/view?usp=sharing **Problem Statement: Cooling of Apples in a Freezer**A few apples are placed in a freezer set at \(-15^\circ C\) to quickly cool them for guests. Initially, the apples are uniformly at a temperature of \(20^\circ C\). The convective heat transfer coefficient is given as \(h = 8 \, \text{W/m}^2\cdot \text{K}\).**Apple Specifications:**- Consider the apples as spheres with a diameter of \(9.0 \, \text{cm}\).- Density (\(\rho\)) of each apple is \(840 \, \text{kg/m}^3\).- Heat capacity (\(C_p\)) is \(3.81 \, \text{kJ/kg}\cdot \text{K}\).- Thermal conductivity (\(k\)) is \(0.418 \, \text{W/m}\cdot \text{K}\).**Objective:**Determine the center and surface temperatures of the apples after 1 hour of cooling.This problem requires calculating how the temperature of the apples changes over time due to convection and conduction, focusing on both the surface and the center of the apples.

Given: Initial temperature of apples (To)= 20 oC Temperature of freezer (T∞) = -15 oC Convective heat transfer coefficient in freezer (h)= 8 Wm2K Diameter of each apple (D) = 0.09 m Diameter of each apple (𝞺)= 840 kgm3 Heat capacity (Cp) =3.81 kJkg.K Thermal conductivity (k)= 0.418 Wm.K It is required to find the center temperature (Tc) and surface temperature (Ts) of the apples after 1 hour (t=3600 s) Biot number is given as Bi=hx1kwhere x1= characteristic length= Volume of the sphereSurface area of the spherex1=π6*D3πD2⇒x1=D6=0.015 mBi=8 Wm2K*0.0960.418 Wm.K⇒Bi=0.287 >0.1 (lumped parameter analysis cannot be used) Thermal diffusivity α=k𝞺Cp=0.418 Wm.K840 kgm3*3.81 1000*Jkg.K⇒α=1.306*10-7 m2s Considering appendix F in text book for sphere m=khx1=0.418 Wm.K8 Wm2K*0.015 m⇒m=3.48333X=αtx12=1.306*10-7 m2s*3600 s(0.015 m)2⇒X=2.0896 From the above X and m values Yc=Tc-T∞To-T∞≃0.2Tc-(-15 oC)20 oC-(-15 oC)=0.2⇒Tc=-8 oC The center temperature of the apples after 1 hour is -8 °CAs there is no graph available for finding surface temperature and moreover, the thermal conductivity of apple is very less and heat transfer coefficient of air is large enough. So heat gets transferred easily from the surface but takes time at the center of apple Therefore at the surface of apple after 1 hour the temperature will be almost equal to the temperature of air Ts≈-15°C