A solid sphere has a temperature of 568 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins? The image depicts two objects radiating energy: a sphere and a cube. 1. **Sphere**: - The sphere is shown with arrows emanating outward, representing radiated energy. - It has a radius labeled as \( R \). - The sphere's surface temperature is denoted by \( T_s \).2. **Cube**: - The cube similarly has arrows radiating outward, indicating energy emission. - It has an edge length labeled as \( L \). - The surface temperature of the cube is marked as \( T_c \).This illustration models how objects emit radiation, relevant in physics studies on thermal radiation and heat transfer. The arrows suggest heat flows equally in all directions from the surface of both geometrical shapes.

Explanation We will use Stefan-Boltzmann law to solve the problem. From Stefan-Boltzmann law we get the power of radiation is, P=eσAT4 where e is the emissivity of the material. A is the surface area of the object. T is the temperature of the object. σ is the Stefan's constant. Given Data Temperature of the sphere is Ts= 568 K Radius of the sphere is R Length of each side of the cube is LCalculation If the sphere is melted and recast into a cube then volume of the cube would be equal to the volume of the sphere. So, L3=43πR3 ⇒L=43π13×R Now Area of the cube is, Ac=6×L2 ⇒Ac=6×43π23×R2 Area of the sphere is As=4πR2 From Stefan-Boltzmann law radiant power of the sphere is Ps=eσAsTs4 Radiant power of the cube is Pc=eσAcTc4 As the radiant power of the cube and the sphere are equal we get, Pc=Ps ⇒eσAcTc4=eσAsTs4 ⇒AcTc4=AsTs4 Tc4=AsAc×Ts4 substituting the values of As and Ac we get, Tc4=4πR26×43π23×R2×Ts4 ⇒Tc4=4π6×43π23×Ts4 ⇒Tc4=4π15.59×Ts4 ⇒Tc4=0.806×Ts4 ⇒Tc=0.80614×Ts substituting the values of Ts we get, Tc=0.80614×568 K ⇒Tc=538 K Answer Temperature of the cube is 538 K