A solid metallic sphere has an initial volume of 36 m ^3 . It is heated uniformly such that its temperature increases by 50^ C . If the coefficient of volume expansion of the metal is 6.0 10⁻⁵ , ^ C ⁻¹ , the increase in the surface area of the sphere is:

Step-by-step solution

Let the initial radius of the sphere be R . The initial volume of the sphere is given by: V₀ = 4 3 R^3 Given V₀ = 36 m ^3 , we have: 4 3 R^3 = 36 R^3 = 27 R = 3 m The initial surface area of the sphere is: A₀ = 4 R^2 = 4 (3)^2 = 36 m ^2 The coefficient of areal expansion is related to the coefficient of volume expansion by: = 2 3 = 2 3 6.0 10⁻⁵ = 4.0 10⁻⁵ , ^ C ⁻¹ The increase in surface area A is: A = A₀ T A = (36 ) (4.0 10⁻⁵) 50 A = 36 200 10⁻⁵ = 0.072 m ^2 Converting to cm ^2 ( 1 m ^2 = 10^4 cm ^2 ): A = 0.072 1