A) Use the following formula for the linear expansion, in this case, of the diameter of the spherical ball:
![d=d_o\lbrack1+\alpha(T_2-T_1)\rbrack](https://img.qammunity.org/2023/formulas/physics/college/5ophzb1211vw08j5h1a8ps5ew9hqr49o74.png)
where,
do: initial diameter = 11.21cm
d: final diameter = ?
α: linear expansion coefficient = 24*10^6 /°C
T2: final temperature = 257°C
T1: intial temperature = 33°C
Replace the previous values of the parameters into the formula for d and simplify:
Hence, the diameter of the spherical ball is 11.27cm
B) First, consider what is the change in the radius of the sphere, as follow:
![\Delta V=0.0324V_o](https://img.qammunity.org/2023/formulas/physics/college/bg07ly95syu5panswynhk21hkuszckyymr.png)
Now, use the following formula for the change in volume with the temperature:
![\Delta V=\beta V_o\Delta T=3\alpha V_o\Delta T](https://img.qammunity.org/2023/formulas/physics/college/9g31iuxuf9pl9sev9rs1rld7mimg2ykxz9.png)
where you have used the equivalence β = 3α.
Solve the equation above for the change in temperature and replace the values of the other parameters:
![\Delta T=(\Delta V)/(3\alpha V_o)=(0.0324V_o)/(3(24\cdot10^(-6)(\degree C)^(-1))V)=450\degree C](https://img.qammunity.org/2023/formulas/physics/college/5l27pmom55zqmxj18h0bx1qt0zw0cu5rj6.png)
Hence, the change in temperature is 450°C