A) Use the following formula for the linear expansion, in this case, of the diameter of the spherical ball:
[tex]d=d_o\lbrack1+\alpha(T_2-T_1)\rbrack[/tex]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:
[tex]\Delta V=0.0324V_o[/tex]Now, use the following formula for the change in volume with the temperature:
[tex]\Delta V=\beta V_o\Delta T=3\alpha V_o\Delta T[/tex]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:
[tex]\Delta T=\frac{\Delta V}{3\alpha V_o}=\frac{0.0324V_o}{3(24\cdot10^{-6}(\degree C)^{-1})V}=450\degree C[/tex]Hence, the change in temperature is 450°C
