Answer
a) heated
b) $860^{\circ}C$
Work Step by Step
(a) We know that aluminum expands and contracts at a greater rate as compared to steel, so the system should be heated.
(b) We can find the required temperature as follows:
$\Delta T=\frac{D_{0,st}-D_{0,Al}}{\alpha_{Al}D_{0,Al}-\alpha_{st}D_{0,st}}$
We plug in the known values to obtain:
$\Delta T=\frac{4.040-4}{(2.4\times 10^{-5})(4)-(1.2\times 10^{-5})(4.040)}$
$\Delta T=842^{\circ}C$
Now the final temperature is:
$T=T_{\circ}+\Delta T$
$T=22^{\circ}C+842^{\circ}C=860^{\circ}C$
