Answer
a) $d^{\prime}=1.183cm$
b) $T=-48^{\circ}C$
Work Step by Step
(a) We can find the final diameter as follows:
$\Delta d=d^{\prime}-d=\alpha d\Delta T$
This can be rearranged as:
$d^{\prime}=d+\alpha d\Delta T$
$\implies d^{\prime}=d(1+\alpha \Delta T)$
We plug in the known values to obtain:
$d^{\prime}=1.178[1+(24\times 10^{-6})(199-23)]=1.183cm$
(b) We can find the required temperature as follows:
$\Delta T=T-T_{\circ}=\frac{d^{\prime}-d}{\alpha d}$
This can be rearranged as:
$T=T_{\circ}+\frac{d^{\prime}-d}{\alpha d}$
We plug in the known values to obtain:
$T=23+\frac{1.176-1.178}{(24\times 10^{-6})(1.178)}$
$T=-48^{\circ}C$
