Answer
(a) greater than
(b) $98C^{\circ}$
Work Step by Step
(a) We know that the thermal conductivity of lead is less than the thermal conductivity of copper; hence the temperature at the lead-copper interference should be greater than $54.0C^{\circ}$.
(b) We know that
$\frac{Q}{t}=kr^2(\frac{\Delta _c}{L})$
We plug in the known values to obtain:
$1.41J/s=(395W/m.K^2)(1.50\times 10^{-2})^2(\frac{\Delta T_c}{0.525m})$
$\implies \Delta_c=8.3^{\circ}$
Now $T_{c,1}=T_{c,\circ}-\Delta T_c$
We plug in the known values to obtain:
$T_{c,1}=(106-8.3)C^{\circ}$
$\implies T_{c,1}=98C^{\circ}$
