Answer
a) increase
b) $35.2J$
Work Step by Step
(a) We know that the heat flow in one second will increase because the thermal conductivity of copper is greater than that of lead.
(b) We can find the required heat as follows:
$Q_c=K_A(\frac{\Delta T}{L})t$
We plug in the known values to obtain:
$Q_c=(395)(0.0150)^2(\frac{104}{0.525})(1.00)$
$Q_c=17.6J$
Now the total heat is given as
$Q_{total}=Q_c+Q_c=2Q_c$
$Q_{total}=2(17.6J)=35.2J$
