Answer
(a) $3.1W$
(b) $1.50KW$
Work Step by Step
(a) We know that
$I=\frac{V}{R_{eq}}$
$I=\frac{120V}{9.62\Omega}=12.474A$
Now the power dissipated in the power cord is given as
$P_C=I^2R_C$
We plug in the known values to obtain:
$P_C=(12.474A)^2(0.020\Omega)$
$P_C=3.1W$
(b) The power dissipated in the heating element is given as
$P_{he}=I^2R_{he}$
We plug in the know values to obtain:
$P_{he}=(12.474A)^2(9.6\Omega)$
$P_{he}=1.50KW$
