Answer
(a) $13\%$
(b) $320J$
Work Step by Step
(a) We can determine the required efficiency as follows:
$W_{out}=\frac{1}{2}(300-100)\times 10^3(600-200)\times 10^{-6}=40J$
and $Q_c=100+180=280J$
Now $\eta=\frac{W_{out}}{W_{out}+Q_c}$
We plug in the known values to obtain:
$\eta=\frac{40}{40+280}=13\%$
(b) We can determine the required heat as follows:
$Q_c=W_{out}+Q_c$
We plug in the known values to obtain:
$Q_c=40+280=320J$