Answer
(a) $8.5KJ$
(b) $6.0KJ$
Work Step by Step
(a) We know that
$\eta=\frac{Work\space done}{Heat \space input}=1-\frac{T_{cold}}{T_{hot}}$
We plug in the known values to obtain:
$\frac{2500}{Heat\space input}=1-\frac{290}{410}$
This simplifies to:
$Heat \space provided=8541J=8.5KJ$
(b) As $Heat \space provided=Work\space done+Heat \space discarded$
$\implies Heat\space discarded=Heat\space provided-Work \space done$
We plug in the known values to obtain:
$Heat \space discarded=8541.16-2500=6041J=6.0KJ$
