Answer
(a) $0.47J/K$
(b) $0.40KJ$
(c) Please see the work below.
Work Step by Step
(a) We can calculate the total entropy change as follows:
$\Delta S_{total}=(\frac{Q}{T})_h+(\frac{Q}{T})_c$
We plug in the known values to obtain:
$\Delta S_{total}=\frac{-660J}{810K}+\frac{410J}{320K}$
$\Delta S_{total}=0.47J/K$
(b) The maximum work done by a reversible engine is given as
$W_{max}=(1-\frac{T_c}{T_h})Q_h$
We plug in the known values to obtain:
$W_{max}=(1-\frac{320K}{810K})(660J)$
$W_{max}=399.26J=0.40KJ$
(c) We know that the difference between the non-reversible and reversible engine is given as
$\Delta W=W_{max}-W$
$\implies \Delta W=(1-\frac{T_c}{T_h})Q_h-(Q_h-Q_c)$
$\Delta W=\frac{-T_c}{T_h}Q_h+Q_c$
$\Delta W=T_c(\frac{-Q_h}{T_h}+\frac{Q_c}{T_c})$
$\Delta W=T_c\Delta S_{total}$
