Answer
(a) $Q_C = -3.72 \times 10^3 J$
(b) $|W| = 2.73 \times 10^3 J$
(c) $ \epsilon = 42.3 \%$
Work Step by Step
(a) Heat discarded $Q_C = -Q_H \frac{T_C}{T_H} $
$Q_C = (-6.45 \times 10^3 J) \frac{300K}{520K} $
$Q_C = -3.72 \times 10^3 J$
(b) Work done $ |𝑄_𝐻 | = |W| + |𝑄_𝐶 |$
$|W| = |𝑄_𝐻 | - |𝑄_𝐶 |$
$|W| =6.45 \times 10^3 J − 3.72 \times 10^3 J$
$|W| = 2.73 \times 10^3 J$
(c) The thermal efficiency $ \epsilon = \frac{W}{|Q_H|} \times 100$
$ \epsilon = \frac{2.73 \times 10^3 J}{6.45 \times 10^3 J} \times 100$
$ \epsilon = 42.3 \%$
