Answer
a) 0.88
b) 300 J
c) 8.3
Work Step by Step
(a) We know that
$\eta=\frac{Work\space done}{Heat\space input}$
We plug in the known values to obtain:
$\eta=\frac{2200}{2500}=0.88$
(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:
$Heat \space discarded=2500-2200=300J$
(c) We know that
$\eta=1-\frac{T_{cold}}{T_{hot}}$
We plug in the known values to obtain:
$0.88=1-\frac{T_{cold}}{T_{hot}}$
$\implies \frac{T_{cold}}{T_{hot}} =0.12$
$\frac{T_{hot}}{T_{cold}}=8.33$
