Answer
a) 2.9 KJ
b) $T_c=262.694K=-10.3^{\circ}C$
Work Step by Step
(a) We know that
$Heat \space removed =Heat\space supplied -Work\space done$
We plug in the known values to obtain:
$Heat\space removed=3240-345=2895J=2.895KJ$
(b) As given that the room temperature is $21^{\circ}C$
so $T_h=21+273=294K$
We know that
$\frac{Q_c}{Q_h}=\frac{T_c}{T_h}$
$\implies T_c=\frac{Q_c}{Q_h}\times T_h$
We plug in the known values to obtain:
$T_c=\frac{2895}{3240}\times 294$
$T_c=262.694K=-10.3^{\circ}C$
