A Carnot engine, having an efficiency of $\mathrm{\eta }$ = 1/10 as heat engine, is used as a refrigerator. If the work done. on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

The efficiency of a Carnot cycle is given by
$\mathrm{\eta }=1-\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\frac{1}{10} ....\left(1\right)$
where, Q₁ = heal taken from source
and Q₂ = heat given to sink
Work done = Q₁ - Q₂ = 10   ….(2)
From eq. (1),
$\frac{{\mathrm{Q}}_1-{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\frac{1}{10}$
or    $\frac{10}{{\mathrm{Q}}_1}=\frac{1}{10}$  or    ${\mathrm{Q}}_1=100\mathrm{J}$
From eq. (2)
${\mathrm{Q}}_2=100-10=90\mathrm{J}$.