A Carnot engine having an efficiency of 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 :

90 J

Work done = Q₂(\(\frac{T_1}{T_2}\)-1) -------(i)

η = 1 - \(\frac{T_2}{T_1}\)

\(\frac{1}{10}\) = 1 - \(\frac{T_2}{T_1}\)

\(\frac{T_2}{T_1}\) = 1 - \(\frac{1}{10}\)

\(\frac{T_2}{T_1}\) = \(\frac{9}{10}\) 

⇒ \(\frac{T_1}{T_2}\) = \(\frac{10}{9}\)

Using equation 1 

10 = Q₂(\(\frac{10}{9}\)-1)

Q₂ = 90J