The efficiency of the Carnot engine is $\eta = \frac{1}{10}$. When the same engine is used as a refrigerator, its coefficient of performance ($\beta$) is given by: $\beta = \frac{1 - \eta}{\eta} = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$ The coefficient of performance is also defined as the ratio of the heat absorbed from the cold reservoir ($Q_2$) to the work done on the system ($W$): $\beta = \frac{Q_2}{W}$ Given that the work done on the system is $W = 10\text{ J}$: $9 = \frac{Q_2}{10}$ $Q_2 = 9 \times 10 = 90\text{ J}$ Therefore, the amount of energy absorbed from the reservoir at the lower temperature is $90\text{ J}$.