A Carnot engine, having an efficiency of `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

Options

A. `99 J`
B. `90 J`
C. `1 J`
D. `100 J`

Explanation

To solve the problem, we need to determine the amount of energy absorbed from the reservoir at a lower temperature (denoted as \( Q_2 \)) when a Carnot engine is used as a refrigerator. We are given the efficiency \( \eta \) of the Carnot engine and the work done on the system.

### Step-by-Step Solution:

1. **Understand the Efficiency of the Carnot Engine**:
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]
where \( T_C \) is the temperature of the cold reservoir and \( T_H \) is the temperature of the hot reservoir. In this case, we are given \( \eta = \frac{1}{10} \).

2. **Calculate the Coefficient of Performance (COP)**:
When the Carnot engine is used as a refrigerator, the coefficient of performance \( \beta \) is given by:
\[ \beta = \frac{Q_2}{W} \]
where \( Q_2 \) is the heat absorbed from the cold reservoir and \( W \) is the work input. The relationship between the efficiency and the COP is:
\[ \beta = \frac{1 - \eta}{\eta} \]

3. **Substitute the Efficiency into the COP Formula**:
Substituting \( \eta = \frac{1}{10} \) into the COP formula:
\[ \beta = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = \frac{\frac{9}{10}}{\frac{1}{10}} = 9 \]

4. **Use the Work Done to Find \( Q_2 \)**:
We know the work done on the system \( W = 10 \, J \). Now, using the COP:
\[ Q_2 = \beta \cdot W = 9 \cdot 10 \, J = 90 \, J \]

5. **Conclusion**:
The amount of energy absorbed from the reservoir at lower temperature is:
\[ Q_2 = 90 \, J \]

### Final Answer:
The amount of energy absorbed from the reservoir at lower temperature is **90 Joules**.