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 step by step, we will use the concepts of efficiency of a Carnot engine and the relationship between work done, heat absorbed, and heat rejected.

### Step-by-Step Solution:

1. **Understand the Efficiency of the Carnot Engine:**
The efficiency (η) of a Carnot engine is defined as:
\[ \eta = \frac{W}{Q_1} \]
where:
- \( W \) is the work done by the engine,
- \( Q_1 \) is the heat absorbed from the hot reservoir.

2. **Given Values:**
From the problem, we have:
- Efficiency \( \eta = \frac{1}{10} \)
- Work done \( W = 10 \, \text{J} \)

3. **Calculate the Heat Supplied (Q1):**
Rearranging the efficiency formula to find \( Q_1 \):
\[ Q_1 = \frac{W}{\eta} \]
Substituting the known values:
\[ Q_1 = \frac{10 \, \text{J}}{\frac{1}{10}} = 10 \, \text{J} \times 10 = 100 \, \text{J} \]

4. **Relate Q1, Q2, and W:**
When the Carnot engine is used as a refrigerator, the relationship between the heat absorbed from the cold reservoir (\( Q_2 \)), the heat supplied (\( Q_1 \)), and the work done (\( W \)) is given by:
\[ Q_2 = Q_1 - W \]

5. **Calculate the Heat Absorbed from the Cold Reservoir (Q2):**
Substituting the values we have:
\[ Q_2 = 100 \, \text{J} - 10 \, \text{J} = 90 \, \text{J} \]

6. **Final Answer:**
The amount of energy absorbed from the reservoir at lower temperature is:
\[ Q_2 = 90 \, \text{J} \]

### Summary:
The amount of energy absorbed from the reservoir at lower temperature is **90 J**.