A carnot engine having an efficiency of `(1)/(5)` is being used as a refrigerator. If the work done on the refrigerator is 8 J, the amount of heat absorbed from the reservoir at lower temperature is:

Options

A. (a)9J
B. (b)32J
C. (c)20J
D. (d)15J

Explanation

To solve the problem, we need to find the amount of heat absorbed from the lower temperature reservoir (Q2) when a Carnot engine with an efficiency of \( \frac{1}{5} \) is used as a refrigerator and the work done on the refrigerator is 8 J.

### Step-by-Step Solution:

1. **Understand the Efficiency of the Carnot Engine**:
The efficiency (\( \eta \)) of the Carnot engine is given as:
\[ \eta = \frac{1}{5} \]

2. **Determine the Coefficient of Performance (COP)**:
The coefficient of performance (COP) for a refrigerator is related to the efficiency of the Carnot engine. The formula for COP is:
\[ \text{COP} = \frac{Q_2}{W} \]
where \( Q_2 \) is the heat absorbed from the lower temperature reservoir and \( W \) is the work done on the refrigerator.

3. **Relate COP to Efficiency**:
The COP can also be expressed in terms of the efficiency:
\[ \text{COP} = \frac{1 - \eta}{\eta} \]
Substituting the efficiency:
\[ \text{COP} = \frac{1 - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{4}{5}}{\frac{1}{5}} = 4 \]

4. **Calculate the Heat Absorbed (Q2)**:
Now, we can use the COP to find \( Q_2 \):
\[ \text{COP} = \frac{Q_2}{W} \]
Rearranging gives:
\[ Q_2 = \text{COP} \times W \]
Substituting the values:
\[ Q_2 = 4 \times 8 \, \text{J} = 32 \, \text{J} \]

5. **Final Answer**:
The amount of heat absorbed from the reservoir at lower temperature is:
\[ Q_2 = 32 \, \text{J} \]

### Summary:
The amount of heat absorbed from the reservoir at lower temperature is 32 J.