A carnot engine having an efficiency of `1/4` is being used as a refrigerator. If the work done on the refrigerator is 5 J, the amount of heat absorbed from the reservoir at lower temperature is:

Options

A. 9 J
B. 12 J
C. 20 J
D. 15 J

Explanation

To solve the problem, we need to find the amount of heat absorbed from the reservoir at a lower temperature (Q2) when a Carnot engine with an efficiency of 1/4 is used as a refrigerator and 5 J of work is done on it.

### Step-by-Step Solution:

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

2. **Given Values**:
- Efficiency, \(η = \frac{1}{4}\)
- Work done on the refrigerator, \(W = 5 \, \text{J}\)

3. **Relating Work, Heat Absorbed, and Heat Rejected**:
For a refrigerator, the relationship can be expressed as:
\[ Q_1 = W + Q_2 \]
where \(Q_2\) is the heat absorbed from the lower temperature reservoir.

4. **Substituting Efficiency into the Equation**:
From the efficiency equation:
\[ \frac{1}{4} = \frac{W}{Q_1} \]
Rearranging gives:
\[ Q_1 = 4W \]
Substituting \(W = 5 \, \text{J}\):
\[ Q_1 = 4 \times 5 = 20 \, \text{J} \]

5. **Finding Q2**:
Now, using the relationship \(Q_1 = W + Q_2\):
\[ 20 = 5 + Q_2 \]
Rearranging gives:
\[ Q_2 = 20 - 5 = 15 \, \text{J} \]

### Final Answer:
The amount of heat absorbed from the reservoir at lower temperature is \(Q_2 = 15 \, \text{J}\).