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. 9 J
B. 32 J
C. 20 J
D. 15 J

Explanation

To solve the problem step by step, we will use the concepts of efficiency and the coefficient of performance (COP) of a refrigerator.

### Step 1: Understand the Given Information
- Efficiency of the Carnot engine, \( \eta = \frac{1}{5} \)
- Work done on the refrigerator, \( W = 8 \, \text{J} \)

### Step 2: Relate Efficiency to Temperatures
The efficiency of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the hot reservoir and \( T_2 \) is the temperature of the cold reservoir.

### Step 3: Substitute the Given Efficiency
Substituting the given efficiency into the formula:
\[ \frac{1}{5} = 1 - \frac{T_2}{T_1} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{5} = \frac{4}{5} \]

### Step 4: Relate Coefficient of Performance to Heat Absorbed
The coefficient of performance (COP) for a refrigerator is defined as:
\[ COP = \frac{Q_L}{W} \]
where \( Q_L \) is the heat absorbed from the cold reservoir.

### Step 5: Express \( Q_L \) in Terms of \( W \)
From the COP definition, we can express \( Q_L \):
\[ Q_L = COP \times W \]

### Step 6: Relate COP to Temperatures
We can also express COP in terms of temperatures:
\[ COP = \frac{T_2}{T_1 - T_2} \]
Now we have two expressions for COP. We can equate them:
\[ \frac{Q_L}{W} = \frac{T_2}{T_1 - T_2} \]

### Step 7: Substitute \( \frac{T_2}{T_1} \) into COP
Using \( \frac{T_2}{T_1} = \frac{4}{5} \), we can express \( T_1 \) in terms of \( T_2 \):
\[ T_1 = \frac{5}{4} T_2 \]
Substituting this into the COP equation:
\[ COP = \frac{T_2}{\frac{5}{4}T_2 - T_2} = \frac{T_2}{\frac{5}{4}T_2 - \frac{4}{4}T_2} = \frac{T_2}{\frac{1}{4}T_2} = 4 \]

### Step 8: Calculate \( Q_L \)
Now substituting \( COP \) back into the equation for \( Q_L \):
\[ Q_L = COP \times W = 4 \times 8 \, \text{J} = 32 \, \text{J} \]

### Final Answer
The amount of heat absorbed from the reservoir at lower temperature is:
\[ \boxed{32 \, \text{J}} \]