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

Explanation

To solve the problem step by step, we will follow the concepts of thermodynamics related to Carnot engines and refrigerators.

### Step 1: Understand the Efficiency of the Carnot Engine
The efficiency (η) of a Carnot engine is given by:
\[ η = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the hot reservoir and \(T_2\) is the temperature of the cold reservoir.

Given that the efficiency is \( \frac{1}{4} \), we can set up the equation:
\[ \frac{1}{4} = 1 - \frac{T_2}{T_1} \]

### Step 2: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 3: Coefficient of Performance for the Refrigerator
For a refrigerator, the coefficient of performance (COP) is defined as:
\[ COP = \frac{Q}{W} \]
where \(Q\) is the heat absorbed from the cold reservoir and \(W\) is the work done on the refrigerator.

### Step 4: Relating COP to Temperatures
The COP can also be expressed in terms of the temperatures:
\[ COP = \frac{T_2}{T_1 - T_2} \]

### Step 5: Setting Up the Equation
From the previous steps, we have:
\[ COP = \frac{Q}{W} = \frac{T_2}{T_1 - T_2} \]
Substituting the work done \(W = 5 \, J\) into the equation gives:
\[ 5 = Q \cdot \frac{T_1 - T_2}{T_2} \]

### Step 6: Substitute \(T_1\) in Terms of \(T_2\)
From the ratio \( \frac{T_2}{T_1} = \frac{3}{4} \), we can express \(T_1\) as:
\[ T_1 = \frac{4}{3} T_2 \]

### Step 7: Substitute \(T_1\) into the Work Equation
Substituting \(T_1\) into the work equation gives:
\[ 5 = Q \cdot \frac{\frac{4}{3} T_2 - T_2}{T_2} \]
This simplifies to:
\[ 5 = Q \cdot \frac{\frac{4}{3} T_2 - \frac{3}{3} T_2}{T_2} = Q \cdot \frac{\frac{1}{3} T_2}{T_2} = Q \cdot \frac{1}{3} \]

### Step 8: Solve for \(Q\)
Now, we can solve for \(Q\):
\[ 5 = \frac{Q}{3} \implies Q = 5 \cdot 3 = 15 \, J \]

### Final Answer
The amount of heat absorbed from the reservoir at lower temperature is:
\[ \boxed{15 \, J} \]