A Carnot refrigerator works between two temperatures of 300 K & 600 K. Find the COP of the refrigerator if heat removed from lower temperature in 300 J.

Options

A. (a)`1`
B. (b)`1.2`
C. (c)`0.5`
D. (d)`0.8`

Explanation

To find the Coefficient of Performance (COP) of a Carnot refrigerator working between two temperatures of 300 K and 600 K, and given that the heat removed from the lower temperature is 300 J, we can follow these steps:

### Step 1: Understand the Concept of COP
The Coefficient of Performance (COP) of a refrigerator is defined as the ratio of the heat removed from the cold reservoir (Q_R) to the work input (W) required to remove that heat. Mathematically, it can be expressed as:

\[ COP = \frac{Q_R}{W} \]

### Step 2: Identify the Given Values
From the problem statement, we have:
- Heat removed from the lower temperature (Q_R) = 300 J
- The temperatures of the hot reservoir (T_H) = 600 K and the cold reservoir (T_C) = 300 K.

### Step 3: Calculate the Work Done (W)
For a Carnot refrigerator, the relationship between the heat absorbed from the cold reservoir (Q_R), the heat rejected to the hot reservoir (Q_H), and the work done (W) is given by:

\[ W = Q_H - Q_R \]

To find Q_H, we can use the ratio of the temperatures since for a Carnot cycle:

\[ \frac{Q_H}{Q_R} = \frac{T_H}{T_C} \]

Substituting the known values:

\[ \frac{Q_H}{300} = \frac{600}{300} \]

This simplifies to:

\[ \frac{Q_H}{300} = 2 \implies Q_H = 2 \times 300 = 600 \text{ J} \]

Now we can calculate the work done (W):

\[ W = Q_H - Q_R = 600 \text{ J} - 300 \text{ J} = 300 \text{ J} \]

### Step 4: Calculate the COP
Now that we have both Q_R and W, we can calculate the COP:

\[ COP = \frac{Q_R}{W} = \frac{300 \text{ J}}{300 \text{ J}} = 1 \]

### Final Answer
The Coefficient of Performance (COP) of the Carnot refrigerator is **1**.

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