If the work done by a Carnot engine working between two temperatures 600K & 300K is used as the work input in Carnot refrigerator, working between 200K & 400K, find the heat (in J) removed from the lower temperature by refrigerator? The heat supplied to engine in 500J

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Efficiency of the Carnot Engine
The efficiency (η) of a Carnot engine is given by the formula:

\[ \eta = 1 - \frac{T_C}{T_H} \]

Where:
- \( T_H \) = Temperature of the hot reservoir = 600 K
- \( T_C \) = Temperature of the cold reservoir = 300 K

Substituting the values:

\[ \eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 \]

### Step 2: Calculate the Work Done by the Carnot Engine
The work done (W) by the Carnot engine can be calculated using the formula:

\[ W = \eta \times Q_H \]

Where:
- \( Q_H \) = Heat supplied to the engine = 500 J

Substituting the values:

\[ W = 0.5 \times 500 = 250 \, \text{J} \]

### Step 3: Use the Work Done as Input for the Carnot Refrigerator
The work done by the engine (250 J) is used as the work input for the Carnot refrigerator. The efficiency of a Carnot refrigerator is given by:

\[ \eta_{refrigerator} = \frac{T_C}{T_H - T_C} \]

Where:
- \( T_H \) = Temperature of the hot reservoir of the refrigerator = 400 K
- \( T_C \) = Temperature of the cold reservoir of the refrigerator = 200 K

Substituting the values:

\[ \eta_{refrigerator} = \frac{200}{400 - 200} = \frac{200}{200} = 1 \]

### Step 4: Calculate the Heat Removed by the Refrigerator
The heat removed (Q_C) from the cold reservoir can be calculated using the formula:

\[ Q_C = W + Q_H \]

Where:
- \( W \) = Work input to the refrigerator = 250 J

Since the efficiency of the refrigerator is 1, it means that all the work done is converted into heat removed. Therefore:

\[ Q_C = W = 250 \, \text{J} \]

### Final Answer
The heat removed from the lower temperature by the refrigerator is **250 J**.

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