A Carnot's engine works as refrigerator between `250 K` and 300K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

Options

A. `420 J`
B. `772 J`
C. `2100 J`
D. `2520 J`

Explanation

To solve the problem of finding the amount of work done in each cycle to operate the Carnot refrigerator, we can follow these steps:

### Step 1: Identify the temperatures
The Carnot refrigerator operates between two temperatures:
- T1 (lower temperature) = 250 K
- T2 (higher temperature) = 300 K

### Step 2: Calculate the efficiency of the Carnot refrigerator
The efficiency (η) of a Carnot refrigerator is given by the formula:
\[ \eta = \frac{T2 - T1}{T2} \]
Substituting the values:
\[ \eta = \frac{300 K - 250 K}{300 K} = \frac{50 K}{300 K} = \frac{1}{6} \]

### Step 3: Use the relationship between work done, heat absorbed, and efficiency
The work done (W) by the refrigerator can be calculated using the formula:
\[ \eta = \frac{W}{Q} \]
Where:
- W = work done
- Q = heat absorbed from the lower temperature reservoir

Rearranging the formula gives us:
\[ W = \eta \times Q \]

### Step 4: Convert heat absorbed from calories to joules
The problem states that the refrigerator receives 500 calories of heat. We need to convert this to joules. The conversion factor is:
\[ 1 \text{ calorie} = 4.2 \text{ joules} \]
Thus,
\[ Q = 500 \text{ calories} \times 4.2 \text{ joules/calorie} = 2100 \text{ joules} \]

### Step 5: Calculate the work done
Now, substituting the values of η and Q into the equation for work done:
\[ W = \frac{1}{6} \times 2100 \text{ joules} = 350 \text{ joules} \]

### Final Answer
The amount of work done in each cycle to operate the refrigerator is **350 joules**.

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