The temperature inside and outside a refrigerator are 273 K and 300 K respectively. Assuming that the refrigerator cycle is reversible. For every joule of work done heat delivered to the surrounding will be nearly :-

Options

A. 11 J
B. 22 J
C. 33 J
D. 50 J

Answer

To solve the problem, we will use the principles of thermodynamics, specifically focusing on the operation of a refrigerator and the relationship between heat transfer, work done, and temperature.

### Step-by-Step Solution:

1. **Identify the Temperatures**:
- The temperature inside the refrigerator (TL) = 273 K
- The temperature outside the refrigerator (TH) = 300 K

2. **Understand the Coefficient of Performance (COP)**:
- The coefficient of performance (β) for a refrigerator is given by the formula:
\[ \beta = \frac{QL}{W} \]
where \(QL\) is the heat extracted from the cold reservoir and \(W\) is the work input.

3. **Relate COP to Temperatures**:
- For a reversible refrigerator, the COP can also be expressed in terms of temperatures:
\[ \beta = \frac{TL}{TH - TL} \]

4. **Calculate QL for 1 Joule of Work**:
- We know that for every 1 Joule of work done (W = 1 J), we can find \(QL\):
\[ QL = W \cdot \beta = 1 \cdot \frac{273}{300 - 273} \]
- Calculate the denominator:
\[ 300 - 273 = 27 \]
- Therefore, substituting back:
\[ QL = 1 \cdot \frac{273}{27} \approx 10.1 \text{ J} \]

5. **Calculate Heat Delivered to Surroundings (QH)**:
- The heat delivered to the surroundings (QH) is the sum of the heat extracted from the cold reservoir (QL) and the work done (W):
\[ QH = QL + W \]
- Substituting the known values:
\[ QH = 10.1 + 1 = 11.1 \text{ J} \]

6. **Round Off the Result**:
- Since the problem asks for a whole number, we round off 11.1 J to 11 J.

### Final Answer:
The heat delivered to the surroundings for every joule of work done will be approximately **11 J**.

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