The temperateus of inside and outside of a refrigerator are 273 K and 303 K respectively. Assuming that the refrigerator cycle is reversible, for every joule of workdone, the heat delivered to the surrounding will be nearly

Options

A. 10 J
B. 20 J
C. 30 J
D. 50 J

Answer

To solve the problem, we will use the principles of thermodynamics, specifically the Carnot cycle for a refrigerator. We will calculate the heat delivered to the surroundings based on the temperatures given and the work done.

### Step-by-Step Solution:

1. **Identify the Temperatures:**
- Inside temperature (T2) = 273 K
- Outside temperature (T1) = 303 K

2. **Use the Carnot Refrigerator Efficiency Formula:**
The efficiency of a reversible refrigerator can be expressed as:
\[ \frac{T2}{T1 - T2} = \frac{Q2}{W} \]
where:
- \( Q2 \) = heat absorbed from the cold reservoir (inside the refrigerator)
- \( W \) = work done on the refrigerator

3. **Substituting Known Values:**
Since we are given that the work done \( W = 1 \) Joule, we can substitute the values into the equation:
\[ \frac{273}{303 - 273} = \frac{Q2}{1} \]

4. **Calculate the Difference in Temperatures:**
\[ T1 - T2 = 303 K - 273 K = 30 K \]

5. **Rearranging the Equation:**
Now, substituting the temperature difference:
\[ \frac{273}{30} = Q2 \]
Therefore,
\[ Q2 = \frac{273}{30} \approx 9.1 \text{ Joules} \]

6. **Calculate Heat Delivered to Surroundings (Q1):**
We know from the first law of thermodynamics that:
\[ Q1 = W + Q2 \]
Substituting the values:
\[ Q1 = 1 \text{ Joule} + 9.1 \text{ Joules} = 10.1 \text{ Joules} \]

7. **Final Answer:**
The heat delivered to the surroundings is approximately:
\[ Q1 \approx 10 \text{ Joules} \]

### Conclusion:
For every joule of work done, the heat delivered to the surrounding will be nearly **10 Joules**.