A refrigerator consumes an average 35 W power to operate between temperature - `10^(@)C` to `25^(@)C`. If there is no loss of energy then how much average heat per second does it transfer ?

Options

A. 298 J/s
B. 35 J/s
C. 350 J/s
D. 263 J/s

Answer

To solve the problem, we need to determine the average heat transferred by the refrigerator per second, given its power consumption and the temperatures involved.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Power consumed by the refrigerator (W) = 35 W
- Higher temperature (TH) = 25°C = 25 + 273 = 298 K
- Lower temperature (TC) = -10°C = -10 + 273 = 263 K

2. **Understanding the Refrigeration Cycle**:
- In a refrigeration cycle, the work done (W) is related to the heat absorbed from the cold reservoir (QC) and the heat rejected to the hot reservoir (QH) by the equation:
\[ QH = QC + W \]

3. **Using the Coefficient of Performance (COP)**:
- The coefficient of performance (K) for a refrigerator can be expressed as:
\[ K = \frac{Q_C}{W} \]
- Rearranging gives us:
\[ Q_C = K \cdot W \]

4. **Finding the Coefficient of Performance**:
- The relationship between the heat absorbed (QC), heat rejected (QH), and the temperatures is given by:
\[ \frac{QH}{W} = \frac{TH}{TH - TC} \]
- Plugging in the known values:
\[ \frac{QH}{35} = \frac{298}{298 - 263} = \frac{298}{35} \]
- This simplifies to:
\[ QH = 35 \cdot \frac{298}{35} = 298 \text{ W} \]

5. **Calculating the Heat Extracted (QC)**:
- Now, substituting back to find QC:
\[ QC = QH - W = 298 - 35 = 263 \text{ W} \]

### Final Answer:
The average heat transferred by the refrigerator per second (QC) is **263 W**.