A refrigerator has to transfer an average of 263J of heat per second from temperature `-10^@C` to `25^@C`. Calculate the average power consumed, assuming no enegy losses in the process.

Explanation

To solve the problem of calculating the average power consumed by a refrigerator that transfers heat from a lower temperature to a higher temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Heat transferred per second (Q) = 263 J/s (or Watts)
- Lower temperature (T1) = -10°C = 263 K (since K = °C + 273)
- Higher temperature (T2) = 25°C = 298 K

2. **Use the Coefficient of Performance (COP) Formula:**
The coefficient of performance for a refrigerator is given by the formula:
\[ COP = \frac{T_2}{T_1 - T_2} \]
where \(T_1\) is the absolute temperature of the cold reservoir and \(T_2\) is the absolute temperature of the hot reservoir.

3. **Substitute the Values into the COP Formula:**
\[ COP = \frac{298}{263 - 298} = \frac{298}{-35} \]
Since we are dealing with absolute temperatures, we need to ensure the temperatures are correctly assigned. The correct formula for work done (W) in terms of heat transferred (Q) and COP is:
\[ W = \frac{Q}{COP} \]

4. **Calculate the Work Done (W):**
Rearranging the COP formula gives us:
\[ W = Q \cdot \frac{T_1 - T_2}{T_2} \]
Substituting the values:
\[ W = 263 \cdot \frac{263 - 298}{298} \]
\[ W = 263 \cdot \frac{-35}{298} \]
\[ W = 263 \cdot -0.1174 \approx -30.9 \text{ J/s} \]
Since we are interested in power, we take the absolute value:
\[ P = 30.9 \text{ Watts} \]

5. **Final Result:**
The average power consumed by the refrigerator is approximately **30.9 Watts**.