In a refrigerator, heat from inside at 277K is transferred to a room at 300K. How many joules of heat shall be delivered to the room for each joule of electrical energy consumed ideally?

Explanation

To solve the problem, we will use the concept of the Coefficient of Performance (COP) of a refrigerator. The COP is defined as the ratio of the heat removed from the cold reservoir (Q2) to the work input (W).

### Step-by-Step Solution:

1. **Identify the temperatures**:
- The temperature of the cold reservoir (inside the refrigerator) is \( T_{\text{lower}} = 277 \, K \).
- The temperature of the hot reservoir (the room) is \( T_{\text{higher}} = 300 \, K \).

2. **Write the formula for the Coefficient of Performance (COP)**:
\[ \text{COP} = \frac{Q_2}{W} = \frac{T_{\text{lower}}}{T_{\text{higher}} - T_{\text{lower}}} \]

3. **Substitute the known values into the COP formula**:
\[ \text{COP} = \frac{277}{300 - 277} = \frac{277}{23} \]

4. **Calculate the value of COP**:
\[ \text{COP} = \frac{277}{23} \approx 12.04 \]

5. **Determine the heat delivered to the room (Q1)**:
- Since we are given that \( W = 1 \, \text{joule} \) (the electrical energy consumed), we can find \( Q_2 \):
\[ Q_2 = \text{COP} \times W = 12.04 \times 1 = 12.04 \, \text{joules} \]

6. **Calculate the total heat delivered to the room (Q1)**:
- The heat rejected to the room (Q1) is the sum of the heat absorbed from the cold reservoir (Q2) and the work done (W):
\[ Q_1 = Q_2 + W = 12.04 + 1 = 13.04 \, \text{joules} \]

### Final Answer:
For each joule of electrical energy consumed ideally, approximately **13.04 joules of heat** shall be delivered to the room.