An ideal refrigerator has a freezer at a temperature of `-13^(@)C`. The coefficient of performance of the engine is `5`. The temperature of the air (to which heat is rejected) will be

Options

A. (a)`325^@C`
B. (b)`325 K`
C. (c)`39^@C`
D. (d)`320^@C`

Answer

To solve the problem, we need to find the temperature of the air (T1) to which heat is rejected by the refrigerator. We are given the following information:

- The temperature of the freezer (T2) is -13°C.
- The coefficient of performance (COP) of the engine is 5.

### Step-by-step Solution:

1. **Convert the Freezer Temperature to Kelvin:**
\[ T_2 = -13°C = -13 + 273 = 260 \, K \]

2. **Use the Coefficient of Performance Formula:**
The coefficient of performance (COP) for a refrigerator is given by:
\[ COP = \frac{T_2}{T_1 - T_2} \]
Here, \(COP = 5\) and \(T_2 = 260 \, K\).

3. **Substitute the Known Values into the Formula:**
\[ 5 = \frac{260}{T_1 - 260} \]

4. **Cross Multiply to Solve for \(T_1\):**
\[ 5(T_1 - 260) = 260 \]

5. **Distribute the 5:**
\[ 5T_1 - 1300 = 260 \]

6. **Add 1300 to Both Sides:**
\[ 5T_1 = 260 + 1300 \]
\[ 5T_1 = 1560 \]

7. **Divide by 5 to Find \(T_1\):**
\[ T_1 = \frac{1560}{5} = 312 \, K \]

8. **Convert \(T_1\) Back to Celsius:**
\[ T_1 = 312 - 273 = 39°C \]

### Final Answer:
The temperature of the air to which heat is rejected is **39°C**.