The cofficient of performance of a refrigerator is 5. If the temperature inside freezer is `-20^(@)C`, the temperature of the surroundings to which it rejects heat is :

Options

A. `31^(@)C`
B. `41^(@)C`
C. `11^(@)C`
D. `21^(@)C`

Answer

To solve the problem, we need to find the temperature of the surroundings (T1) to which the refrigerator rejects heat, given the coefficient of performance (COP) and the temperature inside the freezer.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Coefficient of performance (COP) of the refrigerator, \( \alpha = 5 \)
- Temperature inside the freezer, \( T_2 = -20^\circ C \)

2. **Convert Temperature to Kelvin:**
- To convert the temperature from Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- Therefore,
\[ T_2 = -20 + 273 = 253 \, K \]

3. **Use the Formula for Coefficient of Performance:**
- The formula for the coefficient of performance of a refrigerator is given by:
\[ \alpha = \frac{T_2}{T_1 - T_2} \]
- Here, we need to find \( T_1 \).

4. **Substitute the Known Values into the Formula:**
- Substitute \( \alpha = 5 \) and \( T_2 = 253 \, K \):
\[ 5 = \frac{253}{T_1 - 253} \]

5. **Cross-Multiply to Solve for \( T_1 \):**
- Cross-multiplying gives:
\[ 5(T_1 - 253) = 253 \]
- Expanding this:
\[ 5T_1 - 1265 = 253 \]

6. **Rearranging the Equation:**
- Add 1265 to both sides:
\[ 5T_1 = 253 + 1265 \]
\[ 5T_1 = 1518 \]

7. **Divide by 5 to Find \( T_1 \):**
- Now divide both sides by 5:
\[ T_1 = \frac{1518}{5} = 303.6 \, K \]

8. **Convert \( T_1 \) Back to Celsius:**
- To convert from Kelvin back to Celsius:
\[ T(°C) = T(K) - 273 \]
- Therefore,
\[ T_1 = 303.6 - 273 = 30.6^\circ C \]

9. **Final Answer:**
- Rounding off, the temperature of the surroundings is approximately:
\[ T_1 \approx 31^\circ C \]