In a mechanical refrigerator the low temperature coils are at a temperature of `-23^(@)C` and the compressed gas in the condenser has a temperature of `27^(@)C`. The theoretical coefficient of performance is

Options

A. `5`
B. `8`
C. `6`
D. `6.5`

Explanation

To find the theoretical coefficient of performance (COP) of a mechanical refrigerator, we can use the formula:

\[ COP = \frac{T_L}{T_H - T_L} \]

where:
- \( T_L \) is the absolute temperature of the low-temperature coils (in Kelvin),
- \( T_H \) is the absolute temperature of the high-temperature coils (in Kelvin).

### Step 1: Convert the temperatures from Celsius to Kelvin

1. **Low-temperature coil temperature**:
\[ T_L = -23^\circ C = -23 + 273.15 = 250.15 \, K \]

2. **High-temperature coil temperature**:
\[ T_H = 27^\circ C = 27 + 273.15 = 300.15 \, K \]

### Step 2: Substitute the values into the COP formula

Now we can substitute \( T_L \) and \( T_H \) into the COP formula:

\[ COP = \frac{T_L}{T_H - T_L} = \frac{250.15}{300.15 - 250.15} \]

### Step 3: Calculate the denominator

Calculate \( T_H - T_L \):

\[ T_H - T_L = 300.15 - 250.15 = 50 \, K \]

### Step 4: Calculate the COP

Now substitute back into the COP equation:

\[ COP = \frac{250.15}{50} = 5.003 \]

### Final Answer

The theoretical coefficient of performance (COP) of the mechanical refrigerator is approximately:

\[ COP \approx 5.00 \]