If the coefficient of performance of a refrigerator is 5 and operates at the room temperature (`27^(@)` C), find the temperature inside the refrigerator.

Options

A. 240 K
B. 250 K
C. 230 K
D. 260 K

Explanation

To solve the problem, we need to find the temperature inside the refrigerator (T2) given the coefficient of performance (COP) and the room temperature (T1). Here are the steps to find the solution:

### Step 1: Understand the Coefficient of Performance (COP)
The coefficient of performance (COP) for a refrigerator is defined as:
\[ \beta = \frac{T_2}{T_1 - T_2} \]
where:
- \( \beta \) is the coefficient of performance,
- \( T_1 \) is the temperature of the surroundings (room temperature),
- \( T_2 \) is the temperature inside the refrigerator.

### Step 2: Convert Room Temperature to Kelvin
The room temperature is given as \( 27^\circ C \). To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]

### Step 3: Substitute Known Values into the COP Formula
We know that:
- \( \beta = 5 \)
- \( T_1 = 300 \, K \)

Substituting these values into the COP formula:
\[ 5 = \frac{T_2}{300 - T_2} \]

### Step 4: Rearrange the Equation
To solve for \( T_2 \), we can cross-multiply:
\[ 5(300 - T_2) = T_2 \]
Expanding this gives:
\[ 1500 - 5T_2 = T_2 \]

### Step 5: Combine Like Terms
Now, we can combine the terms involving \( T_2 \):
\[ 1500 = T_2 + 5T_2 \]
\[ 1500 = 6T_2 \]

### Step 6: Solve for \( T_2 \)
Now, divide both sides by 6 to find \( T_2 \):
\[ T_2 = \frac{1500}{6} = 250 \, K \]

### Step 7: Convert \( T_2 \) Back to Celsius
To convert \( T_2 \) back to degrees Celsius:
\[ T_2 = 250 - 273 = -23^\circ C \]

### Final Answer
The temperature inside the refrigerator is \( 250 \, K \) or \( -23^\circ C \).

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