For a refrigerator , heat absorbed from source is 800 J and heat supplied to sink is 500 J then find coefficient of performance is :-

Options

A. `(5)/(8)`
B. `(8)/(5)`
C. `(5)/(3)`
D. `(3)/(5)`

Explanation

To solve the problem, we need to find the coefficient of performance (COP) of the refrigerator using the given values of heat absorbed from the source and heat supplied to the sink.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Heat absorbed from the source (Q1) = 800 J
- Heat supplied to the sink (Q2) = 500 J

2. **Understand the Coefficient of Performance (COP)**:
- The coefficient of performance (COP) for a refrigerator is defined as:
\[ COP = \frac{Q2}{Q1 - Q2} \]
- Here, Q2 is the heat removed from the cold reservoir (heat supplied to the sink), and \(Q1 - Q2\) is the work done by the refrigerator.

3. **Substitute the Values into the Formula**:
- Substitute the values of Q1 and Q2 into the COP formula:
\[ COP = \frac{500}{800 - 500} \]

4. **Calculate the Denominator**:
- Calculate \(800 - 500\):
\[ 800 - 500 = 300 \]

5. **Calculate the COP**:
- Now substitute this value back into the formula:
\[ COP = \frac{500}{300} \]

6. **Simplify the Fraction**:
- Simplify \(\frac{500}{300}\):
\[ COP = \frac{5}{3} \]

7. **Final Answer**:
- Therefore, the coefficient of performance (COP) is:
\[ COP = \frac{5}{3} \]

### Conclusion:
The coefficient of performance of the refrigerator is \(\frac{5}{3}\).

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