A refrigerator with `COP= 1//3` release `200 J` at heat to a reservoir. Then the work done on the working substance is

Options

A. `(100)/3 J`
B. `100 J`
C. `(200)/3 J`
D. `150 J` of heat been added to the gas

Explanation

To solve the problem, we need to use the relationship between the Coefficient of Performance (COP) of a refrigerator, the heat released to the hot reservoir (Q2), and the work done (W) on the working substance. The COP is defined as:

\[ COP = \frac{Q_2}{W} \]

Where:
- \( Q_2 \) is the heat released to the hot reservoir.
- \( W \) is the work done on the working substance.

Given:
- \( COP = \frac{1}{3} \)
- \( Q_2 = 200 \, J \)

### Step 1: Use the COP formula to express W in terms of Q2.
From the definition of COP, we can rearrange the formula to find W:

\[ W = \frac{Q_2}{COP} \]

### Step 2: Substitute the given values into the equation.
Now, we substitute the values of \( Q_2 \) and \( COP \):

\[ W = \frac{200 \, J}{\frac{1}{3}} = 200 \, J \times 3 = 600 \, J \]

### Step 3: Conclusion
The work done on the working substance is:

\[ W = 600 \, J \]

### Final Answer
Thus, the work done on the working substance is **600 Joules**.

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