A refrigerator with `COP= 1//3` release `200 J` of 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`

Explanation

To solve the problem, we need to find the work done on the working substance of a refrigerator given its coefficient of performance (COP) and the amount of heat released to a reservoir.

### Step-by-Step Solution:

1. **Understand the Coefficient of Performance (COP)**:
The coefficient of performance (COP) of a refrigerator is defined as:
\[ \text{COP} = \frac{Q_2}{W} \]
where \( Q_2 \) is the heat removed from the cold reservoir and \( W \) is the work done on the working substance.

2. **Identify Given Values**:
- COP \( = \frac{1}{3} \)
- Heat released to the reservoir \( Q_1 = 200 \, \text{J} \)

3. **Apply the First Law of Thermodynamics**:
According to the first law of thermodynamics for a refrigerator:
\[ Q_1 = Q_2 + W \]
Rearranging gives:
\[ Q_2 = Q_1 - W \]

4. **Substitute \( Q_2 \) into the COP Formula**:
Substitute \( Q_2 \) from the previous step into the COP equation:
\[ \text{COP} = \frac{Q_1 - W}{W} \]
Plugging in the values we have:
\[ \frac{1}{3} = \frac{200 - W}{W} \]

5. **Cross Multiply to Solve for \( W \)**:
Cross multiplying gives:
\[ 1 \cdot W = 3(200 - W) \]
Simplifying this:
\[ W = 600 - 3W \]

6. **Rearranging the Equation**:
Combine like terms:
\[ W + 3W = 600 \]
\[ 4W = 600 \]

7. **Solve for \( W \)**:
Divide both sides by 4:
\[ W = \frac{600}{4} = 150 \, \text{J} \]

### Final Answer:
The work done on the working substance is \( W = 150 \, \text{J} \).