A refrigerator absorbs 2000 cal of heat from ice trays. If the coefficient of performance is 4, then work done by the motor is

Options

A. 2100 J
B. 4200 J
C. 8400 J
D. 500 J

Answer

To solve the problem, we need to use the concept of the Coefficient of Performance (COP) of a refrigerator. The COP is defined as the ratio of the heat extracted from the cold reservoir (Q2) to the work done by the refrigerator (W).

### Step-by-Step Solution:

1. **Identify the given values:**
- Heat absorbed from the ice trays (Q2) = 2000 cal
- Coefficient of Performance (COP, β) = 4

2. **Write the formula for Coefficient of Performance:**
\[ \beta = \frac{Q2}{W} \]
where:
- \( Q2 \) is the heat absorbed from the cold reservoir (2000 cal).
- \( W \) is the work done by the refrigerator.

3. **Rearrange the formula to solve for work done (W):**
\[ W = \frac{Q2}{\beta} \]

4. **Substitute the known values into the equation:**
\[ W = \frac{2000 \text{ cal}}{4} \]

5. **Calculate W:**
\[ W = 500 \text{ cal} \]

6. **Convert calories to joules (if necessary):**
Since 1 cal = 4.184 joules,
\[ W = 500 \text{ cal} \times 4.184 \text{ J/cal} = 2092 \text{ J} \]

### Final Answer:
The work done by the motor is **2092 joules**.