A refrigerator transfer 180 joule of energy in one second from temperature –3°C to 27°C. Calculate the average power consumed, assuming no energy losses in the process.

Options

A. 18 W
B. 54 W
C. 20 W
D. 120 W

Answer

To solve the problem of calculating the average power consumed by the refrigerator, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- Heat transferred from the sink (Q2) = 180 Joules
- Time taken (t) = 1 second
- Temperature of the sink (T2) = -3°C = 270 K (in Kelvin)
- Temperature of the source (T1) = 27°C = 300 K (in Kelvin)

### Step 2: Use the Formula for Work Done
The work done (W) by the refrigerator can be calculated using the relationship:
\[ W = Q1 - Q2 \]
Where Q1 is the heat rejected to the source.

### Step 3: Use the Coefficient of Performance (COP) for Refrigerators
The Coefficient of Performance (COP) for a refrigerator is given by:
\[ \text{COP} = \frac{Q2}{W} = \frac{T2}{T1 - T2} \]
From this, we can rearrange to find W:
\[ W = \frac{Q2 \cdot (T1 - T2)}{T2} \]

### Step 4: Substitute the Values into the COP Equation
Substituting the known values:
- Q2 = 180 J
- T1 = 300 K
- T2 = 270 K

We can calculate:
\[ W = \frac{180 \cdot (300 - 270)}{270} \]
\[ W = \frac{180 \cdot 30}{270} \]

### Step 5: Simplify the Calculation
Now, simplifying the expression:
\[ W = \frac{5400}{270} = 20 \text{ Joules} \]

### Step 6: Calculate the Average Power
The average power (P) consumed by the refrigerator is given by:
\[ P = \frac{W}{t} \]
Where t = 1 second.

Thus,
\[ P = \frac{20 \text{ J}}{1 \text{ s}} = 20 \text{ Watts} \]

### Final Answer
The average power consumed by the refrigerator is **20 Watts**.

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