In a refrigerator, heat from inside at 277K is transferred to a room at 300K. How many joules of heat shall be delivered to the room for each joule of electrical energy consumed ideally?

Explanation

To solve the problem, we will use the concept of the coefficient of performance (COP) of a refrigerator, which is defined as the ratio of the heat extracted from the cold reservoir (Q2) to the work input (W). Here are the steps to arrive at the solution:

### Step 1: Identify the temperatures
We are given:
- Temperature of the cold reservoir (inside the refrigerator), T1 = 277 K
- Temperature of the hot reservoir (room), T2 = 300 K

### Step 2: Write the formula for the coefficient of performance (COP)
The coefficient of performance (β) for a refrigerator is given by the formula:
\[ \beta = \frac{Q_2}{W} \]
Where:
- \( Q_2 \) = heat extracted from the cold reservoir
- \( W \) = work input (electrical energy consumed)

### Step 3: Relate the heat delivered to the room (Q1) to the work done (W)
From the first law of thermodynamics for the refrigerator, we have:
\[ Q_1 = W + Q_2 \]
Where:
- \( Q_1 \) = heat delivered to the hot reservoir (room)

### Step 4: Express W in terms of Q2 and Q1
From the equation above, we can express work (W) as:
\[ W = Q_1 - Q_2 \]

### Step 5: Use the Carnot efficiency relation
For an ideal refrigerator operating between two temperatures, the COP can also be expressed as:
\[ \beta = \frac{T_1}{T_2 - T_1} \]
Substituting the given temperatures:
\[ \beta = \frac{277}{300 - 277} = \frac{277}{23} \approx 12.04 \]

### Step 6: Calculate heat delivered (Q2) for 1 joule of work (W)
Given that we are consuming 1 joule of electrical energy (W = 1 J):
\[ Q_2 = \beta \cdot W = 12.04 \cdot 1 \approx 12.04 \text{ joules} \]

### Step 7: Calculate the total heat delivered to the room (Q1)
Now, substituting \( Q_2 \) back into the equation for \( Q_1 \):
\[ Q_1 = W + Q_2 = 1 + 12.04 \approx 13.04 \text{ joules} \]

### Final Answer
The total heat delivered to the room for each joule of electrical energy consumed ideally is approximately **13.04 joules**.

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