The temperature inside a refrigerator is `t_(2)^(@)C` . The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

Options

A. `(t_(2)+273)/(t_(1)-t_(2))`
B. `(t_(1)+t_(2))/(t_(1)-273)`
C. `(t_1)/(t_(1)-t_(2))`
D. `(t_(1)+273)/(t_(1)-t_(2))`

Explanation

To solve the problem, we need to determine the amount of heat delivered to the room for each joule of electrical energy consumed by the refrigerator. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the Coefficient of Performance (COP)**:
The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat removed from the cold reservoir (inside the refrigerator) to the work input (electrical energy consumed). It is given by:
\[ \beta = \frac{Q_2}{W} \]
where \( Q_2 \) is the heat absorbed from the cold space (inside the refrigerator) and \( W \) is the work done (energy consumed).

2. **Relate Heat Delivered to the Room**:
The heat delivered to the room (hot reservoir) can be expressed as:
\[ Q_1 = Q_2 + W \]
where \( Q_1 \) is the heat delivered to the room.

3. **Express Work in Terms of COP**:
From the definition of COP, we can rearrange it to express \( W \):
\[ W = \frac{Q_2}{\beta} \]

4. **Substituting for \( Q_2 \)**:
Now, substituting \( W \) back into the equation for \( Q_1 \):
\[ Q_1 = Q_2 + \frac{Q_2}{\beta} \]
This can be simplified to:
\[ Q_1 = Q_2 \left(1 + \frac{1}{\beta}\right) \]

5. **Finding the COP in Terms of Temperatures**:
The COP of a refrigerator can also be expressed in terms of the temperatures of the cold and hot reservoirs:
\[ \beta = \frac{T_2 + 273}{T_1 - T_2} \]
where \( T_2 \) is the temperature inside the refrigerator in Celsius, and \( T_1 \) is the temperature of the room in Celsius.

6. **Substituting COP into the Heat Equation**:
Substitute the expression for \( \beta \) into the equation for \( Q_1 \):
\[ Q_1 = Q_2 \left(1 + \frac{T_1 - T_2}{T_2 + 273}\right) \]

7. **Final Expression**:
Rearranging gives us the final expression for the heat delivered to the room for each joule of electrical energy consumed:
\[ \frac{Q_1}{W} = 1 + \frac{T_1 - T_2}{T_2 + 273} \]