The temperature inside a refrigerator is t2 ​ C and the room temperature is t1 ​ C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

Options

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

Answer

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. We will use the concept of the Coefficient of Performance (COP) of a refrigerator and the relationship between heat absorbed, work done, and heat rejected.

### Step-by-Step Solution:

1. **Understanding the Coefficient of Performance (COP):**
The Coefficient of Performance (COP) of a refrigerator is defined as:
\[ \text{COP} = \frac{Q_C}{W} \]
where \( Q_C \) is the heat extracted from the cold reservoir (inside the refrigerator) and \( W \) is the work input (electrical energy consumed).

2. **Relating COP to Temperatures:**
For an ideal refrigerator, the COP can also be expressed in terms of the temperatures of the cold and hot reservoirs:
\[ \text{COP} = \frac{T_2 + 273}{T_1 - T_2} \]
Here, \( T_2 \) is the temperature inside the refrigerator (in °C) and \( T_1 \) is the room temperature (in °C). We convert these temperatures to Kelvin by adding 273.

3. **Finding Heat Delivered to the Room:**
The heat delivered to the room \( Q_H \) can be related to the heat extracted and the work done:
\[ Q_H = Q_C + W \]
Dividing both sides by \( W \):
\[ \frac{Q_H}{W} = \frac{Q_C}{W} + 1 \]

4. **Substituting COP into the Equation:**
From the COP expression, we know:
\[ \frac{Q_C}{W} = \frac{T_2 + 273}{T_1 - T_2} \]
Substituting this into the equation for \( \frac{Q_H}{W} \):
\[ \frac{Q_H}{W} = \frac{T_2 + 273}{T_1 - T_2} + 1 \]

5. **Simplifying the Expression:**
To combine the terms, we can express 1 as \( \frac{T_1 - T_2}{T_1 - T_2} \):
\[ \frac{Q_H}{W} = \frac{T_2 + 273 + (T_1 - T_2)}{T_1 - T_2} = \frac{T_1 + 273}{T_1 - T_2} \]

6. **Final Result:**
Thus, the amount of heat delivered to the room for each joule of electrical energy consumed ideally is:
\[ \frac{Q_H}{W} = \frac{T_1 + 273}{T_1 - T_2} \]