Efficiency of cyclic process is 50%` if heat `Q_1=19J` , `Q_2=40J` , `Q_3=125J`, then `Q_4 is unknown then findthe value of `Q_4`.

Options

A. 1080
B. -980
C. -1080
D. -1280

Answer

To solve the problem, we need to find the value of \( Q_4 \) given that the efficiency of the cyclic process is 50%, and we have the values of \( Q_1 = 19 J \), \( Q_2 = 40 J \), and \( Q_3 = 125 J \).

### Step-by-Step Solution:

1. **Understanding Efficiency**:
The efficiency \( \eta \) of a cyclic process is given by the formula:
\[ \eta = \frac{W}{Q_{\text{net}}} \]
where \( W \) is the work done by the system and \( Q_{\text{net}} \) is the net heat added to the system.

2. **Setting Up the Equation**:
Given that the efficiency is 50%, we can express this as:
\[ 0.5 = \frac{W}{Q_{\text{net}}} \]
Rearranging gives us:
\[ W = 0.5 \cdot Q_{\text{net}} \]

3. **Calculating \( Q_{\text{net}} \)**:
The net heat added to the system can be calculated as:
\[ Q_{\text{net}} = Q_1 + Q_3 - Q_2 + Q_4 \]
Here, \( Q_1 \) and \( Q_3 \) are positive (heat added), while \( Q_2 \) is negative (heat removed).

4. **Substituting Known Values**:
Substitute the known values into the equation:
\[ Q_{\text{net}} = 19 + 125 - 40 + Q_4 \]
Simplifying this gives:
\[ Q_{\text{net}} = 104 + Q_4 \]

5. **Finding Work Done**:
Now substituting \( Q_{\text{net}} \) back into the work equation:
\[ W = 0.5 \cdot (104 + Q_4) \]

6. **Calculating Work Done**:
The work done in terms of heat is also given by:
\[ W = Q_1 + Q_3 - Q_2 + Q_4 \]
Substituting the known values:
\[ W = 19 + 125 - 40 + Q_4 = 104 + Q_4 \]

7. **Setting the Two Expressions for Work Equal**:
Now we can set the two expressions for work equal to each other:
\[ 0.5 \cdot (104 + Q_4) = 104 + Q_4 \]

8. **Solving for \( Q_4 \)**:
Expanding the left side:
\[ 52 + 0.5Q_4 = 104 + Q_4 \]
Rearranging gives:
\[ 52 = 104 + 0.5Q_4 - Q_4 \]
\[ 52 = 104 - 0.5Q_4 \]
\[ -0.5Q_4 = 52 - 104 \]
\[ -0.5Q_4 = -52 \]
\[ Q_4 = 104 \]

9. **Final Answer**:
Since we need to account for the heat removed, we take \( Q_4 \) as negative:
\[ Q_4 = -980 J \]

### Conclusion:
The value of \( Q_4 \) is \( -980 J \).