A heat engine is involved with exchange of heat of 1915 J , - 40 J , + 125 J and -QJ , during one cycle achieving an efficiency of `50.0 `% The value of Q is:

Options

A. 980 J
B. 640 J
C. 40 J
D. 400 J

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the heat exchanges
The heat engine has the following heat exchanges during one cycle:
- Heat supplied: \( Q_1 = 1915 \, \text{J} \)
- Heat removed: \( Q_2 = -40 \, \text{J} \)
- Heat added: \( Q_3 = 125 \, \text{J} \)
- Heat removed: \( Q_4 = -Q \, \text{J} \)

### Step 2: Calculate the total heat input
The total heat input \( Q_{in} \) can be calculated as:
\[ Q_{in} = Q_1 + Q_3 = 1915 \, \text{J} + 125 \, \text{J} = 2040 \, \text{J} \]

### Step 3: Calculate the net heat exchange
The net heat exchange \( Q_{net} \) is given by:
\[ Q_{net} = Q_1 + Q_2 + Q_3 + Q_4 = 1915 \, \text{J} - 40 \, \text{J} + 125 \, \text{J} - Q \]
This simplifies to:
\[ Q_{net} = 2000 \, \text{J} - Q \]

### Step 4: Use the efficiency formula
The efficiency \( \eta \) of the heat engine is given by:
\[ \eta = \frac{W_{net}}{Q_{in}} \]
Where \( W_{net} \) is the net work done. We know the efficiency is 50%, or \( 0.5 \). Therefore:
\[ 0.5 = \frac{2000 - Q}{2040} \]

### Step 5: Solve for \( Q \)
Cross-multiplying gives:
\[ 0.5 \times 2040 = 2000 - Q \]
Calculating \( 0.5 \times 2040 \):
\[ 1020 = 2000 - Q \]
Rearranging gives:
\[ Q = 2000 - 1020 = 980 \, \text{J} \]

### Final Answer
Thus, the value of \( Q \) is:
\[ \boxed{980 \, \text{J}} \]